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2020-10-18 14:27:57 -07:00
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docs/chapters/derivatives/content.en-GB.md
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@@ -66,7 +66,7 @@ And that's the first part done: the two components inside the parentheses are ac
... = n \left (
\frac{x!}{y!(x-y)!} t^{y} (1-t)^{x-y} - \frac{x!}{k!(x-k)!} t^k (1-t)^{x-k}
\right )
\ ,\ with\ x=n-1,\ y=k-1
~,~with~x=n-1,~y=k-1
\\
... = n \left ( B_{(n-1),(k-1)}(t) - B_{(n-1),k}(t) \right )
\end{array}
@@ -98,7 +98,7 @@ Two of these terms fall way: the first term falls away because there is no -1<su
\[\begin{array}{lclc}
n \cdot B_{n-1,BLUE[0]}(t) \cdot w_1 &-& n \cdot B_{n-1,BLUE[0]}(t) \cdot w_0 &+ \\
n \cdot B_{n-1,RED[1]}(t) \cdot w_2 &-& \ n \cdot B_{n-1,RED[1]}(t) \cdot w_1 &+ \\
n \cdot B_{n-1,RED[1]}(t) \cdot w_2 &-& ~n \cdot B_{n-1,RED[1]}(t) \cdot w_1 &+ \\
n \cdot B_{n-1,MAGENTA[2]}(t) \cdot w_3 &-& n \cdot B_{n-1,MAGENTA[2]}(t) \cdot w_2 &+ \\
...
\end{array}\]
@@ -109,14 +109,14 @@ And that's just a summation of lower order curves:
Bézier_{n,k}(t) \frac{d}{dt} = n \cdot B_{(n-1),BLUE[0]}(t) \cdot (w_1 - w_0)
+ n \cdot B_{(n-1),RED[1]}(t) \cdot (w_2 - w_1)
+ n \cdot B_{(n-1),MAGENTA[2]}(t) \cdot (w_3 - w_2)
\ + \ ...
~+ ~...
\]
We can rewrite this as a normal summation, and we're done:
\[
Bézier_{n,k}(t) \frac{d}{dt} = \sum_{k=0}^{n-1} n \cdot B_{n-1,k}(t) \cdot (w_{k+1} - w_k)
= \sum_{k=0}^{n-1} B_{n-1,k}(t) \cdot \underset{derivative\ weights}
= \sum_{k=0}^{n-1} B_{n-1,k}(t) \cdot \underset{derivative~weights}
{\underbrace{n \cdot (w_{k+1} - w_k)}}
\]
@@ -126,21 +126,21 @@ Let's rewrite that in a form similar to our original formula, so we can see the
\[
Bézier(n,t) = \sum_{i=0}^{n}
\underset{binomial\ term}{\underbrace{\binom{n}{i}}}
\underset{binomial~term}{\underbrace{\binom{n}{i}}}
\cdot\
\underset{polynomial\ term}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\underset{polynomial~term}{\underbrace{(1-t)^{n-i} \cdot t^{i}}}
\cdot\
\underset{weight}{\underbrace{w_i}}
\]
\[
Bézier'(n,t) = \sum_{i=0}^{k}
\underset{binomial\ term}{\underbrace{\binom{k}{i}}}
\underset{binomial~term}{\underbrace{\binom{k}{i}}}
\cdot\
\underset{polynomial\ term}{\underbrace{(1-t)^{k-i} \cdot t^{i}}}
\underset{polynomial~term}{\underbrace{(1-t)^{k-i} \cdot t^{i}}}
\cdot\
\underset{derivative\ weight}{\underbrace{n \cdot (w_{i+1} - w_i)}}
{\ , \ with \ k=n-1}
\underset{derivative~weight}{\underbrace{n \cdot (w_{i+1} - w_i)}}
{~, ~with ~k=n-1}
\]
@@ -148,8 +148,8 @@ What are the differences? In terms of the actual Bézier curve, virtually nothin
\[ \begin{array}{llll}
B(n,t), & & w = \{A,B,C,D\} \\
B'(n,t), & n = 3, & w' = \{A',B',C'\} &= \{3 \cdot (B-A), {\ } 3 \cdot (C-B), {\ } 3 \cdot (D-C)\} \\
B''(n,t), & n = 2, & w'' = \{A'',B''\} &= \{2 \cdot (B'-A'), {\ } 2 \cdot (C'-B')\} \\
B'(n,t), & n = 3, & w' = \{A',B',C'\} &= \{3 \cdot (B-A), {~} 3 \cdot (C-B), {~} 3 \cdot (D-C)\} \\
B''(n,t), & n = 2, & w'' = \{A'',B''\} &= \{2 \cdot (B'-A'), {~} 2 \cdot (C'-B')\} \\
B'''(n,t), & n = 1, & w''' = \{A'''\} &= \{1 \cdot (B''-A'')\}
\end{array} \]