From cc2a505cbad34d5d8c61b22347689094ae523759 Mon Sep 17 00:00:00 2001
From: bhegerle <bhegerle@gmail.com>
Date: Tue, 27 Aug 2019 16:46:01 +0100
Subject: [PATCH] Section on curve order raising/lowering (#211)

* edit to second curve order-raising derivation

* bug fix for curve order-lowering: could invert MtM because it had extra column of zeroes
---
 components/sections/reordering/content.en-GB.md | 6 +++---
 components/sections/reordering/handler.js       | 2 +-
 2 files changed, 4 insertions(+), 4 deletions(-)

diff --git a/components/sections/reordering/content.en-GB.md b/components/sections/reordering/content.en-GB.md
index 271b931e..2849651e 100644
--- a/components/sections/reordering/content.en-GB.md
+++ b/components/sections/reordering/content.en-GB.md
@@ -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:
 
diff --git a/components/sections/reordering/handler.js b/components/sections/reordering/handler.js
index f65e3dba..0160cfd2 100644
--- a/components/sections/reordering/handler.js
+++ b/components/sections/reordering/handler.js
@@ -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 {