Components and root finding (#190)

components/sections/components/content.en-GB.md

@@ -1,10 +1,10 @@
 # Component functions
 
-One of the first things people run into when they start using Bézier curves in their own programs is "I know how to draw the curve, but how do I determine the bounding box?". It's actually reasonably straight forward to do so, but it requires having some knowledge on exploiting math to get the values we need. For bounding boxes, we aren't actually interested in the curve itself, but only in its "extremities": the minimum and maximum values the curve has for its x- and y-axis values. If you remember your calculus (provided you ever took calculus, otherwise it's going to be hard to remember) we can determine function extremities using the first derivative of that function, but this poses a problem, since our function is parametric: every axis has its own function.
+One of the first things people run into when they start using Bézier curves in their own programs is "I know how to draw the curve, but how do I determine the bounding box?". It's actually reasonably straightforward to do so, but it requires having some knowledge on exploiting math to get the values we need. For bounding boxes, we aren't actually interested in the curve itself, but only in its "extremities": the minimum and maximum values the curve has for its x- and y-axis values. If you remember your calculus (provided you ever took calculus, otherwise it's going to be hard to remember) we can determine function extremities using the first derivative of that function, but this poses a problem, since our function is parametric: every axis has its own function.
 
 The solution: compute the derivative for each axis separately, and then fit them back together in the same way we do for the original.
 
-Let's look at how a parametric Bézier curve "splits up" into two normal functions, one for the x-axis and one for the y-axis. Note the left-most figure is again an interactive curve, without labeled axes (you get coordinates in the graph instead).  The center and right-most figures are the component functions for computing the x-axis value, given a value for <i>t</i> (between 0 and 1 inclusive), and the y-axis value, respectively.
+Let's look at how a parametric Bézier curve "splits up" into two normal functions, one for the x-axis and one for the y-axis. Note the leftmost figure is again an interactive curve, without labeled axes (you get coordinates in the graph instead).  The center and rightmost figures are the component functions for computing the x-axis value, given a value for <i>t</i> (between 0 and 1 inclusive), and the y-axis value, respectively.
 
 If you move points in a curve sideways, you should only see the middle graph change; likely, moving points vertically should only show a change in the right graph.
 

components/sections/extremities/content.en-GB.md

@@ -1,6 +1,6 @@
 # Finding extremities: root finding
 
-Now that we understand (well, superficially anyway) the component functions, we can find the extremities of our Bézier curve by finding maxima and minima on the component functions, by solving the equations B'(t) = 0 and B''(t) = 0. Although, in the case of quadratic curves there is no B''(t), so we only need to compute B'(t) = 0. So, how do we compute the first and second derivatives? Fairly easily, actually, until our derivatives are 4th order or higher... then things get really hard. But let's start simple:
+Now that we understand (well, superficially anyway) the component functions, we can find the extremities of our Bézier curve by finding maxima and minima on the component functions, by solving the equations B'(t) = 0 and B''(t) = 0. That said, in the case of quadratic curves there is no B''(t), so we only need to compute B'(t) = 0. So, how do we compute the first and second derivatives? Fairly easily, actually, until our derivatives are 4th order or higher... then things get really hard. But let's start simple:
 
 ### Quadratic curves: linear derivatives.
 
@@ -58,7 +58,7 @@ This gives us thee coefficients *a*, *b*, and *c* that are expressed in terms of
 \end{aligned}
 \]
 
-Easy peasy. We can now almost trivially find the roots by plugging those values into the quadratic formula. We also note that the second derivative of a cubic curve means computing the first derivative of a quadratic curve, and we just saw how to do that in the section above.
+Easy-peasy. We can now almost trivially find the roots by plugging those values into the quadratic formula. We also note that the second derivative of a cubic curve means computing the first derivative of a quadratic curve, and we just saw how to do that in the section above.
 
 ### Quartic curves: Cardano's algorithm.
 
@@ -72,7 +72,7 @@ generic form. So:
   \end{aligned}
 \]
 
-This is easier because for the "easier formula" we can use [regular calculus](http://www.wolframalpha.com/input/?i=t^3+%2B+pt+%2B+q) to find the roots (as a cubic function, however, it can have up to three roots, but two of those can be complex. For the purpose of Bézier curve extremities, we can completely ignore those complex roots, since our *t* is a plain real number from 0 to 1).
+This is easier because for the "easier formula" we can use [regular calculus](http://www.wolframalpha.com/input/?i=t^3+%2B+pt+%2B+q) to find the roots. (As a cubic function, however, it can have up to three roots, but two of those can be complex. For the purpose of Bézier curve extremities, we can completely ignore those complex roots, since our *t* is a plain real number from 0 to 1.)
 
 So, the trick is to figure out how to turn the first formula into the second formula, and to then work out the maths that gives us the roots. This is explained in detail over at [Ken J. Ward's page](https://trans4mind.com/personal_development/mathematics/polynomials/cubicAlgebra.htm) for solving the cubic equation, so instead of showing the maths, I'm simply going to show the programming code for solving the cubic equation, with the complex roots getting totally ignored.
 
@@ -174,9 +174,9 @@ And that's it. The maths is complicated, but the code is pretty much just "follo
 
 The problem with this is that as the order of the curve goes up, we can't actually solve those equations the normal way. We can't take the function, and then work out what the solutions are. Not to mention that even solving a third order derivative (for a fourth order curve) is already a royal pain in the backside. We need a better solution. We need numerical approaches.
 
-That's a fancy word for saying "rather than solve the function, treat the problem as a sequence of identical operations, the performing of which gets us closer and closer to the real answer". As it turns out, there is a really nice numerical root finding algorithm, called the [Newton-Raphson](http://en.wikipedia.org/wiki/Newton-Raphson) root finding method (yes, after *[that](https://en.wikipedia.org/wiki/Isaac_Newton)* Newton), which we can make use of.
+That's a fancy word for saying "rather than solve the function, treat the problem as a sequence of identical operations, the performing of which gets us closer and closer to the real answer". As it turns out, there is a really nice numerical root-finding algorithm, called the [Newton-Raphson](http://en.wikipedia.org/wiki/Newton-Raphson) root finding method (yes, after *[that](https://en.wikipedia.org/wiki/Isaac_Newton)* Newton), which we can make use of.
 
-The Newton-Raphson approach consists of picking a value *t* (any will do), and getting the corresponding value at that *t* value. For normal functions, we can treat that value as a height. If the height is zero, we're done, we have found a root. If it's not, we take the tangent of the curve at that point, and extend it until it passes the x-axis, which will be at some new point *t*. We then repeat the procedure with this new value, and we keep doing this until we find our root.
+The Newton-Raphson approach consists of picking a value *t* (any value will do), and getting the corresponding value of the function at that *t* value. For normal functions, we can treat that value as a height. If the height is zero, we're done, we have found a root. If it's not, we take the tangent of the curve at that point, and extend it until it passes the x-axis, which will be at some new point *t*. We then repeat the procedure with this new value, and we keep doing this until we find our root.
 
 Mathematically, this means that for some *t*, at step *n=1*, we perform the following calculation until *f<sub>y</sub>*(*t*) is zero, so that the next *t* is the same as the one we already have:
 
@@ -184,7 +184,7 @@ Mathematically, this means that for some *t*, at step *n=1*, we perform the foll
   t_{n+1} = t_n - \frac{f_y(t_n)}{f'_y(t_n)}
 \]
 
-(The wikipedia article has a decent animation for this process, so I'm not adding a sketch for that here unless there are requests for it)
+(The Wikipedia article has a decent animation for this process, so I'm not adding a sketch for that here unless there are requests for it)
 
 Now, this works well only if we can pick good starting points, and our curve is continuously differentiable and doesn't have oscillations. Glossing over the exact meaning of those terms, the curves we're dealing with conform to those constraints, so as long as we pick good starting points, this will work. So the question is: which starting points do we pick?