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Section on curve order raising/lowering (#211)

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* edit to second curve order-raising derivation

* bug fix for curve order-lowering: could invert MtM because it had extra column of zeroes
This commit is contained in:
bhegerle
2019-08-27 16:46:01 +01:00
committed by Pomax
parent 1fcaa1c1bb
commit cc2a505cba
2 changed files with 4 additions and 4 deletions
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6
components/sections/reordering/content.en-GB.md
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@@ -57,13 +57,13 @@ So by using this seemingly silly trick, we can suddenly express part of our n<su
\[
\begin{aligned}
t B^n_i(t) &= t \frac{n!}{(n-i)!i!} (1-t)^{n-i} t^i \\
&= \frac{i+1}{n+1} \frac{(n+1)!}{(n+1)!(i+1)!} (1-t)^{n-i} t^{i+1} \\
&= \frac{i+1}{k} \frac{k!}{k!(i+1)!} (1-t)^{n-i} t^(i+1), \textit{where } k = n + 1 \\
&= \frac{i+1}{n+1} \frac{(n+1)!}{((n+1)-(i+1))!(i+1)!} (1-t)^{(n+1)-(i+1)} t^{i+1} \\
&= \frac{i+1}{k} \frac{k!}{(k-(i+1))!(i+1)!} (1-t)^{k-(i+1)} t^{i+1}, \textit{where } k = n + 1 \\
&= \frac{i+1}{k} B^k_{i+1}(t)
\end{aligned}
\]
So, with both of those changed from an order `n` expression to an order `(n+1)` expression, we can put them back together again. Now, where the order `n` function had a summation from 0 to `n`, the order `n+1` function uses a summation from 0 to `n+1`, but this shouldn't be a problem as long as we can add some new terms that "contribute nothing". If you read the section on derivatives, you may remember that "higher terms than there is a binomial for" and "lower than zero terms" both "contribute nothing". So as long as we can add terms that have the same form as the terms we need, we can just include them in the summation, they'll sit there and do nothing, and the resulting function stays identical to the lower order curve.
So, with both of those changed from an order `n` expression to an order `(n+1)` expression, we can put them back together again. Now, where the order `n` function had a summation from 0 to `n`, the order `n+1` function uses a summation from 0 to `n+1`, but this shouldn't be a problem as long as we can add some new terms that "contribute nothing". In the next section on derivatives, there is a discussion about why "higher terms than there is a binomial for" and "lower than zero terms" both "contribute nothing". So as long as we can add terms that have the same form as the terms we need, we can just include them in the summation, they'll sit there and do nothing, and the resulting function stays identical to the lower order curve.
Let's do this:

2
components/sections/reordering/handler.js
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@@ -29,7 +29,7 @@ var Reordering = {
// build M, which will be (k) rows by (k-1) columns
for(i=0; i<k; i++) {
M[i] = (new Array(k)).fill(0);
M[i] = (new Array(k - 1)).fill(0);
if(i===0) { M[i][0] = 1; }
else if(i===n) { M[i][i-1] = 1; }
else {