diff --git a/docs/index.html b/docs/index.html
index 97848f3a..8696d610 100644
--- a/docs/index.html
+++ b/docs/index.html
@@ -179,7 +179,7 @@
         <a class="btclk" href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu</a>
         or use the QR code on the right, if that's the kind of convenience you prefer =)
       </p>
-    
+
     </div>
     <div class="btcqr">
         <a href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">
@@ -191,7 +191,7 @@
 
 			<br style="clear: both;" />
 
-			<p>— <a href="https://twitter.com/TheRealPomax">Pomax</a></p>
+			<p>— <a href="https://mastodon.social/@TheRealPomax">Pomax</a></p>
 			<noscript>
 				<div class="note">
 					<header>
@@ -566,7 +566,7 @@
 						thus 80% the distance away from the second point) then we can compute that really easily:
 					</p>
 					<!--
- 
+
       ╭        p =  some point       ╮
       │         1                    │
       │        p =  some other point │
@@ -621,7 +621,7 @@ Given │  distance= (p  - p )         │,  our new point = p  +  distance  ·
 						kind of number manipulation:
 					</p>
 					<!--
- 
+
 f(x) = cos (x)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/0cc876c56200446c60114c1b0eeeb2cc.svg" width="96px" height="17px" loading="lazy" />
@@ -631,7 +631,7 @@ f(x) = cos (x)
 					</p>
 					<p>So far, so good. Now, let's look at parametric functions, and how they cheat. Let's take the following two functions:</p>
 					<!--
- 
+
 f(a) = cos (a)
 f(b) = sin (b)
 -->
@@ -642,9 +642,9 @@ f(b) = sin (b)
 						function. Parametric functions cheat by changing that. In a parametric function all the different functions share a variable, like this:
 					</p>
 					<!--
- 
+
 ╭ f (t) = cos (t)
-╡  a              
+╡  a
 │ f (t) = sin (t)
 ╰  b
 -->
@@ -662,8 +662,8 @@ f(b) = sin (b)
 						obvious:
 					</p>
 					<!--
- 
-{ x = cos (t) 
+
+{ x = cos (t)
   y = sin (t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/6914ba615733c387251682db7a3db045.svg" width="77px" height="40px" loading="lazy" />
@@ -691,7 +691,7 @@ f(b) = sin (b)
 					</p>
 					<p>You may remember polynomials from high school. They're those sums that look like this:</p>
 					<!--
- 
+
              3         2
 f(x) = a  · x  + b  · x  + c  · x + d
 -->
@@ -712,12 +712,12 @@ f(x) = a  · x  + b  · x  + c  · x + d
 						values:
 					</p>
 					<!--
- 
-linear= (1-t) + t                                        
+
+linear= (1-t) + t
              2                      2
-square= (1-t)  + 2  · (1-t)  · t + t                     
+square= (1-t)  + 2  · (1-t)  · t + t
              3             2                       2    3
- cubic= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+ cubic= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img
 						class="LaTeX SVG"
@@ -731,10 +731,10 @@ square= (1-t)  + 2  · (1-t)  · t + t
 						easy. Check out these binomial terms:
 					</p>
 					<!--
- 
- linear=  1 + 1           
- square=  1 + 2 + 1       
-  cubic=  1 + 3 + 3 + 1   
+
+ linear=  1 + 1
+ square=  1 + 2 + 1
+  cubic=  1 + 3 + 3 + 1
 quartic= 1 + 4 + 6 + 4 + 1
 -->
 					<img
@@ -755,11 +755,11 @@ quartic= 1 + 4 + 6 + 4 + 1
 					</p>
 					<!--
 
-linear= \colorblue a + \colorred b                                                                                                                   
-square= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b                                                      
+linear= \colorblue a + \colorred b
+square= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b
  cubic= \colorblue a  · \colorblue a  · \colorblue a + \colorblue a  · \colorblue a  · \colorred b + \colorblue a  · \colorred b  · \colorred b + \co
-                                                                                             
-                                                                                             
+
+
                                                        lorred b  · \colorred b  · \colorred b
 -->
 					<img
@@ -775,9 +775,9 @@ square= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred
 						generic function for this:
 					</p>
 					<!--
- 
+
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t
               ‾‾ i=0
 -->
 					<img
@@ -1070,9 +1070,9 @@ function Bezier(3,t):
 					<!--
 
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset
               ‾‾ i=0
-                                                                weight\underbracew 
+                                                                weight\underbracew
                                                                                   i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/c2f2fe0ef5d0089d9dd8e5e3999405cb.svg" width="336px" height="55px" loading="lazy" />
@@ -1083,11 +1083,11 @@ Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \unders
 						(210,30), we use this Bézier curve:
 					</p>
 					<!--
- 
+
 ╭                               3                                   2                                             2                       3
-╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t  
+╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t
 │                               3                                    2                                             2                      3
-╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t 
+╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/be73034ac382b54863c7a18c2932cbbc.svg" width="473px" height="40px" loading="lazy" />
 					<p>Which gives us the curve we saw at the top of the article:</p>
@@ -1208,9 +1208,9 @@ function Bezier(3,t,w[]):
 					</p>
 					<p>Adding these ratio values to the regular Bézier curve function is fairly easy. Where the regular function is the following:</p>
 					<!--
- 
+
               __ n                    n-i     i
-Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w 
+Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
               ‾‾ i=0                                i
 -->
 					<img
@@ -1222,13 +1222,13 @@ Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
 					/>
 					<p>The function for rational Bézier curves has two more terms:</p>
 					<!--
- 
+
                         __ n                    n-i     i
-                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio 
+                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio
                         ‾‾ i=0                                i                    i
 Rational  Bézier(n,t) = ────────────────────────────────────────────────────────────
                                      __ n                     n-i      i
-                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio  
+                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio
                                      ‾‾ i=0                                       i
 -->
 					<img
@@ -1409,8 +1409,8 @@ function RationalBezier(3,t,w[],r[]):
 						other values, then the general formula for this is:
 					</p>
 					<!--
- 
-mixture = a  ·  value  + b  ·  value 
+
+mixture = a  ·  value  + b  ·  value
                      1              2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/dfd6ded3f0addcf43e0a1581627a2220.svg" width="212px" height="16px" loading="lazy" />
@@ -1423,8 +1423,8 @@ mixture = a  ·  value  + b  ·  value
 						And that's easy:
 					</p>
 					<!--
- 
-m = a  ·  value  + (1 - a)  ·  value 
+
+m = a  ·  value  + (1 - a)  ·  value
                1                    2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/4e0fa763b173e3a683587acf83733353.svg" width="213px" height="16px" loading="lazy" />
@@ -1494,61 +1494,61 @@ m = a  ·  value  + (1 - a)  ·  value
 						refer to coordinate values "in one or more dimensions":
 					</p>
 					<!--
- 
+
                   3                   2                             2          3
-B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t 
+B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t
         1              2                        3                        4
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/9a9a55f5b0323d9ea88f82fc6be58ad3.svg" width="468px" height="20px" loading="lazy" />
 					<p>Disregarding our actual coordinates for a moment, we have:</p>
 					<!--
- 
+
             3             2                       2    3
-B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/87cfac83cb8a4b0bee68ef006effc611.svg" width="353px" height="19px" loading="lazy" />
 					<p>We can write this as a sum of four expressions:</p>
 					<!--
- 
+
                 3
-... =      (1-t) 
+... =      (1-t)
                 2
     + 3  · (1-t)   · t
                      2
-    + 3  · (1-t)  · t 
+    + 3  · (1-t)  · t
              3
-    +       t  
+    +       t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/cdd88611833f3b178df91278359a4193.svg" width="140px" height="75px" loading="lazy" />
 					<p>And we can expand these expressions:</p>
 					<!--
- 
+
                                    3         2
 ... =  (1-t)  · (1-t)  · (1-t) = -t  + 3  · t  - 3  · t + 1
                                        3         2
     + 3  · (1-t)  · (1-t)  · t = 3  · t  - 6  · t  + 3  · t
                                             3         2
-    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t 
+    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t
                                              3
-    +        t  · t  · t       =            t  
+    +        t  · t  · t       =            t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/ec118f296511c6e9ac8727be3703a7ce.svg" width="397px" height="75px" loading="lazy" />
 					<p>Furthermore, we can make all the 1 and 0 factors explicit:</p>
 					<!--
- 
+
              3         2
 ... = -1  · t  + 3  · t  - 3  · t + 1
              3         2
-    + +3  · t  - 6  · t  + 3  · t + 0 
+    + +3  · t  - 6  · t  + 3  · t + 0
              3         2
     + -3  · t  + 3  · t  + 0  · t + 0
              3         2
-    + +1  · t  + 0  · t  + 0  · t + 0 
+    + +1  · t  + 0  · t  + 0  · t + 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/67a5ea33d6c6558f7d954b18226f4956.svg" width="217px" height="75px" loading="lazy" />
 					<p>And <em>that</em>, we can view as a series of four matrix operations:</p>
 					<!--
- 
+
                  ┌ -1 ┐                    ┌ 3  ┐                    ┌ -3 ┐                    ┌ 1 ┐
 ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ -6 │ + ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ 0 │
 └ t  t  t 1 ┘    │ -3 │   └ t  t  t 1 ┘    │ 3  │   └ t  t  t 1 ┘    │ 0  │   └ t  t  t 1 ┘    │ 0 │
@@ -1557,7 +1557,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/8ecff6b8a37d60385d287ea2b26876db.svg" width="607px" height="72px" loading="lazy" />
 					<p>If we compact this into a single matrix operation, we get:</p>
 					<!--
- 
+
                  ┌ -1  3 -3 1 ┐
 ┌  3  2     ┐  · │  3 -6  3 0 │
 └ t  t  t 1 ┘    │ -3  3  0 0 │
@@ -1569,7 +1569,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 						<code>t</code> matrix horizontally, and our big "mixing" matrix upside down:
 					</p>
 					<!--
- 
+
                  ┌  1  0  0 0 ┐
 ┌      2  3 ┐  · │ -3  3  0 0 │
 └ 1 t t  t  ┘    │  3 -6  3 0 │
@@ -1578,7 +1578,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/1a64ed455c6dd2f8cacca5e5e12bdcc1.svg" width="227px" height="72px" loading="lazy" />
 					<p>And then finally, we can add in our original coordinates as a single third matrix:</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -1591,7 +1591,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 					<img class="LaTeX SVG" src="./images/chapters/matrix/b32cae2dfc47d5f36df0bc3defb7dfa8.svg" width="323px" height="73px" loading="lazy" />
 					<p>We can perform the same trick for the quadratic curve, in which case we end up with:</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2003,7 +2003,7 @@ function drawCurvePoint(points[], t):
 						respectively: (we'll reverse the Bézier coefficients vector for legibility)
 					</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2020,7 +2020,7 @@ B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -2043,7 +2043,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						"point on the curve" information into a new matrix multiplication:
 					</p>
 					<!--
- 
+
                                                   ┌ P  ┐                                              ┌ P  ┐
                                    ┌  1  0 0 ┐    │  1 │                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2060,7 +2060,7 @@ B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌
 					/>
 					<p>and</p>
 					<!--
- 
+
                                                                ┌ P  ┐                                                       ┌ P  ┐
                                                                │  1 │                    ┌ 1 0  0 0  ┐                      │  1 │
                                              ┌  1  0  0 0 ┐    │ P  │                    │ 0 z  0 0  │    ┌  1  0  0 0 ┐    │ P  │
@@ -2090,7 +2090,7 @@ B(t) = ┌                    2         3 ┐  · │ -3  3  0 0 │  · │  2
 							at the quadratic curve first:
 						</p>
 						<!--
- 
+
                                                   ┌ P  ┐
                      ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2106,7 +2106,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                                         ┌ P  ┐
                                                                                         │  1 │
 = ┌      2 ┐  · \underset we turn this...\underbrace\kern 2.25em Z  · M \kern 2.25em  · │ P  │
@@ -2122,7 +2122,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                          ┌ P  ┐
                                                         -1               │  1 │
 = ┌      2 ┐  · \underset into this...\underbrace M  · M    · Z  · M   · │ P  │
@@ -2138,7 +2138,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                                                                    ┌ P  ┐
                                                                                                                    │  1 │
 = ┌      2 ┐  · M \underset ...to get this!\underbrace \kern 1.25em  · \kern 1.25em Q \kern 1.25em  · \kern 1.25em │ P  │
@@ -2165,7 +2165,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							coordinates that represent the curve from <em>t = 0</em> to <em>t = z</em>. So let's get computing:
 						</p>
 						<!--
- 
+
                     ┌ 1 0 0 ┐
      -1             │   1   │    ┌ 1 0 0  ┐    ┌  1  0 0 ┐   ┌     1            0        0  ┐
 Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = │  -(z-1)          z        0  │
@@ -2181,7 +2181,7 @@ Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = 
 						/>
 						<p>Excellent! Now we can form our new quadratic curve:</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭      ┌ P  ┐ ╮
                                │  1 │                      │      │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q  · │ P  │ │
@@ -2197,7 +2197,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                     ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌     1            0        0  ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │  -(z-1)          z        0  │  · │ P  │ │
@@ -2213,7 +2213,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌                        P                          ┐
                                │                         1                         │
                 ┌  1  0 0 ┐    │               z  · P  - (z-1)  · P                │
@@ -2242,7 +2242,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							write it down in a more simple form because of the zero, but what we <em>actually</em> evaluated, making the zero explicit, was:
 						</p>
 						<!--
- 
+
                                                             ┌ P  ┐
                                              ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2258,7 +2258,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                              ┌ P  ┐
                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2275,7 +2275,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 						/>
 						<p>If we want the interval [<em>z</em>,1], we will be evaluating this instead:</p>
 						<!--
- 
+
                                                                     ┌ P  ┐
                                                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2291,7 +2291,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 							loading="lazy"
 						/>
 						<!--
- 
+
                 ┌              2        ┐                   ┌ P  ┐
                 │ 1  z        z         │    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │  · │ P  │
@@ -2311,7 +2311,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 							<code>[M · something]</code>:
 						</p>
 						<!--
- 
+
                       ┌ 1 0 0 ┐    ┌              2        ┐
       -1              │   1   │    │ 1  z        z         │    ┌  1  0 0 ┐   ┌      2                   2 ┐
 Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │ = │ (z-1)  -2  · z  · (z-1) z  │
@@ -2327,7 +2327,7 @@ Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  ·
 						/>
 						<p>So, our final second curve looks like:</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭       ┌ P  ┐ ╮
                                │  1 │                      │       │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q'  · │ P  │ │
@@ -2343,7 +2343,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                   ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌      2                   2 ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │ (z-1)  -2  · z  · (z-1) z  │  · │ P  │ │
@@ -2359,7 +2359,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌  2                                      2       ┐
                                │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
                 ┌  1  0 0 ┐    │        3                       2              1 │
@@ -2389,7 +2389,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						<code>t = z</code>, we get two subcurves that are described as Bézier curves with simple-to-derive coordinates:
 					</p>
 					<!--
- 
+
                                              ┌                        P                          ┐
                                     ┌ P  ┐   │                         1                         │
 ┌     1            0        0  ┐    │  1 │   │               z  · P  - (z-1)  · P                │
@@ -2407,7 +2407,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                            ┌  2                                      2       ┐
                                   ┌ P  ┐   │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
 ┌      2                   2 ┐    │  1 │   │        3                       2              1 │
@@ -2428,7 +2428,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						yourself, though) and simply show you the resulting new coordinate sets:
 					</p>
 					<!--
- 
+
                                                               ┌                                    P                                      ┐
                                                      ┌ P  ┐   │                                     1                                     │
 ┌    1            0                0         0  ┐    │  1 │   │                           z  · P  - (z-1)  · P                            │
@@ -2449,7 +2449,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                                               ┌  3               2                                 2              3       ┐
                                                      ┌ P  ┐   │ z   · P  - 3  · z   · (z-1)  · P  + 3  · z  · (z-1)   · P  - (z-1)   · P  │
 ┌       3           2                      2  3 ┐    │  1 │   │        4                        3                        2              1 │
@@ -2499,7 +2499,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						that the start and end weights are the same as the start and end weights for the old curve):
 					</p>
 					<!--
- 
+
               __ k                                                                                            k-i     i
 Bézier(k,t) = ❯      \underset binomial term\underbrace\binomki  · \ \underset polynomial term\underbrace(1-t)     · t    · \ \underset new
               ‾‾ i=0
@@ -2523,9 +2523,9 @@ Bézier(k,t) = ❯      \underset binomial term\underbrace\binomki  · \ \unders
 					</p>
 					<p>We start by taking the standard Bézier function, and condensing it a little:</p>
 					<!--
- 
+
               __ n       n               n                       n-i     i
-Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t 
+Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t
               ‾‾ i=0  i  i               i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/9fc4ecc087d389dd0111bcba165cd5d0.svg" width="408px" height="41px" loading="lazy" />
@@ -2535,7 +2535,7 @@ Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t
 						<code>t</code> and <code>1-t</code>:
 					</p>
 					<!--
- 
+
 x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/ff224ded6bbbc94b43130f5f8eeb5d29.svg" width="379px" height="16px" loading="lazy" />
@@ -2544,9 +2544,9 @@ x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
 						<code>t</code> component:
 					</p>
 					<!--
- 
-Bézier(n,t)= (1-t) B(n,t) + t B(n,t)                    
-             __ n               n      __ n         n   
+
+Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
+             __ n               n      __ n         n
            = ❯      w  (1 - t) B (t) + ❯      w  t B (t)
              ‾‾ i=0  i          i      ‾‾ i=0  i    i
 -->
@@ -2556,18 +2556,18 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
 						I promise, it's about to make sense. We start with <code>(1-t)</code>:
 					</p>
 					<!--
- 
+
          n              n!         n-i  i
-(1 - t) B (t)= (1-t) ──────── (1-t)    t                  
+(1 - t) B (t)= (1-t) ──────── (1-t)    t
          i           (n-i)!i!
                n+1-i   (n+1)!        n+1-i  i
-             = ───── ────────── (1-t)      t              
-                n+1  (n+1-i)!i!                           
+             = ───── ────────── (1-t)      t
+                n+1  (n+1-i)!i!
                k-i    k!         k-i  i
              = ─── ──────── (1-t)    t ,  where  k = n + 1
                 k  (k-i)!i!
                k-i  k
-             = ─── B (t)                                  
+             = ─── B (t)
                 k   i
 -->
 					<img
@@ -2585,18 +2585,18 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
 						<code>t</code> part, but that's not a problem:
 					</p>
 					<!--
- 
+
    n          n!         n-i  i
-t B (t)= t ──────── (1-t)    t                                    
+t B (t)= t ──────── (1-t)    t
    i       (n-i)!i!
          i+1        (n+1)!             (n+1)-(i+1)  i+1
-       = ─── ──────────────────── (1-t)            t              
-         n+1 ((n+1)-(i+1))!(i+1)!                                 
+       = ─── ──────────────────── (1-t)            t
+         n+1 ((n+1)-(i+1))!(i+1)!
          i+1        k!             k-(i+1)  i+1
        = ─── ──────────────── (1-t)        t   ,  where  k = n + 1
           k  (k-(i+1))!(i+1)!
          i+1  k
-       = ─── B   (t)                                              
+       = ─── B   (t)
           k   i+1
 -->
 					<img
@@ -2616,21 +2616,21 @@ t B (t)= t ──────── (1-t)    t
 					</p>
 					<p>Let's do this:</p>
 					<!--
- 
+
              __ n+1             n      __ n+1       n
-Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)                   
+Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)
              ‾‾ i=0  i          i      ‾‾ i=0  i    i
              __ n+1    k-i  k      __ n+1    i+1  k
            = ❯      w  ─── B (t) + ❯      w  ─── B   (t),  where  k = n + 1
              ‾‾ i=0  i  k   i      ‾‾ i=0  i  k   i+1
              __ n+1    k-i  k      __ n+1      i  k
-           = ❯      w  ─── B (t) + ❯      p    ─ B (t)                     
+           = ❯      w  ─── B (t) + ❯      p    ─ B (t)
              ‾‾ i=0  i  k   i      ‾‾ i=0  i-1 k  i
              __ n+1 ╭    k-i        i ╮  k
-           = ❯      │ w  ─── + p    ─ │ B (t)                              
+           = ❯      │ w  ─── + p    ─ │ B (t)
              ‾‾ i=0 ╰  i  k     i-1 k ╯  i
              __ n+1                      k                 i
-           = ❯      (w  (1-s) + p    s) B (t),  where  s = ─               
+           = ❯      (w  (1-s) + p    s) B (t),  where  s = ─
              ‾‾ i=0   i          i-1     i                 k
 -->
 					<img
@@ -2645,14 +2645,14 @@ Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)
 						Bézier(n+1,t) as a very simple matrix multiplication:
 					</p>
 					<!--
- 
-M B  = B 
+
+M B  = B
    n    k
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/1f5b60d190a1c7099b3411e4cc477291.svg" width="71px" height="16px" loading="lazy" />
 					<p>where the matrix <strong>M</strong> is an <code>n+1</code> by <code>n</code> matrix, and looks like:</p>
 					<!--
- 
+
     ┌ 1  0   .   .   .   .    .    .  ┐
     │ 1 k-1                           │
     │ ─ ───  0   .   .   .    0    .  │
@@ -2694,20 +2694,20 @@ M = │        k   k                    │
 						<strong>A x = b</strong>, we can use it. And as it so happens, that's exactly what we're dealing with, so:
 					</p>
 					<!--
- 
-              M B = B             
+
+              M B = B
                  n   k
            T         T
-         (M  M) B = M  B          
+         (M  M) B = M  B
                  n      k
   T   -1   T          T   -1  T
-(M  M)   (M  M) B = (M  M)   M  B 
+(M  M)   (M  M) B = (M  M)   M  B
                  n               k
                       T   -1  T
-              I B = (M  M)   M  B 
+              I B = (M  M)   M  B
                  n               k
                       T   -1  T
-                B = (M  M)   M  B 
+                B = (M  M)   M  B
                  n               k
 -->
 					<img
@@ -2762,9 +2762,9 @@ M = │        k   k                    │
 					</p>
 					<p>First, let's look at the derivative rule for Bézier curves, which is:</p>
 					<!--
- 
+
                     __ n-1
-Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t) 
+Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
                     ‾‾ i=0   i+1  i                    i
 -->
 					<img
@@ -2779,7 +2779,7 @@ Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
 						is the same as a summation where each term is multiplied by <i>n</i>) as:
 					</p>
 					<!--
- 
+
                __ n-1
 Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
                ‾‾ i=0                i           i+1  i
@@ -2805,7 +2805,7 @@ Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
 							only the derivative of the polynomial basis function. So, let's find that:
 						</p>
 						<!--
- 
+
         d    ╭ n ╮  k      n-k d
 B   (t) ── = │   │ t  (1-t)    ──
  n,k    dt   ╰ k ╯             dt
@@ -2822,7 +2822,7 @@ B   (t) ── = │   │ t  (1-t)    ──
 							<a href="https://en.wikipedia.org/wiki/Chain_rule">chain</a> rules gives us:
 						</p>
 						<!--
- 
+
       ╭ n ╮        k-1      n-k    k         n-k-1
 ... = │   │ (k  · t    (1-t)    + t   · (1-t)       · (n-k)  · -1)
       ╰ k ╯
@@ -2836,9 +2836,9 @@ B   (t) ── = │   │ t  (1-t)    ──
 						/>
 						<p>Which is hard to work with, so let's expand that properly:</p>
 						<!--
- 
+
         kn!     k-1      n-k   (n-k)n!   k      n-1-k
-... = ──────── t    (1-t)    - ──────── t  (1-t)     
+... = ──────── t    (1-t)    - ──────── t  (1-t)
       k!(n-k)!                 k!(n-k)!
 -->
 						<img
@@ -2853,12 +2853,12 @@ B   (t) ── = │   │ t  (1-t)    ──
 							look like "x! over y!(x-y)!". If we can do that in a way that involves terms of <i>n-1</i> and <i>k-1</i>, we'll be on the right track.
 						</p>
 						<!--
- 
+
            n!       k-1      n-k   (n-k)n!   k      n-1-k
-... = ──────────── t    (1-t)    - ──────── t  (1-t)                                   
+... = ──────────── t    (1-t)    - ──────── t  (1-t)
       (k-1)!(n-k)!                 k!(n-k)!
         ╭    (n-1)!     k-1      n-k   (n-k)(n-1)!  k      n-1-k ╮
-... = n │ ──────────── t    (1-t)    - ─────────── t  (1-t)      │                     
+... = n │ ──────────── t    (1-t)    - ─────────── t  (1-t)      │
         ╰ (k-1)!(n-k)!                  k!(n-k)!                 ╯
         ╭        (n-1)!         (k-1)      (n-1)-(k-1)      (n-1)!     k      (n-1)-k ╮
 ... = n │ ──────────────────── t      (1-t)            - ──────────── t  (1-t)        │
@@ -2873,11 +2873,11 @@ B   (t) ── = │   │ t  (1-t)    ──
 						/>
 						<p>And that's the first part done: the two components inside the parentheses are actually regular, lower-order Bézier expressions:</p>
 						<!--
- 
+
         ╭    x!     y      x-y      x!     k      x-k ╮
 ... = n │ ──────── t  (1-t)    - ──────── t  (1-t)    │  , with x=n-1, y=k-1
-        ╰ y!(x-y)!               k!(x-k)!             ╯                     
-... = n (B           (t) - B       (t))                                     
+        ╰ y!(x-y)!               k!(x-k)!             ╯
+... = n (B           (t) - B       (t))
           (n-1),(k-1)       (n-1),k
 -->
 						<img
@@ -2892,19 +2892,19 @@ B   (t) ── = │   │ t  (1-t)    ──
 							through to its derivative:
 						</p>
 						<!--
- 
+
 Bézier   (t)    = B   (t)  · w  + B   (t)  · w  + B   (t)  · w  + B   (t)  · w  + ...
       n,k          n,0        0    n,1        1    n,2        2    n,3        3
              d
-Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +                              
+Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +
       n,k    dt          n-1,-1       n-1,0         0
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,0       n-1,1         1
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,1       n-1,2         2
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,2       n-1,3         3
-                  ...                                                                
+                  ...
 -->
 						<img
 							class="LaTeX SVG"
@@ -2918,18 +2918,18 @@ Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +
 							following:
 						</p>
 						<!--
- 
-n  · B      (t)  · w               +                                     
+
+n  · B      (t)  · w               +
       n-1,-1        0
 n  · B                (t)  · w     - n  · B                (t)  · w     +
       n-1,\colorblue 0        1            n-1,\colorblue 0        0
 n  · B               (t)  · w      - n  · B               (t)  · w      +
-      n-1,\colorred 1        2             n-1,\colorred 1        1      
+      n-1,\colorred 1        2             n-1,\colorred 1        1
 n  · B                   (t)  · w  - n  · B                   (t)  · w  +
       n-1,\colormagenta 2        3         n-1,\colormagenta 2        2
 ...                                - n  · B     (t)  · w                +
                                            n-1,3        3
-...                                                                      
+...
 -->
 						<img
 							class="LaTeX SVG"
@@ -2946,14 +2946,14 @@ n  · B                   (t)  · w  - n  · B                   (t)  · w  +
 							this term would contribute "nothing", so we can ignore it, too. This means we're left with:
 						</p>
 						<!--
- 
+
 n  · B                (t)  · w     - n  · B                (t)  · w     +
       n-1,\colorblue 0        1            n-1,\colorblue 0        0
 n  · B               (t)  · w      -  n  · B               (t)  · w     +
-      n-1,\colorred 1        2              n-1,\colorred 1        1     
+      n-1,\colorred 1        2              n-1,\colorred 1        1
 n  · B                   (t)  · w  - n  · B                   (t)  · w  +
       n-1,\colormagenta 2        3         n-1,\colormagenta 2        2
-...                                
+...
 -->
 						<img
 							class="LaTeX SVG"
@@ -2964,9 +2964,9 @@ n  · B                   (t)  · w  - n  · B                   (t)  · w  +
 						/>
 						<p>And that's just a summation of lower order curves:</p>
 						<!--
- 
+
              d
-Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B                 (t)  · (w  - w ) + n  · B                     (t)  · (w  - w )  
+Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B                 (t)  · (w  - w ) + n  · B                     (t)  · (w  - w )
       n,k    dt         (n-1),\colorblue 0         1    0          (n-1),\colorred 1         2    1          (n-1),\colormagenta 2         3    2
                                                                       +  ...
 -->
@@ -2979,7 +2979,7 @@ Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B
 						/>
 						<p>We can rewrite this as a normal summation, and we're done:</p>
 						<!--
- 
+
              d    __ n-1                                 __ n-1
 Bézier   (t) ── = ❯      n  · B     (t)  · (w    - w ) = ❯      B     (t)  · \underset derivative weights \underbracen  · (w    - w )
       n,k    dt   ‾‾ k=0       n-1,k         k+1    k    ‾‾ k=0  n-1,k                                                      k+1    k
@@ -2998,11 +2998,11 @@ Bézier   (t) ── = ❯      n  · B     (t)  · (w    - w ) = ❯      B
 						Bézier curves, and then the derivative:
 					</p>
 					<!--
- 
+
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset
               ‾‾ i=0
-                                                               weight\underbracew 
+                                                               weight\underbracew
                                                                                  i
 -->
 					<img
@@ -3013,7 +3013,7 @@ Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \unders
 						loading="lazy"
 					/>
 					<!--
- 
+
                __ k                                                                                            k-i     i
 Bézier'(n,t) = ❯      \underset binomial term\underbrace\binomki  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset derivative
                ‾‾ i=0
@@ -3034,11 +3034,11 @@ Bézier'(n,t) = ❯      \underset binomial term\underbrace\binomki  · \ \under
 						derivative one:
 					</p>
 					<!--
- 
-B(n,t),           w = {A,B,C,D}   
+
+B(n,t),           w = {A,B,C,D}
 B'(n,t),   n = 3, w' = {A',B',C'} = {3  · (B-A),   3  · (C-B),   3  · (D-C)}
-B''(n,t),  n = 2, w'' = {A'',B''} = {2  · (B'-A'),   2  · (C'-B')}          
-B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}                        
+B''(n,t),  n = 2, w'' = {A'',B''} = {2  · (B'-A'),   2  · (C'-B')}
+B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}
 -->
 					<img
 						class="LaTeX SVG"
@@ -3065,10 +3065,10 @@ B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}
 						indicates the direction of travel at specific points, and is literally just the first derivative of our curve:
 					</p>
 					<!--
- 
+
 tangent (t) = B' (t)
        x        x
-                    
+
 tangent (t) = B' (t)
        y        y
 -->
@@ -3084,14 +3084,14 @@ tangent (t) = B' (t)
 						point, and then do whatever it is we want to do based on those directions:
 					</p>
 					<!--
- 
+
                            ┌─────────────────┐
                            │      2         2
-    d = \| tangent(t)\| =  │B' (t)  + B' (t)  
+    d = \| tangent(t)\| =  │B' (t)  + B' (t)
                           ⟍│  x         y
 
                            tangent (t)     B' (t)
-^                                 x          x    
+^                                 x          x
 x(t) = \| tangent (t)\| =─────────────── = ──────
                  x       \| tangent(t)\|     d
 
@@ -3113,11 +3113,11 @@ y(t) = \| tangent (t)\| = ─────────────── = ──
 						is typically of length 1.0, so all we have to do is rotate the normalized directional vector and we're done:
 					</p>
 					<!--
- 
+
              ^           \pi   ^           \pi     ^
-normal (t) = x(t)  · cos ─── - y(t)  · sin ─── = - y(t)                                             
+normal (t) = x(t)  · cos ─── - y(t)  · sin ─── = - y(t)
       x                   2                 2
-                                                                                                    
+
                                                            ^           \pi   ^           \pi    ^
 normal (t) = \underset quarter circle rotation \underbrace x(t)  · sin ─── + y(t)  · cos ───  = x(t)
       y                                                                 2                 2
@@ -3140,7 +3140,7 @@ normal (t) = \underset quarter circle rotation \underbrace x(t)  · sin ──
 							To turn any point <i>(x,y)</i> into a rotated point <i>(x',y')</i> (over 0,0) by some angle φ, we apply this nice and easy computation:
 						</p>
 						<!--
- 
+
 x' = x  · cos (\phi) - y  · sin (\phi)
 y' = x  · sin (\phi) + y  · cos (\phi)
 -->
@@ -3153,7 +3153,7 @@ y' = x  · sin (\phi) + y  · cos (\phi)
 						/>
 						<p>Which is the "long" version of the following matrix transformation:</p>
 						<!--
- 
+
 ┌ x' ┐ = ┌ cos (\phi) -sin (\phi) ┐ ┌ x ┐
 └ y' ┘   └ sin (\phi) cos (\phi)  ┘ └ y ┘
 -->
@@ -3616,10 +3616,10 @@ generateRMFrames(steps) -> frames:
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
+
 B'(t) = a(1-t) + b(t)= 0,
           a - at + bt= 0,
-           (b-a)t + a= 0 
+           (b-a)t + a= 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -3630,10 +3630,10 @@ B'(t) = a(1-t) + b(t)= 0,
 					/>
 					<p>And then we turn this into our solution for <code>t</code> using basic arithmetics:</p>
 					<!--
- 
-(b-a)t + a= 0, 
+
+(b-a)t + a= 0,
     (b-a)t= -a,
-            -a 
+            -a
          t= ───
             b-a
 -->
@@ -3656,7 +3656,7 @@ B'(t) = a(1-t) + b(t)= 0,
 						you haven't, it looks like this:
 					</p>
 					<!--
- 
+
                                                    ┌────────┐
                                                    │ 2
                2                              -b ±⟍│b  - 4ac
@@ -3681,9 +3681,9 @@ Given f(t) = at  + bt + c, f(t)=0   when  t = ───────────
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
-B(t)  uses { p ,p ,p ,p  }                                                  
-              1  2  3  4                                                    
+
+B(t)  uses { p ,p ,p ,p  }
+              1  2  3  4
 B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
                1  2  3            1      2  1    2      3  2    3      4  3
 -->
@@ -3696,21 +3696,21 @@ B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
 					/>
 					<p>And then, using these <em>v</em> values, we can find out what our <em>a</em>, <em>b</em>, and <em>c</em> should be:</p>
 					<!--
- 
+
               2                  2
-B'(t)= v (1-t)  + 2v (1-t)t + v t            
+B'(t)= v (1-t)  + 2v (1-t)t + v t
         1           2          3
            2                    2       2
-  ...= v (t  - 2t + 1) + 2v (t-t ) + v t     
+  ...= v (t  - 2t + 1) + 2v (t-t ) + v t
         1                  2          3
           2                          2      2
-  ...= v t  - 2v t + v  + 2v t - 2v t  + v t 
+  ...= v t  - 2v t + v  + 2v t - 2v t  + v t
         1       1     1     2      2      3
           2       2      2
   ...= v t  - 2v t  + v t  - 2v t + v  + 2v t
         1       2      3       1     1     2
                    2
-  ...= (v -2v +v )t  + 2(v -v )t + v         
+  ...= (v -2v +v )t  + 2(v -v )t + v
          1   2  3         2  1      1
 -->
 					<img
@@ -3725,12 +3725,12 @@ B'(t)= v (1-t)  + 2v (1-t)t + v t
 						expressions of our original coordinate values, so we can do some substitution to get:
 					</p>
 					<!--
- 
+
 a= v -2v +v  = 3(-p  + 3p  - 3p  + p )
     1   2  3       1     2     3    4
-b= 2(v -v ) = 6(p  - 2p  + p )        
+b= 2(v -v ) = 6(p  - 2p  + p )
       2  1       1     2    3
-c= v  = 3(p -p )                      
+c= v  = 3(p -p )
     1      2  1
 -->
 					<img
@@ -3757,11 +3757,11 @@ c= v  = 3(p -p )
 						really hard to solve, it can be rewritten to a form for which finding the roots is "easier" (even if not "easy"):
 					</p>
 					<!--
- 
+
                    3     2
 very hard: solve at  + bt  + ct + d = 0
                   3
-   easier: solve t  + pt + q = 0       
+   easier: solve t  + pt + q = 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -4170,7 +4170,7 @@ function getCubicRoots(pa, pb, pc, pd) {
 						<code>f<sub>y</sub>(x)</code> is zero, so that the next <code>t</code> is the same as the one we already have:
 					</p>
 					<!--
- 
+
             f (x )
              y  n
 x    = x  - ───────
@@ -4300,11 +4300,11 @@ x    = x  - ───────
 						function for the y-component to an <em>n-2</em> term function. For instance, if we have a cubic curve such as this:
 					</p>
 					<!--
- 
+
 ╭                            3                               2                                          2                       3
-╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t  
+╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t
 │                            3                                2                                          2                      3
-╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t 
+╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/e5db5e945606d3429e4475ff92283a9c.svg" width="497px" height="40px" loading="lazy" />
 					<p>
@@ -4312,11 +4312,11 @@ x    = x  - ───────
 						by -160, gives us:
 					</p>
 					<!--
- 
+
 ╭                          3                               2                                          2                       3
-╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t  
+╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t
 │                          3                               2                                          2                       3
-╰ y = \colorblue 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t 
+╰ y = \colorblue 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/d412c72ed342c2036965ff251c8fb443.svg" width="481px" height="40px" loading="lazy" />
 					<p>
@@ -4324,20 +4324,20 @@ x    = x  - ───────
 						become:
 					</p>
 					<!--
- 
+
 ╭                          3                               2                                         2                       3
-╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                          3                               2                                          2                     3
-╰ y = \colorblue 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorblue  + 0  · t 
+╰ y = \colorblue 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorblue  + 0  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/e49d7bb45a18d14fbb12df4d91e2c67b.svg" width="473px" height="40px" loading="lazy" />
 					<p>If we drop all the zero-terms, this gives us:</p>
 					<!--
- 
+
 ╭                                   2                                         2                       3
-╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                                   2                                          2
-╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t                        
+╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/016f37843b2af2ae34dc8b2975505f3d.svg" width="408px" height="40px" loading="lazy" />
 					<p>
@@ -4418,7 +4418,7 @@ x    = x  - ───────
 					</p>
 					<p>What we need to do is solve a simple equation:</p>
 					<!--
- 
+
 C(t) = 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/inflections/ed68dcfb203517ca080fe48914769fb0.svg" width="59px" height="16px" loading="lazy" />
@@ -4428,7 +4428,7 @@ C(t) = 0
 						circle being on the other side of the curve. So what does <em>C(t)</em> look like? Actually something that seems not too hard:
 					</p>
 					<!--
- 
+
 C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)  ·   Bézier \prime\prime(t)
                x                     y                          y                     x
 -->
@@ -4457,15 +4457,15 @@ C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)
 							with the first and second derivatives, given our basis functions:
 						</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t                            
+Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t
              1           2            3           4
-      \prime            2                2                                      
+      \prime            2                2
 Bézier      (t) = a(1-t)  + 2b(1-t)t + ct   { a=3(x -x ),b=3(x -x ),c=3(x -x ) }
-                                                   2  1       3  2       4  3   
+                                                   2  1       3  2       4  3
       \prime\prime
-Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\                        
+Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 -->
 						<img
 							class="LaTeX SVG"
@@ -4476,14 +4476,14 @@ Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 						/>
 						<p>And of course the same functions for <em>y</em>:</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t 
+Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t
              1           2            3           4
-      \prime            2                2           
-Bézier      (t) = d(1-t)  + 2e(1-t)t + ft            
+      \prime            2                2
+Bézier      (t) = d(1-t)  + 2e(1-t)t + ft
       \prime\prime
-Bézier            (t) = w(1-t) + zt                  
+Bézier            (t) = w(1-t) + zt
 -->
 						<img
 							class="LaTeX SVG"
@@ -4497,19 +4497,19 @@ Bézier            (t) = w(1-t) + zt
 							rather overly complicated set of arithmetic expressions:
 						</p>
 						<!--
- 
+
      2             2             2             2             2
--18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y              
+-18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y
         2  1          3  1          4  1          1  2          3  2
      2             2             2             2             2
-+36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y              
++36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y
         4  2          1  3          2  3          4  3          1  4
-     2             2                                                             
--36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y 
+     2             2
+-36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y
         2  4          3  4         2  1          3  1          4  1          1  2
-+54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y 
++54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y
        3  2          4  2          1  3          2  3          1  4          2  4
--18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y   
+-18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y
      2  1          3  1          1  2          3  2          1  3          2  3
 -->
 						<img
@@ -4530,7 +4530,7 @@ Bézier            (t) = w(1-t) + zt
 						<code>t</code> values, we end up with the following simple term function for <em>C(t)</em>:
 					</p>
 					<!--
- 
+
                                  2
 (3 x  y +2 x  y +3 x  y -x  y ) t  + (3 x  y -x  y -3 x  y ) t + (x  y -x  y )
     3  2    4  2    2  3  4  3           3  2  4  2    2  3        2  3  3  2
@@ -4547,7 +4547,7 @@ Bézier            (t) = w(1-t) + zt
 						simplification:
 					</p>
 					<!--
- 
+
 a = x   · y  ╮
      3     2 │
 b = x   · y  │                             2
@@ -4566,7 +4566,7 @@ d = x   · y  │
 					/>
 					<p>This is a plain quadratic curve, and we know how to solve <em>C(t) = 0</em>; we use the quadratic formula:</p>
 					<!--
- 
+
                                                   ┌──────────┐
                                                   │ 2
 x = -3a + 2b + 3c - d ╮                      -y ±⟍│y  - 4 x z
@@ -4646,7 +4646,7 @@ z =       c - a       ╯                             2x
 								This edge is described by the function:
 							</p>
 							<!--
- 
+
       2
     -x  + 2x + 3
 y = ────────────, { x ≤1 }
@@ -4666,7 +4666,7 @@ y = ────────────, { x ≤1 }
 								the function:
 							</p>
 							<!--
- 
+
      ┌──────────┐
      │        2
     ⟍│3(4x - x )  - x
@@ -4684,7 +4684,7 @@ y = ─────────────────, { 0 ≤x ≤1 }
 						<li>
 							<p>...on the top (blue) edge, the curve's start point touches the curve, forming a loop. This edge is described by the function:</p>
 							<!--
- 
+
       2
     -x  + 3x
 y = ────────, { x ≤0 }
@@ -4751,7 +4751,7 @@ y = ────────, { x ≤0 }
 						<em>z</em> coordinate that is always 1, then we can suddenly add arbitrary values. For example:
 					</p>
 					<!--
- 
+
 ┌ 1 0 a ┐    ┌  x  ┐   ┌ 1  · x + 0  · y + a  · z ┐   ┌ x + a  · 1 ┐   ┌ x + a ┐
 │ 0 1 b │  · │  y  │ = │ 0  · x + 1  · y + b  · z │ = │ y + b  · 1 │ = │ y + b │
 └ 0 0 1 ┘    └ z=1 ┘   └ 0  · x + 0  · y + 1  · z ┘   └   1  · z   ┘   └  z=1  ┘
@@ -4762,7 +4762,7 @@ y = ────────, { x ≤0 }
 						want to subtract p1.x and p1.y, we use:
 					</p>
 					<!--
- 
+
      ┌ 1 0 -P   ┐            ┌ 1  · x + 0  · y - P    · 1 ┐   ┌ x - P   ┐
      │       1x │    ┌ x ┐   │                    1x      │   │      1x │
 T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = │ y - P   │
@@ -4778,7 +4778,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						transformation looks like this:
 					</p>
 					<!--
- 
+
 ┌ 1 S 0 ┐    ┌ x ┐   ┌ x + S  · y ┐
 │ 0 1 0 │  · │ y │ = │     y      │
 └ 0 0 1 ┘    └ 1 ┘   └     1      ┘
@@ -4789,7 +4789,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						simply <em>-x/y</em>, because <em>-x/y \</em> y = -x*. Done:
 					</p>
 					<!--
- 
+
      ┌    U     ┐
      │     2x   │
      │ 1 -─── 0 │
@@ -4807,7 +4807,7 @@ T  = │    U     │
 						x-scaling by 1/point3<sub>x</sub>, and y-scaling by point2<sub>y</sub>. This is really easy:
 					</p>
 					<!--
- 
+
      ┌  1        ┐
      │ ───  0  0 │
      │ V         │
@@ -4828,7 +4828,7 @@ T  = │      1    │
 						offset, in this case 1:
 					</p>
 					<!--
- 
+
      ┌    1    0 0 ┐
      │ 1 - W       │
      │      3y     │
@@ -4844,7 +4844,7 @@ T  = │ ─────── 1 0 │
 						be:
 					</p>
 					<!--
- 
+
                 ╭                           (-x +x )(-y +y )                ╮
                 │                              1  2    1  4                 │
                 │                -x  + x  - ────────────────                │
@@ -4879,7 +4879,7 @@ mapped  = (x) = │                                1  2                       
 						that leave?
 					</p>
 					<!--
- 
+
       ╭                 x   · y       x   · y              ╮
       │                  2     4       2     3             │
       │            x  - ──────── / x -────────             │   ╭         x           ╮
@@ -4899,11 +4899,11 @@ mapped  = (x) = │                                1  2                       
 						factors to see just how simple this is:
 					</p>
 					<!--
- 
-      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y 
+
+      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y
       │          43         │                 │  4    2     42 │            4                3
 ... = │                     │,   where  x   = │ ────────────── │,    y   = ──,   and  y   = ──
-      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y 
+      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y
       ╰  42    43       32  ╯                 ╰  3    2     32 ╯            2                2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/canonical/88e3fae7aeef6d7614290587422542c9.svg" width="563px" height="55px" loading="lazy" />
@@ -4981,16 +4981,16 @@ mapped  = (x) = │                                1  2                       
 					</p>
 					<p>First, let's look at the function for x(t):</p>
 					<!--
- 
+
              3          2            2     3
-x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt 
+x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt
 -->
 					<img class="LaTeX SVG" src="./images/chapters/yforx/378d0fd8cefa688d530ac38930d66844.svg" width="335px" height="19px" loading="lazy" />
 					<p>
 						We can rewrite this to a plain polynomial form, by just fully writing out the expansion and then collecting the polynomial factors, as:
 					</p>
 					<!--
- 
+
                          3                  2
 x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 -->
@@ -5001,7 +5001,7 @@ x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 						known constants, we can trivially rewrite this (by moving the <code>x(t)</code> across the equal sign) as:
 					</p>
 					<!--
- 
+
                    3                  2
 (-a + 3b - 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + (a-x) = 0
 -->
@@ -5085,7 +5085,7 @@ y = curve.get(t).y</textarea
 						following seemingly straight forward (if a bit overwhelming) formula:
 					</p>
 					<!--
- 
+
     ┌───────────────┐
 ╭ z │      2       2
 |   │f '(t) +f '(t)   dt
@@ -5094,7 +5094,7 @@ y = curve.get(t).y</textarea
 					<img class="LaTeX SVG" src="./images/chapters/arclength/59ebc3a7c3547a50998d1ea3664fb688.svg" width="140px" height="33px" loading="lazy" />
 					<p>or, more commonly written using Leibnitz notation as:</p>
 					<!--
- 
+
               ┌─────────────────┐
           ╭ z │       2        2
 length =  |  ⟍│(dx/dt) +(dy/dt)   dt
@@ -5133,7 +5133,7 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 |   ⟍│(dx/dt) +(dy/dt)   dt =  |   f(t) dt  ≃ ┌ \underset strip 1 \underbrace C   · f(t )   + ...  + \underset strip n \underbrace C   · f(t )  ┐ =
 ╯ -1                           ╯ -1           └                                1       1                                            n       n   ┘
                                                                                     __ n
-                                           \underset strips 1 through n \underbrace ❯      C   · f(t )   
+                                           \underset strips 1 through n \underbrace ❯      C   · f(t )
                                                                                     ‾‾ i=1  i       i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/arclength/b76753476ad6ecfe4b8f39bcf9432980.svg" width="623px" height="71px" loading="lazy" />
@@ -5197,16 +5197,16 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							shifting and scaling the inputs. Doing so, we get the following:
 						</p>
 						<!--
- 
+
      ┌─────────────────┐
  ╭ z │       2        2
- |  ⟍│(dx/dt) +(dy/dt)   dt                                        
+ |  ⟍│(dx/dt) +(dy/dt)   dt
  ╯ 0
      z    ┌        ╭ z         z ╮                ╭ z         z ╮ ┐
  ≃ \ ─  · │ C   · f│ ─  · t  + ─ │ + ... + C   · f│ ─  · t  + ─ │ │
      2    └  1     ╰ 2     1   2 ╯          n     ╰ 2     n   2 ╯ ┘
     z    __ n         ╭ z         z ╮
-= \ ─  · ❯     C   · f│ ─  · t  + ─ │                              
+= \ ─  · ❯     C   · f│ ─  · t  + ─ │
     2    ‾‾ i=1 i     ╰ 2     i   2 ╯
 -->
 						<img
@@ -5228,10 +5228,10 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							<a href="./legendre-gauss.html">these tables</a> would tell us that for <em>n=2</em> we must use the following values:
 						</p>
 						<!--
- 
-C  = 1     
+
+C  = 1
  1
-C  = 1     
+C  = 1
  2
         1
 t  = - ────
@@ -5247,7 +5247,7 @@ t  = + ────
 							Which means that in order for us to approximate the integral, we must plug these values into the approximate function, which gives us:
 						</p>
 						<!--
- 
+
     ┌─────────────────┐
 ╭ z │       2        2       z    ┌  ╭ z     -1    z ╮    ╭ z     1     z ╮ ┐
 |  ⟍│(dx/dt) +(dy/dt)   dt ≃ ─  · │ f│ ─  · ──── + ─ │ + f│ ─  · ──── + ─ │ │
@@ -5396,13 +5396,13 @@ t  = + ────
 					</p>
 					<p>So, what does the function look like? This:</p>
 					<!--
- 
+
          x'y'' - x''y'
 \kappa = ─────────────
                    3
                    ─
              2   2 2
-          (x' +y' ) 
+          (x' +y' )
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/060acd6ff0a050fe4d98a7802a2b3a3f.svg" width="113px" height="47px" loading="lazy" />
 					<p>
@@ -5410,14 +5410,14 @@ t  = + ────
 						tiny bit:
 					</p>
 					<!--
- 
+
             B '(t)B ''(t) - B ''(t)B '(t)
              x     y         x      y
 \kappa(t) = ─────────────────────────────
                                   3
                                   ─
                         2       2 2
-                 (B '(t) +B '(t) ) 
+                 (B '(t) +B '(t) )
                    x       y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/afd8cb8b0fe291ff703752c1c9cc33d4.svg" width="239px" height="55px" loading="lazy" />
@@ -5494,7 +5494,7 @@ function kappa(t, B):
 						circle that "fits" the curve's curvature at any point, using what is possibly the simplest bit of maths found in this entire primer:
 					</p>
 					<!--
- 
+
            1
 R(t) = ─────────
        \kappa(t)
@@ -5898,8 +5898,8 @@ lli = function(line1, line2):
 						points, so logically <code>C</code> will have a function that interpolates between those two coordinates:
 					</p>
 					<!--
- 
-C = u(t)  · P       + (1-u(t))  · P    
+
+C = u(t)  · P       + (1-u(t))  · P
               start                 end
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8c6662f605722fb2ff6cd7f65243a126.svg" width="229px" height="16px" loading="lazy" />
@@ -5912,22 +5912,22 @@ C = u(t)  · P       + (1-u(t))  · P
 						(with thanks to Boris Zbarsky) shows us the following two formulae:
 					</p>
 					<!--
- 
+
                         2
-                   (1-t) 
+                   (1-t)
 u(t)           = ───────────
      quadratic    2        2
-                 t  + (1-t) 
+                 t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8e7cfee39c98f2ddf9b635a914066cf6.svg" width="183px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                     3
-               (1-t) 
+               (1-t)
 u(t)       = ───────────
      cubic    3        3
-             t  + (1-t) 
+             t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/a0b99054cc82ca1fb147f077e175ef10.svg" width="161px" height="41px" loading="lazy" />
 					<p>
@@ -5938,7 +5938,7 @@ u(t)       = ───────────
 					</p>
 					<p>We start by observing that, given <code>A</code>, <code>B</code>, and <code>C</code>, the following always holds:</p>
 					<!--
- 
+
             distance(B,C)
 ratio(t) = ────────────── =  Constant
             distance(A,B)
@@ -5946,22 +5946,22 @@ ratio(t) = ────────────── =  Constant
 					<img class="LaTeX SVG" src="./images/chapters/abc/131454dcbac04e567f322979f4af80c6.svg" width="245px" height="39px" loading="lazy" />
 					<p>Working out the maths for this, we see the following two formulae for quadratic and cubic curves:</p>
 					<!--
- 
+
                        2        2
                       t  + (1-t)  - 1
 ratio(t)           = |───────────────|
          quadratic       2        2
-                        t  + (1-t) 
+                        t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/5924e162b50272c40c842fad14b8fa48.svg" width="239px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                    3        3
                   t  + (1-t)  - 1
 ratio(t)       = |───────────────|
          cubic       3        3
-                    t  + (1-t) 
+                    t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8cd992c1ceaae2e67695285beef23a24.svg" width="219px" height="41px" loading="lazy" />
 					<p>
@@ -5971,7 +5971,7 @@ ratio(t)       = |───────────────|
 						<code>ratio(t)</code> function to find <code>A</code>:
 					</p>
 					<!--
- 
+
           C - B           B - C
 A = B - ───────── = B + ─────────
          ratio(t)        ratio(t)
@@ -5986,11 +5986,11 @@ A = B - ───────── = B + ─────────
 						pair, after which we can reverse engineer <code>v1</code> and <code>v2</code>:
 					</p>
 					<!--
- 
+
                                  ╭     e  - t  · A
                                  │      1
-╭ e = (1-t)  · v  + t  · A       │ v = ───────────    
-╡  1            1            ==> ╡  1      1-t         
+╭ e = (1-t)  · v  + t  · A       │ v = ───────────
+╡  1            1            ==> ╡  1      1-t
 │ e = (1-t)  · A + t  · v        │     e  - (1-t)  · A
 ╰  2                     2       │      2
                                  │ v = ───────────────
@@ -5999,14 +5999,14 @@ A = B - ───────── = B + ─────────
 					<img class="LaTeX SVG" src="./images/chapters/abc/3166afa345aec1abda432c39b68d39a0.svg" width="339px" height="73px" loading="lazy" />
 					<p>And then reverse engineer the curve's control points:</p>
 					<!--
- 
+
                                       ╭     v  - (1-t)  ·  start
                                       │      1
 ╭ v = (1-t)  ·  start + t  · C        │ C = ────────────────────
-╡  1                          1   ==> ╡  1           t           
+╡  1                          1   ==> ╡  1           t
 │ v = (1-t)  · C  + t  ·  end         │     v  - t  ·  end
 ╰  2            2                     │      2
-                                      │ C = ──────────────      
+                                      │ C = ──────────────
                                       ╰  2       1-t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8bd3e6fed5bf8d871d30221ae400fd93.svg" width="383px" height="75px" loading="lazy" />
@@ -6034,15 +6034,15 @@ A = B - ───────── = B + ─────────
 						<code>B</code> point, and <code>B</code> and end point as a ratio, using
 					</p>
 					<!--
- 
+
 ╭ d = || Start - B||
 │  1
-│ d = || End - B||  
+│ d = || End - B||
 │  2
-╡      d             
+╡      d
 │       1
-│  t= ─────         
-│     d +d 
+│  t= ─────
+│     d +d
 ╰      1  2
 -->
 					<img
@@ -6090,9 +6090,9 @@ A = B - ───────── = B + ─────────
 						<code>e2</code> coordinates must obey the standard de Casteljau rule for linear interpolation:
 					</p>
 					<!--
- 
-╭ e = B + t  · d    
-╡  1                 
+
+╭ e = B + t  · d
+╡  1
 │ e = B - (1-t)  · d
 ╰  2
 -->
@@ -6112,7 +6112,7 @@ A = B - ───────── = B + ─────────
 						curve with a loop in it. To do this, we can use the <a href="https://en.wikipedia.org/wiki/Atan2">atan2</a> function:
 					</p>
 					<!--
- 
+
 \phi = (atan2(E -S , E -S ) - atan2(B -S , B -S ) + 2 \pi)   mod  2 \pi
                y  y   x  x           y  y   x  x
 -->
@@ -6129,8 +6129,8 @@ A = B - ───────── = B + ─────────
 						<code>d</code>:
 					</p>
 					<!--
- 
-d = {  d  if  0 ≤\phi ≤\pi             
+
+d = {  d  if  0 ≤\phi ≤\pi
       -d  if  \phi < 0 \lor \phi > \pi
 -->
 					<img
@@ -6290,7 +6290,7 @@ for (coordinate, index) in LUT:
 						maths, that means we're trying to solve:
 					</p>
 					<!--
- 
+
 dist(B(t), c) = r
 -->
 					<img
@@ -6302,15 +6302,15 @@ dist(B(t), c) = r
 					/>
 					<p>Which seems simple enough. Unfortunately, when we expand that <code>dist</code> function, things get a lot more problematic:</p>
 					<!--
- 
-r=  dist(B(t), c)                                                                                                              
+
+r=  dist(B(t), c)
     ┌─────────────────────────┐
     │          2             2
- =  │(B t - c )  + (B t - c )                                                                                                  
-   ⟍│  x     x       y     y                                                                                                   
+ =  │(B t - c )  + (B t - c )
+   ⟍│  x     x       y     y
     ┌─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┐
     │╭         3             2                 2       3      ╮2   ╭         3             2                 2       3      ╮2
- =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │  
+ =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │
    ⟍│╰  1             2               3             4       x ╯    ╰  1             2               3             4       y ╯
 -->
 					<img
@@ -6573,17 +6573,17 @@ findClosest(start, p, r, LUT):
 						<code>B</code>, we can compute the associated <code>C</code>:
 					</p>
 					<!--
- 
+
 C = u(t)   ·  Start + (1-u(t) )  ·  End
         q                    q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/48887d68a861a0acdf8313e23fb19880.svg" width="247px" height="24px" loading="lazy" />
 					<p>And then the associated <code>A</code>:</p>
 					<!--
- 
+
           C - B            B - C
 A = B - ────────── = B + ──────────
-         ratio(t)         ratio(t) 
+         ratio(t)         ratio(t)
                  q                q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/2e65bc9c934380c2de6a24bcd5c1c7b7.svg" width="228px" height="41px" loading="lazy" />
@@ -6708,12 +6708,12 @@ A = B - ────────── = B + ──────────
 							form, where each line corresponds to a power of t, and each column is for a specific coefficient. First, we expand the function:
 						</p>
 						<!--
- 
+
                     2                    2
-B          = a (1-t)  + 2 b (1-t) t + c t    
-  quadratic                                  
+B          = a (1-t)  + 2 b (1-t) t + c t
+  quadratic
                          2            2     2
-           = a - 2at + at  + 2bt - 2bt  + ct 
+           = a - 2at + at  + 2bt - 2bt  + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -6724,12 +6724,12 @@ B          = a (1-t)  + 2 b (1-t) t + c t
 						/>
 						<p>And then we (trivially) rearrange the terms across multiple lines:</p>
 						<!--
- 
-B          =a               
+
+B          =a
   quadratic
-            - 2at+ 2bt      
+            - 2at+ 2bt
                 2     2    2
-            + at - 2bt + ct 
+            + at - 2bt + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -6745,7 +6745,7 @@ B          =a
 						</p>
 						<p>With that arrangement, we can easily decompose this as a matrix multiplication:</p>
 						<!--
- 
+
                                          ┌  a  +  0  + 0 ┐                 ┌  1  0 0 ┐    ┌ a ┐
 B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌      2 ┐  · │ -2  2 0 │  · │ b │
   quadratic                └ 1 t t  ┘    └  a  + -2b + c ┘   └ 1 t t  ┘    └  1 -2 1 ┘    └ c ┘
@@ -6759,9 +6759,9 @@ B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌
 						/>
 						<p>We can do the same for the cubic curve, of course. We know the base function for cubics:</p>
 						<!--
- 
+
               3          2             2     3
-B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt 
+B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
   cubic
 -->
 						<img
@@ -6773,14 +6773,14 @@ B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
 						/>
 						<p>So we write out the expansion and rearrange:</p>
 						<!--
- 
-B      =a                     
+
+B      =a
   cubic
-        - 3at + 3bt           
-             2     2    2     
-        + 3at - 6bt +3ct      
+        - 3at + 3bt
+             2     2    2
+        + 3at - 6bt +3ct
             3      3    3    3
-        - at  + 3bt -3ct + dt 
+        - at  + 3bt -3ct + dt
 -->
 						<img
 							class="LaTeX SVG"
@@ -6791,7 +6791,7 @@ B      =a
 						/>
 						<p>Which we can then decompose:</p>
 						<!--
- 
+
                                         ┌  1  0  0 0 ┐    ┌ a ┐
 B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
   cubic                └ 1 t t  t  ┘    │  3 -6  3 0 │    │ c │
@@ -6806,7 +6806,7 @@ B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
 						/>
 						<p>And, of course, we can do this for quartic curves too (skipping the expansion step):</p>
 						<!--
- 
+
                                              ┌  1  0   0   0 0 ┐    ┌ a ┐
                                              │ -4  4   0   0 0 │    │ b │
 B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  · │ c │
@@ -6832,7 +6832,7 @@ B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  ·
 						we want to do curve fitting:
 					</p>
 					<!--
- 
+
     ┌ p   ┐
     │  1  │
     │ p   │
@@ -6870,9 +6870,9 @@ P = │  2  │
 					</p>
 					<p>To get these values, we first compute the general "distance along the polygon" matrix:</p>
 					<!--
- 
+
                                ╭            d  = 0
-D = ┌ d  d  ... d  ┐,   where  ╡             1                 
+D = ┌ d  d  ... d  ┐,   where  ╡             1
     └  1  2      n ┘           │ d  = d    +  length(p   , p )
                                ╰  i    i-1            i-1   i
 -->
@@ -6889,10 +6889,10 @@ D = ┌ d  d  ... d  ┐,   where  ╡             1
 						<code>i=n</code> fall in the [0,1] interval, so we need to scale all values down by whatever the total length of the polygon is:
 					</p>
 					<!--
- 
+
                                ╭    s  = 0
                                │     1
-S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d  
+S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
     └  1  2      n ┘           │  i    i    n
                                │    s  = 1
                                ╰     n
@@ -6914,9 +6914,9 @@ S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
 						to for the associated <code>t</code> value", which we'll square to get rid of any pesky negative signs:
 					</p>
 					<!--
- 
+
                            2
-E(C)  = (p  -   Bézier(s )) 
+E(C)  = (p  -   Bézier(s ))
     i     i             i
 -->
 					<img
@@ -6931,9 +6931,9 @@ E(C)  = (p  -   Bézier(s ))
 						function. So, we literally just do that; the total error function is simply the sum of all these individual errors:
 					</p>
 					<!--
- 
+
        __ n                      2
-E(C) = ❯      (p  -   Bézier(s )) 
+E(C) = ❯      (p  -   Bézier(s ))
        ‾‾ i=1   i             i
 -->
 					<img
@@ -6950,9 +6950,9 @@ E(C) = ❯      (p  -   Bézier(s ))
 						<strong>T x M x C</strong> we talked about before, which gives us:
 					</p>
 					<!--
- 
+
                 2
-E(C) = (P - TMC) 
+E(C) = (P - TMC)
 -->
 					<img
 						class="LaTeX SVG"
@@ -6963,7 +6963,7 @@ E(C) = (P - TMC)
 					/>
 					<p>In which we can replace the rather cumbersome "squaring" operation with a more conventional matrix equivalent:</p>
 					<!--
- 
+
                 T
 E(C) = (P - TMC)  (P - TMC)
 -->
@@ -6985,7 +6985,7 @@ E(C) = (P - TMC)  (P - TMC)
 						those, and then use that 𝕋 instead of <strong>T</strong> in our error function:
 					</p>
 					<!--
- 
+
     ┌    0    1      n-2   n-1  ┐
     │   s    s  ... s     s     │
     │    1    1      1     1    │
@@ -7005,7 +7005,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Which, because of the first and last values in <strong>S</strong>, means:</p>
 					<!--
- 
+
     ┌    1     0  ...   0    0    ┐
     │                  n-2   n-1  │
     │    1    s       s     s     │
@@ -7025,7 +7025,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Now we can properly write out the error function as matrix operations:</p>
 					<!--
- 
+
                 T
 E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 -->
@@ -7042,7 +7042,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						taking the derivative, and determining where that derivative is zero:
 					</p>
 					<!--
- 
+
 ∂E          T
 ── = 0 = -2𝕋  (P - 𝕋MC)
 ∂C
@@ -7081,7 +7081,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						for <strong>C</strong>:
 					</p>
 					<!--
- 
+
      -1   T   -1  T
 C = M   (𝕋  𝕋)   𝕋  P
 -->
@@ -7164,13 +7164,13 @@ C = M   (𝕋  𝕋)   𝕋  P
 						previous and next point:
 					</p>
 					<!--
- 
+
                ┌  V  = P       ┐
                │   1    2      │
 ┌ P  ┐         │  V  = P       │
 │  1 │         │   2    3      │
 │ P  │         │       P  - P  │
-│  2 │       = │        3    1 │              
+│  2 │       = │        3    1 │
 │ P  │points   │ V'  = ─────── │ point-tangent
 │  3 │         │   1      2    │
 │ P  │         │       P  - P  │
@@ -7303,7 +7303,7 @@ for p = 1 to points.length-3 (inclusive):
 						so how different is the matrix form for Catmull-Rom curves?:
 					</p>
 					<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7340,7 +7340,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							we'll get to that:
 						</p>
 						<!--
- 
+
                         ┌   P     ┐
                         │    2    │
 ┌ V   ┐        ┌ P  ┐   │   P     │
@@ -7368,7 +7368,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						</p>
 						<p>Computing <em>T</em> is really more "arranging the numbers":</p>
 						<!--
- 
+
               ┌   P     ┐
               │    2    │
      ┌ P  ┐   │   P     │   ┌  0  · P1 + 1  · P2 + 0  · P3 + 0  · P4 ┐   ┌  0  1 0 0 ┐    ┌ P  ┐
@@ -7391,7 +7391,7 @@ T  · │  2 │ = │  3    1 │ = │ ──  · P1 + 0  · P2 + ─  · P3 +
 						/>
 						<p>Thus:</p>
 						<!--
- 
+
     ┌  0  1 0 0 ┐
     │  0  0 1 0 │
     │ -1    1   │
@@ -7415,7 +7415,7 @@ T = │ ──  0 ─ 0 │
 							factor into both our coordinate vector representation, and mapping matrix <em>T</em>:
 						</p>
 						<!--
- 
+
           ┌   P     ┐
           │    2    │
 ┌ V   ┐   │   P     │        ┌  0  1  0 0  ┐
@@ -7441,7 +7441,7 @@ T = │ ──  0 ─ 0 │
 							coordinates, and see what we end up with:
 						</p>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7460,7 +7460,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						/>
 						<p>Replace point/tangent vector with the expression for all-coordinates:</p>
 						<!--
- 
+
                                                     ┌  0  1  0 0  ┐    ┌ P  ┐
                                                     │  0  0  1 0  │    │  1 │
                                  ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -7479,7 +7479,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │ ──  0 ─
 						/>
 						<p>and merge the matrices:</p>
 						<!--
- 
+
                                  ┌  0    1      0   0  ┐
                                  │ -1          1       │    ┌ P  ┐
                                  │ ──    0     ──   0  │    │  1 │
@@ -7500,7 +7500,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  1 1           1 -1 │  · │  2 │
 						/>
 						<p>This looks a lot like the Bézier matrix form, which as we saw in the chapter on Bézier curves, should look like this:</p>
 						<!--
- 
+
                                                ┌ P  ┐
                                                │  1 │
                              ┌  1  0  0 0 ┐    │ P  │
@@ -7519,7 +7519,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>So, if we want to express a Catmull-Rom curve using a Bézier curve, we'll need to turn this Catmull-Rom bit:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │
@@ -7540,7 +7540,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Into something that looks like this:</p>
 						<!--
- 
+
                   ┌ P  ┐
                   │  1 │
 ┌  1  0  0 0 ┐    │ P  │
@@ -7559,7 +7559,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And the way we do that is with a fairly straight forward bit of matrix rewriting. We start with the equality we need to ensure:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐                          ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │                          │  1 │
@@ -7580,7 +7580,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we remove the coordinate vector from both sides without affecting the equality:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │
 │ ──    0     ──   0  │
@@ -7601,7 +7601,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we can "get rid of" the Bézier matrix on the right by left-multiply both with the inverse of the Bézier matrix:</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -7625,7 +7625,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							outright remove any matrix/inverse pair:
 						</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -7646,7 +7646,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And now we're <em>basically</em> done. We just multiply those two matrices and we know what <em>V</em> is:</p>
 						<!--
- 
+
 ┌  0  1  0 0  ┐
 │ -1    1     │
 │ ──  1 ── 0  │
@@ -7668,7 +7668,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							<li>Start with the Catmull-Rom function:</li>
 						</ol>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7689,7 +7689,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite to pure coordinate form:</li>
 						</ol>
 						<!--
- 
+
                                       ┌   P     ┐
                                       │    2    │
                                       │   P     │
@@ -7714,7 +7714,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for "normal" coordinate vector:</li>
 						</ol>
 						<!--
- 
+
                                       ┌  0  1  0 0  ┐    ┌ P  ┐
                                       │  0  0  1 0  │    │  1 │
                    ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -7735,7 +7735,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>merge the inner matrices:</li>
 						</ol>
 						<!--
- 
+
                    ┌  0    1      0   0  ┐
                    │ -1          1       │    ┌ P  ┐
                    │ ──    0     ──   0  │    │  1 │
@@ -7758,7 +7758,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for Bézier matrix form:</li>
 						</ol>
 						<!--
- 
+
                                      ┌  0  1  0 0  ┐    ┌ P  ┐
                                      │ -1    1     │    │  1 │
                    ┌  1  0  0 0 ┐    │ ──  1 ── 0  │    │ P  │
@@ -7779,7 +7779,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>and transform the coordinates so we have a "pure" Bézier expression:</li>
 						</ol>
 						<!--
- 
+
                                      ┌     P       ┐
                                      │      2      │
                                      │      P -P   │
@@ -7808,13 +7808,13 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						curve that has the vector:
 					</p>
 					<!--
- 
+
                        ┌     P       ┐
                        │      2      │
 ┌ P  ┐                 │      P -P   │
 │  1 │                 │       3  1  │
 │ P  │                 │ P  + ────── │
-│  2 │             ==> │  2   6  · τ │        
+│  2 │             ==> │  2   6  · τ │
 │ P  │ CatmullRom      │      P -P   │  Bézier
 │  3 │                 │       4  2  │
 │ P  │                 │ P  - ────── │
@@ -7834,11 +7834,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						tension Catmull-Rom curve with the following coordinate values:
 					</p>
 					<!--
- 
+
 ┌ P  ┐              ┌       P         ┐
 │  1 │              │        1        │
 │ P  │              │       P         │
-│  2 │          ==> │        4        │           
+│  2 │          ==> │        4        │
 │ P  │  Bézier      │ P  + 3(P  - P ) │ CatmullRom
 │  3 │              │  4      1    2  │
 │ P  │              │ P  + 3(P  - P ) │
@@ -7853,11 +7853,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 					/>
 					<p>Or, if your API allows you to specify Catmull-Rom curves using plain coordinates:</p>
 					<!--
- 
+
 ┌ P  ┐              ┌ P  + 6(P  - P ) ┐
 │  1 │              │  4      1    2  │
 │ P  │              │       P         │
-│  2 │          ==> │        1        │           
+│  2 │          ==> │        1        │
 │ P  │  Bézier      │       P         │ CatmullRom
 │  3 │              │        4        │
 │ P  │              │ P  + 6(P  - P ) │
@@ -7940,8 +7940,8 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						the same as it's "outgoing" derivative:
 					</p>
 					<!--
- 
-B'(1)  = B'(0)   
+
+B'(1)  = B'(0)
      n        n+1
 -->
 					<img class="LaTeX SVG" src="./images/chapters/polybezier/a37252ff55837b918d9d64078ae92ae7.svg" width="124px" height="17px" loading="lazy" />
@@ -7951,7 +7951,7 @@ B'(1)  = B'(0)
 						last control point through the last on-curve point. And mirroring any point A through any point B is really simple:
 					</p>
 					<!--
- 
+
            ┌ B  + (B  - A ) ┐   ┌ 2B  - A  ┐
 Mirrored = │  x     x    x  │ = │   x    x │
            │ B  + (B  - A ) │   │ 2B  - A  │
@@ -8113,7 +8113,7 @@ Mirrored = │  x     x    x  │ = │   x    x │
 							any point on <code>O(t)</code> is a fixed distance <code>d</code> away from coordinate <code>B(t)</code>. So let's math that:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d
 -->
 						<img
@@ -8130,7 +8130,7 @@ O(t) = B(t) + d
 							at <code>B(t)</code>. Easy enough:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d  · N(t)
 -->
 						<img
@@ -8148,7 +8148,7 @@ O(t) = B(t) + d  · N(t)
 							<code>N(t)</code> always has length 1, which we can achieve by dividing <code>B'(t)</code> by its magnitude:
 						</p>
 						<!--
- 
+
           ╭   B'(t)    ╮
 N(t) \bot │ ────────── │
           ╰ || B'(t)|| ╯
@@ -8166,10 +8166,10 @@ N(t) \bot │ ────────── │
 							denoted with double vertical bars:
 						</p>
 						<!--
- 
+
                     ┌───────────┐
                ╭ b  │   2      2
-|| f(x,y)|| =  |    │f '  + f '  
+|| f(x,y)|| =  |    │f '  + f '
                ╯ a ⟍│ x      y
 -->
 						<img
@@ -8183,10 +8183,10 @@ N(t) \bot │ ────────── │
 							So if we want the length of the tangent, we plug in <code>B'(t)</code>, with <code>t = 0</code> as start and <code>t = 1</code> as end:
 						</p>
 						<!--
- 
+
                    ┌───────────────────┐
               ╭ 1  │       2          2
-|| B'(t)|| =  |    │B ''(t)  + B ''(t)  
+|| B'(t)|| =  |    │B ''(t)  + B ''(t)
               ╯ 0 ⟍│ x          y
 -->
 						<img
@@ -8383,7 +8383,7 @@ N(t) \bot │ ────────── │
 						some angle <em>φ</em>:
 					</p>
 					<!--
- 
+
              S = \begin{pmatrix} 1
 0 \end{pmatrix}  ,  \ E = \begin{pmatrix} cos(φ)
               sin(φ) \end{pmatrix}
@@ -8391,7 +8391,7 @@ N(t) \bot │ ────────── │
 					<img class="LaTeX SVG" src="./images/chapters/circles/7ab3da0922477af4cc09f5852100976b.svg" width="175px" height="40px" loading="lazy" />
 					<p>What we want to find is the intersection of the tangents, so we want a point C such that:</p>
 					<!--
- 
+
               C = S + a  · \begin{pmatrix} 0
 1 \end{pmatrix}  ,  \ C = E + b  · \begin{pmatrix} -sin(φ)
                    cos(φ) \end{pmatrix}
@@ -8402,22 +8402,22 @@ N(t) \bot │ ────────── │
 						through E (at some distance <em>b</em> from E). Solving this gives us:
 					</p>
 					<!--
- 
+
 ╭ C  = 1 = cos(φ) + b  · -sin(φ)
-╡  x                             
-│ C  = a = sin(φ) + b  · cos(φ) 
+╡  x
+│ C  = a = sin(φ) + b  · cos(φ)
 ╰  y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/8237af1396bb567d70c8b5e4dd7f8115.svg" width="219px" height="40px" loading="lazy" />
 					<p>First we solve for <em>b</em>:</p>
 					<!--
- 
+
 1 = cos(φ) + b  · -sin(φ)  → \ 1 - cos(φ) = -b  · sin(φ)  → \ -1 + cos(φ) = b  · sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/1829e42ea956ee4df0e45d9ac5334ef7.svg" width="560px" height="17px" loading="lazy" />
 					<p>which yields:</p>
 					<!--
- 
+
     cos(φ)-1
 b = ────────
      sin(φ)
@@ -8425,21 +8425,21 @@ b = ────────
 					<img class="LaTeX SVG" src="./images/chapters/circles/06369b00338310df0a810c592485aa0a.svg" width="101px" height="39px" loading="lazy" />
 					<p>which we can then substitute in the expression for <em>a</em>:</p>
 					<!--
- 
- a= sin(φ) + b  · cos(φ)          
+
+ a= sin(φ) + b  · cos(φ)
              -1 + cos(φ)
 ..= sin(φ) + ───────────  · cos(φ)
                sin(φ)
                           2
              -cos(φ) + cos (φ)
-..= sin(φ) + ─────────────────    
-                  sin(φ)          
+..= sin(φ) + ─────────────────
+                  sin(φ)
        2         2
     sin (φ) + cos (φ) - cos(φ)
-..= ──────────────────────────    
+..= ──────────────────────────
               sin(φ)
     1 - cos(φ)
- a= ──────────                    
+ a= ──────────
       sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/5a23f3bc298c85540c6dd18e304d9224.svg" width="231px" height="195px" loading="lazy" />
@@ -8450,7 +8450,7 @@ b = ────────
 						polynomial, and we know where the circle arc's actual point P is for angle φ/2:
 					</p>
 					<!--
- 
+
          φ                 φ
 P  = cos(─)  ,  \ P  = sin(─)
  x       2         y       2
@@ -8458,7 +8458,7 @@ P  = cos(─)  ,  \ P  = sin(─)
 					<img class="LaTeX SVG" src="./images/chapters/circles/8b4e1d0a62380ed011f27c645ed13b28.svg" width="188px" height="32px" loading="lazy" />
 					<p>We compute T, observing that if <em>t=0.5</em>, the polynomial values (1-t)², 2(1-t)t, and t² are 0.25, 0.5, and 0.25 respectively:</p>
 					<!--
- 
+
     1    2    1    1
 T = ─S + ─C + ─E = ─(S + 2C + E)
     4    4    4    4
@@ -8466,10 +8466,10 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/a04bd1558a76e60b8ca6e1fe4fa38c00.svg" width="252px" height="35px" loading="lazy" />
 					<p>Which, worked out for the x and y components, gives:</p>
 					<!--
- 
+
 ╭     1
-│ T = ─(3 + cos(φ))                                    
-╡  x  4                                                 
+│ T = ─(3 + cos(φ))
+╡  x  4
 │     1╭ 2-2cos(φ)          ╮   1╭     ╭ φ ╮          ╮
 │ T = ─│ ───────── + sin(φ) │ = ─│ 2tan│ ─ │ + sin(φ) │
 ╰  y  4╰  sin(φ)            ╯   4╰     ╰ 2 ╯          ╯
@@ -8477,17 +8477,17 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/986ae9104e0bc52e95689ae7ae4504db.svg" width="408px" height="77px" loading="lazy" />
 					<p>And the distance between these two is the standard Euclidean distance:</p>
 					<!--
- 
+
                  1                   φ        4╭ φ ╮
 d (φ)= T  - P  = ─(3 + cos(φ)) - cos(─) = 2sin │ ─ │  ,
  x      x    x   4                   2         ╰ 4 ╯
                  1╭     ╭ φ ╮          ╮       φ
-d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,   
+d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
  y      y    y   4╰     ╰ 2 ╯          ╯       2
-     ⇓                                                 
+     ⇓
         ┌───────┐                     ┌───────┐
         │ 2    2                 4 φ  │   1
- d(φ)=  │d  + d   =  ...   = 2sin (─) │───────         
+ d(φ)=  │d  + d   =  ...   = 2sin (─) │───────
        ⟍│ x    y                   4  │   2 φ
                                       │cos (─)
                                      ⟍│     2
@@ -8534,7 +8534,7 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 						precision:
 					</p>
 					<!--
- 
+
                 ╭  ┌─────────────┐ ╮
                 │  │     ┌──────┐  │
                 │ ⟍│2+ε-⟍│ε(2+ε)   │
@@ -8585,14 +8585,14 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 					</p>
 					<p>So let's again formally describe this:</p>
 					<!--
- 
-P = (1, 0)                     
+
+P = (1, 0)
  1
-P = (1, k)                     
- 2                             
+P = (1, k)
+ 2
 P = P  + k  · (sin(θ), -cos(θ))
  3   4
-P = (cos(θ), sin(θ))           
+P = (cos(θ), sin(θ))
  4
 -->
 					<img
@@ -8616,29 +8616,29 @@ P = (cos(θ), sin(θ))
 						P<sub>2</sub>" (which is "half height" P<sub>2</sub>), and so forth:
 					</p>
 					<!--
- 
-     P   + P  
-      2     3 
+
+     P   + P
+      2     3
        y     y   k + sin(θ) - k  · cos(θ)
- A = ───────── = ────────────────────────                                 
+ A = ───────── = ────────────────────────
   y      2                  2
           1
-     A  + ─P     k + sin(θ) - k  · cos(θ)   
+     A  + ─P     k + sin(θ) - k  · cos(θ)
       y   2 2    ──────────────────────── + ─
              y              2               2   2k + sin(θ) + k  · cos(θ)
-e  = ───────── = ──────────────────────────── = ───────────────────────── 
+e  = ───────── = ──────────────────────────── = ─────────────────────────
  1       2                    2                             4
-  y                                                                       
+  y
                                        k
      A  +  mid(P , P )   A  + sin(θ) - ─ cos(θ)
       y         4   3     y            2          k + 3sin(θ) 2k  · cos(θ)
 e  = ───────────────── = ────────────────────── = ────────────────────────
  2           2                     2                         4
   y
-     e   + e  
-      1     2 
+     e   + e
+      1     2
        y     y   3k + 4sin(θ) - 3k  · cos(θ)
- B = ───────── = ───────────────────────────                              
+ B = ───────── = ───────────────────────────
   y      2                    8
 -->
 					<img
@@ -8662,15 +8662,15 @@ e  = ───────────────── = ───────
 							<a href="https://www.wolframalpha.com/">Wolfram Alpha</a> is definitely the way to go. That said, let's get going:
 						</p>
 						<!--
- 
+
 3k + 4sin(θ)) - 3k  · cos(θ)      θ
-────────────────────────────= sin(─)                  
+────────────────────────────= sin(─)
              8                    2
                                   ╭ θ ╮
-3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │               
+3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │
                                   ╰ 2 ╯
                                   ╭ θ ╮
-           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)     
+           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)
                                   ╰ 2 ╯
                                 ╭     ╭ θ ╮          ╮
              3k (1 - cos(θ))= 4 │ 2sin│ ─ │ - sin(θ) │
@@ -8678,12 +8678,12 @@ e  = ───────────────── = ───────
                                         θ
                                    2sin(─) - sin(θ)
                                         2
-                          3k= 4  · ────────────────   
+                          3k= 4  · ────────────────
                                       1 - cos(θ)
                                        ╭ θ ╮
                                    2sin│ ─ │ - sin(θ)
                               4        ╰ 2 ╯
-                           k= ─  · ────────────────── 
+                           k= ─  · ──────────────────
                               3        1 - cos(θ)
 -->
 						<img
@@ -8698,11 +8698,11 @@ e  = ───────────────── = ───────
 							<em>k</em>:
 						</p>
 						<!--
- 
+
             ╭ θ ╮
         2sin│ ─ │ - sin(θ)
    4        ╰ 2 ╯
-k= ─  · ──────────────────         
+k= ─  · ──────────────────
    3        1 - cos(θ)
         ╭     ╭ θ ╮               ╮
         │ 2sin│ ─ │               │
@@ -8710,10 +8710,10 @@ k= ─  · ──────────────────
 k= ─  · │ ────────── - ────────── │
    3    ╰ 1 - cos(θ)   1 - cos(θ) ╯
    4    ╭    ╭ θ ╮      ╭ θ ╮ ╮
-k= ─  · │ csc│ ─ │ - cot│ ─ │ │    
+k= ─  · │ csc│ ─ │ - cot│ ─ │ │
    3    ╰    ╰ 2 ╯      ╰ 2 ╯ ╯
    4       ╭ θ ╮
-k= ─  · tan│ ─ │                   
+k= ─  · tan│ ─ │
    3       ╰ 4 ╯
 -->
 						<img
@@ -8731,7 +8731,7 @@ k= ─  · tan│ ─ │
 						involving the circular arc's angle:
 					</p>
 					<!--
- 
+
            4    ╭ θ ╮
 k = f(θ) = ─ tan│ ─ │
            3    ╰ 4 ╯
@@ -8747,13 +8747,13 @@ k = f(θ) = ─ tan│ ─ │
 						Which means that for any circular arc with angle θ and radius <em>r</em>, our Bézier approximation based on three points of incidence is:
 					</p>
 					<!--
- 
-    start= (r, 0)                                          
-control  = (r, k)                                          
-        1                                                  
+
+    start= (r, 0)
+control  = (r, k)
+        1
 control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
         2
-      end= r  · (cos(θ), sin(θ))                           
+      end= r  · (cos(θ), sin(θ))
 -->
 					<img
 						class="LaTeX SVG"
@@ -8764,7 +8764,7 @@ control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
 					/>
 					<p>Which also gives us the commonly found value of 0.55228 for quarter circles, based on them having an angle of half π:</p>
 					<!--
- 
+
  ╭ \pi ╮   4       ╭ \pi ╮   4  ┌─┐
 f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.55228474983[...]
  ╰  2  ╯   3       ╰  8  ╯   3
@@ -8778,13 +8778,13 @@ f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.552284
 					/>
 					<p>And thus giving us the following Bézier coordinates for a quarter circle of radius <em>r</em>:</p>
 					<!--
- 
-    start= (r, 0)           
+
+    start= (r, 0)
 control  = (r, 0.55228  · r)
-        1                   
+        1
 control  = (0.55228  · r, r)
         2
-      end= (0, r)           
+      end= (0, r)
 -->
 					<img
 						class="LaTeX SVG"
@@ -8928,7 +8928,7 @@ getMostWrongT(radius, bezier, start, end, epsilon=1e-15):
 					</p>
 					<p>First, let's express the total error (given circular arc angle θ, and some <em>k</em>) using standard calculus notation:</p>
 					<!--
- 
+
                     ┌───────────────────────┐
               ╭ 1   │         2            2
 erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
@@ -8951,7 +8951,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						expression looks like:
 					</p>
 					<!--
- 
+
 ╭        ┌───────┐         ╮
 │  ╭ 1   │ 2    2          │ d
 │  |  \| │B  + B   - r\|dt │ ── = 0
@@ -8970,7 +8970,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						are each other's inverse operations, so they cancel out, leaving us with our original function:
 					</p>
 					<!--
- 
+
    ┌───────┐
    │ 2    2
 \| │B  + B   - r\| = 0,    t ∈[0,1]
@@ -8985,7 +8985,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 					/>
 					<p>And now we just solve for that... oh wait. We've seen this before. In order to solve this, we'd end up needing to solve this:</p>
 					<!--
- 
+
  2    2
 B  + B  = r
  x    y
@@ -9482,7 +9482,7 @@ radialError(radius, points[]):
 						end, just like for Bézier curves), by evaluating the following function:
 					</p>
 					<!--
- 
+
            __ n
 Point(t) = ❯      P   · N   (t)
            ‾‾ i=0  i     i,k
@@ -9503,7 +9503,7 @@ Point(t) = ❯      P   · N   (t)
 					</p>
 					<p>Then the N() function itself. What does it look like?</p>
 					<!--
- 
+
           ╭       t- knot         ╮                ╭       knot     -t       ╮
           │              i        │                │           (i+k)         │
 N   (t) = │ ───────────────────── │  · N     (t) + │ ─────────────────────── │  · N       (t)
@@ -9518,9 +9518,9 @@ N   (t) = │ ───────────────────── 
 						some point; obviously, it does, and specifically it does so for the following <code>i</code>/<code>k</code> values:
 					</p>
 					<!--
- 
+
           ╭ 1  if  t ∈[ knot , knot   )
-N   (t) = ╡                 i      i+1  
+N   (t) = ╡                 i      i+1
  i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/2514e1aa0565840e33fde0b146e3efe2.svg" width="237px" height="40px" loading="lazy" />
@@ -9541,7 +9541,7 @@ N   (t) = ╡                 i      i+1
 						P(t), we can compute this point by evaluating <em>d(t)</em> on a curve section between knots <code>i</code> and <code>i+1</code>:
 					</p>
 					<!--
- 
+
  k               k-1                   k-1
 d (t) = α     · d   (t) + (1-α   )  · d   (t)
  i       i,k     i            i,k      i-1
@@ -9552,7 +9552,7 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 						by:
 					</p>
 					<!--
- 
+
               t -  knots[i]
 α    = ───────────────────────────
  i,k    knots[i+1+n-k] -  knots[i]
@@ -9566,9 +9566,9 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 					</p>
 					<p>Of course, the recursion does need a stop condition:</p>
 					<!--
- 
+
  k           0                ╭ 1  if  t ∈[ knot , knot   )
-d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1  
+d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
  0           i       i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/49af474c33ce0ee0733626ea3d988570.svg" width="367px" height="40px" loading="lazy" />
@@ -9582,34 +9582,34 @@ d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
 						recursion diagram:
 					</p>
 					<!--
- 
+
      ╭                                ╭                                ╭          1     0           0
-     │                                │                                │         α   × d ,   with  d    either 0 or 1 
+     │                                │                                │         α   × d ,   with  d    either 0 or 1
      │                                │          2     1           1   │          3     3           3
-     │                                │         α   × d ,   with  d  = ╡                 +                              
+     │                                │         α   × d ,   with  d  = ╡                 +
      │                                │          3     3           3   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      │          3     2           2   │                                ╰ ╰      3 ╯     2           2
-     │         α   × d ,   with  d  = ╡                 +                                                                
+     │         α   × d ,   with  d  = ╡                 +
      │          3     3           3   │                                ╭           1     0
-     │                                │                                │          α   × d                             
+     │                                │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           2     2
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
  3   │                                │ ╰      3 ╯     2           2   │ ╭      1 ╮     0           0
-d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1     
+d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
  3   │                                ╰                                ╰ ╰      2 ╯     1           1
-     │                 +                                                                                                 
+     │                 +
      │                                ╭           2     1
-     │                                │          α   × d                                                                
+     │                                │          α   × d
      │                                │           2     2
-     │                                │                                                                                 
-     │ ╭      3 ╮     2           2   │                 +                                                               
-     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0                               
-     │ ╰      3 ╯     2           2   │                                │          α   × d 
+     │                                │
+     │ ╭      3 ╮     2           2   │                 +
+     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0
+     │ ╰      3 ╯     2           2   │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           1     1
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
      │                                │ ╰      2 ╯     1           1   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      ╰                                ╰                                ╰ ╰      1 ╯     0           0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/20e910bbea2e6eff511cb13cef18ef3b.svg" width="641px" height="336px" loading="lazy" />
diff --git a/docs/ko-KR/index.html b/docs/ko-KR/index.html
index 7ec8933f..4d4abe7e 100644
--- a/docs/ko-KR/index.html
+++ b/docs/ko-KR/index.html
@@ -183,7 +183,7 @@
         <a class="btclk" href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu</a>
         or use the QR code on the right, if that's the kind of convenience you prefer =)
       </p>
-    
+
     </div>
     <div class="btcqr">
         <a href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">
@@ -195,7 +195,7 @@
 
 			<br style="clear: both;" />
 
-			<p>— <a href="https://twitter.com/TheRealPomax">Pomax</a></p>
+			<p>— <a href="https://mastodon.social/@TheRealPomax">Pomax</a></p>
 			<noscript>
 				<div class="note">
 					<header>
@@ -577,7 +577,7 @@
 					</p>
 					<!--
 \setmainfont[Ligatures=TeX]TeX Gyre Pagella \setmathfontTeX Gyre Pagella Math
- 
+
                                            ╭  p =  한 점      ╮
                                            │   1            │
                                            │  p =  다른 점     │
@@ -636,7 +636,7 @@
 						값(<i>x</i>라고 합시다)을 어떤 연산을 통해 다른 값과 대응시키는 함수가 있다고 합시다.
 					</p>
 					<!--
- 
+
 f(x) = cos (x)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/0cc876c56200446c60114c1b0eeeb2cc.svg" width="96px" height="17px" loading="lazy" />
@@ -646,7 +646,7 @@ f(x) = cos (x)
 					</p>
 					<p>여기까지는 좋습니다. 이제 매개변수 함수를 보고, 어떤 꼼수를 쓰는지 확인해 봅시다. 아래의 두 함수를 생각해 봅시다.</p>
 					<!--
- 
+
 f(a) = cos (a)
 f(b) = sin (b)
 -->
@@ -657,9 +657,9 @@ f(b) = sin (b)
 						꼼수입니다. 매개변수 함수에서는 아래와 같이 여러 가지 함수가 같은 변수를 공유합니다.
 					</p>
 					<!--
- 
+
 ╭ f (t) = cos (t)
-╡  a              
+╡  a
 │ f (t) = sin (t)
 ╰  b
 -->
@@ -676,8 +676,8 @@ f(b) = sin (b)
 						매개변수 곡선을 만들 때 보통 의미하는 바를 나타내도록 바꾸면 이 식이 눈에 더 잘 들어올지도 모르겠습니다.
 					</p>
 					<!--
- 
-{ x = cos (t) 
+
+{ x = cos (t)
   y = sin (t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/6914ba615733c387251682db7a3db045.svg" width="77px" height="40px" loading="lazy" />
@@ -711,7 +711,7 @@ f(b) = sin (b)
 					</p>
 					<p>중·고등학교에서 다항식을 배운 기억이 나시나요? 이렇게 생긴 합을 다항식이라고 합니다.</p>
 					<!--
- 
+
              3         2
 f(x) = a  · x  + b  · x  + c  · x + d
 -->
@@ -731,12 +731,12 @@ f(x) = a  · x  + b  · x  + c  · x + d
 						"이항계수"의 형태를 띱니다. 이항계수라고 하면 어려워 보이지만 사실은 두 값을 섞는 것을 간단하게 나타낸 것입니다.
 					</p>
 					<!--
- 
-1차= (1-t) + t                                        
+
+1차= (1-t) + t
          2                      2
-2차= (1-t)  + 2  · (1-t)  · t + t                     
+2차= (1-t)  + 2  · (1-t)  · t + t
          3             2                       2    3
-3차= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+3차= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img
 						class="LaTeX SVG"
@@ -750,10 +750,10 @@ f(x) = a  · x  + b  · x  + c  · x + d
 						간단하게 바뀝니다. 밑에 늘어놓은 이항계수를 보세요.
 					</p>
 					<!--
- 
-1차=  1 + 1           
-2차=  1 + 2 + 1       
-3차=  1 + 3 + 3 + 1   
+
+1차=  1 + 1
+2차=  1 + 2 + 1
+3차=  1 + 3 + 3 + 1
 4차= 1 + 4 + 6 + 4 + 1
 -->
 					<img
@@ -775,11 +775,11 @@ f(x) = a  · x  + b  · x  + c  · x + d
 					</p>
 					<!--
 
-1차= \colorblue a + \colorred b                                                                                                                       
-2차= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b                                                          
+1차= \colorblue a + \colorred b
+2차= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b
 3차= \colorblue a  · \colorblue a  · \colorblue a + \colorblue a  · \colorblue a  · \colorred b + \colorblue a  · \colorred b  · \colorred b + \colorr
-                                                                                           
-                                                                                           
+
+
                                                          ed b  · \colorred b  · \colorred b
 -->
 					<img
@@ -794,9 +794,9 @@ f(x) = a  · x  + b  · x  + c  · x + d
 						이쪽도 그렇게 어렵지 않았네요. 이것으로 이항 다항식을 모두 배웠고, 혹시 필요할 경우를 위해 아래에 일반항의 식을 추가합니다.
 					</p>
 					<!--
- 
+
               __ n                                                                            n-i     i
-Bézier(n,t) = ❯      \underset 이항계수 부분\underbrace\binomni  · \ \underset 문자 부분\underbrace(1-t)     · t 
+Bézier(n,t) = ❯      \underset 이항계수 부분\underbrace\binomni  · \ \underset 문자 부분\underbrace(1-t)     · t
               ‾‾ i=0
 -->
 					<img
@@ -1111,9 +1111,9 @@ function Bezier(3,t):
 					<!--
 
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset
               ‾‾ i=0
-                                                                weight\underbracew 
+                                                                weight\underbracew
                                                                                   i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/c2f2fe0ef5d0089d9dd8e5e3999405cb.svg" width="336px" height="55px" loading="lazy" />
@@ -1124,11 +1124,11 @@ Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \unders
 						(210,30), we use this Bézier curve:
 					</p>
 					<!--
- 
+
 ╭                               3                                   2                                             2                       3
-╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t  
+╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t
 │                               3                                    2                                             2                      3
-╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t 
+╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/be73034ac382b54863c7a18c2932cbbc.svg" width="473px" height="40px" loading="lazy" />
 					<p>Which gives us the curve we saw at the top of the article:</p>
@@ -1256,9 +1256,9 @@ function Bezier(3,t,w[]):
 					</p>
 					<p>Adding these ratio values to the regular Bézier curve function is fairly easy. Where the regular function is the following:</p>
 					<!--
- 
+
               __ n                    n-i     i
-Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w 
+Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
               ‾‾ i=0                                i
 -->
 					<img
@@ -1270,13 +1270,13 @@ Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
 					/>
 					<p>The function for rational Bézier curves has two more terms:</p>
 					<!--
- 
+
                         __ n                    n-i     i
-                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio 
+                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio
                         ‾‾ i=0                                i                    i
 Rational  Bézier(n,t) = ────────────────────────────────────────────────────────────
                                      __ n                     n-i      i
-                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio  
+                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio
                                      ‾‾ i=0                                       i
 -->
 					<img
@@ -1464,8 +1464,8 @@ function RationalBezier(3,t,w[],r[]):
 						other values, then the general formula for this is:
 					</p>
 					<!--
- 
-mixture = a  ·  value  + b  ·  value 
+
+mixture = a  ·  value  + b  ·  value
                      1              2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/dfd6ded3f0addcf43e0a1581627a2220.svg" width="212px" height="16px" loading="lazy" />
@@ -1478,8 +1478,8 @@ mixture = a  ·  value  + b  ·  value
 						And that's easy:
 					</p>
 					<!--
- 
-m = a  ·  value  + (1 - a)  ·  value 
+
+m = a  ·  value  + (1 - a)  ·  value
                1                    2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/4e0fa763b173e3a683587acf83733353.svg" width="213px" height="16px" loading="lazy" />
@@ -1553,61 +1553,61 @@ m = a  ·  value  + (1 - a)  ·  value
 						refer to coordinate values "in one or more dimensions":
 					</p>
 					<!--
- 
+
                   3                   2                             2          3
-B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t 
+B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t
         1              2                        3                        4
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/9a9a55f5b0323d9ea88f82fc6be58ad3.svg" width="468px" height="20px" loading="lazy" />
 					<p>Disregarding our actual coordinates for a moment, we have:</p>
 					<!--
- 
+
             3             2                       2    3
-B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/87cfac83cb8a4b0bee68ef006effc611.svg" width="353px" height="19px" loading="lazy" />
 					<p>We can write this as a sum of four expressions:</p>
 					<!--
- 
+
                 3
-... =      (1-t) 
+... =      (1-t)
                 2
     + 3  · (1-t)   · t
                      2
-    + 3  · (1-t)  · t 
+    + 3  · (1-t)  · t
              3
-    +       t  
+    +       t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/cdd88611833f3b178df91278359a4193.svg" width="140px" height="75px" loading="lazy" />
 					<p>And we can expand these expressions:</p>
 					<!--
- 
+
                                    3         2
 ... =  (1-t)  · (1-t)  · (1-t) = -t  + 3  · t  - 3  · t + 1
                                        3         2
     + 3  · (1-t)  · (1-t)  · t = 3  · t  - 6  · t  + 3  · t
                                             3         2
-    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t 
+    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t
                                              3
-    +        t  · t  · t       =            t  
+    +        t  · t  · t       =            t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/ec118f296511c6e9ac8727be3703a7ce.svg" width="397px" height="75px" loading="lazy" />
 					<p>Furthermore, we can make all the 1 and 0 factors explicit:</p>
 					<!--
- 
+
              3         2
 ... = -1  · t  + 3  · t  - 3  · t + 1
              3         2
-    + +3  · t  - 6  · t  + 3  · t + 0 
+    + +3  · t  - 6  · t  + 3  · t + 0
              3         2
     + -3  · t  + 3  · t  + 0  · t + 0
              3         2
-    + +1  · t  + 0  · t  + 0  · t + 0 
+    + +1  · t  + 0  · t  + 0  · t + 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/67a5ea33d6c6558f7d954b18226f4956.svg" width="217px" height="75px" loading="lazy" />
 					<p>And <em>that</em>, we can view as a series of four matrix operations:</p>
 					<!--
- 
+
                  ┌ -1 ┐                    ┌ 3  ┐                    ┌ -3 ┐                    ┌ 1 ┐
 ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ -6 │ + ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ 0 │
 └ t  t  t 1 ┘    │ -3 │   └ t  t  t 1 ┘    │ 3  │   └ t  t  t 1 ┘    │ 0  │   └ t  t  t 1 ┘    │ 0 │
@@ -1616,7 +1616,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/8ecff6b8a37d60385d287ea2b26876db.svg" width="607px" height="72px" loading="lazy" />
 					<p>If we compact this into a single matrix operation, we get:</p>
 					<!--
- 
+
                  ┌ -1  3 -3 1 ┐
 ┌  3  2     ┐  · │  3 -6  3 0 │
 └ t  t  t 1 ┘    │ -3  3  0 0 │
@@ -1628,7 +1628,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 						<code>t</code> matrix horizontally, and our big "mixing" matrix upside down:
 					</p>
 					<!--
- 
+
                  ┌  1  0  0 0 ┐
 ┌      2  3 ┐  · │ -3  3  0 0 │
 └ 1 t t  t  ┘    │  3 -6  3 0 │
@@ -1637,7 +1637,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/1a64ed455c6dd2f8cacca5e5e12bdcc1.svg" width="227px" height="72px" loading="lazy" />
 					<p>And then finally, we can add in our original coordinates as a single third matrix:</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -1650,7 +1650,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 					<img class="LaTeX SVG" src="./images/chapters/matrix/b32cae2dfc47d5f36df0bc3defb7dfa8.svg" width="323px" height="73px" loading="lazy" />
 					<p>We can perform the same trick for the quadratic curve, in which case we end up with:</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2083,7 +2083,7 @@ function drawCurvePoint(points[], t):
 						respectively: (we'll reverse the Bézier coefficients vector for legibility)
 					</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2100,7 +2100,7 @@ B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -2123,7 +2123,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						"point on the curve" information into a new matrix multiplication:
 					</p>
 					<!--
- 
+
                                                   ┌ P  ┐                                              ┌ P  ┐
                                    ┌  1  0 0 ┐    │  1 │                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2140,7 +2140,7 @@ B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌
 					/>
 					<p>and</p>
 					<!--
- 
+
                                                                ┌ P  ┐                                                       ┌ P  ┐
                                                                │  1 │                    ┌ 1 0  0 0  ┐                      │  1 │
                                              ┌  1  0  0 0 ┐    │ P  │                    │ 0 z  0 0  │    ┌  1  0  0 0 ┐    │ P  │
@@ -2170,7 +2170,7 @@ B(t) = ┌                    2         3 ┐  · │ -3  3  0 0 │  · │  2
 							at the quadratic curve first:
 						</p>
 						<!--
- 
+
                                                   ┌ P  ┐
                      ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2186,7 +2186,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                                         ┌ P  ┐
                                                                                         │  1 │
 = ┌      2 ┐  · \underset we turn this...\underbrace\kern 2.25em Z  · M \kern 2.25em  · │ P  │
@@ -2202,7 +2202,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                          ┌ P  ┐
                                                         -1               │  1 │
 = ┌      2 ┐  · \underset into this...\underbrace M  · M    · Z  · M   · │ P  │
@@ -2218,7 +2218,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                                                                    ┌ P  ┐
                                                                                                                    │  1 │
 = ┌      2 ┐  · M \underset ...to get this!\underbrace \kern 1.25em  · \kern 1.25em Q \kern 1.25em  · \kern 1.25em │ P  │
@@ -2245,7 +2245,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							coordinates that represent the curve from <em>t = 0</em> to <em>t = z</em>. So let's get computing:
 						</p>
 						<!--
- 
+
                     ┌ 1 0 0 ┐
      -1             │   1   │    ┌ 1 0 0  ┐    ┌  1  0 0 ┐   ┌     1            0        0  ┐
 Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = │  -(z-1)          z        0  │
@@ -2261,7 +2261,7 @@ Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = 
 						/>
 						<p>Excellent! Now we can form our new quadratic curve:</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭      ┌ P  ┐ ╮
                                │  1 │                      │      │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q  · │ P  │ │
@@ -2277,7 +2277,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                     ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌     1            0        0  ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │  -(z-1)          z        0  │  · │ P  │ │
@@ -2293,7 +2293,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌                        P                          ┐
                                │                         1                         │
                 ┌  1  0 0 ┐    │               z  · P  - (z-1)  · P                │
@@ -2322,7 +2322,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							write it down in a more simple form because of the zero, but what we <em>actually</em> evaluated, making the zero explicit, was:
 						</p>
 						<!--
- 
+
                                                             ┌ P  ┐
                                              ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2338,7 +2338,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                              ┌ P  ┐
                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2355,7 +2355,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 						/>
 						<p>If we want the interval [<em>z</em>,1], we will be evaluating this instead:</p>
 						<!--
- 
+
                                                                     ┌ P  ┐
                                                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2371,7 +2371,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 							loading="lazy"
 						/>
 						<!--
- 
+
                 ┌              2        ┐                   ┌ P  ┐
                 │ 1  z        z         │    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │  · │ P  │
@@ -2391,7 +2391,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 							<code>[M · something]</code>:
 						</p>
 						<!--
- 
+
                       ┌ 1 0 0 ┐    ┌              2        ┐
       -1              │   1   │    │ 1  z        z         │    ┌  1  0 0 ┐   ┌      2                   2 ┐
 Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │ = │ (z-1)  -2  · z  · (z-1) z  │
@@ -2407,7 +2407,7 @@ Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  ·
 						/>
 						<p>So, our final second curve looks like:</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭       ┌ P  ┐ ╮
                                │  1 │                      │       │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q'  · │ P  │ │
@@ -2423,7 +2423,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                   ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌      2                   2 ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │ (z-1)  -2  · z  · (z-1) z  │  · │ P  │ │
@@ -2439,7 +2439,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌  2                                      2       ┐
                                │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
                 ┌  1  0 0 ┐    │        3                       2              1 │
@@ -2469,7 +2469,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						<code>t = z</code>, we get two subcurves that are described as Bézier curves with simple-to-derive coordinates:
 					</p>
 					<!--
- 
+
                                              ┌                        P                          ┐
                                     ┌ P  ┐   │                         1                         │
 ┌     1            0        0  ┐    │  1 │   │               z  · P  - (z-1)  · P                │
@@ -2487,7 +2487,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                            ┌  2                                      2       ┐
                                   ┌ P  ┐   │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
 ┌      2                   2 ┐    │  1 │   │        3                       2              1 │
@@ -2508,7 +2508,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						yourself, though) and simply show you the resulting new coordinate sets:
 					</p>
 					<!--
- 
+
                                                               ┌                                    P                                      ┐
                                                      ┌ P  ┐   │                                     1                                     │
 ┌    1            0                0         0  ┐    │  1 │   │                           z  · P  - (z-1)  · P                            │
@@ -2529,7 +2529,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                                               ┌  3               2                                 2              3       ┐
                                                      ┌ P  ┐   │ z   · P  - 3  · z   · (z-1)  · P  + 3  · z  · (z-1)   · P  - (z-1)   · P  │
 ┌       3           2                      2  3 ┐    │  1 │   │        4                        3                        2              1 │
@@ -2581,7 +2581,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						that the start and end weights are the same as the start and end weights for the old curve):
 					</p>
 					<!--
- 
+
               __ k                                                                                            k-i     i
 Bézier(k,t) = ❯      \underset binomial term\underbrace\binomki  · \ \underset polynomial term\underbrace(1-t)     · t    · \ \underset new
               ‾‾ i=0
@@ -2605,9 +2605,9 @@ Bézier(k,t) = ❯      \underset binomial term\underbrace\binomki  · \ \unders
 					</p>
 					<p>We start by taking the standard Bézier function, and condensing it a little:</p>
 					<!--
- 
+
               __ n       n               n                       n-i     i
-Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t 
+Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t
               ‾‾ i=0  i  i               i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/9fc4ecc087d389dd0111bcba165cd5d0.svg" width="408px" height="41px" loading="lazy" />
@@ -2617,7 +2617,7 @@ Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t
 						<code>t</code> and <code>1-t</code>:
 					</p>
 					<!--
- 
+
 x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/ff224ded6bbbc94b43130f5f8eeb5d29.svg" width="379px" height="16px" loading="lazy" />
@@ -2626,9 +2626,9 @@ x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
 						<code>t</code> component:
 					</p>
 					<!--
- 
-Bézier(n,t)= (1-t) B(n,t) + t B(n,t)                    
-             __ n               n      __ n         n   
+
+Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
+             __ n               n      __ n         n
            = ❯      w  (1 - t) B (t) + ❯      w  t B (t)
              ‾‾ i=0  i          i      ‾‾ i=0  i    i
 -->
@@ -2638,18 +2638,18 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
 						I promise, it's about to make sense. We start with <code>(1-t)</code>:
 					</p>
 					<!--
- 
+
          n              n!         n-i  i
-(1 - t) B (t)= (1-t) ──────── (1-t)    t                  
+(1 - t) B (t)= (1-t) ──────── (1-t)    t
          i           (n-i)!i!
                n+1-i   (n+1)!        n+1-i  i
-             = ───── ────────── (1-t)      t              
-                n+1  (n+1-i)!i!                           
+             = ───── ────────── (1-t)      t
+                n+1  (n+1-i)!i!
                k-i    k!         k-i  i
              = ─── ──────── (1-t)    t ,  where  k = n + 1
                 k  (k-i)!i!
                k-i  k
-             = ─── B (t)                                  
+             = ─── B (t)
                 k   i
 -->
 					<img
@@ -2667,18 +2667,18 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
 						<code>t</code> part, but that's not a problem:
 					</p>
 					<!--
- 
+
    n          n!         n-i  i
-t B (t)= t ──────── (1-t)    t                                    
+t B (t)= t ──────── (1-t)    t
    i       (n-i)!i!
          i+1        (n+1)!             (n+1)-(i+1)  i+1
-       = ─── ──────────────────── (1-t)            t              
-         n+1 ((n+1)-(i+1))!(i+1)!                                 
+       = ─── ──────────────────── (1-t)            t
+         n+1 ((n+1)-(i+1))!(i+1)!
          i+1        k!             k-(i+1)  i+1
        = ─── ──────────────── (1-t)        t   ,  where  k = n + 1
           k  (k-(i+1))!(i+1)!
          i+1  k
-       = ─── B   (t)                                              
+       = ─── B   (t)
           k   i+1
 -->
 					<img
@@ -2698,21 +2698,21 @@ t B (t)= t ──────── (1-t)    t
 					</p>
 					<p>Let's do this:</p>
 					<!--
- 
+
              __ n+1             n      __ n+1       n
-Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)                   
+Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)
              ‾‾ i=0  i          i      ‾‾ i=0  i    i
              __ n+1    k-i  k      __ n+1    i+1  k
            = ❯      w  ─── B (t) + ❯      w  ─── B   (t),  where  k = n + 1
              ‾‾ i=0  i  k   i      ‾‾ i=0  i  k   i+1
              __ n+1    k-i  k      __ n+1      i  k
-           = ❯      w  ─── B (t) + ❯      p    ─ B (t)                     
+           = ❯      w  ─── B (t) + ❯      p    ─ B (t)
              ‾‾ i=0  i  k   i      ‾‾ i=0  i-1 k  i
              __ n+1 ╭    k-i        i ╮  k
-           = ❯      │ w  ─── + p    ─ │ B (t)                              
+           = ❯      │ w  ─── + p    ─ │ B (t)
              ‾‾ i=0 ╰  i  k     i-1 k ╯  i
              __ n+1                      k                 i
-           = ❯      (w  (1-s) + p    s) B (t),  where  s = ─               
+           = ❯      (w  (1-s) + p    s) B (t),  where  s = ─
              ‾‾ i=0   i          i-1     i                 k
 -->
 					<img
@@ -2727,14 +2727,14 @@ Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)
 						Bézier(n+1,t) as a very simple matrix multiplication:
 					</p>
 					<!--
- 
-M B  = B 
+
+M B  = B
    n    k
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/1f5b60d190a1c7099b3411e4cc477291.svg" width="71px" height="16px" loading="lazy" />
 					<p>where the matrix <strong>M</strong> is an <code>n+1</code> by <code>n</code> matrix, and looks like:</p>
 					<!--
- 
+
     ┌ 1  0   .   .   .   .    .    .  ┐
     │ 1 k-1                           │
     │ ─ ───  0   .   .   .    0    .  │
@@ -2776,20 +2776,20 @@ M = │        k   k                    │
 						<strong>A x = b</strong>, we can use it. And as it so happens, that's exactly what we're dealing with, so:
 					</p>
 					<!--
- 
-              M B = B             
+
+              M B = B
                  n   k
            T         T
-         (M  M) B = M  B          
+         (M  M) B = M  B
                  n      k
   T   -1   T          T   -1  T
-(M  M)   (M  M) B = (M  M)   M  B 
+(M  M)   (M  M) B = (M  M)   M  B
                  n               k
                       T   -1  T
-              I B = (M  M)   M  B 
+              I B = (M  M)   M  B
                  n               k
                       T   -1  T
-                B = (M  M)   M  B 
+                B = (M  M)   M  B
                  n               k
 -->
 					<img
@@ -2853,9 +2853,9 @@ M = │        k   k                    │
 					</p>
 					<p>First, let's look at the derivative rule for Bézier curves, which is:</p>
 					<!--
- 
+
                     __ n-1
-Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t) 
+Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
                     ‾‾ i=0   i+1  i                    i
 -->
 					<img
@@ -2870,7 +2870,7 @@ Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
 						is the same as a summation where each term is multiplied by <i>n</i>) as:
 					</p>
 					<!--
- 
+
                __ n-1
 Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
                ‾‾ i=0                i           i+1  i
@@ -2896,7 +2896,7 @@ Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
 							only the derivative of the polynomial basis function. So, let's find that:
 						</p>
 						<!--
- 
+
         d    ╭ n ╮  k      n-k d
 B   (t) ── = │   │ t  (1-t)    ──
  n,k    dt   ╰ k ╯             dt
@@ -2913,7 +2913,7 @@ B   (t) ── = │   │ t  (1-t)    ──
 							<a href="https://en.wikipedia.org/wiki/Chain_rule">chain</a> rules gives us:
 						</p>
 						<!--
- 
+
       ╭ n ╮        k-1      n-k    k         n-k-1
 ... = │   │ (k  · t    (1-t)    + t   · (1-t)       · (n-k)  · -1)
       ╰ k ╯
@@ -2927,9 +2927,9 @@ B   (t) ── = │   │ t  (1-t)    ──
 						/>
 						<p>Which is hard to work with, so let's expand that properly:</p>
 						<!--
- 
+
         kn!     k-1      n-k   (n-k)n!   k      n-1-k
-... = ──────── t    (1-t)    - ──────── t  (1-t)     
+... = ──────── t    (1-t)    - ──────── t  (1-t)
       k!(n-k)!                 k!(n-k)!
 -->
 						<img
@@ -2944,12 +2944,12 @@ B   (t) ── = │   │ t  (1-t)    ──
 							look like "x! over y!(x-y)!". If we can do that in a way that involves terms of <i>n-1</i> and <i>k-1</i>, we'll be on the right track.
 						</p>
 						<!--
- 
+
            n!       k-1      n-k   (n-k)n!   k      n-1-k
-... = ──────────── t    (1-t)    - ──────── t  (1-t)                                   
+... = ──────────── t    (1-t)    - ──────── t  (1-t)
       (k-1)!(n-k)!                 k!(n-k)!
         ╭    (n-1)!     k-1      n-k   (n-k)(n-1)!  k      n-1-k ╮
-... = n │ ──────────── t    (1-t)    - ─────────── t  (1-t)      │                     
+... = n │ ──────────── t    (1-t)    - ─────────── t  (1-t)      │
         ╰ (k-1)!(n-k)!                  k!(n-k)!                 ╯
         ╭        (n-1)!         (k-1)      (n-1)-(k-1)      (n-1)!     k      (n-1)-k ╮
 ... = n │ ──────────────────── t      (1-t)            - ──────────── t  (1-t)        │
@@ -2964,11 +2964,11 @@ B   (t) ── = │   │ t  (1-t)    ──
 						/>
 						<p>And that's the first part done: the two components inside the parentheses are actually regular, lower-order Bézier expressions:</p>
 						<!--
- 
+
         ╭    x!     y      x-y      x!     k      x-k ╮
 ... = n │ ──────── t  (1-t)    - ──────── t  (1-t)    │  , with x=n-1, y=k-1
-        ╰ y!(x-y)!               k!(x-k)!             ╯                     
-... = n (B           (t) - B       (t))                                     
+        ╰ y!(x-y)!               k!(x-k)!             ╯
+... = n (B           (t) - B       (t))
           (n-1),(k-1)       (n-1),k
 -->
 						<img
@@ -2983,19 +2983,19 @@ B   (t) ── = │   │ t  (1-t)    ──
 							through to its derivative:
 						</p>
 						<!--
- 
+
 Bézier   (t)    = B   (t)  · w  + B   (t)  · w  + B   (t)  · w  + B   (t)  · w  + ...
       n,k          n,0        0    n,1        1    n,2        2    n,3        3
              d
-Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +                              
+Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +
       n,k    dt          n-1,-1       n-1,0         0
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,0       n-1,1         1
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,1       n-1,2         2
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,2       n-1,3         3
-                  ...                                                                
+                  ...
 -->
 						<img
 							class="LaTeX SVG"
@@ -3009,18 +3009,18 @@ Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +
 							following:
 						</p>
 						<!--
- 
-n  · B      (t)  · w               +                                     
+
+n  · B      (t)  · w               +
       n-1,-1        0
 n  · B                (t)  · w     - n  · B                (t)  · w     +
       n-1,\colorblue 0        1            n-1,\colorblue 0        0
 n  · B               (t)  · w      - n  · B               (t)  · w      +
-      n-1,\colorred 1        2             n-1,\colorred 1        1      
+      n-1,\colorred 1        2             n-1,\colorred 1        1
 n  · B                   (t)  · w  - n  · B                   (t)  · w  +
       n-1,\colormagenta 2        3         n-1,\colormagenta 2        2
 ...                                - n  · B     (t)  · w                +
                                            n-1,3        3
-...                                                                      
+...
 -->
 						<img
 							class="LaTeX SVG"
@@ -3037,14 +3037,14 @@ n  · B                   (t)  · w  - n  · B                   (t)  · w  +
 							this term would contribute "nothing", so we can ignore it, too. This means we're left with:
 						</p>
 						<!--
- 
+
 n  · B                (t)  · w     - n  · B                (t)  · w     +
       n-1,\colorblue 0        1            n-1,\colorblue 0        0
 n  · B               (t)  · w      -  n  · B               (t)  · w     +
-      n-1,\colorred 1        2              n-1,\colorred 1        1     
+      n-1,\colorred 1        2              n-1,\colorred 1        1
 n  · B                   (t)  · w  - n  · B                   (t)  · w  +
       n-1,\colormagenta 2        3         n-1,\colormagenta 2        2
-...                                
+...
 -->
 						<img
 							class="LaTeX SVG"
@@ -3055,9 +3055,9 @@ n  · B                   (t)  · w  - n  · B                   (t)  · w  +
 						/>
 						<p>And that's just a summation of lower order curves:</p>
 						<!--
- 
+
              d
-Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B                 (t)  · (w  - w ) + n  · B                     (t)  · (w  - w )  
+Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B                 (t)  · (w  - w ) + n  · B                     (t)  · (w  - w )
       n,k    dt         (n-1),\colorblue 0         1    0          (n-1),\colorred 1         2    1          (n-1),\colormagenta 2         3    2
                                                                       +  ...
 -->
@@ -3070,7 +3070,7 @@ Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B
 						/>
 						<p>We can rewrite this as a normal summation, and we're done:</p>
 						<!--
- 
+
              d    __ n-1                                 __ n-1
 Bézier   (t) ── = ❯      n  · B     (t)  · (w    - w ) = ❯      B     (t)  · \underset derivative weights \underbracen  · (w    - w )
       n,k    dt   ‾‾ k=0       n-1,k         k+1    k    ‾‾ k=0  n-1,k                                                      k+1    k
@@ -3089,11 +3089,11 @@ Bézier   (t) ── = ❯      n  · B     (t)  · (w    - w ) = ❯      B
 						Bézier curves, and then the derivative:
 					</p>
 					<!--
- 
+
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset
               ‾‾ i=0
-                                                               weight\underbracew 
+                                                               weight\underbracew
                                                                                  i
 -->
 					<img
@@ -3104,7 +3104,7 @@ Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \unders
 						loading="lazy"
 					/>
 					<!--
- 
+
                __ k                                                                                            k-i     i
 Bézier'(n,t) = ❯      \underset binomial term\underbrace\binomki  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset derivative
                ‾‾ i=0
@@ -3125,11 +3125,11 @@ Bézier'(n,t) = ❯      \underset binomial term\underbrace\binomki  · \ \under
 						derivative one:
 					</p>
 					<!--
- 
-B(n,t),           w = {A,B,C,D}   
+
+B(n,t),           w = {A,B,C,D}
 B'(n,t),   n = 3, w' = {A',B',C'} = {3  · (B-A),   3  · (C-B),   3  · (D-C)}
-B''(n,t),  n = 2, w'' = {A'',B''} = {2  · (B'-A'),   2  · (C'-B')}          
-B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}                        
+B''(n,t),  n = 2, w'' = {A'',B''} = {2  · (B'-A'),   2  · (C'-B')}
+B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}
 -->
 					<img
 						class="LaTeX SVG"
@@ -3158,10 +3158,10 @@ B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}
 						indicates the direction of travel at specific points, and is literally just the first derivative of our curve:
 					</p>
 					<!--
- 
+
 tangent (t) = B' (t)
        x        x
-                    
+
 tangent (t) = B' (t)
        y        y
 -->
@@ -3177,14 +3177,14 @@ tangent (t) = B' (t)
 						point, and then do whatever it is we want to do based on those directions:
 					</p>
 					<!--
- 
+
                            ┌─────────────────┐
                            │      2         2
-    d = \| tangent(t)\| =  │B' (t)  + B' (t)  
+    d = \| tangent(t)\| =  │B' (t)  + B' (t)
                           ⟍│  x         y
 
                            tangent (t)     B' (t)
-^                                 x          x    
+^                                 x          x
 x(t) = \| tangent (t)\| =─────────────── = ──────
                  x       \| tangent(t)\|     d
 
@@ -3206,11 +3206,11 @@ y(t) = \| tangent (t)\| = ─────────────── = ──
 						is typically of length 1.0, so all we have to do is rotate the normalized directional vector and we're done:
 					</p>
 					<!--
- 
+
              ^           \pi   ^           \pi     ^
-normal (t) = x(t)  · cos ─── - y(t)  · sin ─── = - y(t)                                             
+normal (t) = x(t)  · cos ─── - y(t)  · sin ─── = - y(t)
       x                   2                 2
-                                                                                                    
+
                                                            ^           \pi   ^           \pi    ^
 normal (t) = \underset quarter circle rotation \underbrace x(t)  · sin ─── + y(t)  · cos ───  = x(t)
       y                                                                 2                 2
@@ -3233,7 +3233,7 @@ normal (t) = \underset quarter circle rotation \underbrace x(t)  · sin ──
 							To turn any point <i>(x,y)</i> into a rotated point <i>(x',y')</i> (over 0,0) by some angle φ, we apply this nice and easy computation:
 						</p>
 						<!--
- 
+
 x' = x  · cos (\phi) - y  · sin (\phi)
 y' = x  · sin (\phi) + y  · cos (\phi)
 -->
@@ -3246,7 +3246,7 @@ y' = x  · sin (\phi) + y  · cos (\phi)
 						/>
 						<p>Which is the "long" version of the following matrix transformation:</p>
 						<!--
- 
+
 ┌ x' ┐ = ┌ cos (\phi) -sin (\phi) ┐ ┌ x ┐
 └ y' ┘   └ sin (\phi) cos (\phi)  ┘ └ y ┘
 -->
@@ -3737,10 +3737,10 @@ generateRMFrames(steps) -> frames:
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
+
 B'(t) = a(1-t) + b(t)= 0,
           a - at + bt= 0,
-           (b-a)t + a= 0 
+           (b-a)t + a= 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -3751,10 +3751,10 @@ B'(t) = a(1-t) + b(t)= 0,
 					/>
 					<p>And then we turn this into our solution for <code>t</code> using basic arithmetics:</p>
 					<!--
- 
-(b-a)t + a= 0, 
+
+(b-a)t + a= 0,
     (b-a)t= -a,
-            -a 
+            -a
          t= ───
             b-a
 -->
@@ -3777,7 +3777,7 @@ B'(t) = a(1-t) + b(t)= 0,
 						you haven't, it looks like this:
 					</p>
 					<!--
- 
+
                                                    ┌────────┐
                                                    │ 2
                2                              -b ±⟍│b  - 4ac
@@ -3802,9 +3802,9 @@ Given f(t) = at  + bt + c, f(t)=0   when  t = ───────────
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
-B(t)  uses { p ,p ,p ,p  }                                                  
-              1  2  3  4                                                    
+
+B(t)  uses { p ,p ,p ,p  }
+              1  2  3  4
 B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
                1  2  3            1      2  1    2      3  2    3      4  3
 -->
@@ -3817,21 +3817,21 @@ B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
 					/>
 					<p>And then, using these <em>v</em> values, we can find out what our <em>a</em>, <em>b</em>, and <em>c</em> should be:</p>
 					<!--
- 
+
               2                  2
-B'(t)= v (1-t)  + 2v (1-t)t + v t            
+B'(t)= v (1-t)  + 2v (1-t)t + v t
         1           2          3
            2                    2       2
-  ...= v (t  - 2t + 1) + 2v (t-t ) + v t     
+  ...= v (t  - 2t + 1) + 2v (t-t ) + v t
         1                  2          3
           2                          2      2
-  ...= v t  - 2v t + v  + 2v t - 2v t  + v t 
+  ...= v t  - 2v t + v  + 2v t - 2v t  + v t
         1       1     1     2      2      3
           2       2      2
   ...= v t  - 2v t  + v t  - 2v t + v  + 2v t
         1       2      3       1     1     2
                    2
-  ...= (v -2v +v )t  + 2(v -v )t + v         
+  ...= (v -2v +v )t  + 2(v -v )t + v
          1   2  3         2  1      1
 -->
 					<img
@@ -3846,12 +3846,12 @@ B'(t)= v (1-t)  + 2v (1-t)t + v t
 						expressions of our original coordinate values, so we can do some substitution to get:
 					</p>
 					<!--
- 
+
 a= v -2v +v  = 3(-p  + 3p  - 3p  + p )
     1   2  3       1     2     3    4
-b= 2(v -v ) = 6(p  - 2p  + p )        
+b= 2(v -v ) = 6(p  - 2p  + p )
       2  1       1     2    3
-c= v  = 3(p -p )                      
+c= v  = 3(p -p )
     1      2  1
 -->
 					<img
@@ -3878,11 +3878,11 @@ c= v  = 3(p -p )
 						really hard to solve, it can be rewritten to a form for which finding the roots is "easier" (even if not "easy"):
 					</p>
 					<!--
- 
+
                    3     2
 very hard: solve at  + bt  + ct + d = 0
                   3
-   easier: solve t  + pt + q = 0       
+   easier: solve t  + pt + q = 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -4291,7 +4291,7 @@ function getCubicRoots(pa, pb, pc, pd) {
 						<code>f<sub>y</sub>(x)</code> is zero, so that the next <code>t</code> is the same as the one we already have:
 					</p>
 					<!--
- 
+
             f (x )
              y  n
 x    = x  - ───────
@@ -4437,11 +4437,11 @@ x    = x  - ───────
 						function for the y-component to an <em>n-2</em> term function. For instance, if we have a cubic curve such as this:
 					</p>
 					<!--
- 
+
 ╭                            3                               2                                          2                       3
-╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t  
+╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t
 │                            3                                2                                          2                      3
-╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t 
+╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/e5db5e945606d3429e4475ff92283a9c.svg" width="497px" height="40px" loading="lazy" />
 					<p>
@@ -4449,11 +4449,11 @@ x    = x  - ───────
 						by -160, gives us:
 					</p>
 					<!--
- 
+
 ╭                          3                               2                                          2                       3
-╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t  
+╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t
 │                          3                               2                                          2                       3
-╰ y = \colorblue 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t 
+╰ y = \colorblue 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/d412c72ed342c2036965ff251c8fb443.svg" width="481px" height="40px" loading="lazy" />
 					<p>
@@ -4461,20 +4461,20 @@ x    = x  - ───────
 						become:
 					</p>
 					<!--
- 
+
 ╭                          3                               2                                         2                       3
-╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                          3                               2                                          2                     3
-╰ y = \colorblue 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorblue  + 0  · t 
+╰ y = \colorblue 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorblue  + 0  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/e49d7bb45a18d14fbb12df4d91e2c67b.svg" width="473px" height="40px" loading="lazy" />
 					<p>If we drop all the zero-terms, this gives us:</p>
 					<!--
- 
+
 ╭                                   2                                         2                       3
-╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                                   2                                          2
-╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t                        
+╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/016f37843b2af2ae34dc8b2975505f3d.svg" width="408px" height="40px" loading="lazy" />
 					<p>
@@ -4581,7 +4581,7 @@ x    = x  - ───────
 					</p>
 					<p>What we need to do is solve a simple equation:</p>
 					<!--
- 
+
 C(t) = 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/inflections/ed68dcfb203517ca080fe48914769fb0.svg" width="59px" height="16px" loading="lazy" />
@@ -4591,7 +4591,7 @@ C(t) = 0
 						circle being on the other side of the curve. So what does <em>C(t)</em> look like? Actually something that seems not too hard:
 					</p>
 					<!--
- 
+
 C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)  ·   Bézier \prime\prime(t)
                x                     y                          y                     x
 -->
@@ -4620,15 +4620,15 @@ C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)
 							with the first and second derivatives, given our basis functions:
 						</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t                            
+Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t
              1           2            3           4
-      \prime            2                2                                      
+      \prime            2                2
 Bézier      (t) = a(1-t)  + 2b(1-t)t + ct   { a=3(x -x ),b=3(x -x ),c=3(x -x ) }
-                                                   2  1       3  2       4  3   
+                                                   2  1       3  2       4  3
       \prime\prime
-Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\                        
+Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 -->
 						<img
 							class="LaTeX SVG"
@@ -4639,14 +4639,14 @@ Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 						/>
 						<p>And of course the same functions for <em>y</em>:</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t 
+Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t
              1           2            3           4
-      \prime            2                2           
-Bézier      (t) = d(1-t)  + 2e(1-t)t + ft            
+      \prime            2                2
+Bézier      (t) = d(1-t)  + 2e(1-t)t + ft
       \prime\prime
-Bézier            (t) = w(1-t) + zt                  
+Bézier            (t) = w(1-t) + zt
 -->
 						<img
 							class="LaTeX SVG"
@@ -4660,19 +4660,19 @@ Bézier            (t) = w(1-t) + zt
 							rather overly complicated set of arithmetic expressions:
 						</p>
 						<!--
- 
+
      2             2             2             2             2
--18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y              
+-18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y
         2  1          3  1          4  1          1  2          3  2
      2             2             2             2             2
-+36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y              
++36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y
         4  2          1  3          2  3          4  3          1  4
-     2             2                                                             
--36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y 
+     2             2
+-36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y
         2  4          3  4         2  1          3  1          4  1          1  2
-+54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y 
++54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y
        3  2          4  2          1  3          2  3          1  4          2  4
--18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y   
+-18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y
      2  1          3  1          1  2          3  2          1  3          2  3
 -->
 						<img
@@ -4693,7 +4693,7 @@ Bézier            (t) = w(1-t) + zt
 						<code>t</code> values, we end up with the following simple term function for <em>C(t)</em>:
 					</p>
 					<!--
- 
+
                                  2
 (3 x  y +2 x  y +3 x  y -x  y ) t  + (3 x  y -x  y -3 x  y ) t + (x  y -x  y )
     3  2    4  2    2  3  4  3           3  2  4  2    2  3        2  3  3  2
@@ -4710,7 +4710,7 @@ Bézier            (t) = w(1-t) + zt
 						simplification:
 					</p>
 					<!--
- 
+
 a = x   · y  ╮
      3     2 │
 b = x   · y  │                             2
@@ -4729,7 +4729,7 @@ d = x   · y  │
 					/>
 					<p>This is a plain quadratic curve, and we know how to solve <em>C(t) = 0</em>; we use the quadratic formula:</p>
 					<!--
- 
+
                                                   ┌──────────┐
                                                   │ 2
 x = -3a + 2b + 3c - d ╮                      -y ±⟍│y  - 4 x z
@@ -4823,7 +4823,7 @@ z =       c - a       ╯                             2x
 								This edge is described by the function:
 							</p>
 							<!--
- 
+
       2
     -x  + 2x + 3
 y = ────────────, { x ≤1 }
@@ -4843,7 +4843,7 @@ y = ────────────, { x ≤1 }
 								the function:
 							</p>
 							<!--
- 
+
      ┌──────────┐
      │        2
     ⟍│3(4x - x )  - x
@@ -4861,7 +4861,7 @@ y = ─────────────────, { 0 ≤x ≤1 }
 						<li>
 							<p>...on the top (blue) edge, the curve's start point touches the curve, forming a loop. This edge is described by the function:</p>
 							<!--
- 
+
       2
     -x  + 3x
 y = ────────, { x ≤0 }
@@ -4928,7 +4928,7 @@ y = ────────, { x ≤0 }
 						<em>z</em> coordinate that is always 1, then we can suddenly add arbitrary values. For example:
 					</p>
 					<!--
- 
+
 ┌ 1 0 a ┐    ┌  x  ┐   ┌ 1  · x + 0  · y + a  · z ┐   ┌ x + a  · 1 ┐   ┌ x + a ┐
 │ 0 1 b │  · │  y  │ = │ 0  · x + 1  · y + b  · z │ = │ y + b  · 1 │ = │ y + b │
 └ 0 0 1 ┘    └ z=1 ┘   └ 0  · x + 0  · y + 1  · z ┘   └   1  · z   ┘   └  z=1  ┘
@@ -4939,7 +4939,7 @@ y = ────────, { x ≤0 }
 						want to subtract p1.x and p1.y, we use:
 					</p>
 					<!--
- 
+
      ┌ 1 0 -P   ┐            ┌ 1  · x + 0  · y - P    · 1 ┐   ┌ x - P   ┐
      │       1x │    ┌ x ┐   │                    1x      │   │      1x │
 T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = │ y - P   │
@@ -4955,7 +4955,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						transformation looks like this:
 					</p>
 					<!--
- 
+
 ┌ 1 S 0 ┐    ┌ x ┐   ┌ x + S  · y ┐
 │ 0 1 0 │  · │ y │ = │     y      │
 └ 0 0 1 ┘    └ 1 ┘   └     1      ┘
@@ -4966,7 +4966,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						simply <em>-x/y</em>, because <em>-x/y \</em> y = -x*. Done:
 					</p>
 					<!--
- 
+
      ┌    U     ┐
      │     2x   │
      │ 1 -─── 0 │
@@ -4984,7 +4984,7 @@ T  = │    U     │
 						x-scaling by 1/point3<sub>x</sub>, and y-scaling by point2<sub>y</sub>. This is really easy:
 					</p>
 					<!--
- 
+
      ┌  1        ┐
      │ ───  0  0 │
      │ V         │
@@ -5005,7 +5005,7 @@ T  = │      1    │
 						offset, in this case 1:
 					</p>
 					<!--
- 
+
      ┌    1    0 0 ┐
      │ 1 - W       │
      │      3y     │
@@ -5021,7 +5021,7 @@ T  = │ ─────── 1 0 │
 						be:
 					</p>
 					<!--
- 
+
                 ╭                           (-x +x )(-y +y )                ╮
                 │                              1  2    1  4                 │
                 │                -x  + x  - ────────────────                │
@@ -5056,7 +5056,7 @@ mapped  = (x) = │                                1  2                       
 						that leave?
 					</p>
 					<!--
- 
+
       ╭                 x   · y       x   · y              ╮
       │                  2     4       2     3             │
       │            x  - ──────── / x -────────             │   ╭         x           ╮
@@ -5076,11 +5076,11 @@ mapped  = (x) = │                                1  2                       
 						factors to see just how simple this is:
 					</p>
 					<!--
- 
-      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y 
+
+      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y
       │          43         │                 │  4    2     42 │            4                3
 ... = │                     │,   where  x   = │ ────────────── │,    y   = ──,   and  y   = ──
-      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y 
+      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y
       ╰  42    43       32  ╯                 ╰  3    2     32 ╯            2                2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/canonical/88e3fae7aeef6d7614290587422542c9.svg" width="563px" height="55px" loading="lazy" />
@@ -5167,16 +5167,16 @@ mapped  = (x) = │                                1  2                       
 					</p>
 					<p>First, let's look at the function for x(t):</p>
 					<!--
- 
+
              3          2            2     3
-x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt 
+x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt
 -->
 					<img class="LaTeX SVG" src="./images/chapters/yforx/378d0fd8cefa688d530ac38930d66844.svg" width="335px" height="19px" loading="lazy" />
 					<p>
 						We can rewrite this to a plain polynomial form, by just fully writing out the expansion and then collecting the polynomial factors, as:
 					</p>
 					<!--
- 
+
                          3                  2
 x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 -->
@@ -5187,7 +5187,7 @@ x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 						known constants, we can trivially rewrite this (by moving the <code>x(t)</code> across the equal sign) as:
 					</p>
 					<!--
- 
+
                    3                  2
 (-a + 3b - 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + (a-x) = 0
 -->
@@ -5278,7 +5278,7 @@ y = curve.get(t).y</textarea
 						following seemingly straight forward (if a bit overwhelming) formula:
 					</p>
 					<!--
- 
+
     ┌───────────────┐
 ╭ z │      2       2
 |   │f '(t) +f '(t)   dt
@@ -5287,7 +5287,7 @@ y = curve.get(t).y</textarea
 					<img class="LaTeX SVG" src="./images/chapters/arclength/59ebc3a7c3547a50998d1ea3664fb688.svg" width="140px" height="33px" loading="lazy" />
 					<p>or, more commonly written using Leibnitz notation as:</p>
 					<!--
- 
+
               ┌─────────────────┐
           ╭ z │       2        2
 length =  |  ⟍│(dx/dt) +(dy/dt)   dt
@@ -5326,7 +5326,7 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 |   ⟍│(dx/dt) +(dy/dt)   dt =  |   f(t) dt  ≃ ┌ \underset strip 1 \underbrace C   · f(t )   + ...  + \underset strip n \underbrace C   · f(t )  ┐ =
 ╯ -1                           ╯ -1           └                                1       1                                            n       n   ┘
                                                                                     __ n
-                                           \underset strips 1 through n \underbrace ❯      C   · f(t )   
+                                           \underset strips 1 through n \underbrace ❯      C   · f(t )
                                                                                     ‾‾ i=1  i       i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/arclength/b76753476ad6ecfe4b8f39bcf9432980.svg" width="623px" height="71px" loading="lazy" />
@@ -5408,16 +5408,16 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							shifting and scaling the inputs. Doing so, we get the following:
 						</p>
 						<!--
- 
+
      ┌─────────────────┐
  ╭ z │       2        2
- |  ⟍│(dx/dt) +(dy/dt)   dt                                        
+ |  ⟍│(dx/dt) +(dy/dt)   dt
  ╯ 0
      z    ┌        ╭ z         z ╮                ╭ z         z ╮ ┐
  ≃ \ ─  · │ C   · f│ ─  · t  + ─ │ + ... + C   · f│ ─  · t  + ─ │ │
      2    └  1     ╰ 2     1   2 ╯          n     ╰ 2     n   2 ╯ ┘
     z    __ n         ╭ z         z ╮
-= \ ─  · ❯     C   · f│ ─  · t  + ─ │                              
+= \ ─  · ❯     C   · f│ ─  · t  + ─ │
     2    ‾‾ i=1 i     ╰ 2     i   2 ╯
 -->
 						<img
@@ -5439,10 +5439,10 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							<a href="./legendre-gauss.html">these tables</a> would tell us that for <em>n=2</em> we must use the following values:
 						</p>
 						<!--
- 
-C  = 1     
+
+C  = 1
  1
-C  = 1     
+C  = 1
  2
         1
 t  = - ────
@@ -5458,7 +5458,7 @@ t  = + ────
 							Which means that in order for us to approximate the integral, we must plug these values into the approximate function, which gives us:
 						</p>
 						<!--
- 
+
     ┌─────────────────┐
 ╭ z │       2        2       z    ┌  ╭ z     -1    z ╮    ╭ z     1     z ╮ ┐
 |  ⟍│(dx/dt) +(dy/dt)   dt ≃ ─  · │ f│ ─  · ──── + ─ │ + f│ ─  · ──── + ─ │ │
@@ -5620,13 +5620,13 @@ t  = + ────
 					</p>
 					<p>So, what does the function look like? This:</p>
 					<!--
- 
+
          x'y'' - x''y'
 \kappa = ─────────────
                    3
                    ─
              2   2 2
-          (x' +y' ) 
+          (x' +y' )
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/060acd6ff0a050fe4d98a7802a2b3a3f.svg" width="113px" height="47px" loading="lazy" />
 					<p>
@@ -5634,14 +5634,14 @@ t  = + ────
 						tiny bit:
 					</p>
 					<!--
- 
+
             B '(t)B ''(t) - B ''(t)B '(t)
              x     y         x      y
 \kappa(t) = ─────────────────────────────
                                   3
                                   ─
                         2       2 2
-                 (B '(t) +B '(t) ) 
+                 (B '(t) +B '(t) )
                    x       y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/afd8cb8b0fe291ff703752c1c9cc33d4.svg" width="239px" height="55px" loading="lazy" />
@@ -5720,7 +5720,7 @@ function kappa(t, B):
 						circle that "fits" the curve's curvature at any point, using what is possibly the simplest bit of maths found in this entire primer:
 					</p>
 					<!--
- 
+
            1
 R(t) = ─────────
        \kappa(t)
@@ -6168,8 +6168,8 @@ lli = function(line1, line2):
 						points, so logically <code>C</code> will have a function that interpolates between those two coordinates:
 					</p>
 					<!--
- 
-C = u(t)  · P       + (1-u(t))  · P    
+
+C = u(t)  · P       + (1-u(t))  · P
               start                 end
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8c6662f605722fb2ff6cd7f65243a126.svg" width="229px" height="16px" loading="lazy" />
@@ -6182,22 +6182,22 @@ C = u(t)  · P       + (1-u(t))  · P
 						(with thanks to Boris Zbarsky) shows us the following two formulae:
 					</p>
 					<!--
- 
+
                         2
-                   (1-t) 
+                   (1-t)
 u(t)           = ───────────
      quadratic    2        2
-                 t  + (1-t) 
+                 t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8e7cfee39c98f2ddf9b635a914066cf6.svg" width="183px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                     3
-               (1-t) 
+               (1-t)
 u(t)       = ───────────
      cubic    3        3
-             t  + (1-t) 
+             t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/a0b99054cc82ca1fb147f077e175ef10.svg" width="161px" height="41px" loading="lazy" />
 					<p>
@@ -6208,7 +6208,7 @@ u(t)       = ───────────
 					</p>
 					<p>We start by observing that, given <code>A</code>, <code>B</code>, and <code>C</code>, the following always holds:</p>
 					<!--
- 
+
             distance(B,C)
 ratio(t) = ────────────── =  Constant
             distance(A,B)
@@ -6216,22 +6216,22 @@ ratio(t) = ────────────── =  Constant
 					<img class="LaTeX SVG" src="./images/chapters/abc/131454dcbac04e567f322979f4af80c6.svg" width="245px" height="39px" loading="lazy" />
 					<p>Working out the maths for this, we see the following two formulae for quadratic and cubic curves:</p>
 					<!--
- 
+
                        2        2
                       t  + (1-t)  - 1
 ratio(t)           = |───────────────|
          quadratic       2        2
-                        t  + (1-t) 
+                        t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/5924e162b50272c40c842fad14b8fa48.svg" width="239px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                    3        3
                   t  + (1-t)  - 1
 ratio(t)       = |───────────────|
          cubic       3        3
-                    t  + (1-t) 
+                    t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8cd992c1ceaae2e67695285beef23a24.svg" width="219px" height="41px" loading="lazy" />
 					<p>
@@ -6241,7 +6241,7 @@ ratio(t)       = |───────────────|
 						<code>ratio(t)</code> function to find <code>A</code>:
 					</p>
 					<!--
- 
+
           C - B           B - C
 A = B - ───────── = B + ─────────
          ratio(t)        ratio(t)
@@ -6256,11 +6256,11 @@ A = B - ───────── = B + ─────────
 						pair, after which we can reverse engineer <code>v1</code> and <code>v2</code>:
 					</p>
 					<!--
- 
+
                                  ╭     e  - t  · A
                                  │      1
-╭ e = (1-t)  · v  + t  · A       │ v = ───────────    
-╡  1            1            ==> ╡  1      1-t         
+╭ e = (1-t)  · v  + t  · A       │ v = ───────────
+╡  1            1            ==> ╡  1      1-t
 │ e = (1-t)  · A + t  · v        │     e  - (1-t)  · A
 ╰  2                     2       │      2
                                  │ v = ───────────────
@@ -6269,14 +6269,14 @@ A = B - ───────── = B + ─────────
 					<img class="LaTeX SVG" src="./images/chapters/abc/3166afa345aec1abda432c39b68d39a0.svg" width="339px" height="73px" loading="lazy" />
 					<p>And then reverse engineer the curve's control points:</p>
 					<!--
- 
+
                                       ╭     v  - (1-t)  ·  start
                                       │      1
 ╭ v = (1-t)  ·  start + t  · C        │ C = ────────────────────
-╡  1                          1   ==> ╡  1           t           
+╡  1                          1   ==> ╡  1           t
 │ v = (1-t)  · C  + t  ·  end         │     v  - t  ·  end
 ╰  2            2                     │      2
-                                      │ C = ──────────────      
+                                      │ C = ──────────────
                                       ╰  2       1-t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8bd3e6fed5bf8d871d30221ae400fd93.svg" width="383px" height="75px" loading="lazy" />
@@ -6304,15 +6304,15 @@ A = B - ───────── = B + ─────────
 						<code>B</code> point, and <code>B</code> and end point as a ratio, using
 					</p>
 					<!--
- 
+
 ╭ d = || Start - B||
 │  1
-│ d = || End - B||  
+│ d = || End - B||
 │  2
-╡      d             
+╡      d
 │       1
-│  t= ─────         
-│     d +d 
+│  t= ─────
+│     d +d
 ╰      1  2
 -->
 					<img
@@ -6374,9 +6374,9 @@ A = B - ───────── = B + ─────────
 						<code>e2</code> coordinates must obey the standard de Casteljau rule for linear interpolation:
 					</p>
 					<!--
- 
-╭ e = B + t  · d    
-╡  1                 
+
+╭ e = B + t  · d
+╡  1
 │ e = B - (1-t)  · d
 ╰  2
 -->
@@ -6396,7 +6396,7 @@ A = B - ───────── = B + ─────────
 						curve with a loop in it. To do this, we can use the <a href="https://en.wikipedia.org/wiki/Atan2">atan2</a> function:
 					</p>
 					<!--
- 
+
 \phi = (atan2(E -S , E -S ) - atan2(B -S , B -S ) + 2 \pi)   mod  2 \pi
                y  y   x  x           y  y   x  x
 -->
@@ -6413,8 +6413,8 @@ A = B - ───────── = B + ─────────
 						<code>d</code>:
 					</p>
 					<!--
- 
-d = {  d  if  0 ≤\phi ≤\pi             
+
+d = {  d  if  0 ≤\phi ≤\pi
       -d  if  \phi < 0 \lor \phi > \pi
 -->
 					<img
@@ -6592,7 +6592,7 @@ for (coordinate, index) in LUT:
 						maths, that means we're trying to solve:
 					</p>
 					<!--
- 
+
 dist(B(t), c) = r
 -->
 					<img
@@ -6604,15 +6604,15 @@ dist(B(t), c) = r
 					/>
 					<p>Which seems simple enough. Unfortunately, when we expand that <code>dist</code> function, things get a lot more problematic:</p>
 					<!--
- 
-r=  dist(B(t), c)                                                                                                              
+
+r=  dist(B(t), c)
     ┌─────────────────────────┐
     │          2             2
- =  │(B t - c )  + (B t - c )                                                                                                  
-   ⟍│  x     x       y     y                                                                                                   
+ =  │(B t - c )  + (B t - c )
+   ⟍│  x     x       y     y
     ┌─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┐
     │╭         3             2                 2       3      ╮2   ╭         3             2                 2       3      ╮2
- =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │  
+ =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │
    ⟍│╰  1             2               3             4       x ╯    ╰  1             2               3             4       y ╯
 -->
 					<img
@@ -6884,17 +6884,17 @@ findClosest(start, p, r, LUT):
 						<code>B</code>, we can compute the associated <code>C</code>:
 					</p>
 					<!--
- 
+
 C = u(t)   ·  Start + (1-u(t) )  ·  End
         q                    q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/48887d68a861a0acdf8313e23fb19880.svg" width="247px" height="24px" loading="lazy" />
 					<p>And then the associated <code>A</code>:</p>
 					<!--
- 
+
           C - B            B - C
 A = B - ────────── = B + ──────────
-         ratio(t)         ratio(t) 
+         ratio(t)         ratio(t)
                  q                q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/2e65bc9c934380c2de6a24bcd5c1c7b7.svg" width="228px" height="41px" loading="lazy" />
@@ -7031,12 +7031,12 @@ A = B - ────────── = B + ──────────
 							form, where each line corresponds to a power of t, and each column is for a specific coefficient. First, we expand the function:
 						</p>
 						<!--
- 
+
                     2                    2
-B          = a (1-t)  + 2 b (1-t) t + c t    
-  quadratic                                  
+B          = a (1-t)  + 2 b (1-t) t + c t
+  quadratic
                          2            2     2
-           = a - 2at + at  + 2bt - 2bt  + ct 
+           = a - 2at + at  + 2bt - 2bt  + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -7047,12 +7047,12 @@ B          = a (1-t)  + 2 b (1-t) t + c t
 						/>
 						<p>And then we (trivially) rearrange the terms across multiple lines:</p>
 						<!--
- 
-B          =a               
+
+B          =a
   quadratic
-            - 2at+ 2bt      
+            - 2at+ 2bt
                 2     2    2
-            + at - 2bt + ct 
+            + at - 2bt + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -7068,7 +7068,7 @@ B          =a
 						</p>
 						<p>With that arrangement, we can easily decompose this as a matrix multiplication:</p>
 						<!--
- 
+
                                          ┌  a  +  0  + 0 ┐                 ┌  1  0 0 ┐    ┌ a ┐
 B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌      2 ┐  · │ -2  2 0 │  · │ b │
   quadratic                └ 1 t t  ┘    └  a  + -2b + c ┘   └ 1 t t  ┘    └  1 -2 1 ┘    └ c ┘
@@ -7082,9 +7082,9 @@ B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌
 						/>
 						<p>We can do the same for the cubic curve, of course. We know the base function for cubics:</p>
 						<!--
- 
+
               3          2             2     3
-B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt 
+B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
   cubic
 -->
 						<img
@@ -7096,14 +7096,14 @@ B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
 						/>
 						<p>So we write out the expansion and rearrange:</p>
 						<!--
- 
-B      =a                     
+
+B      =a
   cubic
-        - 3at + 3bt           
-             2     2    2     
-        + 3at - 6bt +3ct      
+        - 3at + 3bt
+             2     2    2
+        + 3at - 6bt +3ct
             3      3    3    3
-        - at  + 3bt -3ct + dt 
+        - at  + 3bt -3ct + dt
 -->
 						<img
 							class="LaTeX SVG"
@@ -7114,7 +7114,7 @@ B      =a
 						/>
 						<p>Which we can then decompose:</p>
 						<!--
- 
+
                                         ┌  1  0  0 0 ┐    ┌ a ┐
 B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
   cubic                └ 1 t t  t  ┘    │  3 -6  3 0 │    │ c │
@@ -7129,7 +7129,7 @@ B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
 						/>
 						<p>And, of course, we can do this for quartic curves too (skipping the expansion step):</p>
 						<!--
- 
+
                                              ┌  1  0   0   0 0 ┐    ┌ a ┐
                                              │ -4  4   0   0 0 │    │ b │
 B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  · │ c │
@@ -7155,7 +7155,7 @@ B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  ·
 						we want to do curve fitting:
 					</p>
 					<!--
- 
+
     ┌ p   ┐
     │  1  │
     │ p   │
@@ -7193,9 +7193,9 @@ P = │  2  │
 					</p>
 					<p>To get these values, we first compute the general "distance along the polygon" matrix:</p>
 					<!--
- 
+
                                ╭            d  = 0
-D = ┌ d  d  ... d  ┐,   where  ╡             1                 
+D = ┌ d  d  ... d  ┐,   where  ╡             1
     └  1  2      n ┘           │ d  = d    +  length(p   , p )
                                ╰  i    i-1            i-1   i
 -->
@@ -7212,10 +7212,10 @@ D = ┌ d  d  ... d  ┐,   where  ╡             1
 						<code>i=n</code> fall in the [0,1] interval, so we need to scale all values down by whatever the total length of the polygon is:
 					</p>
 					<!--
- 
+
                                ╭    s  = 0
                                │     1
-S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d  
+S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
     └  1  2      n ┘           │  i    i    n
                                │    s  = 1
                                ╰     n
@@ -7237,9 +7237,9 @@ S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
 						to for the associated <code>t</code> value", which we'll square to get rid of any pesky negative signs:
 					</p>
 					<!--
- 
+
                            2
-E(C)  = (p  -   Bézier(s )) 
+E(C)  = (p  -   Bézier(s ))
     i     i             i
 -->
 					<img
@@ -7254,9 +7254,9 @@ E(C)  = (p  -   Bézier(s ))
 						function. So, we literally just do that; the total error function is simply the sum of all these individual errors:
 					</p>
 					<!--
- 
+
        __ n                      2
-E(C) = ❯      (p  -   Bézier(s )) 
+E(C) = ❯      (p  -   Bézier(s ))
        ‾‾ i=1   i             i
 -->
 					<img
@@ -7273,9 +7273,9 @@ E(C) = ❯      (p  -   Bézier(s ))
 						<strong>T x M x C</strong> we talked about before, which gives us:
 					</p>
 					<!--
- 
+
                 2
-E(C) = (P - TMC) 
+E(C) = (P - TMC)
 -->
 					<img
 						class="LaTeX SVG"
@@ -7286,7 +7286,7 @@ E(C) = (P - TMC)
 					/>
 					<p>In which we can replace the rather cumbersome "squaring" operation with a more conventional matrix equivalent:</p>
 					<!--
- 
+
                 T
 E(C) = (P - TMC)  (P - TMC)
 -->
@@ -7308,7 +7308,7 @@ E(C) = (P - TMC)  (P - TMC)
 						those, and then use that 𝕋 instead of <strong>T</strong> in our error function:
 					</p>
 					<!--
- 
+
     ┌    0    1      n-2   n-1  ┐
     │   s    s  ... s     s     │
     │    1    1      1     1    │
@@ -7328,7 +7328,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Which, because of the first and last values in <strong>S</strong>, means:</p>
 					<!--
- 
+
     ┌    1     0  ...   0    0    ┐
     │                  n-2   n-1  │
     │    1    s       s     s     │
@@ -7348,7 +7348,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Now we can properly write out the error function as matrix operations:</p>
 					<!--
- 
+
                 T
 E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 -->
@@ -7365,7 +7365,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						taking the derivative, and determining where that derivative is zero:
 					</p>
 					<!--
- 
+
 ∂E          T
 ── = 0 = -2𝕋  (P - 𝕋MC)
 ∂C
@@ -7404,7 +7404,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						for <strong>C</strong>:
 					</p>
 					<!--
- 
+
      -1   T   -1  T
 C = M   (𝕋  𝕋)   𝕋  P
 -->
@@ -7503,13 +7503,13 @@ C = M   (𝕋  𝕋)   𝕋  P
 						previous and next point:
 					</p>
 					<!--
- 
+
                ┌  V  = P       ┐
                │   1    2      │
 ┌ P  ┐         │  V  = P       │
 │  1 │         │   2    3      │
 │ P  │         │       P  - P  │
-│  2 │       = │        3    1 │              
+│  2 │       = │        3    1 │
 │ P  │points   │ V'  = ─────── │ point-tangent
 │  3 │         │   1      2    │
 │ P  │         │       P  - P  │
@@ -7642,7 +7642,7 @@ for p = 1 to points.length-3 (inclusive):
 						so how different is the matrix form for Catmull-Rom curves?:
 					</p>
 					<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7679,7 +7679,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							we'll get to that:
 						</p>
 						<!--
- 
+
                         ┌   P     ┐
                         │    2    │
 ┌ V   ┐        ┌ P  ┐   │   P     │
@@ -7707,7 +7707,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						</p>
 						<p>Computing <em>T</em> is really more "arranging the numbers":</p>
 						<!--
- 
+
               ┌   P     ┐
               │    2    │
      ┌ P  ┐   │   P     │   ┌  0  · P1 + 1  · P2 + 0  · P3 + 0  · P4 ┐   ┌  0  1 0 0 ┐    ┌ P  ┐
@@ -7730,7 +7730,7 @@ T  · │  2 │ = │  3    1 │ = │ ──  · P1 + 0  · P2 + ─  · P3 +
 						/>
 						<p>Thus:</p>
 						<!--
- 
+
     ┌  0  1 0 0 ┐
     │  0  0 1 0 │
     │ -1    1   │
@@ -7754,7 +7754,7 @@ T = │ ──  0 ─ 0 │
 							factor into both our coordinate vector representation, and mapping matrix <em>T</em>:
 						</p>
 						<!--
- 
+
           ┌   P     ┐
           │    2    │
 ┌ V   ┐   │   P     │        ┌  0  1  0 0  ┐
@@ -7780,7 +7780,7 @@ T = │ ──  0 ─ 0 │
 							coordinates, and see what we end up with:
 						</p>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7799,7 +7799,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						/>
 						<p>Replace point/tangent vector with the expression for all-coordinates:</p>
 						<!--
- 
+
                                                     ┌  0  1  0 0  ┐    ┌ P  ┐
                                                     │  0  0  1 0  │    │  1 │
                                  ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -7818,7 +7818,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │ ──  0 ─
 						/>
 						<p>and merge the matrices:</p>
 						<!--
- 
+
                                  ┌  0    1      0   0  ┐
                                  │ -1          1       │    ┌ P  ┐
                                  │ ──    0     ──   0  │    │  1 │
@@ -7839,7 +7839,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  1 1           1 -1 │  · │  2 │
 						/>
 						<p>This looks a lot like the Bézier matrix form, which as we saw in the chapter on Bézier curves, should look like this:</p>
 						<!--
- 
+
                                                ┌ P  ┐
                                                │  1 │
                              ┌  1  0  0 0 ┐    │ P  │
@@ -7858,7 +7858,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>So, if we want to express a Catmull-Rom curve using a Bézier curve, we'll need to turn this Catmull-Rom bit:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │
@@ -7879,7 +7879,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Into something that looks like this:</p>
 						<!--
- 
+
                   ┌ P  ┐
                   │  1 │
 ┌  1  0  0 0 ┐    │ P  │
@@ -7898,7 +7898,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And the way we do that is with a fairly straight forward bit of matrix rewriting. We start with the equality we need to ensure:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐                          ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │                          │  1 │
@@ -7919,7 +7919,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we remove the coordinate vector from both sides without affecting the equality:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │
 │ ──    0     ──   0  │
@@ -7940,7 +7940,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we can "get rid of" the Bézier matrix on the right by left-multiply both with the inverse of the Bézier matrix:</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -7964,7 +7964,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							outright remove any matrix/inverse pair:
 						</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -7985,7 +7985,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And now we're <em>basically</em> done. We just multiply those two matrices and we know what <em>V</em> is:</p>
 						<!--
- 
+
 ┌  0  1  0 0  ┐
 │ -1    1     │
 │ ──  1 ── 0  │
@@ -8007,7 +8007,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							<li>Start with the Catmull-Rom function:</li>
 						</ol>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -8028,7 +8028,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite to pure coordinate form:</li>
 						</ol>
 						<!--
- 
+
                                       ┌   P     ┐
                                       │    2    │
                                       │   P     │
@@ -8053,7 +8053,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for "normal" coordinate vector:</li>
 						</ol>
 						<!--
- 
+
                                       ┌  0  1  0 0  ┐    ┌ P  ┐
                                       │  0  0  1 0  │    │  1 │
                    ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -8074,7 +8074,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>merge the inner matrices:</li>
 						</ol>
 						<!--
- 
+
                    ┌  0    1      0   0  ┐
                    │ -1          1       │    ┌ P  ┐
                    │ ──    0     ──   0  │    │  1 │
@@ -8097,7 +8097,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for Bézier matrix form:</li>
 						</ol>
 						<!--
- 
+
                                      ┌  0  1  0 0  ┐    ┌ P  ┐
                                      │ -1    1     │    │  1 │
                    ┌  1  0  0 0 ┐    │ ──  1 ── 0  │    │ P  │
@@ -8118,7 +8118,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>and transform the coordinates so we have a "pure" Bézier expression:</li>
 						</ol>
 						<!--
- 
+
                                      ┌     P       ┐
                                      │      2      │
                                      │      P -P   │
@@ -8147,13 +8147,13 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						curve that has the vector:
 					</p>
 					<!--
- 
+
                        ┌     P       ┐
                        │      2      │
 ┌ P  ┐                 │      P -P   │
 │  1 │                 │       3  1  │
 │ P  │                 │ P  + ────── │
-│  2 │             ==> │  2   6  · τ │        
+│  2 │             ==> │  2   6  · τ │
 │ P  │ CatmullRom      │      P -P   │  Bézier
 │  3 │                 │       4  2  │
 │ P  │                 │ P  - ────── │
@@ -8173,11 +8173,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						tension Catmull-Rom curve with the following coordinate values:
 					</p>
 					<!--
- 
+
 ┌ P  ┐              ┌       P         ┐
 │  1 │              │        1        │
 │ P  │              │       P         │
-│  2 │          ==> │        4        │           
+│  2 │          ==> │        4        │
 │ P  │  Bézier      │ P  + 3(P  - P ) │ CatmullRom
 │  3 │              │  4      1    2  │
 │ P  │              │ P  + 3(P  - P ) │
@@ -8192,11 +8192,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 					/>
 					<p>Or, if your API allows you to specify Catmull-Rom curves using plain coordinates:</p>
 					<!--
- 
+
 ┌ P  ┐              ┌ P  + 6(P  - P ) ┐
 │  1 │              │  4      1    2  │
 │ P  │              │       P         │
-│  2 │          ==> │        1        │           
+│  2 │          ==> │        1        │
 │ P  │  Bézier      │       P         │ CatmullRom
 │  3 │              │        4        │
 │ P  │              │ P  + 6(P  - P ) │
@@ -8287,8 +8287,8 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						the same as it's "outgoing" derivative:
 					</p>
 					<!--
- 
-B'(1)  = B'(0)   
+
+B'(1)  = B'(0)
      n        n+1
 -->
 					<img class="LaTeX SVG" src="./images/chapters/polybezier/a37252ff55837b918d9d64078ae92ae7.svg" width="124px" height="17px" loading="lazy" />
@@ -8298,7 +8298,7 @@ B'(1)  = B'(0)
 						last control point through the last on-curve point. And mirroring any point A through any point B is really simple:
 					</p>
 					<!--
- 
+
            ┌ B  + (B  - A ) ┐   ┌ 2B  - A  ┐
 Mirrored = │  x     x    x  │ = │   x    x │
            │ B  + (B  - A ) │   │ 2B  - A  │
@@ -8474,7 +8474,7 @@ Mirrored = │  x     x    x  │ = │   x    x │
 							any point on <code>O(t)</code> is a fixed distance <code>d</code> away from coordinate <code>B(t)</code>. So let's math that:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d
 -->
 						<img
@@ -8491,7 +8491,7 @@ O(t) = B(t) + d
 							at <code>B(t)</code>. Easy enough:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d  · N(t)
 -->
 						<img
@@ -8509,7 +8509,7 @@ O(t) = B(t) + d  · N(t)
 							<code>N(t)</code> always has length 1, which we can achieve by dividing <code>B'(t)</code> by its magnitude:
 						</p>
 						<!--
- 
+
           ╭   B'(t)    ╮
 N(t) \bot │ ────────── │
           ╰ || B'(t)|| ╯
@@ -8527,10 +8527,10 @@ N(t) \bot │ ────────── │
 							denoted with double vertical bars:
 						</p>
 						<!--
- 
+
                     ┌───────────┐
                ╭ b  │   2      2
-|| f(x,y)|| =  |    │f '  + f '  
+|| f(x,y)|| =  |    │f '  + f '
                ╯ a ⟍│ x      y
 -->
 						<img
@@ -8544,10 +8544,10 @@ N(t) \bot │ ────────── │
 							So if we want the length of the tangent, we plug in <code>B'(t)</code>, with <code>t = 0</code> as start and <code>t = 1</code> as end:
 						</p>
 						<!--
- 
+
                    ┌───────────────────┐
               ╭ 1  │       2          2
-|| B'(t)|| =  |    │B ''(t)  + B ''(t)  
+|| B'(t)|| =  |    │B ''(t)  + B ''(t)
               ╯ 0 ⟍│ x          y
 -->
 						<img
@@ -8761,7 +8761,7 @@ N(t) \bot │ ────────── │
 						some angle <em>φ</em>:
 					</p>
 					<!--
- 
+
              S = \begin{pmatrix} 1
 0 \end{pmatrix}  ,  \ E = \begin{pmatrix} cos(φ)
               sin(φ) \end{pmatrix}
@@ -8769,7 +8769,7 @@ N(t) \bot │ ────────── │
 					<img class="LaTeX SVG" src="./images/chapters/circles/7ab3da0922477af4cc09f5852100976b.svg" width="175px" height="40px" loading="lazy" />
 					<p>What we want to find is the intersection of the tangents, so we want a point C such that:</p>
 					<!--
- 
+
               C = S + a  · \begin{pmatrix} 0
 1 \end{pmatrix}  ,  \ C = E + b  · \begin{pmatrix} -sin(φ)
                    cos(φ) \end{pmatrix}
@@ -8780,22 +8780,22 @@ N(t) \bot │ ────────── │
 						through E (at some distance <em>b</em> from E). Solving this gives us:
 					</p>
 					<!--
- 
+
 ╭ C  = 1 = cos(φ) + b  · -sin(φ)
-╡  x                             
-│ C  = a = sin(φ) + b  · cos(φ) 
+╡  x
+│ C  = a = sin(φ) + b  · cos(φ)
 ╰  y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/8237af1396bb567d70c8b5e4dd7f8115.svg" width="219px" height="40px" loading="lazy" />
 					<p>First we solve for <em>b</em>:</p>
 					<!--
- 
+
 1 = cos(φ) + b  · -sin(φ)  → \ 1 - cos(φ) = -b  · sin(φ)  → \ -1 + cos(φ) = b  · sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/1829e42ea956ee4df0e45d9ac5334ef7.svg" width="560px" height="17px" loading="lazy" />
 					<p>which yields:</p>
 					<!--
- 
+
     cos(φ)-1
 b = ────────
      sin(φ)
@@ -8803,21 +8803,21 @@ b = ────────
 					<img class="LaTeX SVG" src="./images/chapters/circles/06369b00338310df0a810c592485aa0a.svg" width="101px" height="39px" loading="lazy" />
 					<p>which we can then substitute in the expression for <em>a</em>:</p>
 					<!--
- 
- a= sin(φ) + b  · cos(φ)          
+
+ a= sin(φ) + b  · cos(φ)
              -1 + cos(φ)
 ..= sin(φ) + ───────────  · cos(φ)
                sin(φ)
                           2
              -cos(φ) + cos (φ)
-..= sin(φ) + ─────────────────    
-                  sin(φ)          
+..= sin(φ) + ─────────────────
+                  sin(φ)
        2         2
     sin (φ) + cos (φ) - cos(φ)
-..= ──────────────────────────    
+..= ──────────────────────────
               sin(φ)
     1 - cos(φ)
- a= ──────────                    
+ a= ──────────
       sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/5a23f3bc298c85540c6dd18e304d9224.svg" width="231px" height="195px" loading="lazy" />
@@ -8828,7 +8828,7 @@ b = ────────
 						polynomial, and we know where the circle arc's actual point P is for angle φ/2:
 					</p>
 					<!--
- 
+
          φ                 φ
 P  = cos(─)  ,  \ P  = sin(─)
  x       2         y       2
@@ -8836,7 +8836,7 @@ P  = cos(─)  ,  \ P  = sin(─)
 					<img class="LaTeX SVG" src="./images/chapters/circles/8b4e1d0a62380ed011f27c645ed13b28.svg" width="188px" height="32px" loading="lazy" />
 					<p>We compute T, observing that if <em>t=0.5</em>, the polynomial values (1-t)², 2(1-t)t, and t² are 0.25, 0.5, and 0.25 respectively:</p>
 					<!--
- 
+
     1    2    1    1
 T = ─S + ─C + ─E = ─(S + 2C + E)
     4    4    4    4
@@ -8844,10 +8844,10 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/a04bd1558a76e60b8ca6e1fe4fa38c00.svg" width="252px" height="35px" loading="lazy" />
 					<p>Which, worked out for the x and y components, gives:</p>
 					<!--
- 
+
 ╭     1
-│ T = ─(3 + cos(φ))                                    
-╡  x  4                                                 
+│ T = ─(3 + cos(φ))
+╡  x  4
 │     1╭ 2-2cos(φ)          ╮   1╭     ╭ φ ╮          ╮
 │ T = ─│ ───────── + sin(φ) │ = ─│ 2tan│ ─ │ + sin(φ) │
 ╰  y  4╰  sin(φ)            ╯   4╰     ╰ 2 ╯          ╯
@@ -8855,17 +8855,17 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/986ae9104e0bc52e95689ae7ae4504db.svg" width="408px" height="77px" loading="lazy" />
 					<p>And the distance between these two is the standard Euclidean distance:</p>
 					<!--
- 
+
                  1                   φ        4╭ φ ╮
 d (φ)= T  - P  = ─(3 + cos(φ)) - cos(─) = 2sin │ ─ │  ,
  x      x    x   4                   2         ╰ 4 ╯
                  1╭     ╭ φ ╮          ╮       φ
-d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,   
+d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
  y      y    y   4╰     ╰ 2 ╯          ╯       2
-     ⇓                                                 
+     ⇓
         ┌───────┐                     ┌───────┐
         │ 2    2                 4 φ  │   1
- d(φ)=  │d  + d   =  ...   = 2sin (─) │───────         
+ d(φ)=  │d  + d   =  ...   = 2sin (─) │───────
        ⟍│ x    y                   4  │   2 φ
                                       │cos (─)
                                      ⟍│     2
@@ -8912,7 +8912,7 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 						precision:
 					</p>
 					<!--
- 
+
                 ╭  ┌─────────────┐ ╮
                 │  │     ┌──────┐  │
                 │ ⟍│2+ε-⟍│ε(2+ε)   │
@@ -8972,14 +8972,14 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 					</p>
 					<p>So let's again formally describe this:</p>
 					<!--
- 
-P = (1, 0)                     
+
+P = (1, 0)
  1
-P = (1, k)                     
- 2                             
+P = (1, k)
+ 2
 P = P  + k  · (sin(θ), -cos(θ))
  3   4
-P = (cos(θ), sin(θ))           
+P = (cos(θ), sin(θ))
  4
 -->
 					<img
@@ -9003,29 +9003,29 @@ P = (cos(θ), sin(θ))
 						P<sub>2</sub>" (which is "half height" P<sub>2</sub>), and so forth:
 					</p>
 					<!--
- 
-     P   + P  
-      2     3 
+
+     P   + P
+      2     3
        y     y   k + sin(θ) - k  · cos(θ)
- A = ───────── = ────────────────────────                                 
+ A = ───────── = ────────────────────────
   y      2                  2
           1
-     A  + ─P     k + sin(θ) - k  · cos(θ)   
+     A  + ─P     k + sin(θ) - k  · cos(θ)
       y   2 2    ──────────────────────── + ─
              y              2               2   2k + sin(θ) + k  · cos(θ)
-e  = ───────── = ──────────────────────────── = ───────────────────────── 
+e  = ───────── = ──────────────────────────── = ─────────────────────────
  1       2                    2                             4
-  y                                                                       
+  y
                                        k
      A  +  mid(P , P )   A  + sin(θ) - ─ cos(θ)
       y         4   3     y            2          k + 3sin(θ) 2k  · cos(θ)
 e  = ───────────────── = ────────────────────── = ────────────────────────
  2           2                     2                         4
   y
-     e   + e  
-      1     2 
+     e   + e
+      1     2
        y     y   3k + 4sin(θ) - 3k  · cos(θ)
- B = ───────── = ───────────────────────────                              
+ B = ───────── = ───────────────────────────
   y      2                    8
 -->
 					<img
@@ -9049,15 +9049,15 @@ e  = ───────────────── = ───────
 							<a href="https://www.wolframalpha.com/">Wolfram Alpha</a> is definitely the way to go. That said, let's get going:
 						</p>
 						<!--
- 
+
 3k + 4sin(θ)) - 3k  · cos(θ)      θ
-────────────────────────────= sin(─)                  
+────────────────────────────= sin(─)
              8                    2
                                   ╭ θ ╮
-3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │               
+3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │
                                   ╰ 2 ╯
                                   ╭ θ ╮
-           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)     
+           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)
                                   ╰ 2 ╯
                                 ╭     ╭ θ ╮          ╮
              3k (1 - cos(θ))= 4 │ 2sin│ ─ │ - sin(θ) │
@@ -9065,12 +9065,12 @@ e  = ───────────────── = ───────
                                         θ
                                    2sin(─) - sin(θ)
                                         2
-                          3k= 4  · ────────────────   
+                          3k= 4  · ────────────────
                                       1 - cos(θ)
                                        ╭ θ ╮
                                    2sin│ ─ │ - sin(θ)
                               4        ╰ 2 ╯
-                           k= ─  · ────────────────── 
+                           k= ─  · ──────────────────
                               3        1 - cos(θ)
 -->
 						<img
@@ -9085,11 +9085,11 @@ e  = ───────────────── = ───────
 							<em>k</em>:
 						</p>
 						<!--
- 
+
             ╭ θ ╮
         2sin│ ─ │ - sin(θ)
    4        ╰ 2 ╯
-k= ─  · ──────────────────         
+k= ─  · ──────────────────
    3        1 - cos(θ)
         ╭     ╭ θ ╮               ╮
         │ 2sin│ ─ │               │
@@ -9097,10 +9097,10 @@ k= ─  · ──────────────────
 k= ─  · │ ────────── - ────────── │
    3    ╰ 1 - cos(θ)   1 - cos(θ) ╯
    4    ╭    ╭ θ ╮      ╭ θ ╮ ╮
-k= ─  · │ csc│ ─ │ - cot│ ─ │ │    
+k= ─  · │ csc│ ─ │ - cot│ ─ │ │
    3    ╰    ╰ 2 ╯      ╰ 2 ╯ ╯
    4       ╭ θ ╮
-k= ─  · tan│ ─ │                   
+k= ─  · tan│ ─ │
    3       ╰ 4 ╯
 -->
 						<img
@@ -9118,7 +9118,7 @@ k= ─  · tan│ ─ │
 						involving the circular arc's angle:
 					</p>
 					<!--
- 
+
            4    ╭ θ ╮
 k = f(θ) = ─ tan│ ─ │
            3    ╰ 4 ╯
@@ -9134,13 +9134,13 @@ k = f(θ) = ─ tan│ ─ │
 						Which means that for any circular arc with angle θ and radius <em>r</em>, our Bézier approximation based on three points of incidence is:
 					</p>
 					<!--
- 
-    start= (r, 0)                                          
-control  = (r, k)                                          
-        1                                                  
+
+    start= (r, 0)
+control  = (r, k)
+        1
 control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
         2
-      end= r  · (cos(θ), sin(θ))                           
+      end= r  · (cos(θ), sin(θ))
 -->
 					<img
 						class="LaTeX SVG"
@@ -9151,7 +9151,7 @@ control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
 					/>
 					<p>Which also gives us the commonly found value of 0.55228 for quarter circles, based on them having an angle of half π:</p>
 					<!--
- 
+
  ╭ \pi ╮   4       ╭ \pi ╮   4  ┌─┐
 f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.55228474983[...]
  ╰  2  ╯   3       ╰  8  ╯   3
@@ -9165,13 +9165,13 @@ f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.552284
 					/>
 					<p>And thus giving us the following Bézier coordinates for a quarter circle of radius <em>r</em>:</p>
 					<!--
- 
-    start= (r, 0)           
+
+    start= (r, 0)
 control  = (r, 0.55228  · r)
-        1                   
+        1
 control  = (0.55228  · r, r)
         2
-      end= (0, r)           
+      end= (0, r)
 -->
 					<img
 						class="LaTeX SVG"
@@ -9315,7 +9315,7 @@ getMostWrongT(radius, bezier, start, end, epsilon=1e-15):
 					</p>
 					<p>First, let's express the total error (given circular arc angle θ, and some <em>k</em>) using standard calculus notation:</p>
 					<!--
- 
+
                     ┌───────────────────────┐
               ╭ 1   │         2            2
 erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
@@ -9338,7 +9338,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						expression looks like:
 					</p>
 					<!--
- 
+
 ╭        ┌───────┐         ╮
 │  ╭ 1   │ 2    2          │ d
 │  |  \| │B  + B   - r\|dt │ ── = 0
@@ -9357,7 +9357,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						are each other's inverse operations, so they cancel out, leaving us with our original function:
 					</p>
 					<!--
- 
+
    ┌───────┐
    │ 2    2
 \| │B  + B   - r\| = 0,    t ∈[0,1]
@@ -9372,7 +9372,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 					/>
 					<p>And now we just solve for that... oh wait. We've seen this before. In order to solve this, we'd end up needing to solve this:</p>
 					<!--
- 
+
  2    2
 B  + B  = r
  x    y
@@ -9902,7 +9902,7 @@ radialError(radius, points[]):
 						end, just like for Bézier curves), by evaluating the following function:
 					</p>
 					<!--
- 
+
            __ n
 Point(t) = ❯      P   · N   (t)
            ‾‾ i=0  i     i,k
@@ -9923,7 +9923,7 @@ Point(t) = ❯      P   · N   (t)
 					</p>
 					<p>Then the N() function itself. What does it look like?</p>
 					<!--
- 
+
           ╭       t- knot         ╮                ╭       knot     -t       ╮
           │              i        │                │           (i+k)         │
 N   (t) = │ ───────────────────── │  · N     (t) + │ ─────────────────────── │  · N       (t)
@@ -9938,9 +9938,9 @@ N   (t) = │ ───────────────────── 
 						some point; obviously, it does, and specifically it does so for the following <code>i</code>/<code>k</code> values:
 					</p>
 					<!--
- 
+
           ╭ 1  if  t ∈[ knot , knot   )
-N   (t) = ╡                 i      i+1  
+N   (t) = ╡                 i      i+1
  i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/2514e1aa0565840e33fde0b146e3efe2.svg" width="237px" height="40px" loading="lazy" />
@@ -9961,7 +9961,7 @@ N   (t) = ╡                 i      i+1
 						P(t), we can compute this point by evaluating <em>d(t)</em> on a curve section between knots <code>i</code> and <code>i+1</code>:
 					</p>
 					<!--
- 
+
  k               k-1                   k-1
 d (t) = α     · d   (t) + (1-α   )  · d   (t)
  i       i,k     i            i,k      i-1
@@ -9972,7 +9972,7 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 						by:
 					</p>
 					<!--
- 
+
               t -  knots[i]
 α    = ───────────────────────────
  i,k    knots[i+1+n-k] -  knots[i]
@@ -9986,9 +9986,9 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 					</p>
 					<p>Of course, the recursion does need a stop condition:</p>
 					<!--
- 
+
  k           0                ╭ 1  if  t ∈[ knot , knot   )
-d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1  
+d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
  0           i       i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/49af474c33ce0ee0733626ea3d988570.svg" width="367px" height="40px" loading="lazy" />
@@ -10002,34 +10002,34 @@ d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
 						recursion diagram:
 					</p>
 					<!--
- 
+
      ╭                                ╭                                ╭          1     0           0
-     │                                │                                │         α   × d ,   with  d    either 0 or 1 
+     │                                │                                │         α   × d ,   with  d    either 0 or 1
      │                                │          2     1           1   │          3     3           3
-     │                                │         α   × d ,   with  d  = ╡                 +                              
+     │                                │         α   × d ,   with  d  = ╡                 +
      │                                │          3     3           3   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      │          3     2           2   │                                ╰ ╰      3 ╯     2           2
-     │         α   × d ,   with  d  = ╡                 +                                                                
+     │         α   × d ,   with  d  = ╡                 +
      │          3     3           3   │                                ╭           1     0
-     │                                │                                │          α   × d                             
+     │                                │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           2     2
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
  3   │                                │ ╰      3 ╯     2           2   │ ╭      1 ╮     0           0
-d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1     
+d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
  3   │                                ╰                                ╰ ╰      2 ╯     1           1
-     │                 +                                                                                                 
+     │                 +
      │                                ╭           2     1
-     │                                │          α   × d                                                                
+     │                                │          α   × d
      │                                │           2     2
-     │                                │                                                                                 
-     │ ╭      3 ╮     2           2   │                 +                                                               
-     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0                               
-     │ ╰      3 ╯     2           2   │                                │          α   × d 
+     │                                │
+     │ ╭      3 ╮     2           2   │                 +
+     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0
+     │ ╰      3 ╯     2           2   │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           1     1
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
      │                                │ ╰      2 ╯     1           1   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      ╰                                ╰                                ╰ ╰      1 ╯     0           0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/20e910bbea2e6eff511cb13cef18ef3b.svg" width="641px" height="336px" loading="lazy" />
diff --git a/docs/news/2020-09-18.html b/docs/news/2020-09-18.html
index 4ad3f20e..7b6615c2 100644
--- a/docs/news/2020-09-18.html
+++ b/docs/news/2020-09-18.html
@@ -176,15 +176,15 @@
 			</ul>
 			<blockquote>
 				<!--
- 
+
            __ n=3                  n-k k
-B(t)     = ❯     \ P  \binomnk(1-t)   t                
-    cubic  ‾‾ k=0    
+B(t)     = ❯     \ P  \binomnk(1-t)   t
+    cubic  ‾‾ k=0
                     k
-                                                       
+
                    3            2             2       3
-         = P  (1-t)  + 3 P (1-t) t + 3P (1-t)t  + P  t 
-                                                    
+         = P  (1-t)  + 3 P (1-t) t + 3P (1-t)t  + P  t
+
             0             1            2           3
 -->
 				<img class="LaTeX SVG" src="./images/news/2020-09-18.html/15225da473048d8c7b5b473b89de0b66.svg" width="401px" height="97px" loading="lazy" />
@@ -271,7 +271,7 @@ draw() {
 				<a href="https://github.com/Pomax/BezierInfo-2/issues">you know where to go</a>.
 			</p>
 			<p>See you in the next post!</p>
-			<p>— <a href="https://twitter.com/TheRealPomax">Pomax</a></p>
+			<p>— <a href="https://mastodon.social/@TheRealPomaxTheRealPomax">Pomax</a></p>
 		</main>
 
 		<hr />
diff --git a/docs/news/2020-09-18.md b/docs/news/2020-09-18.md
index 2c50de26..45c2231a 100644
--- a/docs/news/2020-09-18.md
+++ b/docs/news/2020-09-18.md
@@ -65,4 +65,4 @@ Enjoy [The new Primer on Bézier Curves](https://pomax.github.io/bezierinfo), an
 
 See you in the next post!
 
-— [Pomax](https://twitter.com/TheRealPomax)
+— [Pomax](https://mastodon.social/@TheRealPomax)
diff --git a/docs/news/2020-11-22.html b/docs/news/2020-11-22.html
index a40155b7..75da0c68 100644
--- a/docs/news/2020-11-22.html
+++ b/docs/news/2020-11-22.html
@@ -133,7 +133,7 @@
 				the point is the circle's center, and where the projection distance actually needs to match the circle radius, so:
 				<a href="https://pomax.github.io/bezierinfo/#circleintersection">let's see how to do that</a>!
 			</p>
-			<p>— <a href="https://twitter.com/TheRealPomax">Pomax</a></p>
+			<p>— <a href="https://mastodon.social/@TheRealPomax">Pomax</a></p>
 		</main>
 
 		<hr />
diff --git a/docs/news/2020-11-22.md b/docs/news/2020-11-22.md
index a22cb0b1..e4d59822 100644
--- a/docs/news/2020-11-22.md
+++ b/docs/news/2020-11-22.md
@@ -5,4 +5,4 @@ While the primer covered line/line, line/curve, and curve/curve intersections, t
 It is, in fact, rather similar to [projecting a point onto a bezier curve](https://pomax.github.io/bezierinfo/#projections) where the point is the circle's center, and where the projection distance actually needs to match the circle radius, so: [let's see how to do that](https://pomax.github.io/bezierinfo/#circleintersection)!
 
 
-— [Pomax](https://twitter.com/TheRealPomax)
+— [Pomax](https://mastodon.social/@TheRealPomax)
diff --git a/docs/news/rss.xml b/docs/news/rss.xml
index 9f41eeb5..bfd59b4a 100644
--- a/docs/news/rss.xml
+++ b/docs/news/rss.xml
@@ -17,10 +17,10 @@
     <title>Curve-circle intersections</title>
     <link>https://pomax.github.io/bezierinfo/news/2020-11-22.html</link>
     <description>
-    
+
 &lt;p&gt;While the primer covered line/line, line/curve, and curve/curve intersections, there was one other obvious intersection conspicuously missing: circle/curve intersections. You&#39;d think those were just an extension on the maths used for the other three, but unfortunately, this is not the case. Rather than using calculus, the only real way to determine where a polynomial curve intersects it is to sample the curve at a resolution high enough to find you intervals on the curve where there likely is an intersection, then refining that interval until you find actual intersections.&lt;/p&gt;
 &lt;p&gt;It is, in fact, rather similar to &lt;a href=&quot;https://pomax.github.io/bezierinfo/#projections&quot;&gt;projecting a point onto a bezier curve&lt;/a&gt; where the point is the circle&#39;s center, and where the projection distance actually needs to match the circle radius, so: &lt;a href=&quot;https://pomax.github.io/bezierinfo/#circleintersection&quot;&gt;let&#39;s see how to do that&lt;/a&gt;!&lt;/p&gt;
-&lt;p&gt;— &lt;a href=&quot;https://twitter.com/TheRealPomax&quot;&gt;Pomax&lt;/a&gt;&lt;/p&gt;
+&lt;p&gt;— &lt;a href=&quot;https://mastodon.social/@TheRealPomax&quot;&gt;Pomax&lt;/a&gt;&lt;/p&gt;
 
     </description>
     <pubDate>Sun Nov 22 2020 00:00:00 +00:00</pubDate>
@@ -29,7 +29,7 @@
     <title>Rewriting the tech stack</title>
     <link>https://pomax.github.io/bezierinfo/news/2020-09-18.html</link>
     <description>
-    
+
 &lt;p&gt;Once upon a time, I needed to draw some Bezier curves because I was trying to create a Japanese kanji composition system that turned strokes into outlines, and that required knowing how to offset Bezier curves and... at the time (2011, time flies) there was no good single source of information for Bezier curves on the web. So I made one. Sure it started small, but it turns out that if you just keep adding bits to something, several years later you have quite the monster, and a single HTML file becomes intractible.&lt;/p&gt;
 &lt;p&gt;So, in 2016, when &lt;a href=&quot;https://reactjs.org/&quot;&gt;React.js&lt;/a&gt; exploded onto the scene, I rewrote the primer as a React app, and it became a lot easier to maintain. Like, &lt;em&gt;a lot&lt;/em&gt; a lot. However, there was a downside: no JS meant no content. Sure, server-side rendering sort of existed, but not really, and because the Primer is hosted through github, there was no &quot;server&quot; to run. Plus, trying to rehydrate an app the size of the Primer from a giant HTML file had truly &lt;em&gt;dire&lt;/em&gt; performance.&lt;/p&gt;
 &lt;p&gt;So I left it a regular React app, and every time I thought &quot;wouldn&#39;t it be nice if it was just... a web page again?&quot; the browser landscape just hadn&#39;t caught up. Finally, in 2020, things are different: with a global pandemic, and some vacation time, and something random causing me to look up the state of HTML custom elements, everything was pointing at it being time to finally, &lt;em&gt;finally&lt;/em&gt;, turn the Primer back into a normal web page.&lt;/p&gt;
@@ -57,15 +57,15 @@
 &lt;/ul&gt;
 &lt;blockquote&gt;
 &lt;!--
- 
+
            __ n=3                  n-k k
-B(t)     = ❯     \ P  \binomnk(1-t)   t                
-    cubic  ‾‾ k=0    
+B(t)     = ❯     \ P  \binomnk(1-t)   t
+    cubic  ‾‾ k=0
                     k
-                                                       
+
                    3            2             2       3
-         = P  (1-t)  + 3 P (1-t) t + 3P (1-t)t  + P  t 
-                                                    
+         = P  (1-t)  + 3 P (1-t) t + 3P (1-t)t  + P  t
+
             0             1            2           3
 --&gt;
 &lt;img class=&quot;LaTeX SVG&quot; src=&quot;./images/news/2020-09-18.html/15225da473048d8c7b5b473b89de0b66.svg&quot; width=&quot;401px&quot; height=&quot;97px&quot; loading=&quot;lazy&quot;&gt;
@@ -105,7 +105,7 @@ draw() {
 &lt;tr&gt;&lt;td&gt;11&lt;/td&gt;&lt;/tr&gt;
 &lt;tr&gt;&lt;td&gt;12&lt;/td&gt;&lt;/tr&gt;
 &lt;tr&gt;&lt;td&gt;13&lt;/td&gt;&lt;/tr&gt;&lt;/table&gt;
-  
+
 &lt;ul&gt;
 &lt;li&gt;Responsive CSS, so the content intelligently reflows where possible.&lt;/li&gt;
 &lt;li&gt;A &quot;Live build&quot; setup for working on the content and code.&lt;/li&gt;
@@ -115,7 +115,7 @@ draw() {
 &lt;p&gt;It&#39;s going to take me a while to detail the entire tech stack, but ultimately what matters is that you get a Primer that is a normal &quot;vanilla&quot; HTML, CSS, and JS page again, that &quot;just works&quot; even with JS disabled.&lt;/p&gt;
 &lt;p&gt;Enjoy &lt;a href=&quot;https://pomax.github.io/bezierinfo&quot;&gt;The new Primer on Bézier Curves&lt;/a&gt;, and if you find any problems, &lt;a href=&quot;https://github.com/Pomax/BezierInfo-2/issues&quot;&gt;you know where to go&lt;/a&gt;.&lt;/p&gt;
 &lt;p&gt;See you in the next post!&lt;/p&gt;
-&lt;p&gt;— &lt;a href=&quot;https://twitter.com/TheRealPomax&quot;&gt;Pomax&lt;/a&gt;&lt;/p&gt;
+&lt;p&gt;— &lt;a href=&quot;https://mastodon.social/@TheRealPomax&quot;&gt;Pomax&lt;/a&gt;&lt;/p&gt;
 
     </description>
     <pubDate>Fri Sep 18 2020 00:00:00 +00:00</pubDate>
diff --git a/docs/ru-RU/index.html b/docs/ru-RU/index.html
index 36232714..6be66690 100644
--- a/docs/ru-RU/index.html
+++ b/docs/ru-RU/index.html
@@ -174,7 +174,7 @@
         <a class="btclk" href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu</a>
         либо используя QR-код справа, если вы предпочитаете такой вид =)
       </p>
-    
+
     </div>
     <div class="btcqr">
         <a href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">
@@ -186,7 +186,7 @@
 
 			<br style="clear: both;" />
 
-			<p>— <a href="https://twitter.com/TheRealPomax">Pomax</a></p>
+			<p>— <a href="https://mastodon.social/@TheRealPomax">Pomax</a></p>
 			<noscript>
 				<div class="note">
 					<header>
@@ -590,7 +590,7 @@
 						80ти % до второй, вычислить результат можно следующим образом:
 					</p>
 					<!--
- 
+
      ╭           p =  неикая точка        ╮
      │            1                       │
      │           p =  неикая другая точка │
@@ -655,7 +655,7 @@
 						другое значение следуя определенной логике
 					</p>
 					<!--
- 
+
 f(x) = cos (x)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/0cc876c56200446c60114c1b0eeeb2cc.svg" width="96px" height="17px" loading="lazy" />
@@ -668,7 +668,7 @@ f(x) = cos (x)
 						внимаю следующий пример:
 					</p>
 					<!--
- 
+
 f(a) = cos (a)
 f(b) = sin (b)
 -->
@@ -680,9 +680,9 @@ f(b) = sin (b)
 						Как здесь:
 					</p>
 					<!--
- 
+
 ╭ f (t) = cos (t)
-╡  a              
+╡  a
 │ f (t) = sin (t)
 ╰  b
 -->
@@ -699,8 +699,8 @@ f(b) = sin (b)
 						наших функций:
 					</p>
 					<!--
- 
-{ x = cos (t) 
+
+{ x = cos (t)
   y = sin (t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/6914ba615733c387251682db7a3db045.svg" width="77px" height="40px" loading="lazy" />
@@ -736,7 +736,7 @@ f(b) = sin (b)
 					</p>
 					<p>Возможно, вы помните полиномы из школьной программы. Они выглядят следующим образом:</p>
 					<!--
- 
+
              3         2
 f(x) = a  · x  + b  · x  + c  · x + d
 -->
@@ -756,12 +756,12 @@ f(x) = a  · x  + b  · x  + c  · x + d
 						<i>b</i> и т.д. принимающими "биноминальную" форму. Звучит сложно, но на практике выглядит понятнее:
 					</p>
 					<!--
- 
-  линийный= (1-t) + t                                        
+
+  линийный= (1-t) + t
                  2                      2
- квадратый= (1-t)  + 2  · (1-t)  · t + t                     
+ квадратый= (1-t)  + 2  · (1-t)  · t + t
                  3             2                       2    3
-кубический= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+кубический= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img
 						class="LaTeX SVG"
@@ -775,10 +775,10 @@ f(x) = a  · x  + b  · x  + c  · x + d
 						ясно. Примите ко внимаю такие биномиальные термины:
 					</p>
 					<!--
- 
-    линийный=  1 + 1           
-   квадратый=  1 + 2 + 1       
-  кубический=  1 + 3 + 3 + 1   
+
+    линийный=  1 + 1
+   квадратый=  1 + 2 + 1
+  кубический=  1 + 3 + 3 + 1
 квартический= 1 + 4 + 6 + 4 + 1
 -->
 					<img
@@ -804,11 +804,11 @@ f(x) = a  · x  + b  · x  + c  · x + d
 					</p>
 					<!--
 
-  линийный= \colorblue a + \colorred b                                                                                                               
- квадратый= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b                                                  
+  линийный= \colorblue a + \colorred b
+ квадратый= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b
 кубический= \colorblue a  · \colorblue a  · \colorblue a + \colorblue a  · \colorblue a  · \colorred b + \colorblue a  · \colorred b  · \colorred b +
-                                                                                               
-                                                                                               
+
+
                                                       \colorred b  · \colorred b  · \colorred b
 -->
 					<img
@@ -824,9 +824,9 @@ f(x) = a  · x  + b  · x  + c  · x + d
 						функцию:
 					</p>
 					<!--
- 
+
               __ n                                                                                                          n-i     i
-Bézier(n,t) = ❯      \underset биноминальный термин\underbrace\binomni  · \ \underset полиноминальный термин\underbrace(1-t)     · t 
+Bézier(n,t) = ❯      \underset биноминальный термин\underbrace\binomni  · \ \underset полиноминальный термин\underbrace(1-t)     · t
               ‾‾ i=0
 -->
 					<img
@@ -1146,9 +1146,9 @@ function Bezier(3,t):
 					<!--
 
               __ n                                                                                                          n-i     i
-Bézier(n,t) = ❯      \underset биноминальный термин\underbrace\binomni  · \ \underset полиноминальный термин\underbrace(1-t)     · t   · \ \underset 
+Bézier(n,t) = ❯      \underset биноминальный термин\underbrace\binomni  · \ \underset полиноминальный термин\underbrace(1-t)     · t   · \ \underset
               ‾‾ i=0
-                                                                  вес\underbracew 
+                                                                  вес\underbracew
                                                                                  i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/a337e3f97387b52d387fc01605314497.svg" width="421px" height="63px" loading="lazy" />
@@ -1159,11 +1159,11 @@ Bézier(n,t) = ❯      \underset биноминальный термин\underb
 						(25,190) и (210,250) заканчиваясь на (210,30), мы запишем это следующим образом:
 					</p>
 					<!--
- 
+
 ╭                               3                                   2                                             2                       3
-╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t  
+╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t
 │                               3                                    2                                             2                      3
-╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t 
+╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/be73034ac382b54863c7a18c2932cbbc.svg" width="473px" height="40px" loading="lazy" />
 					<p>Это производит кривую, график которой мы видели в первой главе этой статьи.</p>
@@ -1296,9 +1296,9 @@ function Bezier(3,t,w[]):
 					</p>
 					<p>Как и прочее, воплощение этого коэффициента не должно составить нам особого труда. Тогда как обычная функция:</p>
 					<!--
- 
+
               __ n                    n-i     i
-Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w 
+Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
               ‾‾ i=0                                i
 -->
 					<img
@@ -1310,13 +1310,13 @@ Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
 					/>
 					<p>Функция для соотносительных кривых Безье имеет два дополнительных термина:</p>
 					<!--
- 
+
                         __ n                    n-i     i
-                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio 
+                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio
                         ‾‾ i=0                                i                    i
 Rational  Bézier(n,t) = ────────────────────────────────────────────────────────────
                                      __ n                     n-i      i
-                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio  
+                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio
                                      ‾‾ i=0                                       i
 -->
 					<img
@@ -1505,8 +1505,8 @@ function RationalBezier(3,t,w[],r[]):
 						сочетание двух других значений, тогда общая формула для этого будет:
 					</p>
 					<!--
- 
-mixture = a  ·  value  + b  ·  value 
+
+mixture = a  ·  value  + b  ·  value
                      1              2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/dfd6ded3f0addcf43e0a1581627a2220.svg" width="212px" height="16px" loading="lazy" />
@@ -1519,8 +1519,8 @@ mixture = a  ·  value  + b  ·  value
 						суму более 100% на выводе, что можно записать как:
 					</p>
 					<!--
- 
-m = a  ·  value  + (1 - a)  ·  value 
+
+m = a  ·  value  + (1 - a)  ·  value
                1                    2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/4e0fa763b173e3a683587acf83733353.svg" width="213px" height="16px" loading="lazy" />
@@ -1598,61 +1598,61 @@ m = a  ·  value  + (1 - a)  ·  value
 						P<sub>...</sub> для обозначения координат в "одном или более пространстве".
 					</p>
 					<!--
- 
+
                   3                   2                             2          3
-B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t 
+B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t
         1              2                        3                        4
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/9a9a55f5b0323d9ea88f82fc6be58ad3.svg" width="468px" height="20px" loading="lazy" />
 					<p>Обобщив, игнорируя конкретные значения, мы получим:</p>
 					<!--
- 
+
             3             2                       2    3
-B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/87cfac83cb8a4b0bee68ef006effc611.svg" width="353px" height="19px" loading="lazy" />
 					<p>Это, в свою очередь, может быть записано как:</p>
 					<!--
- 
+
                 3
-... =      (1-t) 
+... =      (1-t)
                 2
     + 3  · (1-t)   · t
                      2
-    + 3  · (1-t)  · t 
+    + 3  · (1-t)  · t
              3
-    +       t  
+    +       t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/cdd88611833f3b178df91278359a4193.svg" width="140px" height="75px" loading="lazy" />
 					<p>Последнее мы можем раскрыть, записав как:</p>
 					<!--
- 
+
                                    3         2
 ... =  (1-t)  · (1-t)  · (1-t) = -t  + 3  · t  - 3  · t + 1
                                        3         2
     + 3  · (1-t)  · (1-t)  · t = 3  · t  - 6  · t  + 3  · t
                                             3         2
-    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t 
+    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t
                                              3
-    +        t  · t  · t       =            t  
+    +        t  · t  · t       =            t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/ec118f296511c6e9ac8727be3703a7ce.svg" width="397px" height="75px" loading="lazy" />
 					<p>Более того можно записать с коэффициентами 1 и 0, включив нивелированные термины:</p>
 					<!--
- 
+
              3         2
 ... = -1  · t  + 3  · t  - 3  · t + 1
              3         2
-    + +3  · t  - 6  · t  + 3  · t + 0 
+    + +3  · t  - 6  · t  + 3  · t + 0
              3         2
     + -3  · t  + 3  · t  + 0  · t + 0
              3         2
-    + +1  · t  + 0  · t  + 0  · t + 0 
+    + +1  · t  + 0  · t  + 0  · t + 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/67a5ea33d6c6558f7d954b18226f4956.svg" width="217px" height="75px" loading="lazy" />
 					<p>И уже это можно рассматривать как серию четырех матричных операций:</p>
 					<!--
- 
+
                  ┌ -1 ┐                    ┌ 3  ┐                    ┌ -3 ┐                    ┌ 1 ┐
 ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ -6 │ + ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ 0 │
 └ t  t  t 1 ┘    │ -3 │   └ t  t  t 1 ┘    │ 3  │   └ t  t  t 1 ┘    │ 0  │   └ t  t  t 1 ┘    │ 0 │
@@ -1661,7 +1661,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/8ecff6b8a37d60385d287ea2b26876db.svg" width="607px" height="72px" loading="lazy" />
 					<p>Скомбинировав в единую матричную операцию, получим:</p>
 					<!--
- 
+
                  ┌ -1  3 -3 1 ┐
 ┌  3  2     ┐  · │  3 -6  3 0 │
 └ t  t  t 1 ┘    │ -3  3  0 0 │
@@ -1673,7 +1673,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 						<code>t</code> матрицу горизонтально, а нашу "большую" матрицу — вертикально:
 					</p>
 					<!--
- 
+
                  ┌  1  0  0 0 ┐
 ┌      2  3 ┐  · │ -3  3  0 0 │
 └ 1 t t  t  ┘    │  3 -6  3 0 │
@@ -1682,7 +1682,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/1a64ed455c6dd2f8cacca5e5e12bdcc1.svg" width="227px" height="72px" loading="lazy" />
 					<p>Наконец, мы можем добавить оригинальные координаты единой третьей матрицей:</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -1695,7 +1695,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 					<img class="LaTeX SVG" src="./images/chapters/matrix/b32cae2dfc47d5f36df0bc3defb7dfa8.svg" width="323px" height="73px" loading="lazy" />
 					<p>Такой же фокус может быть проделан с квадратной кривой, в коем случае мы получаем:</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2141,7 +2141,7 @@ function drawCurvePoint(points[], t):
 						respectively: (we'll reverse the Bézier coefficients vector for legibility)
 					</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2158,7 +2158,7 @@ B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -2181,7 +2181,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						"point on the curve" information into a new matrix multiplication:
 					</p>
 					<!--
- 
+
                                                   ┌ P  ┐                                              ┌ P  ┐
                                    ┌  1  0 0 ┐    │  1 │                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2198,7 +2198,7 @@ B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌
 					/>
 					<p>and</p>
 					<!--
- 
+
                                                                ┌ P  ┐                                                       ┌ P  ┐
                                                                │  1 │                    ┌ 1 0  0 0  ┐                      │  1 │
                                              ┌  1  0  0 0 ┐    │ P  │                    │ 0 z  0 0  │    ┌  1  0  0 0 ┐    │ P  │
@@ -2228,7 +2228,7 @@ B(t) = ┌                    2         3 ┐  · │ -3  3  0 0 │  · │  2
 							at the quadratic curve first:
 						</p>
 						<!--
- 
+
                                                   ┌ P  ┐
                      ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2244,7 +2244,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                                         ┌ P  ┐
                                                                                         │  1 │
 = ┌      2 ┐  · \underset we turn this...\underbrace\kern 2.25em Z  · M \kern 2.25em  · │ P  │
@@ -2260,7 +2260,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                          ┌ P  ┐
                                                         -1               │  1 │
 = ┌      2 ┐  · \underset into this...\underbrace M  · M    · Z  · M   · │ P  │
@@ -2276,7 +2276,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                                                                    ┌ P  ┐
                                                                                                                    │  1 │
 = ┌      2 ┐  · M \underset ...to get this!\underbrace \kern 1.25em  · \kern 1.25em Q \kern 1.25em  · \kern 1.25em │ P  │
@@ -2303,7 +2303,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							coordinates that represent the curve from <em>t = 0</em> to <em>t = z</em>. So let's get computing:
 						</p>
 						<!--
- 
+
                     ┌ 1 0 0 ┐
      -1             │   1   │    ┌ 1 0 0  ┐    ┌  1  0 0 ┐   ┌     1            0        0  ┐
 Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = │  -(z-1)          z        0  │
@@ -2319,7 +2319,7 @@ Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = 
 						/>
 						<p>Excellent! Now we can form our new quadratic curve:</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭      ┌ P  ┐ ╮
                                │  1 │                      │      │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q  · │ P  │ │
@@ -2335,7 +2335,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                     ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌     1            0        0  ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │  -(z-1)          z        0  │  · │ P  │ │
@@ -2351,7 +2351,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌                        P                          ┐
                                │                         1                         │
                 ┌  1  0 0 ┐    │               z  · P  - (z-1)  · P                │
@@ -2380,7 +2380,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							write it down in a more simple form because of the zero, but what we <em>actually</em> evaluated, making the zero explicit, was:
 						</p>
 						<!--
- 
+
                                                             ┌ P  ┐
                                              ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2396,7 +2396,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                              ┌ P  ┐
                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2413,7 +2413,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 						/>
 						<p>If we want the interval [<em>z</em>,1], we will be evaluating this instead:</p>
 						<!--
- 
+
                                                                     ┌ P  ┐
                                                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2429,7 +2429,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 							loading="lazy"
 						/>
 						<!--
- 
+
                 ┌              2        ┐                   ┌ P  ┐
                 │ 1  z        z         │    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │  · │ P  │
@@ -2449,7 +2449,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 							<code>[M · something]</code>:
 						</p>
 						<!--
- 
+
                       ┌ 1 0 0 ┐    ┌              2        ┐
       -1              │   1   │    │ 1  z        z         │    ┌  1  0 0 ┐   ┌      2                   2 ┐
 Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │ = │ (z-1)  -2  · z  · (z-1) z  │
@@ -2465,7 +2465,7 @@ Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  ·
 						/>
 						<p>So, our final second curve looks like:</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭       ┌ P  ┐ ╮
                                │  1 │                      │       │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q'  · │ P  │ │
@@ -2481,7 +2481,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                   ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌      2                   2 ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │ (z-1)  -2  · z  · (z-1) z  │  · │ P  │ │
@@ -2497,7 +2497,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌  2                                      2       ┐
                                │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
                 ┌  1  0 0 ┐    │        3                       2              1 │
@@ -2527,7 +2527,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						<code>t = z</code>, we get two subcurves that are described as Bézier curves with simple-to-derive coordinates:
 					</p>
 					<!--
- 
+
                                              ┌                        P                          ┐
                                     ┌ P  ┐   │                         1                         │
 ┌     1            0        0  ┐    │  1 │   │               z  · P  - (z-1)  · P                │
@@ -2545,7 +2545,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                            ┌  2                                      2       ┐
                                   ┌ P  ┐   │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
 ┌      2                   2 ┐    │  1 │   │        3                       2              1 │
@@ -2566,7 +2566,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						yourself, though) and simply show you the resulting new coordinate sets:
 					</p>
 					<!--
- 
+
                                                               ┌                                    P                                      ┐
                                                      ┌ P  ┐   │                                     1                                     │
 ┌    1            0                0         0  ┐    │  1 │   │                           z  · P  - (z-1)  · P                            │
@@ -2587,7 +2587,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                                               ┌  3               2                                 2              3       ┐
                                                      ┌ P  ┐   │ z   · P  - 3  · z   · (z-1)  · P  + 3  · z  · (z-1)   · P  - (z-1)   · P  │
 ┌       3           2                      2  3 ┐    │  1 │   │        4                        3                        2              1 │
@@ -2639,7 +2639,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						that the start and end weights are the same as the start and end weights for the old curve):
 					</p>
 					<!--
- 
+
               __ k                                                                                            k-i     i
 Bézier(k,t) = ❯      \underset binomial term\underbrace\binomki  · \ \underset polynomial term\underbrace(1-t)     · t    · \ \underset new
               ‾‾ i=0
@@ -2663,9 +2663,9 @@ Bézier(k,t) = ❯      \underset binomial term\underbrace\binomki  · \ \unders
 					</p>
 					<p>We start by taking the standard Bézier function, and condensing it a little:</p>
 					<!--
- 
+
               __ n       n               n                       n-i     i
-Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t 
+Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t
               ‾‾ i=0  i  i               i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/9fc4ecc087d389dd0111bcba165cd5d0.svg" width="408px" height="41px" loading="lazy" />
@@ -2675,7 +2675,7 @@ Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t
 						<code>t</code> and <code>1-t</code>:
 					</p>
 					<!--
- 
+
 x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/ff224ded6bbbc94b43130f5f8eeb5d29.svg" width="379px" height="16px" loading="lazy" />
@@ -2684,9 +2684,9 @@ x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
 						<code>t</code> component:
 					</p>
 					<!--
- 
-Bézier(n,t)= (1-t) B(n,t) + t B(n,t)                    
-             __ n               n      __ n         n   
+
+Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
+             __ n               n      __ n         n
            = ❯      w  (1 - t) B (t) + ❯      w  t B (t)
              ‾‾ i=0  i          i      ‾‾ i=0  i    i
 -->
@@ -2696,18 +2696,18 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
 						I promise, it's about to make sense. We start with <code>(1-t)</code>:
 					</p>
 					<!--
- 
+
          n              n!         n-i  i
-(1 - t) B (t)= (1-t) ──────── (1-t)    t                  
+(1 - t) B (t)= (1-t) ──────── (1-t)    t
          i           (n-i)!i!
                n+1-i   (n+1)!        n+1-i  i
-             = ───── ────────── (1-t)      t              
-                n+1  (n+1-i)!i!                           
+             = ───── ────────── (1-t)      t
+                n+1  (n+1-i)!i!
                k-i    k!         k-i  i
              = ─── ──────── (1-t)    t ,  where  k = n + 1
                 k  (k-i)!i!
                k-i  k
-             = ─── B (t)                                  
+             = ─── B (t)
                 k   i
 -->
 					<img
@@ -2725,18 +2725,18 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
 						<code>t</code> part, but that's not a problem:
 					</p>
 					<!--
- 
+
    n          n!         n-i  i
-t B (t)= t ──────── (1-t)    t                                    
+t B (t)= t ──────── (1-t)    t
    i       (n-i)!i!
          i+1        (n+1)!             (n+1)-(i+1)  i+1
-       = ─── ──────────────────── (1-t)            t              
-         n+1 ((n+1)-(i+1))!(i+1)!                                 
+       = ─── ──────────────────── (1-t)            t
+         n+1 ((n+1)-(i+1))!(i+1)!
          i+1        k!             k-(i+1)  i+1
        = ─── ──────────────── (1-t)        t   ,  where  k = n + 1
           k  (k-(i+1))!(i+1)!
          i+1  k
-       = ─── B   (t)                                              
+       = ─── B   (t)
           k   i+1
 -->
 					<img
@@ -2756,21 +2756,21 @@ t B (t)= t ──────── (1-t)    t
 					</p>
 					<p>Let's do this:</p>
 					<!--
- 
+
              __ n+1             n      __ n+1       n
-Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)                   
+Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)
              ‾‾ i=0  i          i      ‾‾ i=0  i    i
              __ n+1    k-i  k      __ n+1    i+1  k
            = ❯      w  ─── B (t) + ❯      w  ─── B   (t),  where  k = n + 1
              ‾‾ i=0  i  k   i      ‾‾ i=0  i  k   i+1
              __ n+1    k-i  k      __ n+1      i  k
-           = ❯      w  ─── B (t) + ❯      p    ─ B (t)                     
+           = ❯      w  ─── B (t) + ❯      p    ─ B (t)
              ‾‾ i=0  i  k   i      ‾‾ i=0  i-1 k  i
              __ n+1 ╭    k-i        i ╮  k
-           = ❯      │ w  ─── + p    ─ │ B (t)                              
+           = ❯      │ w  ─── + p    ─ │ B (t)
              ‾‾ i=0 ╰  i  k     i-1 k ╯  i
              __ n+1                      k                 i
-           = ❯      (w  (1-s) + p    s) B (t),  where  s = ─               
+           = ❯      (w  (1-s) + p    s) B (t),  where  s = ─
              ‾‾ i=0   i          i-1     i                 k
 -->
 					<img
@@ -2785,14 +2785,14 @@ Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)
 						Bézier(n+1,t) as a very simple matrix multiplication:
 					</p>
 					<!--
- 
-M B  = B 
+
+M B  = B
    n    k
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/1f5b60d190a1c7099b3411e4cc477291.svg" width="71px" height="16px" loading="lazy" />
 					<p>where the matrix <strong>M</strong> is an <code>n+1</code> by <code>n</code> matrix, and looks like:</p>
 					<!--
- 
+
     ┌ 1  0   .   .   .   .    .    .  ┐
     │ 1 k-1                           │
     │ ─ ───  0   .   .   .    0    .  │
@@ -2834,20 +2834,20 @@ M = │        k   k                    │
 						<strong>A x = b</strong>, we can use it. And as it so happens, that's exactly what we're dealing with, so:
 					</p>
 					<!--
- 
-              M B = B             
+
+              M B = B
                  n   k
            T         T
-         (M  M) B = M  B          
+         (M  M) B = M  B
                  n      k
   T   -1   T          T   -1  T
-(M  M)   (M  M) B = (M  M)   M  B 
+(M  M)   (M  M) B = (M  M)   M  B
                  n               k
                       T   -1  T
-              I B = (M  M)   M  B 
+              I B = (M  M)   M  B
                  n               k
                       T   -1  T
-                B = (M  M)   M  B 
+                B = (M  M)   M  B
                  n               k
 -->
 					<img
@@ -2911,9 +2911,9 @@ M = │        k   k                    │
 					</p>
 					<p>Для начала рассмотрим правило производных для кривых Безье:</p>
 					<!--
- 
+
                     __ n-1
-Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t) 
+Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
                     ‾‾ i=0   i+1  i                    i
 -->
 					<img
@@ -2928,7 +2928,7 @@ Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
 						— то-же что функция сумы, с каждой составляющей умноженной на <i>n</i>) как:
 					</p>
 					<!--
- 
+
                __ n-1
 Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
                ‾‾ i=0                i           i+1  i
@@ -2954,7 +2954,7 @@ Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
 							общей формулы функции, потому производная кривой затрагивает только производные полиноминалов базовой функции. Найдем сперва это:
 						</p>
 						<!--
- 
+
         d    ╭ n ╮  k      n-k d
 B   (t) ── = │   │ t  (1-t)    ──
  n,k    dt   ╰ k ╯             dt
@@ -2973,7 +2973,7 @@ B   (t) ── = │   │ t  (1-t)    ──
 							следующее:
 						</p>
 						<!--
- 
+
       ╭ n ╮        k-1      n-k    k         n-k-1
 ... = │   │ (k  · t    (1-t)    + t   · (1-t)       · (n-k)  · -1)
       ╰ k ╯
@@ -2987,9 +2987,9 @@ B   (t) ── = │   │ t  (1-t)    ──
 						/>
 						<p>В таком виде с этой формулой работать сложно, потому раскроем:</p>
 						<!--
- 
+
         kn!     k-1      n-k   (n-k)n!   k      n-1-k
-... = ──────── t    (1-t)    - ──────── t  (1-t)     
+... = ──────── t    (1-t)    - ──────── t  (1-t)
       k!(n-k)!                 k!(n-k)!
 -->
 						<img
@@ -3005,12 +3005,12 @@ B   (t) ── = │   │ t  (1-t)    ──
 							<i>k-1</i> — мы на верном пути.
 						</p>
 						<!--
- 
+
            n!       k-1      n-k   (n-k)n!   k      n-1-k
-... = ──────────── t    (1-t)    - ──────── t  (1-t)                                   
+... = ──────────── t    (1-t)    - ──────── t  (1-t)
       (k-1)!(n-k)!                 k!(n-k)!
         ╭    (n-1)!     k-1      n-k   (n-k)(n-1)!  k      n-1-k ╮
-... = n │ ──────────── t    (1-t)    - ─────────── t  (1-t)      │                     
+... = n │ ──────────── t    (1-t)    - ─────────── t  (1-t)      │
         ╰ (k-1)!(n-k)!                  k!(n-k)!                 ╯
         ╭        (n-1)!         (k-1)      (n-1)-(k-1)      (n-1)!     k      (n-1)-k ╮
 ... = n │ ──────────────────── t      (1-t)            - ──────────── t  (1-t)        │
@@ -3025,11 +3025,11 @@ B   (t) ── = │   │ t  (1-t)    ──
 						/>
 						<p>Первая часть готова: компоненты в скобках на самом то деле обычные выражения кривых Безье на порядок ниже:</p>
 						<!--
- 
+
         ╭    x!     y      x-y      x!     k      x-k ╮
 ... = n │ ──────── t  (1-t)    - ──────── t  (1-t)    │  , где x=n-1, y=k-1
-        ╰ y!(x-y)!               k!(x-k)!             ╯                    
-... = n (B           (t) - B       (t))                                    
+        ╰ y!(x-y)!               k!(x-k)!             ╯
+... = n (B           (t) - B       (t))
           (n-1),(k-1)       (n-1),k
 -->
 						<img
@@ -3043,19 +3043,19 @@ B   (t) ── = │   │ t  (1-t)    ──
 							Теперь применим это к записям наших формул с "весами". Начнем с формулы кривой Безье приведенной выше и пройдемся по ее производным.
 						</p>
 						<!--
- 
+
 Bézier   (t)    = B   (t)  · w  + B   (t)  · w  + B   (t)  · w  + B   (t)  · w  + ...
       n,k          n,0        0    n,1        1    n,2        2    n,3        3
              d
-Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +                              
+Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +
       n,k    dt          n-1,-1       n-1,0         0
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,0       n-1,1         1
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,1       n-1,2         2
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,2       n-1,3         3
-                  ...                                                                
+                  ...
 -->
 						<img
 							class="LaTeX SVG"
@@ -3069,18 +3069,18 @@ Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +
 							следующее:
 						</p>
 						<!--
- 
-n  · B      (t)  · w               +                                     
+
+n  · B      (t)  · w               +
       n-1,-1        0
 n  · B                (t)  · w     - n  · B                (t)  · w     +
       n-1,\colorblue 0        1            n-1,\colorblue 0        0
 n  · B               (t)  · w      - n  · B               (t)  · w      +
-      n-1,\colorred 1        2             n-1,\colorred 1        1      
+      n-1,\colorred 1        2             n-1,\colorred 1        1
 n  · B                   (t)  · w  - n  · B                   (t)  · w  +
       n-1,\colormagenta 2        3         n-1,\colormagenta 2        2
 ...                                - n  · B     (t)  · w                +
                                            n-1,3        3
-...                                                                      
+...
 -->
 						<img
 							class="LaTeX SVG"
@@ -3097,14 +3097,14 @@ n  · B                   (t)  · w  - n  · B                   (t)  · w  +
 							привносит в результат, затем также может быть проигнорирован. Это значит, нам остается:
 						</p>
 						<!--
- 
+
 n  · B                (t)  · w     - n  · B                (t)  · w     +
       n-1,\colorblue 0        1            n-1,\colorblue 0        0
 n  · B               (t)  · w      -  n  · B               (t)  · w     +
-      n-1,\colorred 1        2              n-1,\colorred 1        1     
+      n-1,\colorred 1        2              n-1,\colorred 1        1
 n  · B                   (t)  · w  - n  · B                   (t)  · w  +
       n-1,\colormagenta 2        3         n-1,\colormagenta 2        2
-...                                
+...
 -->
 						<img
 							class="LaTeX SVG"
@@ -3115,9 +3115,9 @@ n  · B                   (t)  · w  - n  · B                   (t)  · w  +
 						/>
 						<p>И это по сути формула функции сумы на 1 порядок ниже:</p>
 						<!--
- 
+
              d
-Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B                 (t)  · (w  - w ) + n  · B                     (t)  · (w  - w )  
+Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B                 (t)  · (w  - w ) + n  · B                     (t)  · (w  - w )
       n,k    dt         (n-1),\colorblue 0         1    0          (n-1),\colorred 1         2    1          (n-1),\colormagenta 2         3    2
                                                                       +  ...
 -->
@@ -3130,7 +3130,7 @@ Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B
 						/>
 						<p>Можно переписать по стандартной форме сумы, и готово:</p>
 						<!--
- 
+
              d    __ n-1                                 __ n-1
 Bézier   (t) ── = ❯      n  · B     (t)  · (w    - w ) = ❯      B     (t)  · \underset вес производной \underbracen  · (w    - w )
       n,k    dt   ‾‾ k=0       n-1,k         k+1    k    ‾‾ k=0  n-1,k                                                   k+1    k
@@ -3149,11 +3149,11 @@ Bézier   (t) ── = ❯      n  · B     (t)  · (w    - w ) = ❯      B
 						производная:
 					</p>
 					<!--
- 
+
               __ n                                                                                                          n-i     i
-Bézier(n,t) = ❯      \underset биноминальный термин\underbrace\binomni  · \ \underset полиноминальный термин\underbrace(1-t)     · t   · \ \underset 
+Bézier(n,t) = ❯      \underset биноминальный термин\underbrace\binomni  · \ \underset полиноминальный термин\underbrace(1-t)     · t   · \ \underset
               ‾‾ i=0
-                                                                 вес\underbracew 
+                                                                 вес\underbracew
                                                                                 i
 -->
 					<img
@@ -3164,7 +3164,7 @@ Bézier(n,t) = ❯      \underset биноминальный термин\underb
 						loading="lazy"
 					/>
 					<!--
- 
+
                __ k                                                                                                          k-i     i
 Bézier'(n,t) = ❯      \underset биноминальный термин\underbrace\binomki  · \ \underset полиноминальный термин\underbrace(1-t)     · t   · \
                ‾‾ i=0
@@ -3184,11 +3184,11 @@ Bézier'(n,t) = ❯      \underset биноминальный термин\under
 						исходя из 4-х контрольных точек A, B, C и D, первая производная получит 3 точки, вторая — 2, третья — 1:
 					</p>
 					<!--
- 
-B(n,t),           w = {A,B,C,D}   
+
+B(n,t),           w = {A,B,C,D}
 B'(n,t),   n = 3, w' = {A',B',C'} = {3  · (B-A),   3  · (C-B),   3  · (D-C)}
-B''(n,t),  n = 2, w'' = {A'',B''} = {2  · (B'-A'),   2  · (C'-B')}          
-B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}                        
+B''(n,t),  n = 2, w'' = {A'',B''} = {2  · (B'-A'),   2  · (C'-B')}
+B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}
 -->
 					<img
 						class="LaTeX SVG"
@@ -3218,10 +3218,10 @@ B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}
 						indicates the direction of travel at specific points, and is literally just the first derivative of our curve:
 					</p>
 					<!--
- 
+
 tangent (t) = B' (t)
        x        x
-                    
+
 tangent (t) = B' (t)
        y        y
 -->
@@ -3237,14 +3237,14 @@ tangent (t) = B' (t)
 						point, and then do whatever it is we want to do based on those directions:
 					</p>
 					<!--
- 
+
                            ┌─────────────────┐
                            │      2         2
-    d = \| tangent(t)\| =  │B' (t)  + B' (t)  
+    d = \| tangent(t)\| =  │B' (t)  + B' (t)
                           ⟍│  x         y
 
                            tangent (t)     B' (t)
-^                                 x          x    
+^                                 x          x
 x(t) = \| tangent (t)\| =─────────────── = ──────
                  x       \| tangent(t)\|     d
 
@@ -3266,11 +3266,11 @@ y(t) = \| tangent (t)\| = ─────────────── = ──
 						is typically of length 1.0, so all we have to do is rotate the normalized directional vector and we're done:
 					</p>
 					<!--
- 
+
              ^           \pi   ^           \pi     ^
-normal (t) = x(t)  · cos ─── - y(t)  · sin ─── = - y(t)                                             
+normal (t) = x(t)  · cos ─── - y(t)  · sin ─── = - y(t)
       x                   2                 2
-                                                                                                    
+
                                                            ^           \pi   ^           \pi    ^
 normal (t) = \underset quarter circle rotation \underbrace x(t)  · sin ─── + y(t)  · cos ───  = x(t)
       y                                                                 2                 2
@@ -3293,7 +3293,7 @@ normal (t) = \underset quarter circle rotation \underbrace x(t)  · sin ──
 							To turn any point <i>(x,y)</i> into a rotated point <i>(x',y')</i> (over 0,0) by some angle φ, we apply this nice and easy computation:
 						</p>
 						<!--
- 
+
 x' = x  · cos (\phi) - y  · sin (\phi)
 y' = x  · sin (\phi) + y  · cos (\phi)
 -->
@@ -3306,7 +3306,7 @@ y' = x  · sin (\phi) + y  · cos (\phi)
 						/>
 						<p>Which is the "long" version of the following matrix transformation:</p>
 						<!--
- 
+
 ┌ x' ┐ = ┌ cos (\phi) -sin (\phi) ┐ ┌ x ┐
 └ y' ┘   └ sin (\phi) cos (\phi)  ┘ └ y ┘
 -->
@@ -3798,10 +3798,10 @@ generateRMFrames(steps) -> frames:
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
+
 B'(t) = a(1-t) + b(t)= 0,
           a - at + bt= 0,
-           (b-a)t + a= 0 
+           (b-a)t + a= 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -3812,10 +3812,10 @@ B'(t) = a(1-t) + b(t)= 0,
 					/>
 					<p>And then we turn this into our solution for <code>t</code> using basic arithmetics:</p>
 					<!--
- 
-(b-a)t + a= 0, 
+
+(b-a)t + a= 0,
     (b-a)t= -a,
-            -a 
+            -a
          t= ───
             b-a
 -->
@@ -3838,7 +3838,7 @@ B'(t) = a(1-t) + b(t)= 0,
 						you haven't, it looks like this:
 					</p>
 					<!--
- 
+
                                                    ┌────────┐
                                                    │ 2
                2                              -b ±⟍│b  - 4ac
@@ -3863,9 +3863,9 @@ Given f(t) = at  + bt + c, f(t)=0   when  t = ───────────
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
-B(t)  uses { p ,p ,p ,p  }                                                  
-              1  2  3  4                                                    
+
+B(t)  uses { p ,p ,p ,p  }
+              1  2  3  4
 B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
                1  2  3            1      2  1    2      3  2    3      4  3
 -->
@@ -3878,21 +3878,21 @@ B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
 					/>
 					<p>And then, using these <em>v</em> values, we can find out what our <em>a</em>, <em>b</em>, and <em>c</em> should be:</p>
 					<!--
- 
+
               2                  2
-B'(t)= v (1-t)  + 2v (1-t)t + v t            
+B'(t)= v (1-t)  + 2v (1-t)t + v t
         1           2          3
            2                    2       2
-  ...= v (t  - 2t + 1) + 2v (t-t ) + v t     
+  ...= v (t  - 2t + 1) + 2v (t-t ) + v t
         1                  2          3
           2                          2      2
-  ...= v t  - 2v t + v  + 2v t - 2v t  + v t 
+  ...= v t  - 2v t + v  + 2v t - 2v t  + v t
         1       1     1     2      2      3
           2       2      2
   ...= v t  - 2v t  + v t  - 2v t + v  + 2v t
         1       2      3       1     1     2
                    2
-  ...= (v -2v +v )t  + 2(v -v )t + v         
+  ...= (v -2v +v )t  + 2(v -v )t + v
          1   2  3         2  1      1
 -->
 					<img
@@ -3907,12 +3907,12 @@ B'(t)= v (1-t)  + 2v (1-t)t + v t
 						expressions of our original coordinate values, so we can do some substitution to get:
 					</p>
 					<!--
- 
+
 a= v -2v +v  = 3(-p  + 3p  - 3p  + p )
     1   2  3       1     2     3    4
-b= 2(v -v ) = 6(p  - 2p  + p )        
+b= 2(v -v ) = 6(p  - 2p  + p )
       2  1       1     2    3
-c= v  = 3(p -p )                      
+c= v  = 3(p -p )
     1      2  1
 -->
 					<img
@@ -3939,11 +3939,11 @@ c= v  = 3(p -p )
 						really hard to solve, it can be rewritten to a form for which finding the roots is "easier" (even if not "easy"):
 					</p>
 					<!--
- 
+
                    3     2
 very hard: solve at  + bt  + ct + d = 0
                   3
-   easier: solve t  + pt + q = 0       
+   easier: solve t  + pt + q = 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -4352,7 +4352,7 @@ function getCubicRoots(pa, pb, pc, pd) {
 						<code>f<sub>y</sub>(x)</code> is zero, so that the next <code>t</code> is the same as the one we already have:
 					</p>
 					<!--
- 
+
             f (x )
              y  n
 x    = x  - ───────
@@ -4501,38 +4501,38 @@ x    = x  - ───────
 						ниже. Так, если у нас была кубическая ф-ция:
 					</p>
 					<!--
- 
+
 ╭                            3                               2                                          2                       3
-╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t  
+╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t
 │                            3                                2                                          2                      3
-╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t 
+╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/e5db5e945606d3429e4475ff92283a9c.svg" width="497px" height="40px" loading="lazy" />
 					<p>Мы переместим первую точку в начало координат, смещая все значания x на -120 и y на -160:</p>
 					<!--
- 
+
 ╭                         3                               2                                          2                       3
-╡ x = \colorred 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t  
+╡ x = \colorred 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t
 │                         3                               2                                          2                       3
-╰ y = \colorred 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t 
+╰ y = \colorred 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/1564fdc8d17a064fea88b9d508878e62.svg" width="465px" height="41px" loading="lazy" />
 					<p>Затем, повернем кривую до оси X (новые значания x и y округлены в иллюстративных целях):</p>
 					<!--
- 
+
 ╭                         3                               2                                         2                       3
-╡ x = \colorred 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorred 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                          3                               2                                          2                    3
-╰  y = \colorred 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorred  + 0  · t 
+╰  y = \colorred 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorred  + 0  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/64fa084c1df7300f06af13f53681463c.svg" width="457px" height="41px" loading="lazy" />
 					<p>Итак, опуская все нулевые значения, получаем</p>
 					<!--
- 
+
 ╭                                   2                                         2                       3
-╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                                   2                                          2
-╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t                        
+╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/016f37843b2af2ae34dc8b2975505f3d.svg" width="408px" height="40px" loading="lazy" />
 					<p>
@@ -4643,7 +4643,7 @@ x    = x  - ───────
 					</p>
 					<p>What we need to do is solve a simple equation:</p>
 					<!--
- 
+
 C(t) = 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/inflections/ed68dcfb203517ca080fe48914769fb0.svg" width="59px" height="16px" loading="lazy" />
@@ -4653,7 +4653,7 @@ C(t) = 0
 						circle being on the other side of the curve. So what does <em>C(t)</em> look like? Actually something that seems not too hard:
 					</p>
 					<!--
- 
+
 C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)  ·   Bézier \prime\prime(t)
                x                     y                          y                     x
 -->
@@ -4682,15 +4682,15 @@ C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)
 							with the first and second derivatives, given our basis functions:
 						</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t                            
+Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t
              1           2            3           4
-      \prime            2                2                                      
+      \prime            2                2
 Bézier      (t) = a(1-t)  + 2b(1-t)t + ct   { a=3(x -x ),b=3(x -x ),c=3(x -x ) }
-                                                   2  1       3  2       4  3   
+                                                   2  1       3  2       4  3
       \prime\prime
-Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\                        
+Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 -->
 						<img
 							class="LaTeX SVG"
@@ -4701,14 +4701,14 @@ Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 						/>
 						<p>And of course the same functions for <em>y</em>:</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t 
+Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t
              1           2            3           4
-      \prime            2                2           
-Bézier      (t) = d(1-t)  + 2e(1-t)t + ft            
+      \prime            2                2
+Bézier      (t) = d(1-t)  + 2e(1-t)t + ft
       \prime\prime
-Bézier            (t) = w(1-t) + zt                  
+Bézier            (t) = w(1-t) + zt
 -->
 						<img
 							class="LaTeX SVG"
@@ -4722,19 +4722,19 @@ Bézier            (t) = w(1-t) + zt
 							rather overly complicated set of arithmetic expressions:
 						</p>
 						<!--
- 
+
      2             2             2             2             2
--18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y              
+-18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y
         2  1          3  1          4  1          1  2          3  2
      2             2             2             2             2
-+36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y              
++36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y
         4  2          1  3          2  3          4  3          1  4
-     2             2                                                             
--36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y 
+     2             2
+-36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y
         2  4          3  4         2  1          3  1          4  1          1  2
-+54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y 
++54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y
        3  2          4  2          1  3          2  3          1  4          2  4
--18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y   
+-18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y
      2  1          3  1          1  2          3  2          1  3          2  3
 -->
 						<img
@@ -4755,7 +4755,7 @@ Bézier            (t) = w(1-t) + zt
 						<code>t</code> values, we end up with the following simple term function for <em>C(t)</em>:
 					</p>
 					<!--
- 
+
                                  2
 (3 x  y +2 x  y +3 x  y -x  y ) t  + (3 x  y -x  y -3 x  y ) t + (x  y -x  y )
     3  2    4  2    2  3  4  3           3  2  4  2    2  3        2  3  3  2
@@ -4772,7 +4772,7 @@ Bézier            (t) = w(1-t) + zt
 						simplification:
 					</p>
 					<!--
- 
+
 a = x   · y  ╮
      3     2 │
 b = x   · y  │                             2
@@ -4791,7 +4791,7 @@ d = x   · y  │
 					/>
 					<p>This is a plain quadratic curve, and we know how to solve <em>C(t) = 0</em>; we use the quadratic formula:</p>
 					<!--
- 
+
                                                   ┌──────────┐
                                                   │ 2
 x = -3a + 2b + 3c - d ╮                      -y ±⟍│y  - 4 x z
@@ -4887,7 +4887,7 @@ z =       c - a       ╯                             2x
 								This edge is described by the function:
 							</p>
 							<!--
- 
+
       2
     -x  + 2x + 3
 y = ────────────, { x ≤1 }
@@ -4907,7 +4907,7 @@ y = ────────────, { x ≤1 }
 								the function:
 							</p>
 							<!--
- 
+
      ┌──────────┐
      │        2
     ⟍│3(4x - x )  - x
@@ -4925,7 +4925,7 @@ y = ─────────────────, { 0 ≤x ≤1 }
 						<li>
 							<p>...on the top (blue) edge, the curve's start point touches the curve, forming a loop. This edge is described by the function:</p>
 							<!--
- 
+
       2
     -x  + 3x
 y = ────────, { x ≤0 }
@@ -4992,7 +4992,7 @@ y = ────────, { x ≤0 }
 						<em>z</em> coordinate that is always 1, then we can suddenly add arbitrary values. For example:
 					</p>
 					<!--
- 
+
 ┌ 1 0 a ┐    ┌  x  ┐   ┌ 1  · x + 0  · y + a  · z ┐   ┌ x + a  · 1 ┐   ┌ x + a ┐
 │ 0 1 b │  · │  y  │ = │ 0  · x + 1  · y + b  · z │ = │ y + b  · 1 │ = │ y + b │
 └ 0 0 1 ┘    └ z=1 ┘   └ 0  · x + 0  · y + 1  · z ┘   └   1  · z   ┘   └  z=1  ┘
@@ -5003,7 +5003,7 @@ y = ────────, { x ≤0 }
 						want to subtract p1.x and p1.y, we use:
 					</p>
 					<!--
- 
+
      ┌ 1 0 -P   ┐            ┌ 1  · x + 0  · y - P    · 1 ┐   ┌ x - P   ┐
      │       1x │    ┌ x ┐   │                    1x      │   │      1x │
 T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = │ y - P   │
@@ -5019,7 +5019,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						transformation looks like this:
 					</p>
 					<!--
- 
+
 ┌ 1 S 0 ┐    ┌ x ┐   ┌ x + S  · y ┐
 │ 0 1 0 │  · │ y │ = │     y      │
 └ 0 0 1 ┘    └ 1 ┘   └     1      ┘
@@ -5030,7 +5030,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						simply <em>-x/y</em>, because <em>-x/y \</em> y = -x*. Done:
 					</p>
 					<!--
- 
+
      ┌    U     ┐
      │     2x   │
      │ 1 -─── 0 │
@@ -5048,7 +5048,7 @@ T  = │    U     │
 						x-scaling by 1/point3<sub>x</sub>, and y-scaling by point2<sub>y</sub>. This is really easy:
 					</p>
 					<!--
- 
+
      ┌  1        ┐
      │ ───  0  0 │
      │ V         │
@@ -5069,7 +5069,7 @@ T  = │      1    │
 						offset, in this case 1:
 					</p>
 					<!--
- 
+
      ┌    1    0 0 ┐
      │ 1 - W       │
      │      3y     │
@@ -5085,7 +5085,7 @@ T  = │ ─────── 1 0 │
 						be:
 					</p>
 					<!--
- 
+
                 ╭                           (-x +x )(-y +y )                ╮
                 │                              1  2    1  4                 │
                 │                -x  + x  - ────────────────                │
@@ -5120,7 +5120,7 @@ mapped  = (x) = │                                1  2                       
 						that leave?
 					</p>
 					<!--
- 
+
       ╭                 x   · y       x   · y              ╮
       │                  2     4       2     3             │
       │            x  - ──────── / x -────────             │   ╭         x           ╮
@@ -5140,11 +5140,11 @@ mapped  = (x) = │                                1  2                       
 						factors to see just how simple this is:
 					</p>
 					<!--
- 
-      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y 
+
+      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y
       │          43         │                 │  4    2     42 │            4                3
 ... = │                     │,   where  x   = │ ────────────── │,    y   = ──,   and  y   = ──
-      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y 
+      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y
       ╰  42    43       32  ╯                 ╰  3    2     32 ╯            2                2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/canonical/88e3fae7aeef6d7614290587422542c9.svg" width="563px" height="55px" loading="lazy" />
@@ -5233,16 +5233,16 @@ mapped  = (x) = │                                1  2                       
 					</p>
 					<p>First, let's look at the function for x(t):</p>
 					<!--
- 
+
              3          2            2     3
-x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt 
+x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt
 -->
 					<img class="LaTeX SVG" src="./images/chapters/yforx/378d0fd8cefa688d530ac38930d66844.svg" width="335px" height="19px" loading="lazy" />
 					<p>
 						We can rewrite this to a plain polynomial form, by just fully writing out the expansion and then collecting the polynomial factors, as:
 					</p>
 					<!--
- 
+
                          3                  2
 x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 -->
@@ -5253,7 +5253,7 @@ x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 						known constants, we can trivially rewrite this (by moving the <code>x(t)</code> across the equal sign) as:
 					</p>
 					<!--
- 
+
                    3                  2
 (-a + 3b - 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + (a-x) = 0
 -->
@@ -5346,7 +5346,7 @@ y = curve.get(t).y</textarea
 						following seemingly straight forward (if a bit overwhelming) formula:
 					</p>
 					<!--
- 
+
     ┌───────────────┐
 ╭ z │      2       2
 |   │f '(t) +f '(t)   dt
@@ -5355,7 +5355,7 @@ y = curve.get(t).y</textarea
 					<img class="LaTeX SVG" src="./images/chapters/arclength/59ebc3a7c3547a50998d1ea3664fb688.svg" width="140px" height="33px" loading="lazy" />
 					<p>or, more commonly written using Leibnitz notation as:</p>
 					<!--
- 
+
               ┌─────────────────┐
           ╭ z │       2        2
 length =  |  ⟍│(dx/dt) +(dy/dt)   dt
@@ -5394,7 +5394,7 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 |   ⟍│(dx/dt) +(dy/dt)   dt =  |   f(t) dt  ≃ ┌ \underset strip 1 \underbrace C   · f(t )   + ...  + \underset strip n \underbrace C   · f(t )  ┐ =
 ╯ -1                           ╯ -1           └                                1       1                                            n       n   ┘
                                                                                     __ n
-                                           \underset strips 1 through n \underbrace ❯      C   · f(t )   
+                                           \underset strips 1 through n \underbrace ❯      C   · f(t )
                                                                                     ‾‾ i=1  i       i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/arclength/b76753476ad6ecfe4b8f39bcf9432980.svg" width="623px" height="71px" loading="lazy" />
@@ -5476,16 +5476,16 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							shifting and scaling the inputs. Doing so, we get the following:
 						</p>
 						<!--
- 
+
      ┌─────────────────┐
  ╭ z │       2        2
- |  ⟍│(dx/dt) +(dy/dt)   dt                                        
+ |  ⟍│(dx/dt) +(dy/dt)   dt
  ╯ 0
      z    ┌        ╭ z         z ╮                ╭ z         z ╮ ┐
  ≃ \ ─  · │ C   · f│ ─  · t  + ─ │ + ... + C   · f│ ─  · t  + ─ │ │
      2    └  1     ╰ 2     1   2 ╯          n     ╰ 2     n   2 ╯ ┘
     z    __ n         ╭ z         z ╮
-= \ ─  · ❯     C   · f│ ─  · t  + ─ │                              
+= \ ─  · ❯     C   · f│ ─  · t  + ─ │
     2    ‾‾ i=1 i     ╰ 2     i   2 ╯
 -->
 						<img
@@ -5507,10 +5507,10 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							<a href="./legendre-gauss.html">these tables</a> would tell us that for <em>n=2</em> we must use the following values:
 						</p>
 						<!--
- 
-C  = 1     
+
+C  = 1
  1
-C  = 1     
+C  = 1
  2
         1
 t  = - ────
@@ -5526,7 +5526,7 @@ t  = + ────
 							Which means that in order for us to approximate the integral, we must plug these values into the approximate function, which gives us:
 						</p>
 						<!--
- 
+
     ┌─────────────────┐
 ╭ z │       2        2       z    ┌  ╭ z     -1    z ╮    ╭ z     1     z ╮ ┐
 |  ⟍│(dx/dt) +(dy/dt)   dt ≃ ─  · │ f│ ─  · ──── + ─ │ + f│ ─  · ──── + ─ │ │
@@ -5690,13 +5690,13 @@ t  = + ────
 					</p>
 					<p>So, what does the function look like? This:</p>
 					<!--
- 
+
          x'y'' - x''y'
 \kappa = ─────────────
                    3
                    ─
              2   2 2
-          (x' +y' ) 
+          (x' +y' )
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/060acd6ff0a050fe4d98a7802a2b3a3f.svg" width="113px" height="47px" loading="lazy" />
 					<p>
@@ -5704,14 +5704,14 @@ t  = + ────
 						tiny bit:
 					</p>
 					<!--
- 
+
             B '(t)B ''(t) - B ''(t)B '(t)
              x     y         x      y
 \kappa(t) = ─────────────────────────────
                                   3
                                   ─
                         2       2 2
-                 (B '(t) +B '(t) ) 
+                 (B '(t) +B '(t) )
                    x       y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/afd8cb8b0fe291ff703752c1c9cc33d4.svg" width="239px" height="55px" loading="lazy" />
@@ -5790,7 +5790,7 @@ function kappa(t, B):
 						circle that "fits" the curve's curvature at any point, using what is possibly the simplest bit of maths found in this entire primer:
 					</p>
 					<!--
- 
+
            1
 R(t) = ─────────
        \kappa(t)
@@ -6241,8 +6241,8 @@ lli = function(line1, line2):
 						points, so logically <code>C</code> will have a function that interpolates between those two coordinates:
 					</p>
 					<!--
- 
-C = u(t)  · P       + (1-u(t))  · P    
+
+C = u(t)  · P       + (1-u(t))  · P
               start                 end
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8c6662f605722fb2ff6cd7f65243a126.svg" width="229px" height="16px" loading="lazy" />
@@ -6255,22 +6255,22 @@ C = u(t)  · P       + (1-u(t))  · P
 						(with thanks to Boris Zbarsky) shows us the following two formulae:
 					</p>
 					<!--
- 
+
                         2
-                   (1-t) 
+                   (1-t)
 u(t)           = ───────────
      quadratic    2        2
-                 t  + (1-t) 
+                 t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8e7cfee39c98f2ddf9b635a914066cf6.svg" width="183px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                     3
-               (1-t) 
+               (1-t)
 u(t)       = ───────────
      cubic    3        3
-             t  + (1-t) 
+             t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/a0b99054cc82ca1fb147f077e175ef10.svg" width="161px" height="41px" loading="lazy" />
 					<p>
@@ -6281,7 +6281,7 @@ u(t)       = ───────────
 					</p>
 					<p>We start by observing that, given <code>A</code>, <code>B</code>, and <code>C</code>, the following always holds:</p>
 					<!--
- 
+
             distance(B,C)
 ratio(t) = ────────────── =  Constant
             distance(A,B)
@@ -6289,22 +6289,22 @@ ratio(t) = ────────────── =  Constant
 					<img class="LaTeX SVG" src="./images/chapters/abc/131454dcbac04e567f322979f4af80c6.svg" width="245px" height="39px" loading="lazy" />
 					<p>Working out the maths for this, we see the following two formulae for quadratic and cubic curves:</p>
 					<!--
- 
+
                        2        2
                       t  + (1-t)  - 1
 ratio(t)           = |───────────────|
          quadratic       2        2
-                        t  + (1-t) 
+                        t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/5924e162b50272c40c842fad14b8fa48.svg" width="239px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                    3        3
                   t  + (1-t)  - 1
 ratio(t)       = |───────────────|
          cubic       3        3
-                    t  + (1-t) 
+                    t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8cd992c1ceaae2e67695285beef23a24.svg" width="219px" height="41px" loading="lazy" />
 					<p>
@@ -6314,7 +6314,7 @@ ratio(t)       = |───────────────|
 						<code>ratio(t)</code> function to find <code>A</code>:
 					</p>
 					<!--
- 
+
           C - B           B - C
 A = B - ───────── = B + ─────────
          ratio(t)        ratio(t)
@@ -6329,11 +6329,11 @@ A = B - ───────── = B + ─────────
 						pair, after which we can reverse engineer <code>v1</code> and <code>v2</code>:
 					</p>
 					<!--
- 
+
                                  ╭     e  - t  · A
                                  │      1
-╭ e = (1-t)  · v  + t  · A       │ v = ───────────    
-╡  1            1            ==> ╡  1      1-t         
+╭ e = (1-t)  · v  + t  · A       │ v = ───────────
+╡  1            1            ==> ╡  1      1-t
 │ e = (1-t)  · A + t  · v        │     e  - (1-t)  · A
 ╰  2                     2       │      2
                                  │ v = ───────────────
@@ -6342,14 +6342,14 @@ A = B - ───────── = B + ─────────
 					<img class="LaTeX SVG" src="./images/chapters/abc/3166afa345aec1abda432c39b68d39a0.svg" width="339px" height="73px" loading="lazy" />
 					<p>And then reverse engineer the curve's control points:</p>
 					<!--
- 
+
                                       ╭     v  - (1-t)  ·  start
                                       │      1
 ╭ v = (1-t)  ·  start + t  · C        │ C = ────────────────────
-╡  1                          1   ==> ╡  1           t           
+╡  1                          1   ==> ╡  1           t
 │ v = (1-t)  · C  + t  ·  end         │     v  - t  ·  end
 ╰  2            2                     │      2
-                                      │ C = ──────────────      
+                                      │ C = ──────────────
                                       ╰  2       1-t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8bd3e6fed5bf8d871d30221ae400fd93.svg" width="383px" height="75px" loading="lazy" />
@@ -6379,15 +6379,15 @@ A = B - ───────── = B + ─────────
 						<code>B</code> point, and <code>B</code> and end point as a ratio, using
 					</p>
 					<!--
- 
+
 ╭ d = || Start - B||
 │  1
-│ d = || End - B||  
+│ d = || End - B||
 │  2
-╡      d             
+╡      d
 │       1
-│  t= ─────         
-│     d +d 
+│  t= ─────
+│     d +d
 ╰      1  2
 -->
 					<img
@@ -6449,9 +6449,9 @@ A = B - ───────── = B + ─────────
 						<code>e2</code> coordinates must obey the standard de Casteljau rule for linear interpolation:
 					</p>
 					<!--
- 
-╭ e = B + t  · d    
-╡  1                 
+
+╭ e = B + t  · d
+╡  1
 │ e = B - (1-t)  · d
 ╰  2
 -->
@@ -6471,7 +6471,7 @@ A = B - ───────── = B + ─────────
 						curve with a loop in it. To do this, we can use the <a href="https://en.wikipedia.org/wiki/Atan2">atan2</a> function:
 					</p>
 					<!--
- 
+
 \phi = (atan2(E -S , E -S ) - atan2(B -S , B -S ) + 2 \pi)   mod  2 \pi
                y  y   x  x           y  y   x  x
 -->
@@ -6488,8 +6488,8 @@ A = B - ───────── = B + ─────────
 						<code>d</code>:
 					</p>
 					<!--
- 
-d = {  d  if  0 ≤\phi ≤\pi             
+
+d = {  d  if  0 ≤\phi ≤\pi
       -d  if  \phi < 0 \lor \phi > \pi
 -->
 					<img
@@ -6670,7 +6670,7 @@ for (coordinate, index) in LUT:
 						maths, that means we're trying to solve:
 					</p>
 					<!--
- 
+
 dist(B(t), c) = r
 -->
 					<img
@@ -6682,15 +6682,15 @@ dist(B(t), c) = r
 					/>
 					<p>Which seems simple enough. Unfortunately, when we expand that <code>dist</code> function, things get a lot more problematic:</p>
 					<!--
- 
-r=  dist(B(t), c)                                                                                                              
+
+r=  dist(B(t), c)
     ┌─────────────────────────┐
     │          2             2
- =  │(B t - c )  + (B t - c )                                                                                                  
-   ⟍│  x     x       y     y                                                                                                   
+ =  │(B t - c )  + (B t - c )
+   ⟍│  x     x       y     y
     ┌─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┐
     │╭         3             2                 2       3      ╮2   ╭         3             2                 2       3      ╮2
- =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │  
+ =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │
    ⟍│╰  1             2               3             4       x ╯    ╰  1             2               3             4       y ╯
 -->
 					<img
@@ -6963,17 +6963,17 @@ findClosest(start, p, r, LUT):
 						<code>B</code>, we can compute the associated <code>C</code>:
 					</p>
 					<!--
- 
+
 C = u(t)   ·  Start + (1-u(t) )  ·  End
         q                    q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/48887d68a861a0acdf8313e23fb19880.svg" width="247px" height="24px" loading="lazy" />
 					<p>And then the associated <code>A</code>:</p>
 					<!--
- 
+
           C - B            B - C
 A = B - ────────── = B + ──────────
-         ratio(t)         ratio(t) 
+         ratio(t)         ratio(t)
                  q                q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/2e65bc9c934380c2de6a24bcd5c1c7b7.svg" width="228px" height="41px" loading="lazy" />
@@ -7112,12 +7112,12 @@ A = B - ────────── = B + ──────────
 							form, where each line corresponds to a power of t, and each column is for a specific coefficient. First, we expand the function:
 						</p>
 						<!--
- 
+
                     2                    2
-B          = a (1-t)  + 2 b (1-t) t + c t    
-  quadratic                                  
+B          = a (1-t)  + 2 b (1-t) t + c t
+  quadratic
                          2            2     2
-           = a - 2at + at  + 2bt - 2bt  + ct 
+           = a - 2at + at  + 2bt - 2bt  + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -7128,12 +7128,12 @@ B          = a (1-t)  + 2 b (1-t) t + c t
 						/>
 						<p>And then we (trivially) rearrange the terms across multiple lines:</p>
 						<!--
- 
-B          =a               
+
+B          =a
   quadratic
-            - 2at+ 2bt      
+            - 2at+ 2bt
                 2     2    2
-            + at - 2bt + ct 
+            + at - 2bt + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -7149,7 +7149,7 @@ B          =a
 						</p>
 						<p>With that arrangement, we can easily decompose this as a matrix multiplication:</p>
 						<!--
- 
+
                                          ┌  a  +  0  + 0 ┐                 ┌  1  0 0 ┐    ┌ a ┐
 B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌      2 ┐  · │ -2  2 0 │  · │ b │
   quadratic                └ 1 t t  ┘    └  a  + -2b + c ┘   └ 1 t t  ┘    └  1 -2 1 ┘    └ c ┘
@@ -7163,9 +7163,9 @@ B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌
 						/>
 						<p>We can do the same for the cubic curve, of course. We know the base function for cubics:</p>
 						<!--
- 
+
               3          2             2     3
-B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt 
+B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
   cubic
 -->
 						<img
@@ -7177,14 +7177,14 @@ B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
 						/>
 						<p>So we write out the expansion and rearrange:</p>
 						<!--
- 
-B      =a                     
+
+B      =a
   cubic
-        - 3at + 3bt           
-             2     2    2     
-        + 3at - 6bt +3ct      
+        - 3at + 3bt
+             2     2    2
+        + 3at - 6bt +3ct
             3      3    3    3
-        - at  + 3bt -3ct + dt 
+        - at  + 3bt -3ct + dt
 -->
 						<img
 							class="LaTeX SVG"
@@ -7195,7 +7195,7 @@ B      =a
 						/>
 						<p>Which we can then decompose:</p>
 						<!--
- 
+
                                         ┌  1  0  0 0 ┐    ┌ a ┐
 B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
   cubic                └ 1 t t  t  ┘    │  3 -6  3 0 │    │ c │
@@ -7210,7 +7210,7 @@ B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
 						/>
 						<p>And, of course, we can do this for quartic curves too (skipping the expansion step):</p>
 						<!--
- 
+
                                              ┌  1  0   0   0 0 ┐    ┌ a ┐
                                              │ -4  4   0   0 0 │    │ b │
 B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  · │ c │
@@ -7236,7 +7236,7 @@ B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  ·
 						we want to do curve fitting:
 					</p>
 					<!--
- 
+
     ┌ p   ┐
     │  1  │
     │ p   │
@@ -7274,9 +7274,9 @@ P = │  2  │
 					</p>
 					<p>To get these values, we first compute the general "distance along the polygon" matrix:</p>
 					<!--
- 
+
                                ╭            d  = 0
-D = ┌ d  d  ... d  ┐,   where  ╡             1                 
+D = ┌ d  d  ... d  ┐,   where  ╡             1
     └  1  2      n ┘           │ d  = d    +  length(p   , p )
                                ╰  i    i-1            i-1   i
 -->
@@ -7293,10 +7293,10 @@ D = ┌ d  d  ... d  ┐,   where  ╡             1
 						<code>i=n</code> fall in the [0,1] interval, so we need to scale all values down by whatever the total length of the polygon is:
 					</p>
 					<!--
- 
+
                                ╭    s  = 0
                                │     1
-S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d  
+S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
     └  1  2      n ┘           │  i    i    n
                                │    s  = 1
                                ╰     n
@@ -7318,9 +7318,9 @@ S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
 						to for the associated <code>t</code> value", which we'll square to get rid of any pesky negative signs:
 					</p>
 					<!--
- 
+
                            2
-E(C)  = (p  -   Bézier(s )) 
+E(C)  = (p  -   Bézier(s ))
     i     i             i
 -->
 					<img
@@ -7335,9 +7335,9 @@ E(C)  = (p  -   Bézier(s ))
 						function. So, we literally just do that; the total error function is simply the sum of all these individual errors:
 					</p>
 					<!--
- 
+
        __ n                      2
-E(C) = ❯      (p  -   Bézier(s )) 
+E(C) = ❯      (p  -   Bézier(s ))
        ‾‾ i=1   i             i
 -->
 					<img
@@ -7354,9 +7354,9 @@ E(C) = ❯      (p  -   Bézier(s ))
 						<strong>T x M x C</strong> we talked about before, which gives us:
 					</p>
 					<!--
- 
+
                 2
-E(C) = (P - TMC) 
+E(C) = (P - TMC)
 -->
 					<img
 						class="LaTeX SVG"
@@ -7367,7 +7367,7 @@ E(C) = (P - TMC)
 					/>
 					<p>In which we can replace the rather cumbersome "squaring" operation with a more conventional matrix equivalent:</p>
 					<!--
- 
+
                 T
 E(C) = (P - TMC)  (P - TMC)
 -->
@@ -7389,7 +7389,7 @@ E(C) = (P - TMC)  (P - TMC)
 						those, and then use that 𝕋 instead of <strong>T</strong> in our error function:
 					</p>
 					<!--
- 
+
     ┌    0    1      n-2   n-1  ┐
     │   s    s  ... s     s     │
     │    1    1      1     1    │
@@ -7409,7 +7409,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Which, because of the first and last values in <strong>S</strong>, means:</p>
 					<!--
- 
+
     ┌    1     0  ...   0    0    ┐
     │                  n-2   n-1  │
     │    1    s       s     s     │
@@ -7429,7 +7429,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Now we can properly write out the error function as matrix operations:</p>
 					<!--
- 
+
                 T
 E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 -->
@@ -7446,7 +7446,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						taking the derivative, and determining where that derivative is zero:
 					</p>
 					<!--
- 
+
 ∂E          T
 ── = 0 = -2𝕋  (P - 𝕋MC)
 ∂C
@@ -7485,7 +7485,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						for <strong>C</strong>:
 					</p>
 					<!--
- 
+
      -1   T   -1  T
 C = M   (𝕋  𝕋)   𝕋  P
 -->
@@ -7585,13 +7585,13 @@ C = M   (𝕋  𝕋)   𝕋  P
 						previous and next point:
 					</p>
 					<!--
- 
+
                ┌  V  = P       ┐
                │   1    2      │
 ┌ P  ┐         │  V  = P       │
 │  1 │         │   2    3      │
 │ P  │         │       P  - P  │
-│  2 │       = │        3    1 │              
+│  2 │       = │        3    1 │
 │ P  │points   │ V'  = ─────── │ point-tangent
 │  3 │         │   1      2    │
 │ P  │         │       P  - P  │
@@ -7724,7 +7724,7 @@ for p = 1 to points.length-3 (inclusive):
 						so how different is the matrix form for Catmull-Rom curves?:
 					</p>
 					<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7761,7 +7761,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							we'll get to that:
 						</p>
 						<!--
- 
+
                         ┌   P     ┐
                         │    2    │
 ┌ V   ┐        ┌ P  ┐   │   P     │
@@ -7789,7 +7789,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						</p>
 						<p>Computing <em>T</em> is really more "arranging the numbers":</p>
 						<!--
- 
+
               ┌   P     ┐
               │    2    │
      ┌ P  ┐   │   P     │   ┌  0  · P1 + 1  · P2 + 0  · P3 + 0  · P4 ┐   ┌  0  1 0 0 ┐    ┌ P  ┐
@@ -7812,7 +7812,7 @@ T  · │  2 │ = │  3    1 │ = │ ──  · P1 + 0  · P2 + ─  · P3 +
 						/>
 						<p>Thus:</p>
 						<!--
- 
+
     ┌  0  1 0 0 ┐
     │  0  0 1 0 │
     │ -1    1   │
@@ -7836,7 +7836,7 @@ T = │ ──  0 ─ 0 │
 							factor into both our coordinate vector representation, and mapping matrix <em>T</em>:
 						</p>
 						<!--
- 
+
           ┌   P     ┐
           │    2    │
 ┌ V   ┐   │   P     │        ┌  0  1  0 0  ┐
@@ -7862,7 +7862,7 @@ T = │ ──  0 ─ 0 │
 							coordinates, and see what we end up with:
 						</p>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7881,7 +7881,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						/>
 						<p>Replace point/tangent vector with the expression for all-coordinates:</p>
 						<!--
- 
+
                                                     ┌  0  1  0 0  ┐    ┌ P  ┐
                                                     │  0  0  1 0  │    │  1 │
                                  ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -7900,7 +7900,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │ ──  0 ─
 						/>
 						<p>and merge the matrices:</p>
 						<!--
- 
+
                                  ┌  0    1      0   0  ┐
                                  │ -1          1       │    ┌ P  ┐
                                  │ ──    0     ──   0  │    │  1 │
@@ -7921,7 +7921,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  1 1           1 -1 │  · │  2 │
 						/>
 						<p>This looks a lot like the Bézier matrix form, which as we saw in the chapter on Bézier curves, should look like this:</p>
 						<!--
- 
+
                                                ┌ P  ┐
                                                │  1 │
                              ┌  1  0  0 0 ┐    │ P  │
@@ -7940,7 +7940,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>So, if we want to express a Catmull-Rom curve using a Bézier curve, we'll need to turn this Catmull-Rom bit:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │
@@ -7961,7 +7961,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Into something that looks like this:</p>
 						<!--
- 
+
                   ┌ P  ┐
                   │  1 │
 ┌  1  0  0 0 ┐    │ P  │
@@ -7980,7 +7980,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And the way we do that is with a fairly straight forward bit of matrix rewriting. We start with the equality we need to ensure:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐                          ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │                          │  1 │
@@ -8001,7 +8001,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we remove the coordinate vector from both sides without affecting the equality:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │
 │ ──    0     ──   0  │
@@ -8022,7 +8022,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we can "get rid of" the Bézier matrix on the right by left-multiply both with the inverse of the Bézier matrix:</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -8046,7 +8046,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							outright remove any matrix/inverse pair:
 						</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -8067,7 +8067,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And now we're <em>basically</em> done. We just multiply those two matrices and we know what <em>V</em> is:</p>
 						<!--
- 
+
 ┌  0  1  0 0  ┐
 │ -1    1     │
 │ ──  1 ── 0  │
@@ -8089,7 +8089,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							<li>Start with the Catmull-Rom function:</li>
 						</ol>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -8110,7 +8110,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite to pure coordinate form:</li>
 						</ol>
 						<!--
- 
+
                                       ┌   P     ┐
                                       │    2    │
                                       │   P     │
@@ -8135,7 +8135,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for "normal" coordinate vector:</li>
 						</ol>
 						<!--
- 
+
                                       ┌  0  1  0 0  ┐    ┌ P  ┐
                                       │  0  0  1 0  │    │  1 │
                    ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -8156,7 +8156,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>merge the inner matrices:</li>
 						</ol>
 						<!--
- 
+
                    ┌  0    1      0   0  ┐
                    │ -1          1       │    ┌ P  ┐
                    │ ──    0     ──   0  │    │  1 │
@@ -8179,7 +8179,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for Bézier matrix form:</li>
 						</ol>
 						<!--
- 
+
                                      ┌  0  1  0 0  ┐    ┌ P  ┐
                                      │ -1    1     │    │  1 │
                    ┌  1  0  0 0 ┐    │ ──  1 ── 0  │    │ P  │
@@ -8200,7 +8200,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>and transform the coordinates so we have a "pure" Bézier expression:</li>
 						</ol>
 						<!--
- 
+
                                      ┌     P       ┐
                                      │      2      │
                                      │      P -P   │
@@ -8229,13 +8229,13 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						curve that has the vector:
 					</p>
 					<!--
- 
+
                        ┌     P       ┐
                        │      2      │
 ┌ P  ┐                 │      P -P   │
 │  1 │                 │       3  1  │
 │ P  │                 │ P  + ────── │
-│  2 │             ==> │  2   6  · τ │        
+│  2 │             ==> │  2   6  · τ │
 │ P  │ CatmullRom      │      P -P   │  Bézier
 │  3 │                 │       4  2  │
 │ P  │                 │ P  - ────── │
@@ -8255,11 +8255,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						tension Catmull-Rom curve with the following coordinate values:
 					</p>
 					<!--
- 
+
 ┌ P  ┐              ┌       P         ┐
 │  1 │              │        1        │
 │ P  │              │       P         │
-│  2 │          ==> │        4        │           
+│  2 │          ==> │        4        │
 │ P  │  Bézier      │ P  + 3(P  - P ) │ CatmullRom
 │  3 │              │  4      1    2  │
 │ P  │              │ P  + 3(P  - P ) │
@@ -8274,11 +8274,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 					/>
 					<p>Or, if your API allows you to specify Catmull-Rom curves using plain coordinates:</p>
 					<!--
- 
+
 ┌ P  ┐              ┌ P  + 6(P  - P ) ┐
 │  1 │              │  4      1    2  │
 │ P  │              │       P         │
-│  2 │          ==> │        1        │           
+│  2 │          ==> │        1        │
 │ P  │  Bézier      │       P         │ CatmullRom
 │  3 │              │        4        │
 │ P  │              │ P  + 6(P  - P ) │
@@ -8369,8 +8369,8 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						the same as it's "outgoing" derivative:
 					</p>
 					<!--
- 
-B'(1)  = B'(0)   
+
+B'(1)  = B'(0)
      n        n+1
 -->
 					<img class="LaTeX SVG" src="./images/chapters/polybezier/a37252ff55837b918d9d64078ae92ae7.svg" width="124px" height="17px" loading="lazy" />
@@ -8380,7 +8380,7 @@ B'(1)  = B'(0)
 						last control point through the last on-curve point. And mirroring any point A through any point B is really simple:
 					</p>
 					<!--
- 
+
            ┌ B  + (B  - A ) ┐   ┌ 2B  - A  ┐
 Mirrored = │  x     x    x  │ = │   x    x │
            │ B  + (B  - A ) │   │ 2B  - A  │
@@ -8557,7 +8557,7 @@ Mirrored = │  x     x    x  │ = │   x    x │
 							any point on <code>O(t)</code> is a fixed distance <code>d</code> away from coordinate <code>B(t)</code>. So let's math that:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d
 -->
 						<img
@@ -8574,7 +8574,7 @@ O(t) = B(t) + d
 							at <code>B(t)</code>. Easy enough:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d  · N(t)
 -->
 						<img
@@ -8592,7 +8592,7 @@ O(t) = B(t) + d  · N(t)
 							<code>N(t)</code> always has length 1, which we can achieve by dividing <code>B'(t)</code> by its magnitude:
 						</p>
 						<!--
- 
+
           ╭   B'(t)    ╮
 N(t) \bot │ ────────── │
           ╰ || B'(t)|| ╯
@@ -8610,10 +8610,10 @@ N(t) \bot │ ────────── │
 							denoted with double vertical bars:
 						</p>
 						<!--
- 
+
                     ┌───────────┐
                ╭ b  │   2      2
-|| f(x,y)|| =  |    │f '  + f '  
+|| f(x,y)|| =  |    │f '  + f '
                ╯ a ⟍│ x      y
 -->
 						<img
@@ -8627,10 +8627,10 @@ N(t) \bot │ ────────── │
 							So if we want the length of the tangent, we plug in <code>B'(t)</code>, with <code>t = 0</code> as start and <code>t = 1</code> as end:
 						</p>
 						<!--
- 
+
                    ┌───────────────────┐
               ╭ 1  │       2          2
-|| B'(t)|| =  |    │B ''(t)  + B ''(t)  
+|| B'(t)|| =  |    │B ''(t)  + B ''(t)
               ╯ 0 ⟍│ x          y
 -->
 						<img
@@ -8847,7 +8847,7 @@ N(t) \bot │ ────────── │
 						some angle <em>φ</em>:
 					</p>
 					<!--
- 
+
              S = \begin{pmatrix} 1
 0 \end{pmatrix}  ,  \ E = \begin{pmatrix} cos(φ)
               sin(φ) \end{pmatrix}
@@ -8855,7 +8855,7 @@ N(t) \bot │ ────────── │
 					<img class="LaTeX SVG" src="./images/chapters/circles/7ab3da0922477af4cc09f5852100976b.svg" width="175px" height="40px" loading="lazy" />
 					<p>What we want to find is the intersection of the tangents, so we want a point C such that:</p>
 					<!--
- 
+
               C = S + a  · \begin{pmatrix} 0
 1 \end{pmatrix}  ,  \ C = E + b  · \begin{pmatrix} -sin(φ)
                    cos(φ) \end{pmatrix}
@@ -8866,22 +8866,22 @@ N(t) \bot │ ────────── │
 						through E (at some distance <em>b</em> from E). Solving this gives us:
 					</p>
 					<!--
- 
+
 ╭ C  = 1 = cos(φ) + b  · -sin(φ)
-╡  x                             
-│ C  = a = sin(φ) + b  · cos(φ) 
+╡  x
+│ C  = a = sin(φ) + b  · cos(φ)
 ╰  y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/8237af1396bb567d70c8b5e4dd7f8115.svg" width="219px" height="40px" loading="lazy" />
 					<p>First we solve for <em>b</em>:</p>
 					<!--
- 
+
 1 = cos(φ) + b  · -sin(φ)  → \ 1 - cos(φ) = -b  · sin(φ)  → \ -1 + cos(φ) = b  · sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/1829e42ea956ee4df0e45d9ac5334ef7.svg" width="560px" height="17px" loading="lazy" />
 					<p>which yields:</p>
 					<!--
- 
+
     cos(φ)-1
 b = ────────
      sin(φ)
@@ -8889,21 +8889,21 @@ b = ────────
 					<img class="LaTeX SVG" src="./images/chapters/circles/06369b00338310df0a810c592485aa0a.svg" width="101px" height="39px" loading="lazy" />
 					<p>which we can then substitute in the expression for <em>a</em>:</p>
 					<!--
- 
- a= sin(φ) + b  · cos(φ)          
+
+ a= sin(φ) + b  · cos(φ)
              -1 + cos(φ)
 ..= sin(φ) + ───────────  · cos(φ)
                sin(φ)
                           2
              -cos(φ) + cos (φ)
-..= sin(φ) + ─────────────────    
-                  sin(φ)          
+..= sin(φ) + ─────────────────
+                  sin(φ)
        2         2
     sin (φ) + cos (φ) - cos(φ)
-..= ──────────────────────────    
+..= ──────────────────────────
               sin(φ)
     1 - cos(φ)
- a= ──────────                    
+ a= ──────────
       sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/5a23f3bc298c85540c6dd18e304d9224.svg" width="231px" height="195px" loading="lazy" />
@@ -8914,7 +8914,7 @@ b = ────────
 						polynomial, and we know where the circle arc's actual point P is for angle φ/2:
 					</p>
 					<!--
- 
+
          φ                 φ
 P  = cos(─)  ,  \ P  = sin(─)
  x       2         y       2
@@ -8922,7 +8922,7 @@ P  = cos(─)  ,  \ P  = sin(─)
 					<img class="LaTeX SVG" src="./images/chapters/circles/8b4e1d0a62380ed011f27c645ed13b28.svg" width="188px" height="32px" loading="lazy" />
 					<p>We compute T, observing that if <em>t=0.5</em>, the polynomial values (1-t)², 2(1-t)t, and t² are 0.25, 0.5, and 0.25 respectively:</p>
 					<!--
- 
+
     1    2    1    1
 T = ─S + ─C + ─E = ─(S + 2C + E)
     4    4    4    4
@@ -8930,10 +8930,10 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/a04bd1558a76e60b8ca6e1fe4fa38c00.svg" width="252px" height="35px" loading="lazy" />
 					<p>Which, worked out for the x and y components, gives:</p>
 					<!--
- 
+
 ╭     1
-│ T = ─(3 + cos(φ))                                    
-╡  x  4                                                 
+│ T = ─(3 + cos(φ))
+╡  x  4
 │     1╭ 2-2cos(φ)          ╮   1╭     ╭ φ ╮          ╮
 │ T = ─│ ───────── + sin(φ) │ = ─│ 2tan│ ─ │ + sin(φ) │
 ╰  y  4╰  sin(φ)            ╯   4╰     ╰ 2 ╯          ╯
@@ -8941,17 +8941,17 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/986ae9104e0bc52e95689ae7ae4504db.svg" width="408px" height="77px" loading="lazy" />
 					<p>And the distance between these two is the standard Euclidean distance:</p>
 					<!--
- 
+
                  1                   φ        4╭ φ ╮
 d (φ)= T  - P  = ─(3 + cos(φ)) - cos(─) = 2sin │ ─ │  ,
  x      x    x   4                   2         ╰ 4 ╯
                  1╭     ╭ φ ╮          ╮       φ
-d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,   
+d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
  y      y    y   4╰     ╰ 2 ╯          ╯       2
-     ⇓                                                 
+     ⇓
         ┌───────┐                     ┌───────┐
         │ 2    2                 4 φ  │   1
- d(φ)=  │d  + d   =  ...   = 2sin (─) │───────         
+ d(φ)=  │d  + d   =  ...   = 2sin (─) │───────
        ⟍│ x    y                   4  │   2 φ
                                       │cos (─)
                                      ⟍│     2
@@ -8998,7 +8998,7 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 						precision:
 					</p>
 					<!--
- 
+
                 ╭  ┌─────────────┐ ╮
                 │  │     ┌──────┐  │
                 │ ⟍│2+ε-⟍│ε(2+ε)   │
@@ -9058,14 +9058,14 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 					</p>
 					<p>So let's again formally describe this:</p>
 					<!--
- 
-P = (1, 0)                     
+
+P = (1, 0)
  1
-P = (1, k)                     
- 2                             
+P = (1, k)
+ 2
 P = P  + k  · (sin(θ), -cos(θ))
  3   4
-P = (cos(θ), sin(θ))           
+P = (cos(θ), sin(θ))
  4
 -->
 					<img
@@ -9089,29 +9089,29 @@ P = (cos(θ), sin(θ))
 						P<sub>2</sub>" (which is "half height" P<sub>2</sub>), and so forth:
 					</p>
 					<!--
- 
-     P   + P  
-      2     3 
+
+     P   + P
+      2     3
        y     y   k + sin(θ) - k  · cos(θ)
- A = ───────── = ────────────────────────                                 
+ A = ───────── = ────────────────────────
   y      2                  2
           1
-     A  + ─P     k + sin(θ) - k  · cos(θ)   
+     A  + ─P     k + sin(θ) - k  · cos(θ)
       y   2 2    ──────────────────────── + ─
              y              2               2   2k + sin(θ) + k  · cos(θ)
-e  = ───────── = ──────────────────────────── = ───────────────────────── 
+e  = ───────── = ──────────────────────────── = ─────────────────────────
  1       2                    2                             4
-  y                                                                       
+  y
                                        k
      A  +  mid(P , P )   A  + sin(θ) - ─ cos(θ)
       y         4   3     y            2          k + 3sin(θ) 2k  · cos(θ)
 e  = ───────────────── = ────────────────────── = ────────────────────────
  2           2                     2                         4
   y
-     e   + e  
-      1     2 
+     e   + e
+      1     2
        y     y   3k + 4sin(θ) - 3k  · cos(θ)
- B = ───────── = ───────────────────────────                              
+ B = ───────── = ───────────────────────────
   y      2                    8
 -->
 					<img
@@ -9135,15 +9135,15 @@ e  = ───────────────── = ───────
 							<a href="https://www.wolframalpha.com/">Wolfram Alpha</a> is definitely the way to go. That said, let's get going:
 						</p>
 						<!--
- 
+
 3k + 4sin(θ)) - 3k  · cos(θ)      θ
-────────────────────────────= sin(─)                  
+────────────────────────────= sin(─)
              8                    2
                                   ╭ θ ╮
-3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │               
+3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │
                                   ╰ 2 ╯
                                   ╭ θ ╮
-           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)     
+           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)
                                   ╰ 2 ╯
                                 ╭     ╭ θ ╮          ╮
              3k (1 - cos(θ))= 4 │ 2sin│ ─ │ - sin(θ) │
@@ -9151,12 +9151,12 @@ e  = ───────────────── = ───────
                                         θ
                                    2sin(─) - sin(θ)
                                         2
-                          3k= 4  · ────────────────   
+                          3k= 4  · ────────────────
                                       1 - cos(θ)
                                        ╭ θ ╮
                                    2sin│ ─ │ - sin(θ)
                               4        ╰ 2 ╯
-                           k= ─  · ────────────────── 
+                           k= ─  · ──────────────────
                               3        1 - cos(θ)
 -->
 						<img
@@ -9171,11 +9171,11 @@ e  = ───────────────── = ───────
 							<em>k</em>:
 						</p>
 						<!--
- 
+
             ╭ θ ╮
         2sin│ ─ │ - sin(θ)
    4        ╰ 2 ╯
-k= ─  · ──────────────────         
+k= ─  · ──────────────────
    3        1 - cos(θ)
         ╭     ╭ θ ╮               ╮
         │ 2sin│ ─ │               │
@@ -9183,10 +9183,10 @@ k= ─  · ──────────────────
 k= ─  · │ ────────── - ────────── │
    3    ╰ 1 - cos(θ)   1 - cos(θ) ╯
    4    ╭    ╭ θ ╮      ╭ θ ╮ ╮
-k= ─  · │ csc│ ─ │ - cot│ ─ │ │    
+k= ─  · │ csc│ ─ │ - cot│ ─ │ │
    3    ╰    ╰ 2 ╯      ╰ 2 ╯ ╯
    4       ╭ θ ╮
-k= ─  · tan│ ─ │                   
+k= ─  · tan│ ─ │
    3       ╰ 4 ╯
 -->
 						<img
@@ -9204,7 +9204,7 @@ k= ─  · tan│ ─ │
 						involving the circular arc's angle:
 					</p>
 					<!--
- 
+
            4    ╭ θ ╮
 k = f(θ) = ─ tan│ ─ │
            3    ╰ 4 ╯
@@ -9220,13 +9220,13 @@ k = f(θ) = ─ tan│ ─ │
 						Which means that for any circular arc with angle θ and radius <em>r</em>, our Bézier approximation based on three points of incidence is:
 					</p>
 					<!--
- 
-    start= (r, 0)                                          
-control  = (r, k)                                          
-        1                                                  
+
+    start= (r, 0)
+control  = (r, k)
+        1
 control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
         2
-      end= r  · (cos(θ), sin(θ))                           
+      end= r  · (cos(θ), sin(θ))
 -->
 					<img
 						class="LaTeX SVG"
@@ -9237,7 +9237,7 @@ control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
 					/>
 					<p>Which also gives us the commonly found value of 0.55228 for quarter circles, based on them having an angle of half π:</p>
 					<!--
- 
+
  ╭ \pi ╮   4       ╭ \pi ╮   4  ┌─┐
 f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.55228474983[...]
  ╰  2  ╯   3       ╰  8  ╯   3
@@ -9251,13 +9251,13 @@ f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.552284
 					/>
 					<p>And thus giving us the following Bézier coordinates for a quarter circle of radius <em>r</em>:</p>
 					<!--
- 
-    start= (r, 0)           
+
+    start= (r, 0)
 control  = (r, 0.55228  · r)
-        1                   
+        1
 control  = (0.55228  · r, r)
         2
-      end= (0, r)           
+      end= (0, r)
 -->
 					<img
 						class="LaTeX SVG"
@@ -9401,7 +9401,7 @@ getMostWrongT(radius, bezier, start, end, epsilon=1e-15):
 					</p>
 					<p>First, let's express the total error (given circular arc angle θ, and some <em>k</em>) using standard calculus notation:</p>
 					<!--
- 
+
                     ┌───────────────────────┐
               ╭ 1   │         2            2
 erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
@@ -9424,7 +9424,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						expression looks like:
 					</p>
 					<!--
- 
+
 ╭        ┌───────┐         ╮
 │  ╭ 1   │ 2    2          │ d
 │  |  \| │B  + B   - r\|dt │ ── = 0
@@ -9443,7 +9443,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						are each other's inverse operations, so they cancel out, leaving us with our original function:
 					</p>
 					<!--
- 
+
    ┌───────┐
    │ 2    2
 \| │B  + B   - r\| = 0,    t ∈[0,1]
@@ -9458,7 +9458,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 					/>
 					<p>And now we just solve for that... oh wait. We've seen this before. In order to solve this, we'd end up needing to solve this:</p>
 					<!--
- 
+
  2    2
 B  + B  = r
  x    y
@@ -9988,7 +9988,7 @@ radialError(radius, points[]):
 						end, just like for Bézier curves), by evaluating the following function:
 					</p>
 					<!--
- 
+
            __ n
 Point(t) = ❯      P   · N   (t)
            ‾‾ i=0  i     i,k
@@ -10009,7 +10009,7 @@ Point(t) = ❯      P   · N   (t)
 					</p>
 					<p>Then the N() function itself. What does it look like?</p>
 					<!--
- 
+
           ╭       t- knot         ╮                ╭       knot     -t       ╮
           │              i        │                │           (i+k)         │
 N   (t) = │ ───────────────────── │  · N     (t) + │ ─────────────────────── │  · N       (t)
@@ -10024,9 +10024,9 @@ N   (t) = │ ───────────────────── 
 						some point; obviously, it does, and specifically it does so for the following <code>i</code>/<code>k</code> values:
 					</p>
 					<!--
- 
+
           ╭ 1  if  t ∈[ knot , knot   )
-N   (t) = ╡                 i      i+1  
+N   (t) = ╡                 i      i+1
  i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/2514e1aa0565840e33fde0b146e3efe2.svg" width="237px" height="40px" loading="lazy" />
@@ -10047,7 +10047,7 @@ N   (t) = ╡                 i      i+1
 						P(t), we can compute this point by evaluating <em>d(t)</em> on a curve section between knots <code>i</code> and <code>i+1</code>:
 					</p>
 					<!--
- 
+
  k               k-1                   k-1
 d (t) = α     · d   (t) + (1-α   )  · d   (t)
  i       i,k     i            i,k      i-1
@@ -10058,7 +10058,7 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 						by:
 					</p>
 					<!--
- 
+
               t -  knots[i]
 α    = ───────────────────────────
  i,k    knots[i+1+n-k] -  knots[i]
@@ -10072,9 +10072,9 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 					</p>
 					<p>Of course, the recursion does need a stop condition:</p>
 					<!--
- 
+
  k           0                ╭ 1  if  t ∈[ knot , knot   )
-d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1  
+d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
  0           i       i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/49af474c33ce0ee0733626ea3d988570.svg" width="367px" height="40px" loading="lazy" />
@@ -10088,34 +10088,34 @@ d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
 						recursion diagram:
 					</p>
 					<!--
- 
+
      ╭                                ╭                                ╭          1     0           0
-     │                                │                                │         α   × d ,   with  d    either 0 or 1 
+     │                                │                                │         α   × d ,   with  d    either 0 or 1
      │                                │          2     1           1   │          3     3           3
-     │                                │         α   × d ,   with  d  = ╡                 +                              
+     │                                │         α   × d ,   with  d  = ╡                 +
      │                                │          3     3           3   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      │          3     2           2   │                                ╰ ╰      3 ╯     2           2
-     │         α   × d ,   with  d  = ╡                 +                                                                
+     │         α   × d ,   with  d  = ╡                 +
      │          3     3           3   │                                ╭           1     0
-     │                                │                                │          α   × d                             
+     │                                │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           2     2
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
  3   │                                │ ╰      3 ╯     2           2   │ ╭      1 ╮     0           0
-d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1     
+d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
  3   │                                ╰                                ╰ ╰      2 ╯     1           1
-     │                 +                                                                                                 
+     │                 +
      │                                ╭           2     1
-     │                                │          α   × d                                                                
+     │                                │          α   × d
      │                                │           2     2
-     │                                │                                                                                 
-     │ ╭      3 ╮     2           2   │                 +                                                               
-     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0                               
-     │ ╰      3 ╯     2           2   │                                │          α   × d 
+     │                                │
+     │ ╭      3 ╮     2           2   │                 +
+     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0
+     │ ╰      3 ╯     2           2   │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           1     1
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
      │                                │ ╰      2 ╯     1           1   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      ╰                                ╰                                ╰ ╰      1 ╯     0           0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/20e910bbea2e6eff511cb13cef18ef3b.svg" width="641px" height="336px" loading="lazy" />
diff --git a/docs/uk-UA/index.html b/docs/uk-UA/index.html
index 7ffff6ef..ba204a57 100644
--- a/docs/uk-UA/index.html
+++ b/docs/uk-UA/index.html
@@ -173,7 +173,7 @@
         <a class="btclk" href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu</a>
         or use the QR code on the right, if that's the kind of convenience you prefer =)
       </p>
-    
+
     </div>
     <div class="btcqr">
         <a href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">
@@ -185,7 +185,7 @@
 
 			<br style="clear: both;" />
 
-			<p>— <a href="https://twitter.com/TheRealPomax">Pomax</a></p>
+			<p>— <a href="https://mastodon.social/@TheRealPomax">Pomax</a></p>
 			<noscript>
 				<div class="note">
 					<header>
@@ -573,7 +573,7 @@
 						thus 80% the distance away from the second point) then we can compute that really easily:
 					</p>
 					<!--
- 
+
       ╭        p =  some point       ╮
       │         1                    │
       │        p =  some other point │
@@ -637,7 +637,7 @@ Given │  distance= (p  - p )         │,  our new point = p  +  distance  ·
 						kind of number manipulation:
 					</p>
 					<!--
- 
+
 f(x) = cos (x)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/0cc876c56200446c60114c1b0eeeb2cc.svg" width="96px" height="17px" loading="lazy" />
@@ -647,7 +647,7 @@ f(x) = cos (x)
 					</p>
 					<p>So far, so good. Now, let's look at parametric functions, and how they cheat. Let's take the following two functions:</p>
 					<!--
- 
+
 f(a) = cos (a)
 f(b) = sin (b)
 -->
@@ -658,9 +658,9 @@ f(b) = sin (b)
 						function. Parametric functions cheat by changing that. In a parametric function all the different functions share a variable, like this:
 					</p>
 					<!--
- 
+
 ╭ f (t) = cos (t)
-╡  a              
+╡  a
 │ f (t) = sin (t)
 ╰  b
 -->
@@ -678,8 +678,8 @@ f(b) = sin (b)
 						obvious:
 					</p>
 					<!--
- 
-{ x = cos (t) 
+
+{ x = cos (t)
   y = sin (t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/6914ba615733c387251682db7a3db045.svg" width="77px" height="40px" loading="lazy" />
@@ -714,7 +714,7 @@ f(b) = sin (b)
 					</p>
 					<p>You may remember polynomials from high school. They're those sums that look like this:</p>
 					<!--
- 
+
              3         2
 f(x) = a  · x  + b  · x  + c  · x + d
 -->
@@ -735,12 +735,12 @@ f(x) = a  · x  + b  · x  + c  · x + d
 						values:
 					</p>
 					<!--
- 
-linear= (1-t) + t                                        
+
+linear= (1-t) + t
              2                      2
-square= (1-t)  + 2  · (1-t)  · t + t                     
+square= (1-t)  + 2  · (1-t)  · t + t
              3             2                       2    3
- cubic= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+ cubic= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img
 						class="LaTeX SVG"
@@ -754,10 +754,10 @@ square= (1-t)  + 2  · (1-t)  · t + t
 						easy. Check out these binomial terms:
 					</p>
 					<!--
- 
- linear=  1 + 1           
- square=  1 + 2 + 1       
-  cubic=  1 + 3 + 3 + 1   
+
+ linear=  1 + 1
+ square=  1 + 2 + 1
+  cubic=  1 + 3 + 3 + 1
 quartic= 1 + 4 + 6 + 4 + 1
 -->
 					<img
@@ -778,11 +778,11 @@ quartic= 1 + 4 + 6 + 4 + 1
 					</p>
 					<!--
 
-linear= \colorblue a + \colorred b                                                                                                                   
-square= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b                                                      
+linear= \colorblue a + \colorred b
+square= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b
  cubic= \colorblue a  · \colorblue a  · \colorblue a + \colorblue a  · \colorblue a  · \colorred b + \colorblue a  · \colorred b  · \colorred b + \co
-                                                                                             
-                                                                                             
+
+
                                                        lorred b  · \colorred b  · \colorred b
 -->
 					<img
@@ -798,9 +798,9 @@ square= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred
 						generic function for this:
 					</p>
 					<!--
- 
+
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t
               ‾‾ i=0
 -->
 					<img
@@ -1119,9 +1119,9 @@ function Bezier(3,t):
 					<!--
 
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset
               ‾‾ i=0
-                                                                weight\underbracew 
+                                                                weight\underbracew
                                                                                   i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/c2f2fe0ef5d0089d9dd8e5e3999405cb.svg" width="336px" height="55px" loading="lazy" />
@@ -1132,11 +1132,11 @@ Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \unders
 						(210,30), we use this Bézier curve:
 					</p>
 					<!--
- 
+
 ╭                               3                                   2                                             2                       3
-╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t  
+╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t
 │                               3                                    2                                             2                      3
-╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t 
+╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/be73034ac382b54863c7a18c2932cbbc.svg" width="473px" height="40px" loading="lazy" />
 					<p>Which gives us the curve we saw at the top of the article:</p>
@@ -1266,9 +1266,9 @@ function Bezier(3,t,w[]):
 					</p>
 					<p>Adding these ratio values to the regular Bézier curve function is fairly easy. Where the regular function is the following:</p>
 					<!--
- 
+
               __ n                    n-i     i
-Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w 
+Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
               ‾‾ i=0                                i
 -->
 					<img
@@ -1280,13 +1280,13 @@ Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
 					/>
 					<p>The function for rational Bézier curves has two more terms:</p>
 					<!--
- 
+
                         __ n                    n-i     i
-                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio 
+                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio
                         ‾‾ i=0                                i                    i
 Rational  Bézier(n,t) = ────────────────────────────────────────────────────────────
                                      __ n                     n-i      i
-                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio  
+                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio
                                      ‾‾ i=0                                       i
 -->
 					<img
@@ -1476,8 +1476,8 @@ function RationalBezier(3,t,w[],r[]):
 						other values, then the general formula for this is:
 					</p>
 					<!--
- 
-mixture = a  ·  value  + b  ·  value 
+
+mixture = a  ·  value  + b  ·  value
                      1              2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/dfd6ded3f0addcf43e0a1581627a2220.svg" width="212px" height="16px" loading="lazy" />
@@ -1490,8 +1490,8 @@ mixture = a  ·  value  + b  ·  value
 						And that's easy:
 					</p>
 					<!--
- 
-m = a  ·  value  + (1 - a)  ·  value 
+
+m = a  ·  value  + (1 - a)  ·  value
                1                    2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/4e0fa763b173e3a683587acf83733353.svg" width="213px" height="16px" loading="lazy" />
@@ -1567,61 +1567,61 @@ m = a  ·  value  + (1 - a)  ·  value
 						refer to coordinate values "in one or more dimensions":
 					</p>
 					<!--
- 
+
                   3                   2                             2          3
-B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t 
+B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t
         1              2                        3                        4
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/9a9a55f5b0323d9ea88f82fc6be58ad3.svg" width="468px" height="20px" loading="lazy" />
 					<p>Disregarding our actual coordinates for a moment, we have:</p>
 					<!--
- 
+
             3             2                       2    3
-B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/87cfac83cb8a4b0bee68ef006effc611.svg" width="353px" height="19px" loading="lazy" />
 					<p>We can write this as a sum of four expressions:</p>
 					<!--
- 
+
                 3
-... =      (1-t) 
+... =      (1-t)
                 2
     + 3  · (1-t)   · t
                      2
-    + 3  · (1-t)  · t 
+    + 3  · (1-t)  · t
              3
-    +       t  
+    +       t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/cdd88611833f3b178df91278359a4193.svg" width="140px" height="75px" loading="lazy" />
 					<p>And we can expand these expressions:</p>
 					<!--
- 
+
                                    3         2
 ... =  (1-t)  · (1-t)  · (1-t) = -t  + 3  · t  - 3  · t + 1
                                        3         2
     + 3  · (1-t)  · (1-t)  · t = 3  · t  - 6  · t  + 3  · t
                                             3         2
-    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t 
+    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t
                                              3
-    +        t  · t  · t       =            t  
+    +        t  · t  · t       =            t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/ec118f296511c6e9ac8727be3703a7ce.svg" width="397px" height="75px" loading="lazy" />
 					<p>Furthermore, we can make all the 1 and 0 factors explicit:</p>
 					<!--
- 
+
              3         2
 ... = -1  · t  + 3  · t  - 3  · t + 1
              3         2
-    + +3  · t  - 6  · t  + 3  · t + 0 
+    + +3  · t  - 6  · t  + 3  · t + 0
              3         2
     + -3  · t  + 3  · t  + 0  · t + 0
              3         2
-    + +1  · t  + 0  · t  + 0  · t + 0 
+    + +1  · t  + 0  · t  + 0  · t + 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/67a5ea33d6c6558f7d954b18226f4956.svg" width="217px" height="75px" loading="lazy" />
 					<p>And <em>that</em>, we can view as a series of four matrix operations:</p>
 					<!--
- 
+
                  ┌ -1 ┐                    ┌ 3  ┐                    ┌ -3 ┐                    ┌ 1 ┐
 ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ -6 │ + ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ 0 │
 └ t  t  t 1 ┘    │ -3 │   └ t  t  t 1 ┘    │ 3  │   └ t  t  t 1 ┘    │ 0  │   └ t  t  t 1 ┘    │ 0 │
@@ -1630,7 +1630,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/8ecff6b8a37d60385d287ea2b26876db.svg" width="607px" height="72px" loading="lazy" />
 					<p>If we compact this into a single matrix operation, we get:</p>
 					<!--
- 
+
                  ┌ -1  3 -3 1 ┐
 ┌  3  2     ┐  · │  3 -6  3 0 │
 └ t  t  t 1 ┘    │ -3  3  0 0 │
@@ -1642,7 +1642,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 						<code>t</code> matrix horizontally, and our big "mixing" matrix upside down:
 					</p>
 					<!--
- 
+
                  ┌  1  0  0 0 ┐
 ┌      2  3 ┐  · │ -3  3  0 0 │
 └ 1 t t  t  ┘    │  3 -6  3 0 │
@@ -1651,7 +1651,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/1a64ed455c6dd2f8cacca5e5e12bdcc1.svg" width="227px" height="72px" loading="lazy" />
 					<p>And then finally, we can add in our original coordinates as a single third matrix:</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -1664,7 +1664,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 					<img class="LaTeX SVG" src="./images/chapters/matrix/b32cae2dfc47d5f36df0bc3defb7dfa8.svg" width="323px" height="73px" loading="lazy" />
 					<p>We can perform the same trick for the quadratic curve, in which case we end up with:</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2103,7 +2103,7 @@ function drawCurvePoint(points[], t):
 						respectively: (we'll reverse the Bézier coefficients vector for legibility)
 					</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2120,7 +2120,7 @@ B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -2143,7 +2143,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						"point on the curve" information into a new matrix multiplication:
 					</p>
 					<!--
- 
+
                                                   ┌ P  ┐                                              ┌ P  ┐
                                    ┌  1  0 0 ┐    │  1 │                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2160,7 +2160,7 @@ B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌
 					/>
 					<p>and</p>
 					<!--
- 
+
                                                                ┌ P  ┐                                                       ┌ P  ┐
                                                                │  1 │                    ┌ 1 0  0 0  ┐                      │  1 │
                                              ┌  1  0  0 0 ┐    │ P  │                    │ 0 z  0 0  │    ┌  1  0  0 0 ┐    │ P  │
@@ -2190,7 +2190,7 @@ B(t) = ┌                    2         3 ┐  · │ -3  3  0 0 │  · │  2
 							at the quadratic curve first:
 						</p>
 						<!--
- 
+
                                                   ┌ P  ┐
                      ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2206,7 +2206,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                                         ┌ P  ┐
                                                                                         │  1 │
 = ┌      2 ┐  · \underset we turn this...\underbrace\kern 2.25em Z  · M \kern 2.25em  · │ P  │
@@ -2222,7 +2222,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                          ┌ P  ┐
                                                         -1               │  1 │
 = ┌      2 ┐  · \underset into this...\underbrace M  · M    · Z  · M   · │ P  │
@@ -2238,7 +2238,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                                                                                                    ┌ P  ┐
                                                                                                                    │  1 │
 = ┌      2 ┐  · M \underset ...to get this!\underbrace \kern 1.25em  · \kern 1.25em Q \kern 1.25em  · \kern 1.25em │ P  │
@@ -2265,7 +2265,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							coordinates that represent the curve from <em>t = 0</em> to <em>t = z</em>. So let's get computing:
 						</p>
 						<!--
- 
+
                     ┌ 1 0 0 ┐
      -1             │   1   │    ┌ 1 0 0  ┐    ┌  1  0 0 ┐   ┌     1            0        0  ┐
 Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = │  -(z-1)          z        0  │
@@ -2281,7 +2281,7 @@ Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = 
 						/>
 						<p>Excellent! Now we can form our new quadratic curve:</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭      ┌ P  ┐ ╮
                                │  1 │                      │      │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q  · │ P  │ │
@@ -2297,7 +2297,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                     ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌     1            0        0  ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │  -(z-1)          z        0  │  · │ P  │ │
@@ -2313,7 +2313,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌                        P                          ┐
                                │                         1                         │
                 ┌  1  0 0 ┐    │               z  · P  - (z-1)  · P                │
@@ -2342,7 +2342,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							write it down in a more simple form because of the zero, but what we <em>actually</em> evaluated, making the zero explicit, was:
 						</p>
 						<!--
- 
+
                                                             ┌ P  ┐
                                              ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2358,7 +2358,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                              ┌ P  ┐
                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2375,7 +2375,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 						/>
 						<p>If we want the interval [<em>z</em>,1], we will be evaluating this instead:</p>
 						<!--
- 
+
                                                                     ┌ P  ┐
                                                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2391,7 +2391,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 							loading="lazy"
 						/>
 						<!--
- 
+
                 ┌              2        ┐                   ┌ P  ┐
                 │ 1  z        z         │    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │  · │ P  │
@@ -2411,7 +2411,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 							<code>[M · something]</code>:
 						</p>
 						<!--
- 
+
                       ┌ 1 0 0 ┐    ┌              2        ┐
       -1              │   1   │    │ 1  z        z         │    ┌  1  0 0 ┐   ┌      2                   2 ┐
 Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │ = │ (z-1)  -2  · z  · (z-1) z  │
@@ -2427,7 +2427,7 @@ Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  ·
 						/>
 						<p>So, our final second curve looks like:</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭       ┌ P  ┐ ╮
                                │  1 │                      │       │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q'  · │ P  │ │
@@ -2443,7 +2443,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                   ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌      2                   2 ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │ (z-1)  -2  · z  · (z-1) z  │  · │ P  │ │
@@ -2459,7 +2459,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌  2                                      2       ┐
                                │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
                 ┌  1  0 0 ┐    │        3                       2              1 │
@@ -2489,7 +2489,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						<code>t = z</code>, we get two subcurves that are described as Bézier curves with simple-to-derive coordinates:
 					</p>
 					<!--
- 
+
                                              ┌                        P                          ┐
                                     ┌ P  ┐   │                         1                         │
 ┌     1            0        0  ┐    │  1 │   │               z  · P  - (z-1)  · P                │
@@ -2507,7 +2507,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                            ┌  2                                      2       ┐
                                   ┌ P  ┐   │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
 ┌      2                   2 ┐    │  1 │   │        3                       2              1 │
@@ -2528,7 +2528,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						yourself, though) and simply show you the resulting new coordinate sets:
 					</p>
 					<!--
- 
+
                                                               ┌                                    P                                      ┐
                                                      ┌ P  ┐   │                                     1                                     │
 ┌    1            0                0         0  ┐    │  1 │   │                           z  · P  - (z-1)  · P                            │
@@ -2549,7 +2549,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>and</p>
 					<!--
- 
+
                                                               ┌  3               2                                 2              3       ┐
                                                      ┌ P  ┐   │ z   · P  - 3  · z   · (z-1)  · P  + 3  · z  · (z-1)   · P  - (z-1)   · P  │
 ┌       3           2                      2  3 ┐    │  1 │   │        4                        3                        2              1 │
@@ -2601,7 +2601,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						that the start and end weights are the same as the start and end weights for the old curve):
 					</p>
 					<!--
- 
+
               __ k                                                                                            k-i     i
 Bézier(k,t) = ❯      \underset binomial term\underbrace\binomki  · \ \underset polynomial term\underbrace(1-t)     · t    · \ \underset new
               ‾‾ i=0
@@ -2625,9 +2625,9 @@ Bézier(k,t) = ❯      \underset binomial term\underbrace\binomki  · \ \unders
 					</p>
 					<p>We start by taking the standard Bézier function, and condensing it a little:</p>
 					<!--
- 
+
               __ n       n               n                       n-i     i
-Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t 
+Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t
               ‾‾ i=0  i  i               i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/9fc4ecc087d389dd0111bcba165cd5d0.svg" width="408px" height="41px" loading="lazy" />
@@ -2637,7 +2637,7 @@ Bézier(n,t) = ❯      w  B (t)  , where  B (t) = \binomni  · (1-t)     · t
 						<code>t</code> and <code>1-t</code>:
 					</p>
 					<!--
- 
+
 x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/ff224ded6bbbc94b43130f5f8eeb5d29.svg" width="379px" height="16px" loading="lazy" />
@@ -2646,9 +2646,9 @@ x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
 						<code>t</code> component:
 					</p>
 					<!--
- 
-Bézier(n,t)= (1-t) B(n,t) + t B(n,t)                    
-             __ n               n      __ n         n   
+
+Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
+             __ n               n      __ n         n
            = ❯      w  (1 - t) B (t) + ❯      w  t B (t)
              ‾‾ i=0  i          i      ‾‾ i=0  i    i
 -->
@@ -2658,18 +2658,18 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
 						I promise, it's about to make sense. We start with <code>(1-t)</code>:
 					</p>
 					<!--
- 
+
          n              n!         n-i  i
-(1 - t) B (t)= (1-t) ──────── (1-t)    t                  
+(1 - t) B (t)= (1-t) ──────── (1-t)    t
          i           (n-i)!i!
                n+1-i   (n+1)!        n+1-i  i
-             = ───── ────────── (1-t)      t              
-                n+1  (n+1-i)!i!                           
+             = ───── ────────── (1-t)      t
+                n+1  (n+1-i)!i!
                k-i    k!         k-i  i
              = ─── ──────── (1-t)    t ,  where  k = n + 1
                 k  (k-i)!i!
                k-i  k
-             = ─── B (t)                                  
+             = ─── B (t)
                 k   i
 -->
 					<img
@@ -2687,18 +2687,18 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
 						<code>t</code> part, but that's not a problem:
 					</p>
 					<!--
- 
+
    n          n!         n-i  i
-t B (t)= t ──────── (1-t)    t                                    
+t B (t)= t ──────── (1-t)    t
    i       (n-i)!i!
          i+1        (n+1)!             (n+1)-(i+1)  i+1
-       = ─── ──────────────────── (1-t)            t              
-         n+1 ((n+1)-(i+1))!(i+1)!                                 
+       = ─── ──────────────────── (1-t)            t
+         n+1 ((n+1)-(i+1))!(i+1)!
          i+1        k!             k-(i+1)  i+1
        = ─── ──────────────── (1-t)        t   ,  where  k = n + 1
           k  (k-(i+1))!(i+1)!
          i+1  k
-       = ─── B   (t)                                              
+       = ─── B   (t)
           k   i+1
 -->
 					<img
@@ -2718,21 +2718,21 @@ t B (t)= t ──────── (1-t)    t
 					</p>
 					<p>Let's do this:</p>
 					<!--
- 
+
              __ n+1             n      __ n+1       n
-Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)                   
+Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)
              ‾‾ i=0  i          i      ‾‾ i=0  i    i
              __ n+1    k-i  k      __ n+1    i+1  k
            = ❯      w  ─── B (t) + ❯      w  ─── B   (t),  where  k = n + 1
              ‾‾ i=0  i  k   i      ‾‾ i=0  i  k   i+1
              __ n+1    k-i  k      __ n+1      i  k
-           = ❯      w  ─── B (t) + ❯      p    ─ B (t)                     
+           = ❯      w  ─── B (t) + ❯      p    ─ B (t)
              ‾‾ i=0  i  k   i      ‾‾ i=0  i-1 k  i
              __ n+1 ╭    k-i        i ╮  k
-           = ❯      │ w  ─── + p    ─ │ B (t)                              
+           = ❯      │ w  ─── + p    ─ │ B (t)
              ‾‾ i=0 ╰  i  k     i-1 k ╯  i
              __ n+1                      k                 i
-           = ❯      (w  (1-s) + p    s) B (t),  where  s = ─               
+           = ❯      (w  (1-s) + p    s) B (t),  where  s = ─
              ‾‾ i=0   i          i-1     i                 k
 -->
 					<img
@@ -2747,14 +2747,14 @@ Bézier(n,t)= ❯      w  (1 - t) B (t) + ❯      w  t B (t)
 						Bézier(n+1,t) as a very simple matrix multiplication:
 					</p>
 					<!--
- 
-M B  = B 
+
+M B  = B
    n    k
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/1f5b60d190a1c7099b3411e4cc477291.svg" width="71px" height="16px" loading="lazy" />
 					<p>where the matrix <strong>M</strong> is an <code>n+1</code> by <code>n</code> matrix, and looks like:</p>
 					<!--
- 
+
     ┌ 1  0   .   .   .   .    .    .  ┐
     │ 1 k-1                           │
     │ ─ ───  0   .   .   .    0    .  │
@@ -2796,20 +2796,20 @@ M = │        k   k                    │
 						<strong>A x = b</strong>, we can use it. And as it so happens, that's exactly what we're dealing with, so:
 					</p>
 					<!--
- 
-              M B = B             
+
+              M B = B
                  n   k
            T         T
-         (M  M) B = M  B          
+         (M  M) B = M  B
                  n      k
   T   -1   T          T   -1  T
-(M  M)   (M  M) B = (M  M)   M  B 
+(M  M)   (M  M) B = (M  M)   M  B
                  n               k
                       T   -1  T
-              I B = (M  M)   M  B 
+              I B = (M  M)   M  B
                  n               k
                       T   -1  T
-                B = (M  M)   M  B 
+                B = (M  M)   M  B
                  n               k
 -->
 					<img
@@ -2873,9 +2873,9 @@ M = │        k   k                    │
 					</p>
 					<p>First, let's look at the derivative rule for Bézier curves, which is:</p>
 					<!--
- 
+
                     __ n-1
-Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t) 
+Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
                     ‾‾ i=0   i+1  i                    i
 -->
 					<img
@@ -2890,7 +2890,7 @@ Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
 						is the same as a summation where each term is multiplied by <i>n</i>) as:
 					</p>
 					<!--
- 
+
                __ n-1
 Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
                ‾‾ i=0                i           i+1  i
@@ -2916,7 +2916,7 @@ Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
 							only the derivative of the polynomial basis function. So, let's find that:
 						</p>
 						<!--
- 
+
         d    ╭ n ╮  k      n-k d
 B   (t) ── = │   │ t  (1-t)    ──
  n,k    dt   ╰ k ╯             dt
@@ -2933,7 +2933,7 @@ B   (t) ── = │   │ t  (1-t)    ──
 							<a href="https://en.wikipedia.org/wiki/Chain_rule">chain</a> rules gives us:
 						</p>
 						<!--
- 
+
       ╭ n ╮        k-1      n-k    k         n-k-1
 ... = │   │ (k  · t    (1-t)    + t   · (1-t)       · (n-k)  · -1)
       ╰ k ╯
@@ -2947,9 +2947,9 @@ B   (t) ── = │   │ t  (1-t)    ──
 						/>
 						<p>Which is hard to work with, so let's expand that properly:</p>
 						<!--
- 
+
         kn!     k-1      n-k   (n-k)n!   k      n-1-k
-... = ──────── t    (1-t)    - ──────── t  (1-t)     
+... = ──────── t    (1-t)    - ──────── t  (1-t)
       k!(n-k)!                 k!(n-k)!
 -->
 						<img
@@ -2964,12 +2964,12 @@ B   (t) ── = │   │ t  (1-t)    ──
 							look like "x! over y!(x-y)!". If we can do that in a way that involves terms of <i>n-1</i> and <i>k-1</i>, we'll be on the right track.
 						</p>
 						<!--
- 
+
            n!       k-1      n-k   (n-k)n!   k      n-1-k
-... = ──────────── t    (1-t)    - ──────── t  (1-t)                                   
+... = ──────────── t    (1-t)    - ──────── t  (1-t)
       (k-1)!(n-k)!                 k!(n-k)!
         ╭    (n-1)!     k-1      n-k   (n-k)(n-1)!  k      n-1-k ╮
-... = n │ ──────────── t    (1-t)    - ─────────── t  (1-t)      │                     
+... = n │ ──────────── t    (1-t)    - ─────────── t  (1-t)      │
         ╰ (k-1)!(n-k)!                  k!(n-k)!                 ╯
         ╭        (n-1)!         (k-1)      (n-1)-(k-1)      (n-1)!     k      (n-1)-k ╮
 ... = n │ ──────────────────── t      (1-t)            - ──────────── t  (1-t)        │
@@ -2984,11 +2984,11 @@ B   (t) ── = │   │ t  (1-t)    ──
 						/>
 						<p>And that's the first part done: the two components inside the parentheses are actually regular, lower-order Bézier expressions:</p>
 						<!--
- 
+
         ╭    x!     y      x-y      x!     k      x-k ╮
 ... = n │ ──────── t  (1-t)    - ──────── t  (1-t)    │  , with x=n-1, y=k-1
-        ╰ y!(x-y)!               k!(x-k)!             ╯                     
-... = n (B           (t) - B       (t))                                     
+        ╰ y!(x-y)!               k!(x-k)!             ╯
+... = n (B           (t) - B       (t))
           (n-1),(k-1)       (n-1),k
 -->
 						<img
@@ -3003,19 +3003,19 @@ B   (t) ── = │   │ t  (1-t)    ──
 							through to its derivative:
 						</p>
 						<!--
- 
+
 Bézier   (t)    = B   (t)  · w  + B   (t)  · w  + B   (t)  · w  + B   (t)  · w  + ...
       n,k          n,0        0    n,1        1    n,2        2    n,3        3
              d
-Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +                              
+Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +
       n,k    dt          n-1,-1       n-1,0         0
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,0       n-1,1         1
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,1       n-1,2         2
-                  n  · (B     (t) - B     (t))  · w  +                               
+                  n  · (B     (t) - B     (t))  · w  +
                          n-1,2       n-1,3         3
-                  ...                                                                
+                  ...
 -->
 						<img
 							class="LaTeX SVG"
@@ -3029,18 +3029,18 @@ Bézier   (t) ── = n  · (B      (t) - B     (t))  · w  +
 							following:
 						</p>
 						<!--
- 
-n  · B      (t)  · w               +                                     
+
+n  · B      (t)  · w               +
       n-1,-1        0
 n  · B                (t)  · w     - n  · B                (t)  · w     +
       n-1,\colorblue 0        1            n-1,\colorblue 0        0
 n  · B               (t)  · w      - n  · B               (t)  · w      +
-      n-1,\colorred 1        2             n-1,\colorred 1        1      
+      n-1,\colorred 1        2             n-1,\colorred 1        1
 n  · B                   (t)  · w  - n  · B                   (t)  · w  +
       n-1,\colormagenta 2        3         n-1,\colormagenta 2        2
 ...                                - n  · B     (t)  · w                +
                                            n-1,3        3
-...                                                                      
+...
 -->
 						<img
 							class="LaTeX SVG"
@@ -3057,14 +3057,14 @@ n  · B                   (t)  · w  - n  · B                   (t)  · w  +
 							this term would contribute "nothing", so we can ignore it, too. This means we're left with:
 						</p>
 						<!--
- 
+
 n  · B                (t)  · w     - n  · B                (t)  · w     +
       n-1,\colorblue 0        1            n-1,\colorblue 0        0
 n  · B               (t)  · w      -  n  · B               (t)  · w     +
-      n-1,\colorred 1        2              n-1,\colorred 1        1     
+      n-1,\colorred 1        2              n-1,\colorred 1        1
 n  · B                   (t)  · w  - n  · B                   (t)  · w  +
       n-1,\colormagenta 2        3         n-1,\colormagenta 2        2
-...                                
+...
 -->
 						<img
 							class="LaTeX SVG"
@@ -3075,9 +3075,9 @@ n  · B                   (t)  · w  - n  · B                   (t)  · w  +
 						/>
 						<p>And that's just a summation of lower order curves:</p>
 						<!--
- 
+
              d
-Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B                 (t)  · (w  - w ) + n  · B                     (t)  · (w  - w )  
+Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B                 (t)  · (w  - w ) + n  · B                     (t)  · (w  - w )
       n,k    dt         (n-1),\colorblue 0         1    0          (n-1),\colorred 1         2    1          (n-1),\colormagenta 2         3    2
                                                                       +  ...
 -->
@@ -3090,7 +3090,7 @@ Bézier   (t) ── = n  · B                  (t)  · (w  - w ) + n  · B
 						/>
 						<p>We can rewrite this as a normal summation, and we're done:</p>
 						<!--
- 
+
              d    __ n-1                                 __ n-1
 Bézier   (t) ── = ❯      n  · B     (t)  · (w    - w ) = ❯      B     (t)  · \underset derivative weights \underbracen  · (w    - w )
       n,k    dt   ‾‾ k=0       n-1,k         k+1    k    ‾‾ k=0  n-1,k                                                      k+1    k
@@ -3109,11 +3109,11 @@ Bézier   (t) ── = ❯      n  · B     (t)  · (w    - w ) = ❯      B
 						Bézier curves, and then the derivative:
 					</p>
 					<!--
- 
+
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset
               ‾‾ i=0
-                                                               weight\underbracew 
+                                                               weight\underbracew
                                                                                  i
 -->
 					<img
@@ -3124,7 +3124,7 @@ Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \unders
 						loading="lazy"
 					/>
 					<!--
- 
+
                __ k                                                                                            k-i     i
 Bézier'(n,t) = ❯      \underset binomial term\underbrace\binomki  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset derivative
                ‾‾ i=0
@@ -3145,11 +3145,11 @@ Bézier'(n,t) = ❯      \underset binomial term\underbrace\binomki  · \ \under
 						derivative one:
 					</p>
 					<!--
- 
-B(n,t),           w = {A,B,C,D}   
+
+B(n,t),           w = {A,B,C,D}
 B'(n,t),   n = 3, w' = {A',B',C'} = {3  · (B-A),   3  · (C-B),   3  · (D-C)}
-B''(n,t),  n = 2, w'' = {A'',B''} = {2  · (B'-A'),   2  · (C'-B')}          
-B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}                        
+B''(n,t),  n = 2, w'' = {A'',B''} = {2  · (B'-A'),   2  · (C'-B')}
+B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}
 -->
 					<img
 						class="LaTeX SVG"
@@ -3178,10 +3178,10 @@ B'''(n,t), n = 1, w''' = {A'''}   = {1  · (B''-A'')}
 						indicates the direction of travel at specific points, and is literally just the first derivative of our curve:
 					</p>
 					<!--
- 
+
 tangent (t) = B' (t)
        x        x
-                    
+
 tangent (t) = B' (t)
        y        y
 -->
@@ -3197,14 +3197,14 @@ tangent (t) = B' (t)
 						point, and then do whatever it is we want to do based on those directions:
 					</p>
 					<!--
- 
+
                            ┌─────────────────┐
                            │      2         2
-    d = \| tangent(t)\| =  │B' (t)  + B' (t)  
+    d = \| tangent(t)\| =  │B' (t)  + B' (t)
                           ⟍│  x         y
 
                            tangent (t)     B' (t)
-^                                 x          x    
+^                                 x          x
 x(t) = \| tangent (t)\| =─────────────── = ──────
                  x       \| tangent(t)\|     d
 
@@ -3226,11 +3226,11 @@ y(t) = \| tangent (t)\| = ─────────────── = ──
 						is typically of length 1.0, so all we have to do is rotate the normalized directional vector and we're done:
 					</p>
 					<!--
- 
+
              ^           \pi   ^           \pi     ^
-normal (t) = x(t)  · cos ─── - y(t)  · sin ─── = - y(t)                                             
+normal (t) = x(t)  · cos ─── - y(t)  · sin ─── = - y(t)
       x                   2                 2
-                                                                                                    
+
                                                            ^           \pi   ^           \pi    ^
 normal (t) = \underset quarter circle rotation \underbrace x(t)  · sin ─── + y(t)  · cos ───  = x(t)
       y                                                                 2                 2
@@ -3253,7 +3253,7 @@ normal (t) = \underset quarter circle rotation \underbrace x(t)  · sin ──
 							To turn any point <i>(x,y)</i> into a rotated point <i>(x',y')</i> (over 0,0) by some angle φ, we apply this nice and easy computation:
 						</p>
 						<!--
- 
+
 x' = x  · cos (\phi) - y  · sin (\phi)
 y' = x  · sin (\phi) + y  · cos (\phi)
 -->
@@ -3266,7 +3266,7 @@ y' = x  · sin (\phi) + y  · cos (\phi)
 						/>
 						<p>Which is the "long" version of the following matrix transformation:</p>
 						<!--
- 
+
 ┌ x' ┐ = ┌ cos (\phi) -sin (\phi) ┐ ┌ x ┐
 └ y' ┘   └ sin (\phi) cos (\phi)  ┘ └ y ┘
 -->
@@ -3757,10 +3757,10 @@ generateRMFrames(steps) -> frames:
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
+
 B'(t) = a(1-t) + b(t)= 0,
           a - at + bt= 0,
-           (b-a)t + a= 0 
+           (b-a)t + a= 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -3771,10 +3771,10 @@ B'(t) = a(1-t) + b(t)= 0,
 					/>
 					<p>And then we turn this into our solution for <code>t</code> using basic arithmetics:</p>
 					<!--
- 
-(b-a)t + a= 0, 
+
+(b-a)t + a= 0,
     (b-a)t= -a,
-            -a 
+            -a
          t= ───
             b-a
 -->
@@ -3797,7 +3797,7 @@ B'(t) = a(1-t) + b(t)= 0,
 						you haven't, it looks like this:
 					</p>
 					<!--
- 
+
                                                    ┌────────┐
                                                    │ 2
                2                              -b ±⟍│b  - 4ac
@@ -3822,9 +3822,9 @@ Given f(t) = at  + bt + c, f(t)=0   when  t = ───────────
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
-B(t)  uses { p ,p ,p ,p  }                                                  
-              1  2  3  4                                                    
+
+B(t)  uses { p ,p ,p ,p  }
+              1  2  3  4
 B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
                1  2  3            1      2  1    2      3  2    3      4  3
 -->
@@ -3837,21 +3837,21 @@ B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
 					/>
 					<p>And then, using these <em>v</em> values, we can find out what our <em>a</em>, <em>b</em>, and <em>c</em> should be:</p>
 					<!--
- 
+
               2                  2
-B'(t)= v (1-t)  + 2v (1-t)t + v t            
+B'(t)= v (1-t)  + 2v (1-t)t + v t
         1           2          3
            2                    2       2
-  ...= v (t  - 2t + 1) + 2v (t-t ) + v t     
+  ...= v (t  - 2t + 1) + 2v (t-t ) + v t
         1                  2          3
           2                          2      2
-  ...= v t  - 2v t + v  + 2v t - 2v t  + v t 
+  ...= v t  - 2v t + v  + 2v t - 2v t  + v t
         1       1     1     2      2      3
           2       2      2
   ...= v t  - 2v t  + v t  - 2v t + v  + 2v t
         1       2      3       1     1     2
                    2
-  ...= (v -2v +v )t  + 2(v -v )t + v         
+  ...= (v -2v +v )t  + 2(v -v )t + v
          1   2  3         2  1      1
 -->
 					<img
@@ -3866,12 +3866,12 @@ B'(t)= v (1-t)  + 2v (1-t)t + v t
 						expressions of our original coordinate values, so we can do some substitution to get:
 					</p>
 					<!--
- 
+
 a= v -2v +v  = 3(-p  + 3p  - 3p  + p )
     1   2  3       1     2     3    4
-b= 2(v -v ) = 6(p  - 2p  + p )        
+b= 2(v -v ) = 6(p  - 2p  + p )
       2  1       1     2    3
-c= v  = 3(p -p )                      
+c= v  = 3(p -p )
     1      2  1
 -->
 					<img
@@ -3898,11 +3898,11 @@ c= v  = 3(p -p )
 						really hard to solve, it can be rewritten to a form for which finding the roots is "easier" (even if not "easy"):
 					</p>
 					<!--
- 
+
                    3     2
 very hard: solve at  + bt  + ct + d = 0
                   3
-   easier: solve t  + pt + q = 0       
+   easier: solve t  + pt + q = 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -4311,7 +4311,7 @@ function getCubicRoots(pa, pb, pc, pd) {
 						<code>f<sub>y</sub>(x)</code> is zero, so that the next <code>t</code> is the same as the one we already have:
 					</p>
 					<!--
- 
+
             f (x )
              y  n
 x    = x  - ───────
@@ -4459,11 +4459,11 @@ x    = x  - ───────
 						function for the y-component to an <em>n-2</em> term function. For instance, if we have a cubic curve such as this:
 					</p>
 					<!--
- 
+
 ╭                            3                               2                                          2                       3
-╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t  
+╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t
 │                            3                                2                                          2                      3
-╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t 
+╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/e5db5e945606d3429e4475ff92283a9c.svg" width="497px" height="40px" loading="lazy" />
 					<p>
@@ -4471,11 +4471,11 @@ x    = x  - ───────
 						by -160, gives us:
 					</p>
 					<!--
- 
+
 ╭                          3                               2                                          2                       3
-╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t  
+╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t
 │                          3                               2                                          2                       3
-╰ y = \colorblue 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t 
+╰ y = \colorblue 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/d412c72ed342c2036965ff251c8fb443.svg" width="481px" height="40px" loading="lazy" />
 					<p>
@@ -4483,20 +4483,20 @@ x    = x  - ───────
 						become:
 					</p>
 					<!--
- 
+
 ╭                          3                               2                                         2                       3
-╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                          3                               2                                          2                     3
-╰ y = \colorblue 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorblue  + 0  · t 
+╰ y = \colorblue 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorblue  + 0  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/e49d7bb45a18d14fbb12df4d91e2c67b.svg" width="473px" height="40px" loading="lazy" />
 					<p>If we drop all the zero-terms, this gives us:</p>
 					<!--
- 
+
 ╭                                   2                                         2                       3
-╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                                   2                                          2
-╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t                        
+╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/016f37843b2af2ae34dc8b2975505f3d.svg" width="408px" height="40px" loading="lazy" />
 					<p>
@@ -4607,7 +4607,7 @@ x    = x  - ───────
 					</p>
 					<p>What we need to do is solve a simple equation:</p>
 					<!--
- 
+
 C(t) = 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/inflections/ed68dcfb203517ca080fe48914769fb0.svg" width="59px" height="16px" loading="lazy" />
@@ -4617,7 +4617,7 @@ C(t) = 0
 						circle being on the other side of the curve. So what does <em>C(t)</em> look like? Actually something that seems not too hard:
 					</p>
 					<!--
- 
+
 C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)  ·   Bézier \prime\prime(t)
                x                     y                          y                     x
 -->
@@ -4646,15 +4646,15 @@ C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)
 							with the first and second derivatives, given our basis functions:
 						</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t                            
+Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t
              1           2            3           4
-      \prime            2                2                                      
+      \prime            2                2
 Bézier      (t) = a(1-t)  + 2b(1-t)t + ct   { a=3(x -x ),b=3(x -x ),c=3(x -x ) }
-                                                   2  1       3  2       4  3   
+                                                   2  1       3  2       4  3
       \prime\prime
-Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\                        
+Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 -->
 						<img
 							class="LaTeX SVG"
@@ -4665,14 +4665,14 @@ Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 						/>
 						<p>And of course the same functions for <em>y</em>:</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t 
+Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t
              1           2            3           4
-      \prime            2                2           
-Bézier      (t) = d(1-t)  + 2e(1-t)t + ft            
+      \prime            2                2
+Bézier      (t) = d(1-t)  + 2e(1-t)t + ft
       \prime\prime
-Bézier            (t) = w(1-t) + zt                  
+Bézier            (t) = w(1-t) + zt
 -->
 						<img
 							class="LaTeX SVG"
@@ -4686,19 +4686,19 @@ Bézier            (t) = w(1-t) + zt
 							rather overly complicated set of arithmetic expressions:
 						</p>
 						<!--
- 
+
      2             2             2             2             2
--18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y              
+-18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y
         2  1          3  1          4  1          1  2          3  2
      2             2             2             2             2
-+36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y              
++36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y
         4  2          1  3          2  3          4  3          1  4
-     2             2                                                             
--36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y 
+     2             2
+-36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y
         2  4          3  4         2  1          3  1          4  1          1  2
-+54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y 
++54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y
        3  2          4  2          1  3          2  3          1  4          2  4
--18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y   
+-18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y
      2  1          3  1          1  2          3  2          1  3          2  3
 -->
 						<img
@@ -4719,7 +4719,7 @@ Bézier            (t) = w(1-t) + zt
 						<code>t</code> values, we end up with the following simple term function for <em>C(t)</em>:
 					</p>
 					<!--
- 
+
                                  2
 (3 x  y +2 x  y +3 x  y -x  y ) t  + (3 x  y -x  y -3 x  y ) t + (x  y -x  y )
     3  2    4  2    2  3  4  3           3  2  4  2    2  3        2  3  3  2
@@ -4736,7 +4736,7 @@ Bézier            (t) = w(1-t) + zt
 						simplification:
 					</p>
 					<!--
- 
+
 a = x   · y  ╮
      3     2 │
 b = x   · y  │                             2
@@ -4755,7 +4755,7 @@ d = x   · y  │
 					/>
 					<p>This is a plain quadratic curve, and we know how to solve <em>C(t) = 0</em>; we use the quadratic formula:</p>
 					<!--
- 
+
                                                   ┌──────────┐
                                                   │ 2
 x = -3a + 2b + 3c - d ╮                      -y ±⟍│y  - 4 x z
@@ -4851,7 +4851,7 @@ z =       c - a       ╯                             2x
 								This edge is described by the function:
 							</p>
 							<!--
- 
+
       2
     -x  + 2x + 3
 y = ────────────, { x ≤1 }
@@ -4871,7 +4871,7 @@ y = ────────────, { x ≤1 }
 								the function:
 							</p>
 							<!--
- 
+
      ┌──────────┐
      │        2
     ⟍│3(4x - x )  - x
@@ -4889,7 +4889,7 @@ y = ─────────────────, { 0 ≤x ≤1 }
 						<li>
 							<p>...on the top (blue) edge, the curve's start point touches the curve, forming a loop. This edge is described by the function:</p>
 							<!--
- 
+
       2
     -x  + 3x
 y = ────────, { x ≤0 }
@@ -4956,7 +4956,7 @@ y = ────────, { x ≤0 }
 						<em>z</em> coordinate that is always 1, then we can suddenly add arbitrary values. For example:
 					</p>
 					<!--
- 
+
 ┌ 1 0 a ┐    ┌  x  ┐   ┌ 1  · x + 0  · y + a  · z ┐   ┌ x + a  · 1 ┐   ┌ x + a ┐
 │ 0 1 b │  · │  y  │ = │ 0  · x + 1  · y + b  · z │ = │ y + b  · 1 │ = │ y + b │
 └ 0 0 1 ┘    └ z=1 ┘   └ 0  · x + 0  · y + 1  · z ┘   └   1  · z   ┘   └  z=1  ┘
@@ -4967,7 +4967,7 @@ y = ────────, { x ≤0 }
 						want to subtract p1.x and p1.y, we use:
 					</p>
 					<!--
- 
+
      ┌ 1 0 -P   ┐            ┌ 1  · x + 0  · y - P    · 1 ┐   ┌ x - P   ┐
      │       1x │    ┌ x ┐   │                    1x      │   │      1x │
 T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = │ y - P   │
@@ -4983,7 +4983,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						transformation looks like this:
 					</p>
 					<!--
- 
+
 ┌ 1 S 0 ┐    ┌ x ┐   ┌ x + S  · y ┐
 │ 0 1 0 │  · │ y │ = │     y      │
 └ 0 0 1 ┘    └ 1 ┘   └     1      ┘
@@ -4994,7 +4994,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						simply <em>-x/y</em>, because <em>-x/y \</em> y = -x*. Done:
 					</p>
 					<!--
- 
+
      ┌    U     ┐
      │     2x   │
      │ 1 -─── 0 │
@@ -5012,7 +5012,7 @@ T  = │    U     │
 						x-scaling by 1/point3<sub>x</sub>, and y-scaling by point2<sub>y</sub>. This is really easy:
 					</p>
 					<!--
- 
+
      ┌  1        ┐
      │ ───  0  0 │
      │ V         │
@@ -5033,7 +5033,7 @@ T  = │      1    │
 						offset, in this case 1:
 					</p>
 					<!--
- 
+
      ┌    1    0 0 ┐
      │ 1 - W       │
      │      3y     │
@@ -5049,7 +5049,7 @@ T  = │ ─────── 1 0 │
 						be:
 					</p>
 					<!--
- 
+
                 ╭                           (-x +x )(-y +y )                ╮
                 │                              1  2    1  4                 │
                 │                -x  + x  - ────────────────                │
@@ -5084,7 +5084,7 @@ mapped  = (x) = │                                1  2                       
 						that leave?
 					</p>
 					<!--
- 
+
       ╭                 x   · y       x   · y              ╮
       │                  2     4       2     3             │
       │            x  - ──────── / x -────────             │   ╭         x           ╮
@@ -5104,11 +5104,11 @@ mapped  = (x) = │                                1  2                       
 						factors to see just how simple this is:
 					</p>
 					<!--
- 
-      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y 
+
+      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y
       │          43         │                 │  4    2     42 │            4                3
 ... = │                     │,   where  x   = │ ────────────── │,    y   = ──,   and  y   = ──
-      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y 
+      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y
       ╰  42    43       32  ╯                 ╰  3    2     32 ╯            2                2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/canonical/88e3fae7aeef6d7614290587422542c9.svg" width="563px" height="55px" loading="lazy" />
@@ -5197,16 +5197,16 @@ mapped  = (x) = │                                1  2                       
 					</p>
 					<p>First, let's look at the function for x(t):</p>
 					<!--
- 
+
              3          2            2     3
-x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt 
+x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt
 -->
 					<img class="LaTeX SVG" src="./images/chapters/yforx/378d0fd8cefa688d530ac38930d66844.svg" width="335px" height="19px" loading="lazy" />
 					<p>
 						We can rewrite this to a plain polynomial form, by just fully writing out the expansion and then collecting the polynomial factors, as:
 					</p>
 					<!--
- 
+
                          3                  2
 x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 -->
@@ -5217,7 +5217,7 @@ x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 						known constants, we can trivially rewrite this (by moving the <code>x(t)</code> across the equal sign) as:
 					</p>
 					<!--
- 
+
                    3                  2
 (-a + 3b - 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + (a-x) = 0
 -->
@@ -5310,7 +5310,7 @@ y = curve.get(t).y</textarea
 						following seemingly straight forward (if a bit overwhelming) formula:
 					</p>
 					<!--
- 
+
     ┌───────────────┐
 ╭ z │      2       2
 |   │f '(t) +f '(t)   dt
@@ -5319,7 +5319,7 @@ y = curve.get(t).y</textarea
 					<img class="LaTeX SVG" src="./images/chapters/arclength/59ebc3a7c3547a50998d1ea3664fb688.svg" width="140px" height="33px" loading="lazy" />
 					<p>or, more commonly written using Leibnitz notation as:</p>
 					<!--
- 
+
               ┌─────────────────┐
           ╭ z │       2        2
 length =  |  ⟍│(dx/dt) +(dy/dt)   dt
@@ -5358,7 +5358,7 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 |   ⟍│(dx/dt) +(dy/dt)   dt =  |   f(t) dt  ≃ ┌ \underset strip 1 \underbrace C   · f(t )   + ...  + \underset strip n \underbrace C   · f(t )  ┐ =
 ╯ -1                           ╯ -1           └                                1       1                                            n       n   ┘
                                                                                     __ n
-                                           \underset strips 1 through n \underbrace ❯      C   · f(t )   
+                                           \underset strips 1 through n \underbrace ❯      C   · f(t )
                                                                                     ‾‾ i=1  i       i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/arclength/b76753476ad6ecfe4b8f39bcf9432980.svg" width="623px" height="71px" loading="lazy" />
@@ -5440,16 +5440,16 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							shifting and scaling the inputs. Doing so, we get the following:
 						</p>
 						<!--
- 
+
      ┌─────────────────┐
  ╭ z │       2        2
- |  ⟍│(dx/dt) +(dy/dt)   dt                                        
+ |  ⟍│(dx/dt) +(dy/dt)   dt
  ╯ 0
      z    ┌        ╭ z         z ╮                ╭ z         z ╮ ┐
  ≃ \ ─  · │ C   · f│ ─  · t  + ─ │ + ... + C   · f│ ─  · t  + ─ │ │
      2    └  1     ╰ 2     1   2 ╯          n     ╰ 2     n   2 ╯ ┘
     z    __ n         ╭ z         z ╮
-= \ ─  · ❯     C   · f│ ─  · t  + ─ │                              
+= \ ─  · ❯     C   · f│ ─  · t  + ─ │
     2    ‾‾ i=1 i     ╰ 2     i   2 ╯
 -->
 						<img
@@ -5471,10 +5471,10 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							<a href="./legendre-gauss.html">these tables</a> would tell us that for <em>n=2</em> we must use the following values:
 						</p>
 						<!--
- 
-C  = 1     
+
+C  = 1
  1
-C  = 1     
+C  = 1
  2
         1
 t  = - ────
@@ -5490,7 +5490,7 @@ t  = + ────
 							Which means that in order for us to approximate the integral, we must plug these values into the approximate function, which gives us:
 						</p>
 						<!--
- 
+
     ┌─────────────────┐
 ╭ z │       2        2       z    ┌  ╭ z     -1    z ╮    ╭ z     1     z ╮ ┐
 |  ⟍│(dx/dt) +(dy/dt)   dt ≃ ─  · │ f│ ─  · ──── + ─ │ + f│ ─  · ──── + ─ │ │
@@ -5654,13 +5654,13 @@ t  = + ────
 					</p>
 					<p>So, what does the function look like? This:</p>
 					<!--
- 
+
          x'y'' - x''y'
 \kappa = ─────────────
                    3
                    ─
              2   2 2
-          (x' +y' ) 
+          (x' +y' )
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/060acd6ff0a050fe4d98a7802a2b3a3f.svg" width="113px" height="47px" loading="lazy" />
 					<p>
@@ -5668,14 +5668,14 @@ t  = + ────
 						tiny bit:
 					</p>
 					<!--
- 
+
             B '(t)B ''(t) - B ''(t)B '(t)
              x     y         x      y
 \kappa(t) = ─────────────────────────────
                                   3
                                   ─
                         2       2 2
-                 (B '(t) +B '(t) ) 
+                 (B '(t) +B '(t) )
                    x       y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/afd8cb8b0fe291ff703752c1c9cc33d4.svg" width="239px" height="55px" loading="lazy" />
@@ -5754,7 +5754,7 @@ function kappa(t, B):
 						circle that "fits" the curve's curvature at any point, using what is possibly the simplest bit of maths found in this entire primer:
 					</p>
 					<!--
- 
+
            1
 R(t) = ─────────
        \kappa(t)
@@ -6204,8 +6204,8 @@ lli = function(line1, line2):
 						points, so logically <code>C</code> will have a function that interpolates between those two coordinates:
 					</p>
 					<!--
- 
-C = u(t)  · P       + (1-u(t))  · P    
+
+C = u(t)  · P       + (1-u(t))  · P
               start                 end
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8c6662f605722fb2ff6cd7f65243a126.svg" width="229px" height="16px" loading="lazy" />
@@ -6218,22 +6218,22 @@ C = u(t)  · P       + (1-u(t))  · P
 						(with thanks to Boris Zbarsky) shows us the following two formulae:
 					</p>
 					<!--
- 
+
                         2
-                   (1-t) 
+                   (1-t)
 u(t)           = ───────────
      quadratic    2        2
-                 t  + (1-t) 
+                 t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8e7cfee39c98f2ddf9b635a914066cf6.svg" width="183px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                     3
-               (1-t) 
+               (1-t)
 u(t)       = ───────────
      cubic    3        3
-             t  + (1-t) 
+             t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/a0b99054cc82ca1fb147f077e175ef10.svg" width="161px" height="41px" loading="lazy" />
 					<p>
@@ -6244,7 +6244,7 @@ u(t)       = ───────────
 					</p>
 					<p>We start by observing that, given <code>A</code>, <code>B</code>, and <code>C</code>, the following always holds:</p>
 					<!--
- 
+
             distance(B,C)
 ratio(t) = ────────────── =  Constant
             distance(A,B)
@@ -6252,22 +6252,22 @@ ratio(t) = ────────────── =  Constant
 					<img class="LaTeX SVG" src="./images/chapters/abc/131454dcbac04e567f322979f4af80c6.svg" width="245px" height="39px" loading="lazy" />
 					<p>Working out the maths for this, we see the following two formulae for quadratic and cubic curves:</p>
 					<!--
- 
+
                        2        2
                       t  + (1-t)  - 1
 ratio(t)           = |───────────────|
          quadratic       2        2
-                        t  + (1-t) 
+                        t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/5924e162b50272c40c842fad14b8fa48.svg" width="239px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                    3        3
                   t  + (1-t)  - 1
 ratio(t)       = |───────────────|
          cubic       3        3
-                    t  + (1-t) 
+                    t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8cd992c1ceaae2e67695285beef23a24.svg" width="219px" height="41px" loading="lazy" />
 					<p>
@@ -6277,7 +6277,7 @@ ratio(t)       = |───────────────|
 						<code>ratio(t)</code> function to find <code>A</code>:
 					</p>
 					<!--
- 
+
           C - B           B - C
 A = B - ───────── = B + ─────────
          ratio(t)        ratio(t)
@@ -6292,11 +6292,11 @@ A = B - ───────── = B + ─────────
 						pair, after which we can reverse engineer <code>v1</code> and <code>v2</code>:
 					</p>
 					<!--
- 
+
                                  ╭     e  - t  · A
                                  │      1
-╭ e = (1-t)  · v  + t  · A       │ v = ───────────    
-╡  1            1            ==> ╡  1      1-t         
+╭ e = (1-t)  · v  + t  · A       │ v = ───────────
+╡  1            1            ==> ╡  1      1-t
 │ e = (1-t)  · A + t  · v        │     e  - (1-t)  · A
 ╰  2                     2       │      2
                                  │ v = ───────────────
@@ -6305,14 +6305,14 @@ A = B - ───────── = B + ─────────
 					<img class="LaTeX SVG" src="./images/chapters/abc/3166afa345aec1abda432c39b68d39a0.svg" width="339px" height="73px" loading="lazy" />
 					<p>And then reverse engineer the curve's control points:</p>
 					<!--
- 
+
                                       ╭     v  - (1-t)  ·  start
                                       │      1
 ╭ v = (1-t)  ·  start + t  · C        │ C = ────────────────────
-╡  1                          1   ==> ╡  1           t           
+╡  1                          1   ==> ╡  1           t
 │ v = (1-t)  · C  + t  ·  end         │     v  - t  ·  end
 ╰  2            2                     │      2
-                                      │ C = ──────────────      
+                                      │ C = ──────────────
                                       ╰  2       1-t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8bd3e6fed5bf8d871d30221ae400fd93.svg" width="383px" height="75px" loading="lazy" />
@@ -6342,15 +6342,15 @@ A = B - ───────── = B + ─────────
 						<code>B</code> point, and <code>B</code> and end point as a ratio, using
 					</p>
 					<!--
- 
+
 ╭ d = || Start - B||
 │  1
-│ d = || End - B||  
+│ d = || End - B||
 │  2
-╡      d             
+╡      d
 │       1
-│  t= ─────         
-│     d +d 
+│  t= ─────
+│     d +d
 ╰      1  2
 -->
 					<img
@@ -6412,9 +6412,9 @@ A = B - ───────── = B + ─────────
 						<code>e2</code> coordinates must obey the standard de Casteljau rule for linear interpolation:
 					</p>
 					<!--
- 
-╭ e = B + t  · d    
-╡  1                 
+
+╭ e = B + t  · d
+╡  1
 │ e = B - (1-t)  · d
 ╰  2
 -->
@@ -6434,7 +6434,7 @@ A = B - ───────── = B + ─────────
 						curve with a loop in it. To do this, we can use the <a href="https://en.wikipedia.org/wiki/Atan2">atan2</a> function:
 					</p>
 					<!--
- 
+
 \phi = (atan2(E -S , E -S ) - atan2(B -S , B -S ) + 2 \pi)   mod  2 \pi
                y  y   x  x           y  y   x  x
 -->
@@ -6451,8 +6451,8 @@ A = B - ───────── = B + ─────────
 						<code>d</code>:
 					</p>
 					<!--
- 
-d = {  d  if  0 ≤\phi ≤\pi             
+
+d = {  d  if  0 ≤\phi ≤\pi
       -d  if  \phi < 0 \lor \phi > \pi
 -->
 					<img
@@ -6632,7 +6632,7 @@ for (coordinate, index) in LUT:
 						maths, that means we're trying to solve:
 					</p>
 					<!--
- 
+
 dist(B(t), c) = r
 -->
 					<img
@@ -6644,15 +6644,15 @@ dist(B(t), c) = r
 					/>
 					<p>Which seems simple enough. Unfortunately, when we expand that <code>dist</code> function, things get a lot more problematic:</p>
 					<!--
- 
-r=  dist(B(t), c)                                                                                                              
+
+r=  dist(B(t), c)
     ┌─────────────────────────┐
     │          2             2
- =  │(B t - c )  + (B t - c )                                                                                                  
-   ⟍│  x     x       y     y                                                                                                   
+ =  │(B t - c )  + (B t - c )
+   ⟍│  x     x       y     y
     ┌─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┐
     │╭         3             2                 2       3      ╮2   ╭         3             2                 2       3      ╮2
- =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │  
+ =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │
    ⟍│╰  1             2               3             4       x ╯    ╰  1             2               3             4       y ╯
 -->
 					<img
@@ -6925,17 +6925,17 @@ findClosest(start, p, r, LUT):
 						<code>B</code>, we can compute the associated <code>C</code>:
 					</p>
 					<!--
- 
+
 C = u(t)   ·  Start + (1-u(t) )  ·  End
         q                    q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/48887d68a861a0acdf8313e23fb19880.svg" width="247px" height="24px" loading="lazy" />
 					<p>And then the associated <code>A</code>:</p>
 					<!--
- 
+
           C - B            B - C
 A = B - ────────── = B + ──────────
-         ratio(t)         ratio(t) 
+         ratio(t)         ratio(t)
                  q                q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/2e65bc9c934380c2de6a24bcd5c1c7b7.svg" width="228px" height="41px" loading="lazy" />
@@ -7074,12 +7074,12 @@ A = B - ────────── = B + ──────────
 							form, where each line corresponds to a power of t, and each column is for a specific coefficient. First, we expand the function:
 						</p>
 						<!--
- 
+
                     2                    2
-B          = a (1-t)  + 2 b (1-t) t + c t    
-  quadratic                                  
+B          = a (1-t)  + 2 b (1-t) t + c t
+  quadratic
                          2            2     2
-           = a - 2at + at  + 2bt - 2bt  + ct 
+           = a - 2at + at  + 2bt - 2bt  + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -7090,12 +7090,12 @@ B          = a (1-t)  + 2 b (1-t) t + c t
 						/>
 						<p>And then we (trivially) rearrange the terms across multiple lines:</p>
 						<!--
- 
-B          =a               
+
+B          =a
   quadratic
-            - 2at+ 2bt      
+            - 2at+ 2bt
                 2     2    2
-            + at - 2bt + ct 
+            + at - 2bt + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -7111,7 +7111,7 @@ B          =a
 						</p>
 						<p>With that arrangement, we can easily decompose this as a matrix multiplication:</p>
 						<!--
- 
+
                                          ┌  a  +  0  + 0 ┐                 ┌  1  0 0 ┐    ┌ a ┐
 B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌      2 ┐  · │ -2  2 0 │  · │ b │
   quadratic                └ 1 t t  ┘    └  a  + -2b + c ┘   └ 1 t t  ┘    └  1 -2 1 ┘    └ c ┘
@@ -7125,9 +7125,9 @@ B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌
 						/>
 						<p>We can do the same for the cubic curve, of course. We know the base function for cubics:</p>
 						<!--
- 
+
               3          2             2     3
-B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt 
+B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
   cubic
 -->
 						<img
@@ -7139,14 +7139,14 @@ B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
 						/>
 						<p>So we write out the expansion and rearrange:</p>
 						<!--
- 
-B      =a                     
+
+B      =a
   cubic
-        - 3at + 3bt           
-             2     2    2     
-        + 3at - 6bt +3ct      
+        - 3at + 3bt
+             2     2    2
+        + 3at - 6bt +3ct
             3      3    3    3
-        - at  + 3bt -3ct + dt 
+        - at  + 3bt -3ct + dt
 -->
 						<img
 							class="LaTeX SVG"
@@ -7157,7 +7157,7 @@ B      =a
 						/>
 						<p>Which we can then decompose:</p>
 						<!--
- 
+
                                         ┌  1  0  0 0 ┐    ┌ a ┐
 B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
   cubic                └ 1 t t  t  ┘    │  3 -6  3 0 │    │ c │
@@ -7172,7 +7172,7 @@ B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
 						/>
 						<p>And, of course, we can do this for quartic curves too (skipping the expansion step):</p>
 						<!--
- 
+
                                              ┌  1  0   0   0 0 ┐    ┌ a ┐
                                              │ -4  4   0   0 0 │    │ b │
 B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  · │ c │
@@ -7198,7 +7198,7 @@ B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  ·
 						we want to do curve fitting:
 					</p>
 					<!--
- 
+
     ┌ p   ┐
     │  1  │
     │ p   │
@@ -7236,9 +7236,9 @@ P = │  2  │
 					</p>
 					<p>To get these values, we first compute the general "distance along the polygon" matrix:</p>
 					<!--
- 
+
                                ╭            d  = 0
-D = ┌ d  d  ... d  ┐,   where  ╡             1                 
+D = ┌ d  d  ... d  ┐,   where  ╡             1
     └  1  2      n ┘           │ d  = d    +  length(p   , p )
                                ╰  i    i-1            i-1   i
 -->
@@ -7255,10 +7255,10 @@ D = ┌ d  d  ... d  ┐,   where  ╡             1
 						<code>i=n</code> fall in the [0,1] interval, so we need to scale all values down by whatever the total length of the polygon is:
 					</p>
 					<!--
- 
+
                                ╭    s  = 0
                                │     1
-S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d  
+S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
     └  1  2      n ┘           │  i    i    n
                                │    s  = 1
                                ╰     n
@@ -7280,9 +7280,9 @@ S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
 						to for the associated <code>t</code> value", which we'll square to get rid of any pesky negative signs:
 					</p>
 					<!--
- 
+
                            2
-E(C)  = (p  -   Bézier(s )) 
+E(C)  = (p  -   Bézier(s ))
     i     i             i
 -->
 					<img
@@ -7297,9 +7297,9 @@ E(C)  = (p  -   Bézier(s ))
 						function. So, we literally just do that; the total error function is simply the sum of all these individual errors:
 					</p>
 					<!--
- 
+
        __ n                      2
-E(C) = ❯      (p  -   Bézier(s )) 
+E(C) = ❯      (p  -   Bézier(s ))
        ‾‾ i=1   i             i
 -->
 					<img
@@ -7316,9 +7316,9 @@ E(C) = ❯      (p  -   Bézier(s ))
 						<strong>T x M x C</strong> we talked about before, which gives us:
 					</p>
 					<!--
- 
+
                 2
-E(C) = (P - TMC) 
+E(C) = (P - TMC)
 -->
 					<img
 						class="LaTeX SVG"
@@ -7329,7 +7329,7 @@ E(C) = (P - TMC)
 					/>
 					<p>In which we can replace the rather cumbersome "squaring" operation with a more conventional matrix equivalent:</p>
 					<!--
- 
+
                 T
 E(C) = (P - TMC)  (P - TMC)
 -->
@@ -7351,7 +7351,7 @@ E(C) = (P - TMC)  (P - TMC)
 						those, and then use that 𝕋 instead of <strong>T</strong> in our error function:
 					</p>
 					<!--
- 
+
     ┌    0    1      n-2   n-1  ┐
     │   s    s  ... s     s     │
     │    1    1      1     1    │
@@ -7371,7 +7371,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Which, because of the first and last values in <strong>S</strong>, means:</p>
 					<!--
- 
+
     ┌    1     0  ...   0    0    ┐
     │                  n-2   n-1  │
     │    1    s       s     s     │
@@ -7391,7 +7391,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Now we can properly write out the error function as matrix operations:</p>
 					<!--
- 
+
                 T
 E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 -->
@@ -7408,7 +7408,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						taking the derivative, and determining where that derivative is zero:
 					</p>
 					<!--
- 
+
 ∂E          T
 ── = 0 = -2𝕋  (P - 𝕋MC)
 ∂C
@@ -7447,7 +7447,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						for <strong>C</strong>:
 					</p>
 					<!--
- 
+
      -1   T   -1  T
 C = M   (𝕋  𝕋)   𝕋  P
 -->
@@ -7546,13 +7546,13 @@ C = M   (𝕋  𝕋)   𝕋  P
 						previous and next point:
 					</p>
 					<!--
- 
+
                ┌  V  = P       ┐
                │   1    2      │
 ┌ P  ┐         │  V  = P       │
 │  1 │         │   2    3      │
 │ P  │         │       P  - P  │
-│  2 │       = │        3    1 │              
+│  2 │       = │        3    1 │
 │ P  │points   │ V'  = ─────── │ point-tangent
 │  3 │         │   1      2    │
 │ P  │         │       P  - P  │
@@ -7685,7 +7685,7 @@ for p = 1 to points.length-3 (inclusive):
 						so how different is the matrix form for Catmull-Rom curves?:
 					</p>
 					<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7722,7 +7722,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							we'll get to that:
 						</p>
 						<!--
- 
+
                         ┌   P     ┐
                         │    2    │
 ┌ V   ┐        ┌ P  ┐   │   P     │
@@ -7750,7 +7750,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						</p>
 						<p>Computing <em>T</em> is really more "arranging the numbers":</p>
 						<!--
- 
+
               ┌   P     ┐
               │    2    │
      ┌ P  ┐   │   P     │   ┌  0  · P1 + 1  · P2 + 0  · P3 + 0  · P4 ┐   ┌  0  1 0 0 ┐    ┌ P  ┐
@@ -7773,7 +7773,7 @@ T  · │  2 │ = │  3    1 │ = │ ──  · P1 + 0  · P2 + ─  · P3 +
 						/>
 						<p>Thus:</p>
 						<!--
- 
+
     ┌  0  1 0 0 ┐
     │  0  0 1 0 │
     │ -1    1   │
@@ -7797,7 +7797,7 @@ T = │ ──  0 ─ 0 │
 							factor into both our coordinate vector representation, and mapping matrix <em>T</em>:
 						</p>
 						<!--
- 
+
           ┌   P     ┐
           │    2    │
 ┌ V   ┐   │   P     │        ┌  0  1  0 0  ┐
@@ -7823,7 +7823,7 @@ T = │ ──  0 ─ 0 │
 							coordinates, and see what we end up with:
 						</p>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7842,7 +7842,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						/>
 						<p>Replace point/tangent vector with the expression for all-coordinates:</p>
 						<!--
- 
+
                                                     ┌  0  1  0 0  ┐    ┌ P  ┐
                                                     │  0  0  1 0  │    │  1 │
                                  ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -7861,7 +7861,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │ ──  0 ─
 						/>
 						<p>and merge the matrices:</p>
 						<!--
- 
+
                                  ┌  0    1      0   0  ┐
                                  │ -1          1       │    ┌ P  ┐
                                  │ ──    0     ──   0  │    │  1 │
@@ -7882,7 +7882,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  1 1           1 -1 │  · │  2 │
 						/>
 						<p>This looks a lot like the Bézier matrix form, which as we saw in the chapter on Bézier curves, should look like this:</p>
 						<!--
- 
+
                                                ┌ P  ┐
                                                │  1 │
                              ┌  1  0  0 0 ┐    │ P  │
@@ -7901,7 +7901,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>So, if we want to express a Catmull-Rom curve using a Bézier curve, we'll need to turn this Catmull-Rom bit:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │
@@ -7922,7 +7922,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Into something that looks like this:</p>
 						<!--
- 
+
                   ┌ P  ┐
                   │  1 │
 ┌  1  0  0 0 ┐    │ P  │
@@ -7941,7 +7941,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And the way we do that is with a fairly straight forward bit of matrix rewriting. We start with the equality we need to ensure:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐                          ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │                          │  1 │
@@ -7962,7 +7962,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we remove the coordinate vector from both sides without affecting the equality:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │
 │ ──    0     ──   0  │
@@ -7983,7 +7983,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we can "get rid of" the Bézier matrix on the right by left-multiply both with the inverse of the Bézier matrix:</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -8007,7 +8007,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							outright remove any matrix/inverse pair:
 						</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -8028,7 +8028,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And now we're <em>basically</em> done. We just multiply those two matrices and we know what <em>V</em> is:</p>
 						<!--
- 
+
 ┌  0  1  0 0  ┐
 │ -1    1     │
 │ ──  1 ── 0  │
@@ -8050,7 +8050,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							<li>Start with the Catmull-Rom function:</li>
 						</ol>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -8071,7 +8071,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite to pure coordinate form:</li>
 						</ol>
 						<!--
- 
+
                                       ┌   P     ┐
                                       │    2    │
                                       │   P     │
@@ -8096,7 +8096,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for "normal" coordinate vector:</li>
 						</ol>
 						<!--
- 
+
                                       ┌  0  1  0 0  ┐    ┌ P  ┐
                                       │  0  0  1 0  │    │  1 │
                    ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -8117,7 +8117,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>merge the inner matrices:</li>
 						</ol>
 						<!--
- 
+
                    ┌  0    1      0   0  ┐
                    │ -1          1       │    ┌ P  ┐
                    │ ──    0     ──   0  │    │  1 │
@@ -8140,7 +8140,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for Bézier matrix form:</li>
 						</ol>
 						<!--
- 
+
                                      ┌  0  1  0 0  ┐    ┌ P  ┐
                                      │ -1    1     │    │  1 │
                    ┌  1  0  0 0 ┐    │ ──  1 ── 0  │    │ P  │
@@ -8161,7 +8161,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>and transform the coordinates so we have a "pure" Bézier expression:</li>
 						</ol>
 						<!--
- 
+
                                      ┌     P       ┐
                                      │      2      │
                                      │      P -P   │
@@ -8190,13 +8190,13 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						curve that has the vector:
 					</p>
 					<!--
- 
+
                        ┌     P       ┐
                        │      2      │
 ┌ P  ┐                 │      P -P   │
 │  1 │                 │       3  1  │
 │ P  │                 │ P  + ────── │
-│  2 │             ==> │  2   6  · τ │        
+│  2 │             ==> │  2   6  · τ │
 │ P  │ CatmullRom      │      P -P   │  Bézier
 │  3 │                 │       4  2  │
 │ P  │                 │ P  - ────── │
@@ -8216,11 +8216,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						tension Catmull-Rom curve with the following coordinate values:
 					</p>
 					<!--
- 
+
 ┌ P  ┐              ┌       P         ┐
 │  1 │              │        1        │
 │ P  │              │       P         │
-│  2 │          ==> │        4        │           
+│  2 │          ==> │        4        │
 │ P  │  Bézier      │ P  + 3(P  - P ) │ CatmullRom
 │  3 │              │  4      1    2  │
 │ P  │              │ P  + 3(P  - P ) │
@@ -8235,11 +8235,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 					/>
 					<p>Or, if your API allows you to specify Catmull-Rom curves using plain coordinates:</p>
 					<!--
- 
+
 ┌ P  ┐              ┌ P  + 6(P  - P ) ┐
 │  1 │              │  4      1    2  │
 │ P  │              │       P         │
-│  2 │          ==> │        1        │           
+│  2 │          ==> │        1        │
 │ P  │  Bézier      │       P         │ CatmullRom
 │  3 │              │        4        │
 │ P  │              │ P  + 6(P  - P ) │
@@ -8330,8 +8330,8 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						the same as it's "outgoing" derivative:
 					</p>
 					<!--
- 
-B'(1)  = B'(0)   
+
+B'(1)  = B'(0)
      n        n+1
 -->
 					<img class="LaTeX SVG" src="./images/chapters/polybezier/a37252ff55837b918d9d64078ae92ae7.svg" width="124px" height="17px" loading="lazy" />
@@ -8341,7 +8341,7 @@ B'(1)  = B'(0)
 						last control point through the last on-curve point. And mirroring any point A through any point B is really simple:
 					</p>
 					<!--
- 
+
            ┌ B  + (B  - A ) ┐   ┌ 2B  - A  ┐
 Mirrored = │  x     x    x  │ = │   x    x │
            │ B  + (B  - A ) │   │ 2B  - A  │
@@ -8517,7 +8517,7 @@ Mirrored = │  x     x    x  │ = │   x    x │
 							any point on <code>O(t)</code> is a fixed distance <code>d</code> away from coordinate <code>B(t)</code>. So let's math that:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d
 -->
 						<img
@@ -8534,7 +8534,7 @@ O(t) = B(t) + d
 							at <code>B(t)</code>. Easy enough:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d  · N(t)
 -->
 						<img
@@ -8552,7 +8552,7 @@ O(t) = B(t) + d  · N(t)
 							<code>N(t)</code> always has length 1, which we can achieve by dividing <code>B'(t)</code> by its magnitude:
 						</p>
 						<!--
- 
+
           ╭   B'(t)    ╮
 N(t) \bot │ ────────── │
           ╰ || B'(t)|| ╯
@@ -8570,10 +8570,10 @@ N(t) \bot │ ────────── │
 							denoted with double vertical bars:
 						</p>
 						<!--
- 
+
                     ┌───────────┐
                ╭ b  │   2      2
-|| f(x,y)|| =  |    │f '  + f '  
+|| f(x,y)|| =  |    │f '  + f '
                ╯ a ⟍│ x      y
 -->
 						<img
@@ -8587,10 +8587,10 @@ N(t) \bot │ ────────── │
 							So if we want the length of the tangent, we plug in <code>B'(t)</code>, with <code>t = 0</code> as start and <code>t = 1</code> as end:
 						</p>
 						<!--
- 
+
                    ┌───────────────────┐
               ╭ 1  │       2          2
-|| B'(t)|| =  |    │B ''(t)  + B ''(t)  
+|| B'(t)|| =  |    │B ''(t)  + B ''(t)
               ╯ 0 ⟍│ x          y
 -->
 						<img
@@ -8806,7 +8806,7 @@ N(t) \bot │ ────────── │
 						some angle <em>φ</em>:
 					</p>
 					<!--
- 
+
              S = \begin{pmatrix} 1
 0 \end{pmatrix}  ,  \ E = \begin{pmatrix} cos(φ)
               sin(φ) \end{pmatrix}
@@ -8814,7 +8814,7 @@ N(t) \bot │ ────────── │
 					<img class="LaTeX SVG" src="./images/chapters/circles/7ab3da0922477af4cc09f5852100976b.svg" width="175px" height="40px" loading="lazy" />
 					<p>What we want to find is the intersection of the tangents, so we want a point C such that:</p>
 					<!--
- 
+
               C = S + a  · \begin{pmatrix} 0
 1 \end{pmatrix}  ,  \ C = E + b  · \begin{pmatrix} -sin(φ)
                    cos(φ) \end{pmatrix}
@@ -8825,22 +8825,22 @@ N(t) \bot │ ────────── │
 						through E (at some distance <em>b</em> from E). Solving this gives us:
 					</p>
 					<!--
- 
+
 ╭ C  = 1 = cos(φ) + b  · -sin(φ)
-╡  x                             
-│ C  = a = sin(φ) + b  · cos(φ) 
+╡  x
+│ C  = a = sin(φ) + b  · cos(φ)
 ╰  y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/8237af1396bb567d70c8b5e4dd7f8115.svg" width="219px" height="40px" loading="lazy" />
 					<p>First we solve for <em>b</em>:</p>
 					<!--
- 
+
 1 = cos(φ) + b  · -sin(φ)  → \ 1 - cos(φ) = -b  · sin(φ)  → \ -1 + cos(φ) = b  · sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/1829e42ea956ee4df0e45d9ac5334ef7.svg" width="560px" height="17px" loading="lazy" />
 					<p>which yields:</p>
 					<!--
- 
+
     cos(φ)-1
 b = ────────
      sin(φ)
@@ -8848,21 +8848,21 @@ b = ────────
 					<img class="LaTeX SVG" src="./images/chapters/circles/06369b00338310df0a810c592485aa0a.svg" width="101px" height="39px" loading="lazy" />
 					<p>which we can then substitute in the expression for <em>a</em>:</p>
 					<!--
- 
- a= sin(φ) + b  · cos(φ)          
+
+ a= sin(φ) + b  · cos(φ)
              -1 + cos(φ)
 ..= sin(φ) + ───────────  · cos(φ)
                sin(φ)
                           2
              -cos(φ) + cos (φ)
-..= sin(φ) + ─────────────────    
-                  sin(φ)          
+..= sin(φ) + ─────────────────
+                  sin(φ)
        2         2
     sin (φ) + cos (φ) - cos(φ)
-..= ──────────────────────────    
+..= ──────────────────────────
               sin(φ)
     1 - cos(φ)
- a= ──────────                    
+ a= ──────────
       sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/5a23f3bc298c85540c6dd18e304d9224.svg" width="231px" height="195px" loading="lazy" />
@@ -8873,7 +8873,7 @@ b = ────────
 						polynomial, and we know where the circle arc's actual point P is for angle φ/2:
 					</p>
 					<!--
- 
+
          φ                 φ
 P  = cos(─)  ,  \ P  = sin(─)
  x       2         y       2
@@ -8881,7 +8881,7 @@ P  = cos(─)  ,  \ P  = sin(─)
 					<img class="LaTeX SVG" src="./images/chapters/circles/8b4e1d0a62380ed011f27c645ed13b28.svg" width="188px" height="32px" loading="lazy" />
 					<p>We compute T, observing that if <em>t=0.5</em>, the polynomial values (1-t)², 2(1-t)t, and t² are 0.25, 0.5, and 0.25 respectively:</p>
 					<!--
- 
+
     1    2    1    1
 T = ─S + ─C + ─E = ─(S + 2C + E)
     4    4    4    4
@@ -8889,10 +8889,10 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/a04bd1558a76e60b8ca6e1fe4fa38c00.svg" width="252px" height="35px" loading="lazy" />
 					<p>Which, worked out for the x and y components, gives:</p>
 					<!--
- 
+
 ╭     1
-│ T = ─(3 + cos(φ))                                    
-╡  x  4                                                 
+│ T = ─(3 + cos(φ))
+╡  x  4
 │     1╭ 2-2cos(φ)          ╮   1╭     ╭ φ ╮          ╮
 │ T = ─│ ───────── + sin(φ) │ = ─│ 2tan│ ─ │ + sin(φ) │
 ╰  y  4╰  sin(φ)            ╯   4╰     ╰ 2 ╯          ╯
@@ -8900,17 +8900,17 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/986ae9104e0bc52e95689ae7ae4504db.svg" width="408px" height="77px" loading="lazy" />
 					<p>And the distance between these two is the standard Euclidean distance:</p>
 					<!--
- 
+
                  1                   φ        4╭ φ ╮
 d (φ)= T  - P  = ─(3 + cos(φ)) - cos(─) = 2sin │ ─ │  ,
  x      x    x   4                   2         ╰ 4 ╯
                  1╭     ╭ φ ╮          ╮       φ
-d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,   
+d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
  y      y    y   4╰     ╰ 2 ╯          ╯       2
-     ⇓                                                 
+     ⇓
         ┌───────┐                     ┌───────┐
         │ 2    2                 4 φ  │   1
- d(φ)=  │d  + d   =  ...   = 2sin (─) │───────         
+ d(φ)=  │d  + d   =  ...   = 2sin (─) │───────
        ⟍│ x    y                   4  │   2 φ
                                       │cos (─)
                                      ⟍│     2
@@ -8957,7 +8957,7 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 						precision:
 					</p>
 					<!--
- 
+
                 ╭  ┌─────────────┐ ╮
                 │  │     ┌──────┐  │
                 │ ⟍│2+ε-⟍│ε(2+ε)   │
@@ -9017,14 +9017,14 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 					</p>
 					<p>So let's again formally describe this:</p>
 					<!--
- 
-P = (1, 0)                     
+
+P = (1, 0)
  1
-P = (1, k)                     
- 2                             
+P = (1, k)
+ 2
 P = P  + k  · (sin(θ), -cos(θ))
  3   4
-P = (cos(θ), sin(θ))           
+P = (cos(θ), sin(θ))
  4
 -->
 					<img
@@ -9048,29 +9048,29 @@ P = (cos(θ), sin(θ))
 						P<sub>2</sub>" (which is "half height" P<sub>2</sub>), and so forth:
 					</p>
 					<!--
- 
-     P   + P  
-      2     3 
+
+     P   + P
+      2     3
        y     y   k + sin(θ) - k  · cos(θ)
- A = ───────── = ────────────────────────                                 
+ A = ───────── = ────────────────────────
   y      2                  2
           1
-     A  + ─P     k + sin(θ) - k  · cos(θ)   
+     A  + ─P     k + sin(θ) - k  · cos(θ)
       y   2 2    ──────────────────────── + ─
              y              2               2   2k + sin(θ) + k  · cos(θ)
-e  = ───────── = ──────────────────────────── = ───────────────────────── 
+e  = ───────── = ──────────────────────────── = ─────────────────────────
  1       2                    2                             4
-  y                                                                       
+  y
                                        k
      A  +  mid(P , P )   A  + sin(θ) - ─ cos(θ)
       y         4   3     y            2          k + 3sin(θ) 2k  · cos(θ)
 e  = ───────────────── = ────────────────────── = ────────────────────────
  2           2                     2                         4
   y
-     e   + e  
-      1     2 
+     e   + e
+      1     2
        y     y   3k + 4sin(θ) - 3k  · cos(θ)
- B = ───────── = ───────────────────────────                              
+ B = ───────── = ───────────────────────────
   y      2                    8
 -->
 					<img
@@ -9094,15 +9094,15 @@ e  = ───────────────── = ───────
 							<a href="https://www.wolframalpha.com/">Wolfram Alpha</a> is definitely the way to go. That said, let's get going:
 						</p>
 						<!--
- 
+
 3k + 4sin(θ)) - 3k  · cos(θ)      θ
-────────────────────────────= sin(─)                  
+────────────────────────────= sin(─)
              8                    2
                                   ╭ θ ╮
-3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │               
+3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │
                                   ╰ 2 ╯
                                   ╭ θ ╮
-           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)     
+           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)
                                   ╰ 2 ╯
                                 ╭     ╭ θ ╮          ╮
              3k (1 - cos(θ))= 4 │ 2sin│ ─ │ - sin(θ) │
@@ -9110,12 +9110,12 @@ e  = ───────────────── = ───────
                                         θ
                                    2sin(─) - sin(θ)
                                         2
-                          3k= 4  · ────────────────   
+                          3k= 4  · ────────────────
                                       1 - cos(θ)
                                        ╭ θ ╮
                                    2sin│ ─ │ - sin(θ)
                               4        ╰ 2 ╯
-                           k= ─  · ────────────────── 
+                           k= ─  · ──────────────────
                               3        1 - cos(θ)
 -->
 						<img
@@ -9130,11 +9130,11 @@ e  = ───────────────── = ───────
 							<em>k</em>:
 						</p>
 						<!--
- 
+
             ╭ θ ╮
         2sin│ ─ │ - sin(θ)
    4        ╰ 2 ╯
-k= ─  · ──────────────────         
+k= ─  · ──────────────────
    3        1 - cos(θ)
         ╭     ╭ θ ╮               ╮
         │ 2sin│ ─ │               │
@@ -9142,10 +9142,10 @@ k= ─  · ──────────────────
 k= ─  · │ ────────── - ────────── │
    3    ╰ 1 - cos(θ)   1 - cos(θ) ╯
    4    ╭    ╭ θ ╮      ╭ θ ╮ ╮
-k= ─  · │ csc│ ─ │ - cot│ ─ │ │    
+k= ─  · │ csc│ ─ │ - cot│ ─ │ │
    3    ╰    ╰ 2 ╯      ╰ 2 ╯ ╯
    4       ╭ θ ╮
-k= ─  · tan│ ─ │                   
+k= ─  · tan│ ─ │
    3       ╰ 4 ╯
 -->
 						<img
@@ -9163,7 +9163,7 @@ k= ─  · tan│ ─ │
 						involving the circular arc's angle:
 					</p>
 					<!--
- 
+
            4    ╭ θ ╮
 k = f(θ) = ─ tan│ ─ │
            3    ╰ 4 ╯
@@ -9179,13 +9179,13 @@ k = f(θ) = ─ tan│ ─ │
 						Which means that for any circular arc with angle θ and radius <em>r</em>, our Bézier approximation based on three points of incidence is:
 					</p>
 					<!--
- 
-    start= (r, 0)                                          
-control  = (r, k)                                          
-        1                                                  
+
+    start= (r, 0)
+control  = (r, k)
+        1
 control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
         2
-      end= r  · (cos(θ), sin(θ))                           
+      end= r  · (cos(θ), sin(θ))
 -->
 					<img
 						class="LaTeX SVG"
@@ -9196,7 +9196,7 @@ control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
 					/>
 					<p>Which also gives us the commonly found value of 0.55228 for quarter circles, based on them having an angle of half π:</p>
 					<!--
- 
+
  ╭ \pi ╮   4       ╭ \pi ╮   4  ┌─┐
 f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.55228474983[...]
  ╰  2  ╯   3       ╰  8  ╯   3
@@ -9210,13 +9210,13 @@ f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.552284
 					/>
 					<p>And thus giving us the following Bézier coordinates for a quarter circle of radius <em>r</em>:</p>
 					<!--
- 
-    start= (r, 0)           
+
+    start= (r, 0)
 control  = (r, 0.55228  · r)
-        1                   
+        1
 control  = (0.55228  · r, r)
         2
-      end= (0, r)           
+      end= (0, r)
 -->
 					<img
 						class="LaTeX SVG"
@@ -9360,7 +9360,7 @@ getMostWrongT(radius, bezier, start, end, epsilon=1e-15):
 					</p>
 					<p>First, let's express the total error (given circular arc angle θ, and some <em>k</em>) using standard calculus notation:</p>
 					<!--
- 
+
                     ┌───────────────────────┐
               ╭ 1   │         2            2
 erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
@@ -9383,7 +9383,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						expression looks like:
 					</p>
 					<!--
- 
+
 ╭        ┌───────┐         ╮
 │  ╭ 1   │ 2    2          │ d
 │  |  \| │B  + B   - r\|dt │ ── = 0
@@ -9402,7 +9402,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						are each other's inverse operations, so they cancel out, leaving us with our original function:
 					</p>
 					<!--
- 
+
    ┌───────┐
    │ 2    2
 \| │B  + B   - r\| = 0,    t ∈[0,1]
@@ -9417,7 +9417,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 					/>
 					<p>And now we just solve for that... oh wait. We've seen this before. In order to solve this, we'd end up needing to solve this:</p>
 					<!--
- 
+
  2    2
 B  + B  = r
  x    y
@@ -9947,7 +9947,7 @@ radialError(radius, points[]):
 						end, just like for Bézier curves), by evaluating the following function:
 					</p>
 					<!--
- 
+
            __ n
 Point(t) = ❯      P   · N   (t)
            ‾‾ i=0  i     i,k
@@ -9968,7 +9968,7 @@ Point(t) = ❯      P   · N   (t)
 					</p>
 					<p>Then the N() function itself. What does it look like?</p>
 					<!--
- 
+
           ╭       t- knot         ╮                ╭       knot     -t       ╮
           │              i        │                │           (i+k)         │
 N   (t) = │ ───────────────────── │  · N     (t) + │ ─────────────────────── │  · N       (t)
@@ -9983,9 +9983,9 @@ N   (t) = │ ───────────────────── 
 						some point; obviously, it does, and specifically it does so for the following <code>i</code>/<code>k</code> values:
 					</p>
 					<!--
- 
+
           ╭ 1  if  t ∈[ knot , knot   )
-N   (t) = ╡                 i      i+1  
+N   (t) = ╡                 i      i+1
  i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/2514e1aa0565840e33fde0b146e3efe2.svg" width="237px" height="40px" loading="lazy" />
@@ -10006,7 +10006,7 @@ N   (t) = ╡                 i      i+1
 						P(t), we can compute this point by evaluating <em>d(t)</em> on a curve section between knots <code>i</code> and <code>i+1</code>:
 					</p>
 					<!--
- 
+
  k               k-1                   k-1
 d (t) = α     · d   (t) + (1-α   )  · d   (t)
  i       i,k     i            i,k      i-1
@@ -10017,7 +10017,7 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 						by:
 					</p>
 					<!--
- 
+
               t -  knots[i]
 α    = ───────────────────────────
  i,k    knots[i+1+n-k] -  knots[i]
@@ -10031,9 +10031,9 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 					</p>
 					<p>Of course, the recursion does need a stop condition:</p>
 					<!--
- 
+
  k           0                ╭ 1  if  t ∈[ knot , knot   )
-d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1  
+d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
  0           i       i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/49af474c33ce0ee0733626ea3d988570.svg" width="367px" height="40px" loading="lazy" />
@@ -10047,34 +10047,34 @@ d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
 						recursion diagram:
 					</p>
 					<!--
- 
+
      ╭                                ╭                                ╭          1     0           0
-     │                                │                                │         α   × d ,   with  d    either 0 or 1 
+     │                                │                                │         α   × d ,   with  d    either 0 or 1
      │                                │          2     1           1   │          3     3           3
-     │                                │         α   × d ,   with  d  = ╡                 +                              
+     │                                │         α   × d ,   with  d  = ╡                 +
      │                                │          3     3           3   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      │          3     2           2   │                                ╰ ╰      3 ╯     2           2
-     │         α   × d ,   with  d  = ╡                 +                                                                
+     │         α   × d ,   with  d  = ╡                 +
      │          3     3           3   │                                ╭           1     0
-     │                                │                                │          α   × d                             
+     │                                │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           2     2
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
  3   │                                │ ╰      3 ╯     2           2   │ ╭      1 ╮     0           0
-d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1     
+d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
  3   │                                ╰                                ╰ ╰      2 ╯     1           1
-     │                 +                                                                                                 
+     │                 +
      │                                ╭           2     1
-     │                                │          α   × d                                                                
+     │                                │          α   × d
      │                                │           2     2
-     │                                │                                                                                 
-     │ ╭      3 ╮     2           2   │                 +                                                               
-     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0                               
-     │ ╰      3 ╯     2           2   │                                │          α   × d 
+     │                                │
+     │ ╭      3 ╮     2           2   │                 +
+     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0
+     │ ╰      3 ╯     2           2   │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           1     1
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
      │                                │ ╰      2 ╯     1           1   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      ╰                                ╰                                ╰ ╰      1 ╯     0           0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/20e910bbea2e6eff511cb13cef18ef3b.svg" width="641px" height="336px" loading="lazy" />
diff --git a/docs/zh-CN/index.html b/docs/zh-CN/index.html
index 48be3507..c2d459e1 100644
--- a/docs/zh-CN/index.html
+++ b/docs/zh-CN/index.html
@@ -172,7 +172,7 @@
         <a class="btclk" href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu</a>
         或者使用右边的二维码，如果你倾向于这种便利的方式 =)
       </p>
-    
+
     </div>
     <div class="btcqr">
         <a href="bitcoin:3GY1HbQ2cH9V4xBLnRYdEfc42Nd1ZyjLZu?label=Primer%20on%20Bezier%20Curves">
@@ -184,7 +184,7 @@
 
 			<br style="clear: both;" />
 
-			<p>— <a href="https://twitter.com/TheRealPomax">Pomax</a></p>
+			<p>— <a href="https://mastodon.social/@TheRealPomax">Pomax</a></p>
 			<noscript>
 				<div class="note">
 					<header>
@@ -527,7 +527,7 @@
 					</p>
 					<p>如果我们知道两点之间的距离，并想找出离第一个点20%间距的一个新的点(也就是离第二个点80%的间距)，我们可以通过简单的计算来得到：</p>
 					<!--
- 
+
       ╭        p =  some point       ╮
       │         1                    │
       │        p =  some other point │
@@ -572,7 +572,7 @@ Given │  distance= (p  - p )         │,  our new point = p  +  distance  ·
 						贝塞尔曲线是“参数”方程的一种形式。从数学上讲，参数方程作弊了：“方程”实际上是一个从输入到<strong>唯一</strong>输出的、良好定义的映射关系。几个输入进来，一个输出返回。改变输入变量，还是只有一个输出值。参数方程在这里作弊了。它们基本上干了这么件事，“好吧，我们想要更多的输出值，所以我们用了多个方程”。举个例子：假如我们有一个方程，通过一些计算，将假设为<i>x</i>的一些值映射到另外的值:
 					</p>
 					<!--
- 
+
 f(x) = cos (x)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/0cc876c56200446c60114c1b0eeeb2cc.svg" width="96px" height="17px" loading="lazy" />
@@ -581,7 +581,7 @@ f(x) = cos (x)
 					</p>
 					<p>到目前没什么问题。现在，让我们来看一下参数方程，以及它们是怎么作弊的。我们取以下两个方程：</p>
 					<!--
- 
+
 f(a) = cos (a)
 f(b) = sin (b)
 -->
@@ -590,9 +590,9 @@ f(b) = sin (b)
 						这俩方程没什么让人印象深刻的，只不过是正弦函数和余弦函数，但正如你所见，输入变量有两个不同的名字。如果我们改变了<i>a</i>的值，<i>f(b)</i>的输出不会有变化，因为这个方程没有用到<i>a</i>。参数方程通过改变这点来作弊。在参数方程中，所有不同的方程共用一个变量，如下所示：
 					</p>
 					<!--
- 
+
 ╭ f (t) = cos (t)
-╡  a              
+╡  a
 │ f (t) = sin (t)
 ╰  b
 -->
@@ -607,8 +607,8 @@ f(b) = sin (b)
 						多个方程，但只有一个变量。如果我们改变了<i>t</i>的值，<i>f<sub>a</sub>(t)</i>和<i>f<sub>b</sub>(t)</i>的输出都会发生变化。你可能会好奇这有什么用，答案其实很简单：对于参数曲线，如果我们用常用的标记来替代<i>f<sub>a</sub>(t)</i>和<i>f<sub>b</sub>(t)</i>，看起来就有些明朗了：
 					</p>
 					<!--
- 
-{ x = cos (t) 
+
+{ x = cos (t)
   y = sin (t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/explanation/6914ba615733c387251682db7a3db045.svg" width="77px" height="40px" loading="lazy" />
@@ -636,7 +636,7 @@ f(b) = sin (b)
 					</p>
 					<p>你可能记得高中所学的多项式，看起来像这样：</p>
 					<!--
- 
+
              3         2
 f(x) = a  · x  + b  · x  + c  · x + d
 -->
@@ -654,12 +654,12 @@ f(x) = a  · x  + b  · x  + c  · x + d
 						贝塞尔曲线不是x的多项式，它是<i>t</i>的多项式，<i>t</i>的值被限制在0和1之间，并且含有<i>a</i>，<i>b</i>等参数。它采用了二次项的形式，听起来很神奇但实际上就是混合不同值的简单描述：
 					</p>
 					<!--
- 
-linear= (1-t) + t                                        
+
+linear= (1-t) + t
              2                      2
-square= (1-t)  + 2  · (1-t)  · t + t                     
+square= (1-t)  + 2  · (1-t)  · t + t
              3             2                       2    3
- cubic= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+ cubic= (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img
 						class="LaTeX SVG"
@@ -670,10 +670,10 @@ square= (1-t)  + 2  · (1-t)  · t + t
 					/>
 					<p>我明白你在想什么：这看起来并不简单，但如果我们拿掉<i>t</i>并让系数乘以1，事情就会立马简单很多，看看这些二次项：</p>
 					<!--
- 
- linear=  1 + 1           
- square=  1 + 2 + 1       
-  cubic=  1 + 3 + 3 + 1   
+
+ linear=  1 + 1
+ square=  1 + 2 + 1
+  cubic=  1 + 3 + 3 + 1
 quartic= 1 + 4 + 6 + 4 + 1
 -->
 					<img
@@ -691,11 +691,11 @@ quartic= 1 + 4 + 6 + 4 + 1
 					</p>
 					<!--
 
-linear= \colorblue a + \colorred b                                                                                                                   
-square= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b                                                      
+linear= \colorblue a + \colorred b
+square= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred b  · \colorred b
  cubic= \colorblue a  · \colorblue a  · \colorblue a + \colorblue a  · \colorblue a  · \colorred b + \colorblue a  · \colorred b  · \colorred b + \co
-                                                                                             
-                                                                                             
+
+
                                                        lorred b  · \colorred b  · \colorred b
 -->
 					<img
@@ -709,9 +709,9 @@ square= \colorblue a  · \colorblue a + \colorblue a  · \colorred b + \colorred
 						基本上它就是“每个<i>a</i>和<i>b</i>结合项”的和，在每个加号后面逐步的将<i>a</i>换成<i>b</i>。因此这也很简单。现在你已经知道了二次多项式，为了叙述的完整性，我将给出一般方程：
 					</p>
 					<!--
- 
+
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t
               ‾‾ i=0
 -->
 					<img
@@ -1007,9 +1007,9 @@ function Bezier(3,t):
 					<!--
 
               __ n                                                                                            n-i     i
-Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset 
+Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \underset polynomial term\underbrace(1-t)     · t   · \ \underset
               ‾‾ i=0
-                                                                weight\underbracew 
+                                                                weight\underbracew
                                                                                   i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/c2f2fe0ef5d0089d9dd8e5e3999405cb.svg" width="336px" height="55px" loading="lazy" />
@@ -1017,11 +1017,11 @@ Bézier(n,t) = ❯      \underset binomial term\underbrace\binomni  · \ \unders
 						看起来很复杂，但实际上“权重”只是我们想让曲线所拥有的坐标值：对于一条n<sup>th</sup>阶曲线，w<sup>0</sup>是起始坐标，w<sup>n</sup>是终点坐标，中间的所有点都是控制点坐标。假设说一条曲线的起点为（110，150），终点为（210，30），并受点（25，190）和点（210，250）的控制，贝塞尔曲线方程就为：
 					</p>
 					<!--
- 
+
 ╭                               3                                   2                                             2                       3
-╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t  
+╡ x = \colordarkred 110  · (1-t)  + \colordarkgreen 25  · 3  · (1-t)   · t + \colordarkblue 210  · 3  · (1-t)  · t  + \coloramber 210  · t
 │                               3                                    2                                             2                      3
-╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t 
+╰ y = \colordarkred 150  · (1-t)  + \colordarkgreen 190  · 3  · (1-t)   · t + \colordarkblue 250  · 3  · (1-t)  · t  + \coloramber 30  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/control/be73034ac382b54863c7a18c2932cbbc.svg" width="473px" height="40px" loading="lazy" />
 					<p>这就是我们在文章开头看到的曲线：</p>
@@ -1134,9 +1134,9 @@ function Bezier(3,t,w[]):
 					</p>
 					<p>Adding these ratio values to the regular Bézier curve function is fairly easy. Where the regular function is the following:</p>
 					<!--
- 
+
               __ n                    n-i     i
-Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w 
+Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
               ‾‾ i=0                                i
 -->
 					<img
@@ -1148,13 +1148,13 @@ Bézier(n,t) = ❯      \binomni  · (1-t)     · t   · w
 					/>
 					<p>The function for rational Bézier curves has two more terms:</p>
 					<!--
- 
+
                         __ n                    n-i     i
-                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio 
+                        ❯      \binomni  · (1-t)     · t   · w   · \colorblue ratio
                         ‾‾ i=0                                i                    i
 Rational  Bézier(n,t) = ────────────────────────────────────────────────────────────
                                      __ n                     n-i      i
-                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio  
+                        \colorblue   ❯      \binomni  ·  (1-t)     ·  t   ·  ratio
                                      ‾‾ i=0                                       i
 -->
 					<img
@@ -1340,8 +1340,8 @@ function RationalBezier(3,t,w[],r[]):
 					</p>
 					<p>这一切都与我们如何从曲线的“起点”变化到曲线“终点”有关。如果有一个值是另外两个值的混合，一般方程如下：</p>
 					<!--
- 
-mixture = a  ·  value  + b  ·  value 
+
+mixture = a  ·  value  + b  ·  value
                      1              2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/dfd6ded3f0addcf43e0a1581627a2220.svg" width="212px" height="16px" loading="lazy" />
@@ -1351,8 +1351,8 @@ mixture = a  ·  value  + b  ·  value
 						2。另外，我们不想让“a”和“b”是互相独立的:如果它们是互相独立的话，我们可以任意选出自己喜欢的值，并得到混合值，比如说100%的value1和100%的value2。原则上这是可以的，但是对于贝塞尔曲线来说，我们通常想要的是起始值和终点值<em>之间</em>的混合值，所以要确保我们不会设置一些“a”和"b"而导致混合值超过100%。这很简单：
 					</p>
 					<!--
- 
-m = a  ·  value  + (1 - a)  ·  value 
+
+m = a  ·  value  + (1 - a)  ·  value
                1                    2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/extended/4e0fa763b173e3a683587acf83733353.svg" width="213px" height="16px" loading="lazy" />
@@ -1421,61 +1421,61 @@ m = a  ·  value  + (1 - a)  ·  value
 						通过将贝塞尔公式表示成一个多项式基本方程、系数矩阵以及实际的坐标，我们也可以用矩阵运算来表示贝塞尔。让我们看一下这对三次曲线来说有什么含义：
 					</p>
 					<!--
- 
+
                   3                   2                             2          3
-B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t 
+B(t) = P   · (1-t)  + P   · 3  · (1-t)   · t + P   · 3  · (1-t)  · t  + P   · t
         1              2                        3                        4
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/9a9a55f5b0323d9ea88f82fc6be58ad3.svg" width="468px" height="20px" loading="lazy" />
 					<p>暂时不用管我们具体的坐标，现在有：</p>
 					<!--
- 
+
             3             2                       2    3
-B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t 
+B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/87cfac83cb8a4b0bee68ef006effc611.svg" width="353px" height="19px" loading="lazy" />
 					<p>可以将它写成四个表达式之和：</p>
 					<!--
- 
+
                 3
-... =      (1-t) 
+... =      (1-t)
                 2
     + 3  · (1-t)   · t
                      2
-    + 3  · (1-t)  · t 
+    + 3  · (1-t)  · t
              3
-    +       t  
+    +       t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/cdd88611833f3b178df91278359a4193.svg" width="140px" height="75px" loading="lazy" />
 					<p>我们可以扩展这些表达式：</p>
 					<!--
- 
+
                                    3         2
 ... =  (1-t)  · (1-t)  · (1-t) = -t  + 3  · t  - 3  · t + 1
                                        3         2
     + 3  · (1-t)  · (1-t)  · t = 3  · t  - 6  · t  + 3  · t
                                             3         2
-    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t 
+    +   3  · (1-t)  · t  · t   =     -3  · t  + 3  · t
                                              3
-    +        t  · t  · t       =            t  
+    +        t  · t  · t       =            t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/ec118f296511c6e9ac8727be3703a7ce.svg" width="397px" height="75px" loading="lazy" />
 					<p>更进一步，我们可以加上所有的1和0系数，以便看得更清楚：</p>
 					<!--
- 
+
              3         2
 ... = -1  · t  + 3  · t  - 3  · t + 1
              3         2
-    + +3  · t  - 6  · t  + 3  · t + 0 
+    + +3  · t  - 6  · t  + 3  · t + 0
              3         2
     + -3  · t  + 3  · t  + 0  · t + 0
              3         2
-    + +1  · t  + 0  · t  + 0  · t + 0 
+    + +1  · t  + 0  · t  + 0  · t + 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/matrix/67a5ea33d6c6558f7d954b18226f4956.svg" width="217px" height="75px" loading="lazy" />
 					<p><em>现在</em>，我们可以将它看作四个矩阵运算：</p>
 					<!--
- 
+
                  ┌ -1 ┐                    ┌ 3  ┐                    ┌ -3 ┐                    ┌ 1 ┐
 ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ -6 │ + ┌  3  2     ┐  · │ 3  │ + ┌  3  2     ┐  · │ 0 │
 └ t  t  t 1 ┘    │ -3 │   └ t  t  t 1 ┘    │ 3  │   └ t  t  t 1 ┘    │ 0  │   └ t  t  t 1 ┘    │ 0 │
@@ -1484,7 +1484,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/8ecff6b8a37d60385d287ea2b26876db.svg" width="607px" height="72px" loading="lazy" />
 					<p>如果我们将它压缩到一个矩阵操作里，就能得到：</p>
 					<!--
- 
+
                  ┌ -1  3 -3 1 ┐
 ┌  3  2     ┐  · │  3 -6  3 0 │
 └ t  t  t 1 ┘    │ -3  3  0 0 │
@@ -1493,7 +1493,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/b9527f7d5a0f5d2d737eac118d69243e.svg" width="227px" height="72px" loading="lazy" />
 					<p>这种多项式表达式一般是以递增的顺序来写的，所以我们应该将<code>t</code>矩阵水平翻转，并将大的那个“混合”矩阵上下颠倒：</p>
 					<!--
- 
+
                  ┌  1  0  0 0 ┐
 ┌      2  3 ┐  · │ -3  3  0 0 │
 └ 1 t t  t  ┘    │  3 -6  3 0 │
@@ -1502,7 +1502,7 @@ B(t) = (1-t)  + 3  · (1-t)   · t + 3  · (1-t)  · t  + t
 					<img class="LaTeX SVG" src="./images/chapters/matrix/1a64ed455c6dd2f8cacca5e5e12bdcc1.svg" width="227px" height="72px" loading="lazy" />
 					<p>最终，我们可以加入原始的坐标，作为第三个单独矩阵：</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -1515,7 +1515,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 					<img class="LaTeX SVG" src="./images/chapters/matrix/b32cae2dfc47d5f36df0bc3defb7dfa8.svg" width="323px" height="73px" loading="lazy" />
 					<p>我们可以对二次曲线运用相同的技巧，可以得到：</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -1916,7 +1916,7 @@ function drawCurvePoint(points[], t):
 						>已经说明可以用矩阵乘法表示曲线，尤其是以下两种分别表示二次曲线和三次曲线的形式（为提高可读性，贝塞尔曲线的系数向量已被倒转次序）：
 					</p>
 					<!--
- 
+
                                     ┌ P  ┐
                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -1933,7 +1933,7 @@ B(t) = ┌      2 ┐  · │ -2  2 0 │  · │ P  │
 					/>
 					<p>和</p>
 					<!--
- 
+
                                           ┌ P  ┐
                                           │  1 │
                         ┌  1  0  0 0 ┐    │ P  │
@@ -1955,7 +1955,7 @@ B(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						>处分割曲线得到两条新的（且显然更小的）贝塞尔曲线，则可以用矩阵表示和一点线性代数求出这两条贝塞尔曲线的坐标。首先将实际的“线上点”信息分离到新的矩阵乘法中：
 					</p>
 					<!--
- 
+
                                                   ┌ P  ┐                                              ┌ P  ┐
                                    ┌  1  0 0 ┐    │  1 │                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -1972,7 +1972,7 @@ B(t) = ┌                    2 ┐  · │ -2  2 0 │  · │ P  │ = ┌
 					/>
 					<p>以及</p>
 					<!--
- 
+
                                                                ┌ P  ┐                                                       ┌ P  ┐
                                                                │  1 │                    ┌ 1 0  0 0  ┐                      │  1 │
                                              ┌  1  0  0 0 ┐    │ P  │                    │ 0 z  0 0  │    ┌  1  0  0 0 ┐    │ P  │
@@ -1998,7 +1998,7 @@ B(t) = ┌                    2         3 ┐  · │ -3  3  0 0 │  · │  2
 						<h2>推导新的凸包坐标</h2>
 						<p>推导分割曲线后所得两段曲线的坐标要花上几步，而且曲线次数越高，花的工夫越多，因此先看二次曲线：</p>
 						<!--
- 
+
                                                   ┌ P  ┐
                      ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2034,7 +2034,7 @@ B(t) = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
 							<em>M</em> · <em>P</em>]）)和表示从<em>t = 0</em>到<em>t = z</em>的曲线的一系列新坐标。计算如下：
 						</p>
 						<!--
- 
+
                     ┌ 1 0 0 ┐
      -1             │   1   │    ┌ 1 0 0  ┐    ┌  1  0 0 ┐   ┌     1            0        0  ┐
 Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = │  -(z-1)          z        0  │
@@ -2050,7 +2050,7 @@ Q = M    · Z  · M = │ 1 ─ 0 │  · │ 0 z 0  │  · │ -2  2 0 │ = 
 						/>
 						<p>很好！现在得出新的二次曲线：</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭      ┌ P  ┐ ╮
                                │  1 │                      │      │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q  · │ P  │ │
@@ -2066,7 +2066,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                     ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌     1            0        0  ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │  -(z-1)          z        0  │  · │ P  │ │
@@ -2082,7 +2082,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌                        P                          ┐
                                │                         1                         │
                 ┌  1  0 0 ┐    │               z  · P  - (z-1)  · P                │
@@ -2111,7 +2111,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							>的一段需要再算一次。首先注意到之前的计算其实是在一般的区间[0,<em>z</em>]上进行的。之所以可以写成更简单的形式是因为0为端点，但将0显式地写出可知<em>真正</em>所计算的是：
 						</p>
 						<!--
- 
+
                                                             ┌ P  ┐
                                              ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2127,7 +2127,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                              ┌ P  ┐
                 ┌ 1 0 0  ┐    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 z 0  │  · │ -2  2 0 │  · │ P  │
@@ -2144,7 +2144,7 @@ B(t) = ┌                              2 ┐  · │ -2  2 0 │  · │ P  │
 						/>
 						<p>如果需要[<em>z</em>,1]区间，那么计算如下：</p>
 						<!--
- 
+
                                                                     ┌ P  ┐
                                                      ┌  1  0 0 ┐    │  1 │
 B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · │ P  │
@@ -2160,7 +2160,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 							loading="lazy"
 						/>
 						<!--
- 
+
                 ┌              2        ┐                   ┌ P  ┐
                 │ 1  z        z         │    ┌  1  0 0 ┐    │  1 │
 = ┌      2 ┐  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │  · │ P  │
@@ -2177,7 +2177,7 @@ B(t) = ┌                                      2 ┐  · │ -2  2 0 │  · 
 						/>
 						<p>用同样的技巧左乘单位矩阵将<code>[某项 · M]</code>变为<code>[M · 某项]</code>：</p>
 						<!--
- 
+
                       ┌ 1 0 0 ┐    ┌              2        ┐
       -1              │   1   │    │ 1  z        z         │    ┌  1  0 0 ┐   ┌      2                   2 ┐
 Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  · │ -2  2 0 │ = │ (z-1)  -2  · z  · (z-1) z  │
@@ -2193,7 +2193,7 @@ Q' = M    · Z'  · M = │ 1 ─ 0 │  · │ 0 1-z 2  · z  · (1-z) │  ·
 						/>
 						<p>那么最终第二条曲线为：</p>
 						<!--
- 
+
                                ┌ P  ┐                      ╭       ┌ P  ┐ ╮
                                │  1 │                      │       │  1 │ │
 B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │ Q'  · │ P  │ │
@@ -2209,7 +2209,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ╭                                   ┌ P  ┐ ╮
                 ┌  1  0 0 ┐    │ ┌      2                   2 ┐    │  1 │ │
 = ┌      2 ┐  · │ -2  2 0 │  · │ │ (z-1)  -2  · z  · (z-1) z  │  · │ P  │ │
@@ -2225,7 +2225,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 							loading="lazy"
 						/>
 						<!--
- 
+
                                ┌  2                                      2       ┐
                                │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
                 ┌  1  0 0 ┐    │        3                       2              1 │
@@ -2252,7 +2252,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						>处分割一条二次曲线可得两条子曲线，它们均为用容易求得的坐标所描述的贝塞尔曲线：
 					</p>
 					<!--
- 
+
                                              ┌                        P                          ┐
                                     ┌ P  ┐   │                         1                         │
 ┌     1            0        0  ┐    │  1 │   │               z  · P  - (z-1)  · P                │
@@ -2270,7 +2270,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>和</p>
 					<!--
- 
+
                                            ┌  2                                      2       ┐
                                   ┌ P  ┐   │ z   · P  - 2  · z  · (z-1)  · P  + (z-1)   · P  │
 ┌      2                   2 ┐    │  1 │   │        3                       2              1 │
@@ -2288,7 +2288,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>虽然三次曲线可以同理推导，但此处省略实际的推导过程（读者可自行写出）并直接展示所得的新坐标集：</p>
 					<!--
- 
+
                                                               ┌                                    P                                      ┐
                                                      ┌ P  ┐   │                                     1                                     │
 ┌    1            0                0         0  ┐    │  1 │   │                           z  · P  - (z-1)  · P                            │
@@ -2309,7 +2309,7 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 					/>
 					<p>和</p>
 					<!--
- 
+
                                                               ┌  3               2                                 2              3       ┐
                                                      ┌ P  ┐   │ z   · P  - 3  · z   · (z-1)  · P  + 3  · z  · (z-1)   · P  - (z-1)   · P  │
 ┌       3           2                      2  3 ┐    │  1 │   │        4                        3                        2              1 │
@@ -2367,15 +2367,15 @@ B(t) = ┌      2 ┐  · M  · Q  · │ P  │ = ┌      2 ┐  · M  · │
 						然后用一个朴素（其实极其有用）的变形技巧：既然<code>t</code>值总在0到1之间（含端点），且<code>1-t</code>加<code>t</code>恒等于1，那么任何数都可表示为<code>t</code>与<code>1-t</code>的和：
 					</p>
 					<!--
- 
+
 x = 1 x = ((1-t) + t) x = (1-t) x + t x = x (1-t) + x t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/ff224ded6bbbc94b43130f5f8eeb5d29.svg" width="379px" height="16px" loading="lazy" />
 					<p>于是用这一看似平凡的性质可将贝塞尔函数拆分为<code>1-t</code>和<code>t</code>两部分之和：</p>
 					<!--
- 
-Bézier(n,t)= (1-t) B(n,t) + t B(n,t)                    
-             __ n               n      __ n         n   
+
+Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
+             __ n               n      __ n         n
            = ❯      w  (1 - t) B (t) + ❯      w  t B (t)
              ‾‾ i=0  i          i      ‾‾ i=0  i    i
 -->
@@ -2398,14 +2398,14 @@ Bézier(n,t)= (1-t) B(n,t) + t B(n,t)
 					</script>
 					<p>接下来从变形转到线性代数（矩阵）——现在Bézier(n,t)和Bézier(n+1,t)之间的关系可用非常简单的矩阵乘法表示：</p>
 					<!--
- 
-M B  = B 
+
+M B  = B
    n    k
 -->
 					<img class="LaTeX SVG" src="./images/chapters/reordering/1f5b60d190a1c7099b3411e4cc477291.svg" width="71px" height="16px" loading="lazy" />
 					<p>其中矩阵<strong>M</strong>为<code>(n+1)×n</code>阶的矩阵，其形如：</p>
 					<!--
- 
+
     ┌ 1  0   .   .   .   .    .    .  ┐
     │ 1 k-1                           │
     │ ─ ───  0   .   .   .    0    .  │
@@ -2445,20 +2445,20 @@ M = │        k   k                    │
 						>，利用法方程组可以求出使方程两侧之差长度最小的<code>x</code>。既然现在面临的问题即为如此，那么：
 					</p>
 					<!--
- 
-              M B = B             
+
+              M B = B
                  n   k
            T         T
-         (M  M) B = M  B          
+         (M  M) B = M  B
                  n      k
   T   -1   T          T   -1  T
-(M  M)   (M  M) B = (M  M)   M  B 
+(M  M)   (M  M) B = (M  M)   M  B
                  n               k
                       T   -1  T
-              I B = (M  M)   M  B 
+              I B = (M  M)   M  B
                  n               k
                       T   -1  T
-                B = (M  M)   M  B 
+                B = (M  M)   M  B
                  n               k
 -->
 					<img
@@ -2513,9 +2513,9 @@ M = │        k   k                    │
 					</p>
 					<p>首先观察贝塞尔函数的求导法则，即：</p>
 					<!--
- 
+
                     __ n-1
-Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t) 
+Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
                     ‾‾ i=0   i+1  i                    i
 -->
 					<img
@@ -2527,7 +2527,7 @@ Bézier'(n,t) = n  · ❯      (b   -b )  ·   Bézier(n-1,t)
 					/>
 					<p>上式可改写为（注意式中的<em>b</em>即权重<em>w</em>，且<em>n</em>乘以一个和式等于每个求和项乘以<em>n</em>再求和）：</p>
 					<!--
- 
+
                __ n-1
 Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
                ‾‾ i=0                i           i+1  i
@@ -2633,10 +2633,10 @@ Bézier'(n,t) = ❯        Bézier(n-1,t)   · n  · (w   -w )
 						如果要将物体沿曲线移动或者从曲线附近“移向远处”，那么与之最相关的两个向量为曲线的切向量和法向量，而这两者都非常容易求得。切向量用于沿曲线移动或者对准曲线方向，它标志着曲线在指定点的行进方向，而且就是曲线函数的一阶导数：
 					</p>
 					<!--
- 
+
 tangent (t) = B' (t)
        x        x
-                    
+
 tangent (t) = B' (t)
        y        y
 -->
@@ -2649,14 +2649,14 @@ tangent (t) = B' (t)
 					/>
 					<p>此即所需的方向向量。可以在每一点将方向向量规范化后得到单位方向向量（即长度为1.0），再根据这些方向进行所需的操作：</p>
 					<!--
- 
+
                            ┌─────────────────┐
                            │      2         2
-    d = \| tangent(t)\| =  │B' (t)  + B' (t)  
+    d = \| tangent(t)\| =  │B' (t)  + B' (t)
                           ⟍│  x         y
 
                            tangent (t)     B' (t)
-^                                 x          x    
+^                                 x          x
 x(t) = \| tangent (t)\| =─────────────── = ──────
                  x       \| tangent(t)\|     d
 
@@ -2685,7 +2685,7 @@ y(t) = \| tangent (t)\| = ─────────────── = ──
 						</p>
 						<p>为了将点(<em>x</em>,<em>y</em>)（绕(0,0)）旋转<em>φ</em>度得到点(<em>x'</em>,<em>y'</em>)，可以使用以下简洁的计算式：</p>
 						<!--
- 
+
 x' = x  · cos (\phi) - y  · sin (\phi)
 y' = x  · sin (\phi) + y  · cos (\phi)
 -->
@@ -2698,7 +2698,7 @@ y' = x  · sin (\phi) + y  · cos (\phi)
 						/>
 						<p>对应“短”版本的矩阵变换为：</p>
 						<!--
- 
+
 ┌ x' ┐ = ┌ cos (\phi) -sin (\phi) ┐ ┌ x ┐
 └ y' ┘   └ sin (\phi) cos (\phi)  ┘ └ y ┘
 -->
@@ -3083,10 +3083,10 @@ generateRMFrames(steps) -> frames:
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
+
 B'(t) = a(1-t) + b(t)= 0,
           a - at + bt= 0,
-           (b-a)t + a= 0 
+           (b-a)t + a= 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -3097,10 +3097,10 @@ B'(t) = a(1-t) + b(t)= 0,
 					/>
 					<p>And then we turn this into our solution for <code>t</code> using basic arithmetics:</p>
 					<!--
- 
-(b-a)t + a= 0, 
+
+(b-a)t + a= 0,
     (b-a)t= -a,
-            -a 
+            -a
          t= ───
             b-a
 -->
@@ -3123,7 +3123,7 @@ B'(t) = a(1-t) + b(t)= 0,
 						you haven't, it looks like this:
 					</p>
 					<!--
- 
+
                                                    ┌────────┐
                                                    │ 2
                2                              -b ±⟍│b  - 4ac
@@ -3148,9 +3148,9 @@ Given f(t) = at  + bt + c, f(t)=0   when  t = ───────────
 						<a href="#derivatives">derivatives section</a>:
 					</p>
 					<!--
- 
-B(t)  uses { p ,p ,p ,p  }                                                  
-              1  2  3  4                                                    
+
+B(t)  uses { p ,p ,p ,p  }
+              1  2  3  4
 B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
                1  2  3            1      2  1    2      3  2    3      4  3
 -->
@@ -3163,21 +3163,21 @@ B'(t)  uses { v ,v ,v  },  where v  = 3(p -p ), v  = 3(p -p ), v  = 3(p -p )
 					/>
 					<p>And then, using these <em>v</em> values, we can find out what our <em>a</em>, <em>b</em>, and <em>c</em> should be:</p>
 					<!--
- 
+
               2                  2
-B'(t)= v (1-t)  + 2v (1-t)t + v t            
+B'(t)= v (1-t)  + 2v (1-t)t + v t
         1           2          3
            2                    2       2
-  ...= v (t  - 2t + 1) + 2v (t-t ) + v t     
+  ...= v (t  - 2t + 1) + 2v (t-t ) + v t
         1                  2          3
           2                          2      2
-  ...= v t  - 2v t + v  + 2v t - 2v t  + v t 
+  ...= v t  - 2v t + v  + 2v t - 2v t  + v t
         1       1     1     2      2      3
           2       2      2
   ...= v t  - 2v t  + v t  - 2v t + v  + 2v t
         1       2      3       1     1     2
                    2
-  ...= (v -2v +v )t  + 2(v -v )t + v         
+  ...= (v -2v +v )t  + 2(v -v )t + v
          1   2  3         2  1      1
 -->
 					<img
@@ -3192,12 +3192,12 @@ B'(t)= v (1-t)  + 2v (1-t)t + v t
 						expressions of our original coordinate values, so we can do some substitution to get:
 					</p>
 					<!--
- 
+
 a= v -2v +v  = 3(-p  + 3p  - 3p  + p )
     1   2  3       1     2     3    4
-b= 2(v -v ) = 6(p  - 2p  + p )        
+b= 2(v -v ) = 6(p  - 2p  + p )
       2  1       1     2    3
-c= v  = 3(p -p )                      
+c= v  = 3(p -p )
     1      2  1
 -->
 					<img
@@ -3224,11 +3224,11 @@ c= v  = 3(p -p )
 						really hard to solve, it can be rewritten to a form for which finding the roots is "easier" (even if not "easy"):
 					</p>
 					<!--
- 
+
                    3     2
 very hard: solve at  + bt  + ct + d = 0
                   3
-   easier: solve t  + pt + q = 0       
+   easier: solve t  + pt + q = 0
 -->
 					<img
 						class="LaTeX SVG"
@@ -3637,7 +3637,7 @@ function getCubicRoots(pa, pb, pc, pd) {
 						<code>f<sub>y</sub>(x)</code> is zero, so that the next <code>t</code> is the same as the one we already have:
 					</p>
 					<!--
- 
+
             f (x )
              y  n
 x    = x  - ───────
@@ -3785,11 +3785,11 @@ x    = x  - ───────
 						function for the y-component to an <em>n-2</em> term function. For instance, if we have a cubic curve such as this:
 					</p>
 					<!--
- 
+
 ╭                            3                               2                                          2                       3
-╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t  
+╡ x = \colorblue 120  · (1-t)  \colorblue  + 35  · 3  · (1-t)   · t \colorblue  + 220  · 3  · (1-t)  · t  \colorblue  + 220  · t
 │                            3                                2                                          2                      3
-╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t 
+╰ y = \colorblue 160  · (1-t)  \colorblue  + 200  · 3  · (1-t)   · t \colorblue  + 260  · 3  · (1-t)  · t  \colorblue  + 40  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/e5db5e945606d3429e4475ff92283a9c.svg" width="497px" height="40px" loading="lazy" />
 					<p>
@@ -3797,11 +3797,11 @@ x    = x  - ───────
 						by -160, gives us:
 					</p>
 					<!--
- 
+
 ╭                          3                               2                                          2                       3
-╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t  
+╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  + 100  · t
 │                          3                               2                                          2                       3
-╰ y = \colorblue 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t 
+╰ y = \colorblue 0  · (1-t)  \colorblue  + 40  · 3  · (1-t)   · t \colorblue  + 100  · 3  · (1-t)  · t  \colorblue  - 120  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/d412c72ed342c2036965ff251c8fb443.svg" width="481px" height="40px" loading="lazy" />
 					<p>
@@ -3809,20 +3809,20 @@ x    = x  - ───────
 						become:
 					</p>
 					<!--
- 
+
 ╭                          3                               2                                         2                       3
-╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorblue 0  · (1-t)  \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                          3                               2                                          2                     3
-╰ y = \colorblue 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorblue  + 0  · t 
+╰ y = \colorblue 0  · (1-t)  \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t  \colorblue  + 0  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/e49d7bb45a18d14fbb12df4d91e2c67b.svg" width="473px" height="40px" loading="lazy" />
 					<p>If we drop all the zero-terms, this gives us:</p>
 					<!--
- 
+
 ╭                                   2                                         2                       3
-╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t  
+╡ x = \colorblue  - 85  · 3  · (1-t)   · t \colorblue  - 12  · 3  · (1-t)  · t  \colorblue  + 156  · t
 │                                   2                                          2
-╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t                        
+╰ y = \colorblue  - 40  · 3  · (1-t)   · t \colorblue  + 140  · 3  · (1-t)  · t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/aligning/016f37843b2af2ae34dc8b2975505f3d.svg" width="408px" height="40px" loading="lazy" />
 					<p>
@@ -3933,7 +3933,7 @@ x    = x  - ───────
 					</p>
 					<p>What we need to do is solve a simple equation:</p>
 					<!--
- 
+
 C(t) = 0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/inflections/ed68dcfb203517ca080fe48914769fb0.svg" width="59px" height="16px" loading="lazy" />
@@ -3943,7 +3943,7 @@ C(t) = 0
 						circle being on the other side of the curve. So what does <em>C(t)</em> look like? Actually something that seems not too hard:
 					</p>
 					<!--
- 
+
 C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)  ·   Bézier \prime\prime(t)
                x                     y                          y                     x
 -->
@@ -3972,15 +3972,15 @@ C(t) =   Bézier \prime(t)  ·   Bézier \prime\prime(t) -   Bézier \prime(t)
 							with the first and second derivatives, given our basis functions:
 						</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t                            
+Bézier(t) = x (1-t)  + 3x (1-t) t + 3x (1-t)t  + x t
              1           2            3           4
-      \prime            2                2                                      
+      \prime            2                2
 Bézier      (t) = a(1-t)  + 2b(1-t)t + ct   { a=3(x -x ),b=3(x -x ),c=3(x -x ) }
-                                                   2  1       3  2       4  3   
+                                                   2  1       3  2       4  3
       \prime\prime
-Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\                        
+Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 -->
 						<img
 							class="LaTeX SVG"
@@ -3991,14 +3991,14 @@ Bézier            (t) = u(1-t) + vt {u=2(b-a),v=2(c-b)}\
 						/>
 						<p>And of course the same functions for <em>y</em>:</p>
 						<!--
- 
+
                    3           2             2      3
-Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t 
+Bézier(t) = y (1-t)  + 3y (1-t) t + 3y (1-t)t  + y t
              1           2            3           4
-      \prime            2                2           
-Bézier      (t) = d(1-t)  + 2e(1-t)t + ft            
+      \prime            2                2
+Bézier      (t) = d(1-t)  + 2e(1-t)t + ft
       \prime\prime
-Bézier            (t) = w(1-t) + zt                  
+Bézier            (t) = w(1-t) + zt
 -->
 						<img
 							class="LaTeX SVG"
@@ -4012,19 +4012,19 @@ Bézier            (t) = w(1-t) + zt
 							rather overly complicated set of arithmetic expressions:
 						</p>
 						<!--
- 
+
      2             2             2             2             2
--18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y              
+-18 t  x  y  + 36 t  x  y  - 18 t  x  y  + 18 t  x  y  - 54 t  x  y
         2  1          3  1          4  1          1  2          3  2
      2             2             2             2             2
-+36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y              
++36 t  x  y  - 36 t  x  y  + 54 t  x  y  - 18 t  x  y  + 18 t  x  y
         4  2          1  3          2  3          4  3          1  4
-     2             2                                                             
--36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y 
+     2             2
+-36 t  x  y  + 18 t  x  y  + 36 t x  y   - 54 t x  y   + 18 t x  y   - 36 t x  y
         2  4          3  4         2  1          3  1          4  1          1  2
-+54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y 
++54 t x  y   - 18 t x  y   + 54 t x  y   - 54 t x  y   - 18 t x  y   + 18 t x  y
        3  2          4  2          1  3          2  3          1  4          2  4
--18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y   
+-18 x  y     + 18 x  y     + 18 x  y     - 18 x  y     - 18 x  y     + 18 x  y
      2  1          3  1          1  2          3  2          1  3          2  3
 -->
 						<img
@@ -4045,7 +4045,7 @@ Bézier            (t) = w(1-t) + zt
 						<code>t</code> values, we end up with the following simple term function for <em>C(t)</em>:
 					</p>
 					<!--
- 
+
                                  2
 (3 x  y +2 x  y +3 x  y -x  y ) t  + (3 x  y -x  y -3 x  y ) t + (x  y -x  y )
     3  2    4  2    2  3  4  3           3  2  4  2    2  3        2  3  3  2
@@ -4062,7 +4062,7 @@ Bézier            (t) = w(1-t) + zt
 						simplification:
 					</p>
 					<!--
- 
+
 a = x   · y  ╮
      3     2 │
 b = x   · y  │                             2
@@ -4081,7 +4081,7 @@ d = x   · y  │
 					/>
 					<p>This is a plain quadratic curve, and we know how to solve <em>C(t) = 0</em>; we use the quadratic formula:</p>
 					<!--
- 
+
                                                   ┌──────────┐
                                                   │ 2
 x = -3a + 2b + 3c - d ╮                      -y ±⟍│y  - 4 x z
@@ -4175,7 +4175,7 @@ z =       c - a       ╯                             2x
 								This edge is described by the function:
 							</p>
 							<!--
- 
+
       2
     -x  + 2x + 3
 y = ────────────, { x ≤1 }
@@ -4195,7 +4195,7 @@ y = ────────────, { x ≤1 }
 								the function:
 							</p>
 							<!--
- 
+
      ┌──────────┐
      │        2
     ⟍│3(4x - x )  - x
@@ -4213,7 +4213,7 @@ y = ─────────────────, { 0 ≤x ≤1 }
 						<li>
 							<p>...on the top (blue) edge, the curve's start point touches the curve, forming a loop. This edge is described by the function:</p>
 							<!--
- 
+
       2
     -x  + 3x
 y = ────────, { x ≤0 }
@@ -4280,7 +4280,7 @@ y = ────────, { x ≤0 }
 						<em>z</em> coordinate that is always 1, then we can suddenly add arbitrary values. For example:
 					</p>
 					<!--
- 
+
 ┌ 1 0 a ┐    ┌  x  ┐   ┌ 1  · x + 0  · y + a  · z ┐   ┌ x + a  · 1 ┐   ┌ x + a ┐
 │ 0 1 b │  · │  y  │ = │ 0  · x + 1  · y + b  · z │ = │ y + b  · 1 │ = │ y + b │
 └ 0 0 1 ┘    └ z=1 ┘   └ 0  · x + 0  · y + 1  · z ┘   └   1  · z   ┘   └  z=1  ┘
@@ -4291,7 +4291,7 @@ y = ────────, { x ≤0 }
 						want to subtract p1.x and p1.y, we use:
 					</p>
 					<!--
- 
+
      ┌ 1 0 -P   ┐            ┌ 1  · x + 0  · y - P    · 1 ┐   ┌ x - P   ┐
      │       1x │    ┌ x ┐   │                    1x      │   │      1x │
 T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = │ y - P   │
@@ -4307,7 +4307,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						transformation looks like this:
 					</p>
 					<!--
- 
+
 ┌ 1 S 0 ┐    ┌ x ┐   ┌ x + S  · y ┐
 │ 0 1 0 │  · │ y │ = │     y      │
 └ 0 0 1 ┘    └ 1 ┘   └     1      ┘
@@ -4318,7 +4318,7 @@ T  = │ 0 1 -P   │  · │ y │ = │ 0  · x + 1  · y - P    · 1 │ = 
 						simply <em>-x/y</em>, because <em>-x/y \</em> y = -x*. Done:
 					</p>
 					<!--
- 
+
      ┌    U     ┐
      │     2x   │
      │ 1 -─── 0 │
@@ -4336,7 +4336,7 @@ T  = │    U     │
 						x-scaling by 1/point3<sub>x</sub>, and y-scaling by point2<sub>y</sub>. This is really easy:
 					</p>
 					<!--
- 
+
      ┌  1        ┐
      │ ───  0  0 │
      │ V         │
@@ -4357,7 +4357,7 @@ T  = │      1    │
 						offset, in this case 1:
 					</p>
 					<!--
- 
+
      ┌    1    0 0 ┐
      │ 1 - W       │
      │      3y     │
@@ -4373,7 +4373,7 @@ T  = │ ─────── 1 0 │
 						be:
 					</p>
 					<!--
- 
+
                 ╭                           (-x +x )(-y +y )                ╮
                 │                              1  2    1  4                 │
                 │                -x  + x  - ────────────────                │
@@ -4408,7 +4408,7 @@ mapped  = (x) = │                                1  2                       
 						that leave?
 					</p>
 					<!--
- 
+
       ╭                 x   · y       x   · y              ╮
       │                  2     4       2     3             │
       │            x  - ──────── / x -────────             │   ╭         x           ╮
@@ -4428,11 +4428,11 @@ mapped  = (x) = │                                1  2                       
 						factors to see just how simple this is:
 					</p>
 					<!--
- 
-      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y 
+
+      ╭         x           ╮                 ╭ x  - x   · y   ╮           y                y
       │          43         │                 │  4    2     42 │            4                3
 ... = │                     │,   where  x   = │ ────────────── │,    y   = ──,   and  y   = ──
-      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y 
+      │ y   + x   (1 - y  ) │            43   │ x  - x   · y   │      42   y           32   y
       ╰  42    43       32  ╯                 ╰  3    2     32 ╯            2                2
 -->
 					<img class="LaTeX SVG" src="./images/chapters/canonical/88e3fae7aeef6d7614290587422542c9.svg" width="563px" height="55px" loading="lazy" />
@@ -4521,16 +4521,16 @@ mapped  = (x) = │                                1  2                       
 					</p>
 					<p>First, let's look at the function for x(t):</p>
 					<!--
- 
+
              3          2            2     3
-x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt 
+x(t) = a(1-t)  + 3b(1-t) t + 3c(1-t)t  + dt
 -->
 					<img class="LaTeX SVG" src="./images/chapters/yforx/378d0fd8cefa688d530ac38930d66844.svg" width="335px" height="19px" loading="lazy" />
 					<p>
 						We can rewrite this to a plain polynomial form, by just fully writing out the expansion and then collecting the polynomial factors, as:
 					</p>
 					<!--
- 
+
                          3                  2
 x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 -->
@@ -4541,7 +4541,7 @@ x(t) = (-a + 3b- 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + a
 						known constants, we can trivially rewrite this (by moving the <code>x(t)</code> across the equal sign) as:
 					</p>
 					<!--
- 
+
                    3                  2
 (-a + 3b - 3c + d)t  + (3a - 6b + 3c)t  + (-3a + 3b)t + (a-x) = 0
 -->
@@ -4634,7 +4634,7 @@ y = curve.get(t).y</textarea
 						following seemingly straight forward (if a bit overwhelming) formula:
 					</p>
 					<!--
- 
+
     ┌───────────────┐
 ╭ z │      2       2
 |   │f '(t) +f '(t)   dt
@@ -4643,7 +4643,7 @@ y = curve.get(t).y</textarea
 					<img class="LaTeX SVG" src="./images/chapters/arclength/59ebc3a7c3547a50998d1ea3664fb688.svg" width="140px" height="33px" loading="lazy" />
 					<p>or, more commonly written using Leibnitz notation as:</p>
 					<!--
- 
+
               ┌─────────────────┐
           ╭ z │       2        2
 length =  |  ⟍│(dx/dt) +(dy/dt)   dt
@@ -4682,7 +4682,7 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 |   ⟍│(dx/dt) +(dy/dt)   dt =  |   f(t) dt  ≃ ┌ \underset strip 1 \underbrace C   · f(t )   + ...  + \underset strip n \underbrace C   · f(t )  ┐ =
 ╯ -1                           ╯ -1           └                                1       1                                            n       n   ┘
                                                                                     __ n
-                                           \underset strips 1 through n \underbrace ❯      C   · f(t )   
+                                           \underset strips 1 through n \underbrace ❯      C   · f(t )
                                                                                     ‾‾ i=1  i       i
 -->
 					<img class="LaTeX SVG" src="./images/chapters/arclength/b76753476ad6ecfe4b8f39bcf9432980.svg" width="623px" height="71px" loading="lazy" />
@@ -4764,16 +4764,16 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							shifting and scaling the inputs. Doing so, we get the following:
 						</p>
 						<!--
- 
+
      ┌─────────────────┐
  ╭ z │       2        2
- |  ⟍│(dx/dt) +(dy/dt)   dt                                        
+ |  ⟍│(dx/dt) +(dy/dt)   dt
  ╯ 0
      z    ┌        ╭ z         z ╮                ╭ z         z ╮ ┐
  ≃ \ ─  · │ C   · f│ ─  · t  + ─ │ + ... + C   · f│ ─  · t  + ─ │ │
      2    └  1     ╰ 2     1   2 ╯          n     ╰ 2     n   2 ╯ ┘
     z    __ n         ╭ z         z ╮
-= \ ─  · ❯     C   · f│ ─  · t  + ─ │                              
+= \ ─  · ❯     C   · f│ ─  · t  + ─ │
     2    ‾‾ i=1 i     ╰ 2     i   2 ╯
 -->
 						<img
@@ -4795,10 +4795,10 @@ length =  |  ⟍│(dx/dt) +(dy/dt)   dt
 							<a href="./legendre-gauss.html">these tables</a> would tell us that for <em>n=2</em> we must use the following values:
 						</p>
 						<!--
- 
-C  = 1     
+
+C  = 1
  1
-C  = 1     
+C  = 1
  2
         1
 t  = - ────
@@ -4814,7 +4814,7 @@ t  = + ────
 							Which means that in order for us to approximate the integral, we must plug these values into the approximate function, which gives us:
 						</p>
 						<!--
- 
+
     ┌─────────────────┐
 ╭ z │       2        2       z    ┌  ╭ z     -1    z ╮    ╭ z     1     z ╮ ┐
 |  ⟍│(dx/dt) +(dy/dt)   dt ≃ ─  · │ f│ ─  · ──── + ─ │ + f│ ─  · ──── + ─ │ │
@@ -4978,13 +4978,13 @@ t  = + ────
 					</p>
 					<p>So, what does the function look like? This:</p>
 					<!--
- 
+
          x'y'' - x''y'
 \kappa = ─────────────
                    3
                    ─
              2   2 2
-          (x' +y' ) 
+          (x' +y' )
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/060acd6ff0a050fe4d98a7802a2b3a3f.svg" width="113px" height="47px" loading="lazy" />
 					<p>
@@ -4992,14 +4992,14 @@ t  = + ────
 						tiny bit:
 					</p>
 					<!--
- 
+
             B '(t)B ''(t) - B ''(t)B '(t)
              x     y         x      y
 \kappa(t) = ─────────────────────────────
                                   3
                                   ─
                         2       2 2
-                 (B '(t) +B '(t) ) 
+                 (B '(t) +B '(t) )
                    x       y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/curvature/afd8cb8b0fe291ff703752c1c9cc33d4.svg" width="239px" height="55px" loading="lazy" />
@@ -5078,7 +5078,7 @@ function kappa(t, B):
 						circle that "fits" the curve's curvature at any point, using what is possibly the simplest bit of maths found in this entire primer:
 					</p>
 					<!--
- 
+
            1
 R(t) = ─────────
        \kappa(t)
@@ -5526,8 +5526,8 @@ lli = function(line1, line2):
 						points, so logically <code>C</code> will have a function that interpolates between those two coordinates:
 					</p>
 					<!--
- 
-C = u(t)  · P       + (1-u(t))  · P    
+
+C = u(t)  · P       + (1-u(t))  · P
               start                 end
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8c6662f605722fb2ff6cd7f65243a126.svg" width="229px" height="16px" loading="lazy" />
@@ -5540,22 +5540,22 @@ C = u(t)  · P       + (1-u(t))  · P
 						(with thanks to Boris Zbarsky) shows us the following two formulae:
 					</p>
 					<!--
- 
+
                         2
-                   (1-t) 
+                   (1-t)
 u(t)           = ───────────
      quadratic    2        2
-                 t  + (1-t) 
+                 t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8e7cfee39c98f2ddf9b635a914066cf6.svg" width="183px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                     3
-               (1-t) 
+               (1-t)
 u(t)       = ───────────
      cubic    3        3
-             t  + (1-t) 
+             t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/a0b99054cc82ca1fb147f077e175ef10.svg" width="161px" height="41px" loading="lazy" />
 					<p>
@@ -5566,7 +5566,7 @@ u(t)       = ───────────
 					</p>
 					<p>We start by observing that, given <code>A</code>, <code>B</code>, and <code>C</code>, the following always holds:</p>
 					<!--
- 
+
             distance(B,C)
 ratio(t) = ────────────── =  Constant
             distance(A,B)
@@ -5574,22 +5574,22 @@ ratio(t) = ────────────── =  Constant
 					<img class="LaTeX SVG" src="./images/chapters/abc/131454dcbac04e567f322979f4af80c6.svg" width="245px" height="39px" loading="lazy" />
 					<p>Working out the maths for this, we see the following two formulae for quadratic and cubic curves:</p>
 					<!--
- 
+
                        2        2
                       t  + (1-t)  - 1
 ratio(t)           = |───────────────|
          quadratic       2        2
-                        t  + (1-t) 
+                        t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/5924e162b50272c40c842fad14b8fa48.svg" width="239px" height="41px" loading="lazy" />
 					<p>And</p>
 					<!--
- 
+
                    3        3
                   t  + (1-t)  - 1
 ratio(t)       = |───────────────|
          cubic       3        3
-                    t  + (1-t) 
+                    t  + (1-t)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8cd992c1ceaae2e67695285beef23a24.svg" width="219px" height="41px" loading="lazy" />
 					<p>
@@ -5599,7 +5599,7 @@ ratio(t)       = |───────────────|
 						<code>ratio(t)</code> function to find <code>A</code>:
 					</p>
 					<!--
- 
+
           C - B           B - C
 A = B - ───────── = B + ─────────
          ratio(t)        ratio(t)
@@ -5614,11 +5614,11 @@ A = B - ───────── = B + ─────────
 						pair, after which we can reverse engineer <code>v1</code> and <code>v2</code>:
 					</p>
 					<!--
- 
+
                                  ╭     e  - t  · A
                                  │      1
-╭ e = (1-t)  · v  + t  · A       │ v = ───────────    
-╡  1            1            ==> ╡  1      1-t         
+╭ e = (1-t)  · v  + t  · A       │ v = ───────────
+╡  1            1            ==> ╡  1      1-t
 │ e = (1-t)  · A + t  · v        │     e  - (1-t)  · A
 ╰  2                     2       │      2
                                  │ v = ───────────────
@@ -5627,14 +5627,14 @@ A = B - ───────── = B + ─────────
 					<img class="LaTeX SVG" src="./images/chapters/abc/3166afa345aec1abda432c39b68d39a0.svg" width="339px" height="73px" loading="lazy" />
 					<p>And then reverse engineer the curve's control points:</p>
 					<!--
- 
+
                                       ╭     v  - (1-t)  ·  start
                                       │      1
 ╭ v = (1-t)  ·  start + t  · C        │ C = ────────────────────
-╡  1                          1   ==> ╡  1           t           
+╡  1                          1   ==> ╡  1           t
 │ v = (1-t)  · C  + t  ·  end         │     v  - t  ·  end
 ╰  2            2                     │      2
-                                      │ C = ──────────────      
+                                      │ C = ──────────────
                                       ╰  2       1-t
 -->
 					<img class="LaTeX SVG" src="./images/chapters/abc/8bd3e6fed5bf8d871d30221ae400fd93.svg" width="383px" height="75px" loading="lazy" />
@@ -5662,15 +5662,15 @@ A = B - ───────── = B + ─────────
 						<code>B</code> point, and <code>B</code> and end point as a ratio, using
 					</p>
 					<!--
- 
+
 ╭ d = || Start - B||
 │  1
-│ d = || End - B||  
+│ d = || End - B||
 │  2
-╡      d             
+╡      d
 │       1
-│  t= ─────         
-│     d +d 
+│  t= ─────
+│     d +d
 ╰      1  2
 -->
 					<img
@@ -5732,9 +5732,9 @@ A = B - ───────── = B + ─────────
 						<code>e2</code> coordinates must obey the standard de Casteljau rule for linear interpolation:
 					</p>
 					<!--
- 
-╭ e = B + t  · d    
-╡  1                 
+
+╭ e = B + t  · d
+╡  1
 │ e = B - (1-t)  · d
 ╰  2
 -->
@@ -5754,7 +5754,7 @@ A = B - ───────── = B + ─────────
 						curve with a loop in it. To do this, we can use the <a href="https://en.wikipedia.org/wiki/Atan2">atan2</a> function:
 					</p>
 					<!--
- 
+
 \phi = (atan2(E -S , E -S ) - atan2(B -S , B -S ) + 2 \pi)   mod  2 \pi
                y  y   x  x           y  y   x  x
 -->
@@ -5771,8 +5771,8 @@ A = B - ───────── = B + ─────────
 						<code>d</code>:
 					</p>
 					<!--
- 
-d = {  d  if  0 ≤\phi ≤\pi             
+
+d = {  d  if  0 ≤\phi ≤\pi
       -d  if  \phi < 0 \lor \phi > \pi
 -->
 					<img
@@ -5952,7 +5952,7 @@ for (coordinate, index) in LUT:
 						maths, that means we're trying to solve:
 					</p>
 					<!--
- 
+
 dist(B(t), c) = r
 -->
 					<img
@@ -5964,15 +5964,15 @@ dist(B(t), c) = r
 					/>
 					<p>Which seems simple enough. Unfortunately, when we expand that <code>dist</code> function, things get a lot more problematic:</p>
 					<!--
- 
-r=  dist(B(t), c)                                                                                                              
+
+r=  dist(B(t), c)
     ┌─────────────────────────┐
     │          2             2
- =  │(B t - c )  + (B t - c )                                                                                                  
-   ⟍│  x     x       y     y                                                                                                   
+ =  │(B t - c )  + (B t - c )
+   ⟍│  x     x       y     y
     ┌─────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────┐
     │╭         3             2                 2       3      ╮2   ╭         3             2                 2       3      ╮2
- =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │  
+ =  ││ x  (1-t)  + 3 x  (1-t)  t + 2 x  (1-t) t  + x  t  - c  │  + │ y  (1-t)  + 3 y  (1-t)  t + 2 y  (1-t) t  + y  t  - c  │
    ⟍│╰  1             2               3             4       x ╯    ╰  1             2               3             4       y ╯
 -->
 					<img
@@ -6244,17 +6244,17 @@ findClosest(start, p, r, LUT):
 						<code>B</code>, we can compute the associated <code>C</code>:
 					</p>
 					<!--
- 
+
 C = u(t)   ·  Start + (1-u(t) )  ·  End
         q                    q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/48887d68a861a0acdf8313e23fb19880.svg" width="247px" height="24px" loading="lazy" />
 					<p>And then the associated <code>A</code>:</p>
 					<!--
- 
+
           C - B            B - C
 A = B - ────────── = B + ──────────
-         ratio(t)         ratio(t) 
+         ratio(t)         ratio(t)
                  q                q
 -->
 					<img class="LaTeX SVG" src="./images/chapters/molding/2e65bc9c934380c2de6a24bcd5c1c7b7.svg" width="228px" height="41px" loading="lazy" />
@@ -6393,12 +6393,12 @@ A = B - ────────── = B + ──────────
 							form, where each line corresponds to a power of t, and each column is for a specific coefficient. First, we expand the function:
 						</p>
 						<!--
- 
+
                     2                    2
-B          = a (1-t)  + 2 b (1-t) t + c t    
-  quadratic                                  
+B          = a (1-t)  + 2 b (1-t) t + c t
+  quadratic
                          2            2     2
-           = a - 2at + at  + 2bt - 2bt  + ct 
+           = a - 2at + at  + 2bt - 2bt  + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -6409,12 +6409,12 @@ B          = a (1-t)  + 2 b (1-t) t + c t
 						/>
 						<p>And then we (trivially) rearrange the terms across multiple lines:</p>
 						<!--
- 
-B          =a               
+
+B          =a
   quadratic
-            - 2at+ 2bt      
+            - 2at+ 2bt
                 2     2    2
-            + at - 2bt + ct 
+            + at - 2bt + ct
 -->
 						<img
 							class="LaTeX SVG"
@@ -6430,7 +6430,7 @@ B          =a
 						</p>
 						<p>With that arrangement, we can easily decompose this as a matrix multiplication:</p>
 						<!--
- 
+
                                          ┌  a  +  0  + 0 ┐                 ┌  1  0 0 ┐    ┌ a ┐
 B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌      2 ┐  · │ -2  2 0 │  · │ b │
   quadratic                └ 1 t t  ┘    └  a  + -2b + c ┘   └ 1 t t  ┘    └  1 -2 1 ┘    └ c ┘
@@ -6444,9 +6444,9 @@ B          = T  · M  · C = ┌      2 ┐  · │ -2a +  2b + 0 │ = ┌
 						/>
 						<p>We can do the same for the cubic curve, of course. We know the base function for cubics:</p>
 						<!--
- 
+
               3          2             2     3
-B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt 
+B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
   cubic
 -->
 						<img
@@ -6458,14 +6458,14 @@ B      =a(1-t)  + 3b(1-t)  t + 3c(1-t)t  + dt
 						/>
 						<p>So we write out the expansion and rearrange:</p>
 						<!--
- 
-B      =a                     
+
+B      =a
   cubic
-        - 3at + 3bt           
-             2     2    2     
-        + 3at - 6bt +3ct      
+        - 3at + 3bt
+             2     2    2
+        + 3at - 6bt +3ct
             3      3    3    3
-        - at  + 3bt -3ct + dt 
+        - at  + 3bt -3ct + dt
 -->
 						<img
 							class="LaTeX SVG"
@@ -6476,7 +6476,7 @@ B      =a
 						/>
 						<p>Which we can then decompose:</p>
 						<!--
- 
+
                                         ┌  1  0  0 0 ┐    ┌ a ┐
 B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
   cubic                └ 1 t t  t  ┘    │  3 -6  3 0 │    │ c │
@@ -6491,7 +6491,7 @@ B      = T  · M  · C = ┌      2  3 ┐  · │ -3  3  0 0 │  · │ b │
 						/>
 						<p>And, of course, we can do this for quartic curves too (skipping the expansion step):</p>
 						<!--
- 
+
                                              ┌  1  0   0   0 0 ┐    ┌ a ┐
                                              │ -4  4   0   0 0 │    │ b │
 B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  · │ c │
@@ -6517,7 +6517,7 @@ B        = T  · M  · C = ┌      2  3  4 ┐  · │  6 -12  6   0 0 │  ·
 						we want to do curve fitting:
 					</p>
 					<!--
- 
+
     ┌ p   ┐
     │  1  │
     │ p   │
@@ -6555,9 +6555,9 @@ P = │  2  │
 					</p>
 					<p>To get these values, we first compute the general "distance along the polygon" matrix:</p>
 					<!--
- 
+
                                ╭            d  = 0
-D = ┌ d  d  ... d  ┐,   where  ╡             1                 
+D = ┌ d  d  ... d  ┐,   where  ╡             1
     └  1  2      n ┘           │ d  = d    +  length(p   , p )
                                ╰  i    i-1            i-1   i
 -->
@@ -6574,10 +6574,10 @@ D = ┌ d  d  ... d  ┐,   where  ╡             1
 						<code>i=n</code> fall in the [0,1] interval, so we need to scale all values down by whatever the total length of the polygon is:
 					</p>
 					<!--
- 
+
                                ╭    s  = 0
                                │     1
-S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d  
+S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
     └  1  2      n ┘           │  i    i    n
                                │    s  = 1
                                ╰     n
@@ -6599,9 +6599,9 @@ S = ┌ s  s  ... s  ┐,   where  ╡ s  = d  / d
 						to for the associated <code>t</code> value", which we'll square to get rid of any pesky negative signs:
 					</p>
 					<!--
- 
+
                            2
-E(C)  = (p  -   Bézier(s )) 
+E(C)  = (p  -   Bézier(s ))
     i     i             i
 -->
 					<img
@@ -6616,9 +6616,9 @@ E(C)  = (p  -   Bézier(s ))
 						function. So, we literally just do that; the total error function is simply the sum of all these individual errors:
 					</p>
 					<!--
- 
+
        __ n                      2
-E(C) = ❯      (p  -   Bézier(s )) 
+E(C) = ❯      (p  -   Bézier(s ))
        ‾‾ i=1   i             i
 -->
 					<img
@@ -6635,9 +6635,9 @@ E(C) = ❯      (p  -   Bézier(s ))
 						<strong>T x M x C</strong> we talked about before, which gives us:
 					</p>
 					<!--
- 
+
                 2
-E(C) = (P - TMC) 
+E(C) = (P - TMC)
 -->
 					<img
 						class="LaTeX SVG"
@@ -6648,7 +6648,7 @@ E(C) = (P - TMC)
 					/>
 					<p>In which we can replace the rather cumbersome "squaring" operation with a more conventional matrix equivalent:</p>
 					<!--
- 
+
                 T
 E(C) = (P - TMC)  (P - TMC)
 -->
@@ -6670,7 +6670,7 @@ E(C) = (P - TMC)  (P - TMC)
 						those, and then use that 𝕋 instead of <strong>T</strong> in our error function:
 					</p>
 					<!--
- 
+
     ┌    0    1      n-2   n-1  ┐
     │   s    s  ... s     s     │
     │    1    1      1     1    │
@@ -6690,7 +6690,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Which, because of the first and last values in <strong>S</strong>, means:</p>
 					<!--
- 
+
     ┌    1     0  ...   0    0    ┐
     │                  n-2   n-1  │
     │    1    s       s     s     │
@@ -6710,7 +6710,7 @@ E(C) = (P - TMC)  (P - TMC)
 					/>
 					<p>Now we can properly write out the error function as matrix operations:</p>
 					<!--
- 
+
                 T
 E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 -->
@@ -6727,7 +6727,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						taking the derivative, and determining where that derivative is zero:
 					</p>
 					<!--
- 
+
 ∂E          T
 ── = 0 = -2𝕋  (P - 𝕋MC)
 ∂C
@@ -6766,7 +6766,7 @@ E(C) = (P - 𝕋MC)  (P - 𝕋MC)
 						for <strong>C</strong>:
 					</p>
 					<!--
- 
+
      -1   T   -1  T
 C = M   (𝕋  𝕋)   𝕋  P
 -->
@@ -6865,13 +6865,13 @@ C = M   (𝕋  𝕋)   𝕋  P
 						previous and next point:
 					</p>
 					<!--
- 
+
                ┌  V  = P       ┐
                │   1    2      │
 ┌ P  ┐         │  V  = P       │
 │  1 │         │   2    3      │
 │ P  │         │       P  - P  │
-│  2 │       = │        3    1 │              
+│  2 │       = │        3    1 │
 │ P  │points   │ V'  = ─────── │ point-tangent
 │  3 │         │   1      2    │
 │ P  │         │       P  - P  │
@@ -7004,7 +7004,7 @@ for p = 1 to points.length-3 (inclusive):
 						so how different is the matrix form for Catmull-Rom curves?:
 					</p>
 					<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7041,7 +7041,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							we'll get to that:
 						</p>
 						<!--
- 
+
                         ┌   P     ┐
                         │    2    │
 ┌ V   ┐        ┌ P  ┐   │   P     │
@@ -7069,7 +7069,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						</p>
 						<p>Computing <em>T</em> is really more "arranging the numbers":</p>
 						<!--
- 
+
               ┌   P     ┐
               │    2    │
      ┌ P  ┐   │   P     │   ┌  0  · P1 + 1  · P2 + 0  · P3 + 0  · P4 ┐   ┌  0  1 0 0 ┐    ┌ P  ┐
@@ -7092,7 +7092,7 @@ T  · │  2 │ = │  3    1 │ = │ ──  · P1 + 0  · P2 + ─  · P3 +
 						/>
 						<p>Thus:</p>
 						<!--
- 
+
     ┌  0  1 0 0 ┐
     │  0  0 1 0 │
     │ -1    1   │
@@ -7116,7 +7116,7 @@ T = │ ──  0 ─ 0 │
 							factor into both our coordinate vector representation, and mapping matrix <em>T</em>:
 						</p>
 						<!--
- 
+
           ┌   P     ┐
           │    2    │
 ┌ V   ┐   │   P     │        ┌  0  1  0 0  ┐
@@ -7142,7 +7142,7 @@ T = │ ──  0 ─ 0 │
 							coordinates, and see what we end up with:
 						</p>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7161,7 +7161,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						/>
 						<p>Replace point/tangent vector with the expression for all-coordinates:</p>
 						<!--
- 
+
                                                     ┌  0  1  0 0  ┐    ┌ P  ┐
                                                     │  0  0  1 0  │    │  1 │
                                  ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -7180,7 +7180,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │ ──  0 ─
 						/>
 						<p>and merge the matrices:</p>
 						<!--
- 
+
                                  ┌  0    1      0   0  ┐
                                  │ -1          1       │    ┌ P  ┐
                                  │ ──    0     ──   0  │    │  1 │
@@ -7201,7 +7201,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  1 1           1 -1 │  · │  2 │
 						/>
 						<p>This looks a lot like the Bézier matrix form, which as we saw in the chapter on Bézier curves, should look like this:</p>
 						<!--
- 
+
                                                ┌ P  ┐
                                                │  1 │
                              ┌  1  0  0 0 ┐    │ P  │
@@ -7220,7 +7220,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>So, if we want to express a Catmull-Rom curve using a Bézier curve, we'll need to turn this Catmull-Rom bit:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │
@@ -7241,7 +7241,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Into something that looks like this:</p>
 						<!--
- 
+
                   ┌ P  ┐
                   │  1 │
 ┌  1  0  0 0 ┐    │ P  │
@@ -7260,7 +7260,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And the way we do that is with a fairly straight forward bit of matrix rewriting. We start with the equality we need to ensure:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │    ┌ P  ┐                          ┌ P  ┐
 │ ──    0     ──   0  │    │  1 │                          │  1 │
@@ -7281,7 +7281,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we remove the coordinate vector from both sides without affecting the equality:</p>
 						<!--
- 
+
 ┌  0    1      0   0  ┐
 │ -1          1       │
 │ ──    0     ──   0  │
@@ -7302,7 +7302,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>Then we can "get rid of" the Bézier matrix on the right by left-multiply both with the inverse of the Bézier matrix:</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -7326,7 +7326,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							outright remove any matrix/inverse pair:
 						</p>
 						<!--
- 
+
                      ┌  0    1      0   0  ┐
                      │ -1          1       │
                      │ ──    0     ──   0  │
@@ -7347,7 +7347,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 						/>
 						<p>And now we're <em>basically</em> done. We just multiply those two matrices and we know what <em>V</em> is:</p>
 						<!--
- 
+
 ┌  0  1  0 0  ┐
 │ -1    1     │
 │ ──  1 ── 0  │
@@ -7369,7 +7369,7 @@ Bézier(t) = ┌      2  3 ┐  · │ -3  3  0 0 │  · │  2 │
 							<li>Start with the Catmull-Rom function:</li>
 						</ol>
 						<!--
- 
+
                                                     ┌ V   ┐
                                                     │  1  │
                                  ┌  1  0  0 0  ┐    │ V   │
@@ -7390,7 +7390,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite to pure coordinate form:</li>
 						</ol>
 						<!--
- 
+
                                       ┌   P     ┐
                                       │    2    │
                                       │   P     │
@@ -7415,7 +7415,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for "normal" coordinate vector:</li>
 						</ol>
 						<!--
- 
+
                                       ┌  0  1  0 0  ┐    ┌ P  ┐
                                       │  0  0  1 0  │    │  1 │
                    ┌  1  0  0 0  ┐    │ -1    1     │    │ P  │
@@ -7436,7 +7436,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>merge the inner matrices:</li>
 						</ol>
 						<!--
- 
+
                    ┌  0    1      0   0  ┐
                    │ -1          1       │    ┌ P  ┐
                    │ ──    0     ──   0  │    │  1 │
@@ -7459,7 +7459,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>rewrite for Bézier matrix form:</li>
 						</ol>
 						<!--
- 
+
                                      ┌  0  1  0 0  ┐    ┌ P  ┐
                                      │ -1    1     │    │  1 │
                    ┌  1  0  0 0 ┐    │ ──  1 ── 0  │    │ P  │
@@ -7480,7 +7480,7 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 							<li>and transform the coordinates so we have a "pure" Bézier expression:</li>
 						</ol>
 						<!--
- 
+
                                      ┌     P       ┐
                                      │      2      │
                                      │      P -P   │
@@ -7509,13 +7509,13 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						curve that has the vector:
 					</p>
 					<!--
- 
+
                        ┌     P       ┐
                        │      2      │
 ┌ P  ┐                 │      P -P   │
 │  1 │                 │       3  1  │
 │ P  │                 │ P  + ────── │
-│  2 │             ==> │  2   6  · τ │        
+│  2 │             ==> │  2   6  · τ │
 │ P  │ CatmullRom      │      P -P   │  Bézier
 │  3 │                 │       4  2  │
 │ P  │                 │ P  - ────── │
@@ -7535,11 +7535,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						tension Catmull-Rom curve with the following coordinate values:
 					</p>
 					<!--
- 
+
 ┌ P  ┐              ┌       P         ┐
 │  1 │              │        1        │
 │ P  │              │       P         │
-│  2 │          ==> │        4        │           
+│  2 │          ==> │        4        │
 │ P  │  Bézier      │ P  + 3(P  - P ) │ CatmullRom
 │  3 │              │  4      1    2  │
 │ P  │              │ P  + 3(P  - P ) │
@@ -7554,11 +7554,11 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 					/>
 					<p>Or, if your API allows you to specify Catmull-Rom curves using plain coordinates:</p>
 					<!--
- 
+
 ┌ P  ┐              ┌ P  + 6(P  - P ) ┐
 │  1 │              │  4      1    2  │
 │ P  │              │       P         │
-│  2 │          ==> │        1        │           
+│  2 │          ==> │        1        │
 │ P  │  Bézier      │       P         │ CatmullRom
 │  3 │              │        4        │
 │ P  │              │ P  + 6(P  - P ) │
@@ -7649,8 +7649,8 @@ CatmullRom(t) = ┌      2  3 ┐  · │  0  0  1 0  │  · │  2  │
 						the same as it's "outgoing" derivative:
 					</p>
 					<!--
- 
-B'(1)  = B'(0)   
+
+B'(1)  = B'(0)
      n        n+1
 -->
 					<img class="LaTeX SVG" src="./images/chapters/polybezier/a37252ff55837b918d9d64078ae92ae7.svg" width="124px" height="17px" loading="lazy" />
@@ -7660,7 +7660,7 @@ B'(1)  = B'(0)
 						last control point through the last on-curve point. And mirroring any point A through any point B is really simple:
 					</p>
 					<!--
- 
+
            ┌ B  + (B  - A ) ┐   ┌ 2B  - A  ┐
 Mirrored = │  x     x    x  │ = │   x    x │
            │ B  + (B  - A ) │   │ 2B  - A  │
@@ -7836,7 +7836,7 @@ Mirrored = │  x     x    x  │ = │   x    x │
 							any point on <code>O(t)</code> is a fixed distance <code>d</code> away from coordinate <code>B(t)</code>. So let's math that:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d
 -->
 						<img
@@ -7853,7 +7853,7 @@ O(t) = B(t) + d
 							at <code>B(t)</code>. Easy enough:
 						</p>
 						<!--
- 
+
 O(t) = B(t) + d  · N(t)
 -->
 						<img
@@ -7871,7 +7871,7 @@ O(t) = B(t) + d  · N(t)
 							<code>N(t)</code> always has length 1, which we can achieve by dividing <code>B'(t)</code> by its magnitude:
 						</p>
 						<!--
- 
+
           ╭   B'(t)    ╮
 N(t) \bot │ ────────── │
           ╰ || B'(t)|| ╯
@@ -7889,10 +7889,10 @@ N(t) \bot │ ────────── │
 							denoted with double vertical bars:
 						</p>
 						<!--
- 
+
                     ┌───────────┐
                ╭ b  │   2      2
-|| f(x,y)|| =  |    │f '  + f '  
+|| f(x,y)|| =  |    │f '  + f '
                ╯ a ⟍│ x      y
 -->
 						<img
@@ -7906,10 +7906,10 @@ N(t) \bot │ ────────── │
 							So if we want the length of the tangent, we plug in <code>B'(t)</code>, with <code>t = 0</code> as start and <code>t = 1</code> as end:
 						</p>
 						<!--
- 
+
                    ┌───────────────────┐
               ╭ 1  │       2          2
-|| B'(t)|| =  |    │B ''(t)  + B ''(t)  
+|| B'(t)|| =  |    │B ''(t)  + B ''(t)
               ╯ 0 ⟍│ x          y
 -->
 						<img
@@ -8125,7 +8125,7 @@ N(t) \bot │ ────────── │
 						some angle <em>φ</em>:
 					</p>
 					<!--
- 
+
              S = \begin{pmatrix} 1
 0 \end{pmatrix}  ,  \ E = \begin{pmatrix} cos(φ)
               sin(φ) \end{pmatrix}
@@ -8133,7 +8133,7 @@ N(t) \bot │ ────────── │
 					<img class="LaTeX SVG" src="./images/chapters/circles/7ab3da0922477af4cc09f5852100976b.svg" width="175px" height="40px" loading="lazy" />
 					<p>What we want to find is the intersection of the tangents, so we want a point C such that:</p>
 					<!--
- 
+
               C = S + a  · \begin{pmatrix} 0
 1 \end{pmatrix}  ,  \ C = E + b  · \begin{pmatrix} -sin(φ)
                    cos(φ) \end{pmatrix}
@@ -8144,22 +8144,22 @@ N(t) \bot │ ────────── │
 						through E (at some distance <em>b</em> from E). Solving this gives us:
 					</p>
 					<!--
- 
+
 ╭ C  = 1 = cos(φ) + b  · -sin(φ)
-╡  x                             
-│ C  = a = sin(φ) + b  · cos(φ) 
+╡  x
+│ C  = a = sin(φ) + b  · cos(φ)
 ╰  y
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/8237af1396bb567d70c8b5e4dd7f8115.svg" width="219px" height="40px" loading="lazy" />
 					<p>First we solve for <em>b</em>:</p>
 					<!--
- 
+
 1 = cos(φ) + b  · -sin(φ)  → \ 1 - cos(φ) = -b  · sin(φ)  → \ -1 + cos(φ) = b  · sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/1829e42ea956ee4df0e45d9ac5334ef7.svg" width="560px" height="17px" loading="lazy" />
 					<p>which yields:</p>
 					<!--
- 
+
     cos(φ)-1
 b = ────────
      sin(φ)
@@ -8167,21 +8167,21 @@ b = ────────
 					<img class="LaTeX SVG" src="./images/chapters/circles/06369b00338310df0a810c592485aa0a.svg" width="101px" height="39px" loading="lazy" />
 					<p>which we can then substitute in the expression for <em>a</em>:</p>
 					<!--
- 
- a= sin(φ) + b  · cos(φ)          
+
+ a= sin(φ) + b  · cos(φ)
              -1 + cos(φ)
 ..= sin(φ) + ───────────  · cos(φ)
                sin(φ)
                           2
              -cos(φ) + cos (φ)
-..= sin(φ) + ─────────────────    
-                  sin(φ)          
+..= sin(φ) + ─────────────────
+                  sin(φ)
        2         2
     sin (φ) + cos (φ) - cos(φ)
-..= ──────────────────────────    
+..= ──────────────────────────
               sin(φ)
     1 - cos(φ)
- a= ──────────                    
+ a= ──────────
       sin(φ)
 -->
 					<img class="LaTeX SVG" src="./images/chapters/circles/5a23f3bc298c85540c6dd18e304d9224.svg" width="231px" height="195px" loading="lazy" />
@@ -8192,7 +8192,7 @@ b = ────────
 						polynomial, and we know where the circle arc's actual point P is for angle φ/2:
 					</p>
 					<!--
- 
+
          φ                 φ
 P  = cos(─)  ,  \ P  = sin(─)
  x       2         y       2
@@ -8200,7 +8200,7 @@ P  = cos(─)  ,  \ P  = sin(─)
 					<img class="LaTeX SVG" src="./images/chapters/circles/8b4e1d0a62380ed011f27c645ed13b28.svg" width="188px" height="32px" loading="lazy" />
 					<p>We compute T, observing that if <em>t=0.5</em>, the polynomial values (1-t)², 2(1-t)t, and t² are 0.25, 0.5, and 0.25 respectively:</p>
 					<!--
- 
+
     1    2    1    1
 T = ─S + ─C + ─E = ─(S + 2C + E)
     4    4    4    4
@@ -8208,10 +8208,10 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/a04bd1558a76e60b8ca6e1fe4fa38c00.svg" width="252px" height="35px" loading="lazy" />
 					<p>Which, worked out for the x and y components, gives:</p>
 					<!--
- 
+
 ╭     1
-│ T = ─(3 + cos(φ))                                    
-╡  x  4                                                 
+│ T = ─(3 + cos(φ))
+╡  x  4
 │     1╭ 2-2cos(φ)          ╮   1╭     ╭ φ ╮          ╮
 │ T = ─│ ───────── + sin(φ) │ = ─│ 2tan│ ─ │ + sin(φ) │
 ╰  y  4╰  sin(φ)            ╯   4╰     ╰ 2 ╯          ╯
@@ -8219,17 +8219,17 @@ T = ─S + ─C + ─E = ─(S + 2C + E)
 					<img class="LaTeX SVG" src="./images/chapters/circles/986ae9104e0bc52e95689ae7ae4504db.svg" width="408px" height="77px" loading="lazy" />
 					<p>And the distance between these two is the standard Euclidean distance:</p>
 					<!--
- 
+
                  1                   φ        4╭ φ ╮
 d (φ)= T  - P  = ─(3 + cos(φ)) - cos(─) = 2sin │ ─ │  ,
  x      x    x   4                   2         ╰ 4 ╯
                  1╭     ╭ φ ╮          ╮       φ
-d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,   
+d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
  y      y    y   4╰     ╰ 2 ╯          ╯       2
-     ⇓                                                 
+     ⇓
         ┌───────┐                     ┌───────┐
         │ 2    2                 4 φ  │   1
- d(φ)=  │d  + d   =  ...   = 2sin (─) │───────         
+ d(φ)=  │d  + d   =  ...   = 2sin (─) │───────
        ⟍│ x    y                   4  │   2 φ
                                       │cos (─)
                                      ⟍│     2
@@ -8276,7 +8276,7 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 						precision:
 					</p>
 					<!--
- 
+
                 ╭  ┌─────────────┐ ╮
                 │  │     ┌──────┐  │
                 │ ⟍│2+ε-⟍│ε(2+ε)   │
@@ -8336,14 +8336,14 @@ d (φ)= T  - P  = ─│ 2tan│ ─ │ + sin(φ) │ - sin(─)  ,
 					</p>
 					<p>So let's again formally describe this:</p>
 					<!--
- 
-P = (1, 0)                     
+
+P = (1, 0)
  1
-P = (1, k)                     
- 2                             
+P = (1, k)
+ 2
 P = P  + k  · (sin(θ), -cos(θ))
  3   4
-P = (cos(θ), sin(θ))           
+P = (cos(θ), sin(θ))
  4
 -->
 					<img
@@ -8367,29 +8367,29 @@ P = (cos(θ), sin(θ))
 						P<sub>2</sub>" (which is "half height" P<sub>2</sub>), and so forth:
 					</p>
 					<!--
- 
-     P   + P  
-      2     3 
+
+     P   + P
+      2     3
        y     y   k + sin(θ) - k  · cos(θ)
- A = ───────── = ────────────────────────                                 
+ A = ───────── = ────────────────────────
   y      2                  2
           1
-     A  + ─P     k + sin(θ) - k  · cos(θ)   
+     A  + ─P     k + sin(θ) - k  · cos(θ)
       y   2 2    ──────────────────────── + ─
              y              2               2   2k + sin(θ) + k  · cos(θ)
-e  = ───────── = ──────────────────────────── = ───────────────────────── 
+e  = ───────── = ──────────────────────────── = ─────────────────────────
  1       2                    2                             4
-  y                                                                       
+  y
                                        k
      A  +  mid(P , P )   A  + sin(θ) - ─ cos(θ)
       y         4   3     y            2          k + 3sin(θ) 2k  · cos(θ)
 e  = ───────────────── = ────────────────────── = ────────────────────────
  2           2                     2                         4
   y
-     e   + e  
-      1     2 
+     e   + e
+      1     2
        y     y   3k + 4sin(θ) - 3k  · cos(θ)
- B = ───────── = ───────────────────────────                              
+ B = ───────── = ───────────────────────────
   y      2                    8
 -->
 					<img
@@ -8413,15 +8413,15 @@ e  = ───────────────── = ───────
 							<a href="https://www.wolframalpha.com/">Wolfram Alpha</a> is definitely the way to go. That said, let's get going:
 						</p>
 						<!--
- 
+
 3k + 4sin(θ)) - 3k  · cos(θ)      θ
-────────────────────────────= sin(─)                  
+────────────────────────────= sin(─)
              8                    2
                                   ╭ θ ╮
-3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │               
+3k + 4sin(θ)) - 3k  · cos(θ)= 8sin│ ─ │
                                   ╰ 2 ╯
                                   ╭ θ ╮
-           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)     
+           3k - 3k  · cos(θ)= 8sin│ ─ │ - 4sin(θ)
                                   ╰ 2 ╯
                                 ╭     ╭ θ ╮          ╮
              3k (1 - cos(θ))= 4 │ 2sin│ ─ │ - sin(θ) │
@@ -8429,12 +8429,12 @@ e  = ───────────────── = ───────
                                         θ
                                    2sin(─) - sin(θ)
                                         2
-                          3k= 4  · ────────────────   
+                          3k= 4  · ────────────────
                                       1 - cos(θ)
                                        ╭ θ ╮
                                    2sin│ ─ │ - sin(θ)
                               4        ╰ 2 ╯
-                           k= ─  · ────────────────── 
+                           k= ─  · ──────────────────
                               3        1 - cos(θ)
 -->
 						<img
@@ -8449,11 +8449,11 @@ e  = ───────────────── = ───────
 							<em>k</em>:
 						</p>
 						<!--
- 
+
             ╭ θ ╮
         2sin│ ─ │ - sin(θ)
    4        ╰ 2 ╯
-k= ─  · ──────────────────         
+k= ─  · ──────────────────
    3        1 - cos(θ)
         ╭     ╭ θ ╮               ╮
         │ 2sin│ ─ │               │
@@ -8461,10 +8461,10 @@ k= ─  · ──────────────────
 k= ─  · │ ────────── - ────────── │
    3    ╰ 1 - cos(θ)   1 - cos(θ) ╯
    4    ╭    ╭ θ ╮      ╭ θ ╮ ╮
-k= ─  · │ csc│ ─ │ - cot│ ─ │ │    
+k= ─  · │ csc│ ─ │ - cot│ ─ │ │
    3    ╰    ╰ 2 ╯      ╰ 2 ╯ ╯
    4       ╭ θ ╮
-k= ─  · tan│ ─ │                   
+k= ─  · tan│ ─ │
    3       ╰ 4 ╯
 -->
 						<img
@@ -8482,7 +8482,7 @@ k= ─  · tan│ ─ │
 						involving the circular arc's angle:
 					</p>
 					<!--
- 
+
            4    ╭ θ ╮
 k = f(θ) = ─ tan│ ─ │
            3    ╰ 4 ╯
@@ -8498,13 +8498,13 @@ k = f(θ) = ─ tan│ ─ │
 						Which means that for any circular arc with angle θ and radius <em>r</em>, our Bézier approximation based on three points of incidence is:
 					</p>
 					<!--
- 
-    start= (r, 0)                                          
-control  = (r, k)                                          
-        1                                                  
+
+    start= (r, 0)
+control  = (r, k)
+        1
 control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
         2
-      end= r  · (cos(θ), sin(θ))                           
+      end= r  · (cos(θ), sin(θ))
 -->
 					<img
 						class="LaTeX SVG"
@@ -8515,7 +8515,7 @@ control  = r · (cos(θ) + k  · sin(θ), sin(θ) - k  · cos(θ))
 					/>
 					<p>Which also gives us the commonly found value of 0.55228 for quarter circles, based on them having an angle of half π:</p>
 					<!--
- 
+
  ╭ \pi ╮   4       ╭ \pi ╮   4  ┌─┐
 f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.55228474983[...]
  ╰  2  ╯   3       ╰  8  ╯   3
@@ -8529,13 +8529,13 @@ f│ ─── │ = ─  · tan│ ─── │ = ─(⟍│2 -1) ≅ 0.552284
 					/>
 					<p>And thus giving us the following Bézier coordinates for a quarter circle of radius <em>r</em>:</p>
 					<!--
- 
-    start= (r, 0)           
+
+    start= (r, 0)
 control  = (r, 0.55228  · r)
-        1                   
+        1
 control  = (0.55228  · r, r)
         2
-      end= (0, r)           
+      end= (0, r)
 -->
 					<img
 						class="LaTeX SVG"
@@ -8679,7 +8679,7 @@ getMostWrongT(radius, bezier, start, end, epsilon=1e-15):
 					</p>
 					<p>First, let's express the total error (given circular arc angle θ, and some <em>k</em>) using standard calculus notation:</p>
 					<!--
- 
+
                     ┌───────────────────────┐
               ╭ 1   │         2            2
 erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
@@ -8702,7 +8702,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						expression looks like:
 					</p>
 					<!--
- 
+
 ╭        ┌───────┐         ╮
 │  ╭ 1   │ 2    2          │ d
 │  |  \| │B  + B   - r\|dt │ ── = 0
@@ -8721,7 +8721,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 						are each other's inverse operations, so they cancel out, leaving us with our original function:
 					</p>
 					<!--
- 
+
    ┌───────┐
    │ 2    2
 \| │B  + B   - r\| = 0,    t ∈[0,1]
@@ -8736,7 +8736,7 @@ erf (θ, k) =  |  \| │B (t,θ,k)  + B (t,θ,k)   - r\|dt
 					/>
 					<p>And now we just solve for that... oh wait. We've seen this before. In order to solve this, we'd end up needing to solve this:</p>
 					<!--
- 
+
  2    2
 B  + B  = r
  x    y
@@ -9259,7 +9259,7 @@ radialError(radius, points[]):
 						end, just like for Bézier curves), by evaluating the following function:
 					</p>
 					<!--
- 
+
            __ n
 Point(t) = ❯      P   · N   (t)
            ‾‾ i=0  i     i,k
@@ -9280,7 +9280,7 @@ Point(t) = ❯      P   · N   (t)
 					</p>
 					<p>Then the N() function itself. What does it look like?</p>
 					<!--
- 
+
           ╭       t- knot         ╮                ╭       knot     -t       ╮
           │              i        │                │           (i+k)         │
 N   (t) = │ ───────────────────── │  · N     (t) + │ ─────────────────────── │  · N       (t)
@@ -9295,9 +9295,9 @@ N   (t) = │ ───────────────────── 
 						some point; obviously, it does, and specifically it does so for the following <code>i</code>/<code>k</code> values:
 					</p>
 					<!--
- 
+
           ╭ 1  if  t ∈[ knot , knot   )
-N   (t) = ╡                 i      i+1  
+N   (t) = ╡                 i      i+1
  i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/2514e1aa0565840e33fde0b146e3efe2.svg" width="237px" height="40px" loading="lazy" />
@@ -9318,7 +9318,7 @@ N   (t) = ╡                 i      i+1
 						P(t), we can compute this point by evaluating <em>d(t)</em> on a curve section between knots <code>i</code> and <code>i+1</code>:
 					</p>
 					<!--
- 
+
  k               k-1                   k-1
 d (t) = α     · d   (t) + (1-α   )  · d   (t)
  i       i,k     i            i,k      i-1
@@ -9329,7 +9329,7 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 						by:
 					</p>
 					<!--
- 
+
               t -  knots[i]
 α    = ───────────────────────────
  i,k    knots[i+1+n-k] -  knots[i]
@@ -9343,9 +9343,9 @@ d (t) = α     · d   (t) + (1-α   )  · d   (t)
 					</p>
 					<p>Of course, the recursion does need a stop condition:</p>
 					<!--
- 
+
  k           0                ╭ 1  if  t ∈[ knot , knot   )
-d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1  
+d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
  0           i       i,1      ╰ 0         otherwise
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/49af474c33ce0ee0733626ea3d988570.svg" width="367px" height="40px" loading="lazy" />
@@ -9359,34 +9359,34 @@ d (t) = 0,  d (t) = N   (t) = ╡                 i      i+1
 						recursion diagram:
 					</p>
 					<!--
- 
+
      ╭                                ╭                                ╭          1     0           0
-     │                                │                                │         α   × d ,   with  d    either 0 or 1 
+     │                                │                                │         α   × d ,   with  d    either 0 or 1
      │                                │          2     1           1   │          3     3           3
-     │                                │         α   × d ,   with  d  = ╡                 +                              
+     │                                │         α   × d ,   with  d  = ╡                 +
      │                                │          3     3           3   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      │          3     2           2   │                                ╰ ╰      3 ╯     2           2
-     │         α   × d ,   with  d  = ╡                 +                                                                
+     │         α   × d ,   with  d  = ╡                 +
      │          3     3           3   │                                ╭           1     0
-     │                                │                                │          α   × d                             
+     │                                │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           2     2
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
  3   │                                │ ╰      3 ╯     2           2   │ ╭      1 ╮     0           0
-d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1     
+d  = ╡                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
  3   │                                ╰                                ╰ ╰      2 ╯     1           1
-     │                 +                                                                                                 
+     │                 +
      │                                ╭           2     1
-     │                                │          α   × d                                                                
+     │                                │          α   × d
      │                                │           2     2
-     │                                │                                                                                 
-     │ ╭      3 ╮     2           2   │                 +                                                               
-     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0                               
-     │ ╰      3 ╯     2           2   │                                │          α   × d 
+     │                                │
+     │ ╭      3 ╮     2           2   │                 +
+     │ │ 1 - α  │  × d ,   with  d  = ╡                                ╭           1     0
+     │ ╰      3 ╯     2           2   │                                │          α   × d
      │                                │ ╭      2 ╮     1           1   │           1     1
-     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +                              
+     │                                │ │ 1 - α  │  × d ,   with  d  = ╡                 +
      │                                │ ╰      2 ╯     1           1   │ ╭      1 ╮     0           0
-     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1 
+     │                                │                                │ │ 1 - α  │  × d ,   with  d    either 0 or 1
      ╰                                ╰                                ╰ ╰      1 ╯     0           0
 -->
 					<img class="LaTeX SVG" src="./images/chapters/bsplines/20e910bbea2e6eff511cb13cef18ef3b.svg" width="641px" height="336px" loading="lazy" />
diff --git a/src/html/index.template.html b/src/html/index.template.html
index 067faf84..6eb42df0 100644
--- a/src/html/index.template.html
+++ b/src/html/index.template.html
@@ -55,7 +55,7 @@
     {{ longDescription }}
     {% include "./fragments/donations.html" %}
     <p>
-      — <a href="https://twitter.com/TheRealPomax">Pomax</a>
+      — <a href="https://mastodon.social/@TheRealPomax">Pomax</a>
     </p>
     {% include "./fragments/noscript.html" %}
     {% include "./fragments/toc.html" %}