From a4954879d5239ab1d357882804992ae427315ef3 Mon Sep 17 00:00:00 2001
From: Michael Hardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:12:31 +0100
Subject: [PATCH 01/15] First attempt to port Chebyshev documentation.

---
 docs/src/chebyshev.md | 0
 1 file changed, 0 insertions(+), 0 deletions(-)
 create mode 100644 docs/src/chebyshev.md

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
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index 000000000..e69de29bb

From 65c36b0b1df804db89e366185c663e2ed420ba1a Mon Sep 17 00:00:00 2001
From: Michael Hardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:15:54 +0100
Subject: [PATCH 02/15] Add Chebyshev documentation to .md file.

---
 docs/src/chebyshev.md | 251 ++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 251 insertions(+)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index e69de29bb..e5f3b17f2 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -0,0 +1,251 @@
+Chebyshev tranform via Fourier transform
+===============================================
+
+```math
+\begin{equation}
+\end{equation}
+```
+We express a function $f$ as a sum of Chebyshev polynomials
+```math 
+\begin{equation} f(x) = \sum^N_{n=0} a_{n}T_n(x)\label{eq:cheb-expansion}\end{equation}
+```
+The Chebyshev polynomials are defined by 
+```math
+\begin{equation} T_n(\cos \theta) = \cos n \theta, {\rm~with~}x = \cos \theta. \end{equation}
+```
+We can see how to find $\{a_{n}\}$ given $\{f(x_j)\}$ via Fourier transform. 
+The Fourier series representation of $f$ on a uniform grid indexed by $j$ is defined by 
+```math
+\begin{equation} f_j = \sum_{k=0}^{M-1} b_{k}\exp\left[i \frac{2\pi k j}{M}\right].\label{eq:fourier-series}\end{equation}
+```
+
+Gauss-Chebyshev-Lobotto points
+===============================================
+
+We pick points 
+```math
+\begin{equation} x_j = \cos \theta_j, \quad \theta_j = \frac{j \pi}{N} \quad 0 \leq j \leq N.\end{equation}
+```
+Then 
+```math
+\begin{equation} T_n{x_j} = \cos \frac{n j \pi}{N}.\end{equation}
+```
+Assuming that $M = 2N$, with $N$ an integer, and $b_{k} = b_{M-k}$ for $k>0$, we have that 
+```math
+\begin{equation} f_j = b_{0} + b_{N}(-1)^j + \sum_{n=1}^{N-1}
+b_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right).\end{equation}
+```
+Comparing this to the expression for $f(x_j)$ in the Chebyshev representation, 
+using the form of $T_n(x_j)$, 
+```math
+\begin{equation} f_j = a_{0} + a_{N}(-1)^j + \frac{1}{2}\sum_{n=1}^{N-1}
+a_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right),\end{equation}
+```
+we find that the Chebyshev representation on the Chebyshev points is equivalent 
+to the Fourier representation on the uniform grid points, if we identify
+```math
+\begin{equation} b_{0} = a_{0}, \quad  b_{N} = a_{N}, \quad b_{j} = \frac{a_{j}}{2} {\rm~for~} 1 \leq j \leq N-1. \end{equation}
+```
+This fact allows us to carry out the Chebyshev tranform by Fourier transforming the $\{f_j\}$ data
+and carrying out the correct normalisation of the resulting coefficients. 
+
+Gauss-Chebyshev-Radau points
+===============================================
+
+The last subsection dealt with grids which contain both endpoints on the $[-1,1]$ domain. 
+Certain problems require domains which contain a single endpoint, i.e., $x \in (-1,1]$. For 
+these cases we choose the points 
+```math
+\begin{equation} x_j = \cos \theta_j, \quad \theta_j = \frac{2 j \pi}{2 N + 1} \quad 0 \leq j \leq N.\end{equation}
+```
+Writing out the Chebyshev series \eq{eq:cheb-expansion}, 
+we have that 
+```math
+\begin{equation} \begin{split} f(x_j) = & \sum^N_{n=0} a_{n} \cos \frac{2 n j \pi}{2 N + 1} \\ & = a_{0} + \sum^N_{n=1} \frac{a_{n}}{2}\left(\exp\left[i \frac{2\pi n j}{2N +1}\right] + \exp\left[-i \frac{2\pi n j}{2N +1}\right]\right).\end{split} \label{eq:cheb-expansion-radau-points}\end{equation}
+```
+The form of the series \eq{eq:cheb-expansion-radau-points} is identical to the form of 
+a Fourier series on an odd number of points, i.e., taking $M = 2 N + 1$ in equation \eq{eq:fourier-series},
+and assuming $b_{k} = b_{M -k}$ for $k>1$,
+we have that 
+```math
+\begin{equation} f_j = b_{0} + \sum_{k=1}^{N} b_{k}\left(\exp\left[i \frac{2\pi k j}{2N+1}\right] + \exp\left[-i \frac{2\pi k j}{2N+1} \right]\right). \end{equation}
+```
+We can thus take a Chebyshev transform using a Fourier transform on Gauss-Chebyshev-Radau points if we identify 
+```math
+\begin{equation} b_{0} = a_{0}, \quad b_{j} = \frac{a_{j}}{2} {\rm~for~} 1 \leq j \leq N. \end{equation}
+```
+
+Chebyshev coefficients of derivatives of a function
+===============================================
+
+Starting from the expression of $f$ as a sum Chebyshev polynomials, equation \eq{eq:cheb-expansion},
+we can obtain an expression for the derivative
+```math \begin{equation}
+\frac{d f}{d x} = \sum^N_{n=0} a_{n}\frac{d T_{n}}{d x}. \label{eq:derivative-def}\end{equation}
+```
+We note that we must be able to express ${d f}/{d x}$ as a sum 
+of Chebyshev polynomials of up to order $N-1$, i.e.,
+```math
+\begin{equation} \frac{d f}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}. \end{equation}
+```
+We must determine the set $\{d_{n}\}$ in terms of the set $\{a_{n}\}$.
+First, we equate the two expressions to find that 
+```math
+\begin{equation} \sum^N_{k=0} a_{k}\frac{d T_{k}}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}. \label{eq:dn-def}\end{equation}
+```
+We use the Chebyshev polynomials of the second kind $U_n{x}$ to aid us in the calculation of the set $\{d_{n}\}$. 
+These polynomials are defined by 
+```math
+\begin{equation} U_{0}(x) = 1, \quad U_{1}(x) = 2x, \quad U_{n+1} = 2 x U_{n}(x) - U_{n-1}(x).\end{equation}
+```
+Note the useful relations 
+```math
+\begin{equation} \frac{d T_{n}}{d x} = n U_{n-1}, {\rm~for~}n\geq 1, \quad \frac{d T_{0}}{d x} = 0,\end{equation}
+```
+```math
+\begin{equation} T_{n} = \frac{1}{2}\left(U_{n} - U_{n -2}\right), T_{0} = U_{0}\quad, {\rm ~and~} \quad 2 T_{1} = U_{1}. \end{equation}
+```
+Using these identities, which may be obtained from the trigonometric definition of $U_{n}(\cos \theta)$
+```math
+\begin{equation}  U_{n}(\cos \theta) \sin \theta = \sin \left((n+1)\theta\right),\end{equation}
+```
+we find that equation  \eq{eq:dn-def} becomes 
+```math
+\begin{equation} \begin{split}\sum^N_{n=1} a_{n} n U_{n-1}(x) =& \frac{d_{N-1}}{2}U_{N-1}+\frac{d_{N-2}}{2}U_{N-2} 
+\\ & + \sum^{N-3}_{k=1} \frac{d_{k}-d_{k+2}}{2}U_{k} + \left(d_{0} - \frac{d_{2}}{2}\right)U_{0}. \end{split}
+\label{eq:dn-def-U}\end{equation}
+```
+Using the orthogonality relation 
+```math
+\begin{equation} \int^1_{-1} U_{m}(x)U_{n}(x)\sqrt{1-x^2} \; d x = 
+\left\{\begin{tabular}{l} $0 {\rm ~if~} n\neq m $ \\ $\pi/2 {\rm ~if~} n=m$  \end{tabular}\right.,\end{equation}
+```
+we obtain the (unqiuely-determined) relations 
+```math
+\begin{equation} \begin{split} &d_{N-1} = 2Na_{N},\quad d_{N-2} = 2(N-1)a_{N-1}, \\ 
+& d_{k} = 2(k+1) a_{k+1} + d_{k+2}, \quad d_{0} = \frac{d_{2}}{2} + a_{1}.\end{split} \label{eq:dn-result-U}\end{equation}
+```
+Note the lack of a second relation for $d_{0}$, but that otherwise the expressions \eq{eq:dn-result-U}
+agree with those of the last section.       
+
+Clenshaw-Curtis integration weights
+===============================================
+
+We require the integration weights for the set of points $\{x_j\}$ chosen 
+in our numerical scheme. The weights $w_{j}$ are defined implicitly by 
+```math
+\begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{j=0}^N f(x_j) w_{j}. \label{eq:w-sum}\end{equation}
+```
+In the Chebyshev scheme we use the change of variables $x = \cos \theta$
+to write 
+```math
+\begin{equation} \int^{1}_{-1} f(x) \; d x = \int^\pi_0 f(\cos\theta) \sin \theta \; d \theta . \label{eq:change-of-variables-integral} \end{equation}
+```
+ Using the series expansion \eq{eq:cheb-expansion} in equation \eq{eq:change-of-variables-integral}
+ we find that 
+ ```math
+ \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum^N_{n=0} a_{n}\int^\pi_0 \cos (n \theta) \sin \theta \; d \theta
+ . \label{eq:series-integral} \end{equation}
+ ```
+ Note the integral identity
+ ```math
+\begin{equation} \int^\pi_0 \cos(n \theta) \sin \theta \; d \theta = \frac{\cos(n \pi) +1}{1 - n^2} {\rm~for~} n \geq 0.\end{equation}
+```
+ Also note that 
+ ```math
+ \begin{equation} \frac{\cos(n \pi) +1}{1 - n^2} = \left\{\begin{tabular}{l} $0 {\rm ~if~} n = 2 r + 1, ~r \in \mathbb{Z} $ \\ $2/(1 - n^2) {\rm ~if~} n=2r,~r. \in \mathbb{Z}$  \end{tabular}\right. \end{equation}
+ ```
+ We define ```math
+ \begin{equation} \J_{n} = \frac{\cos(n \pi) +1}{1 - n^2}. \end{equation}
+ ```
+ Using this definition, we can write the integral of $f(x)$ can be written 
+ in terms of a sum over of the Chebyshev coefficients:
+ ```math
+ \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{n=0}^N \J_{n} a_{n}. \label{eq:Cheb-sum}\end{equation}
+ ```
+ 
+ To avoid computing the set of coefficients $\{a_{n}\}$ every time we wish to integrate $f(x_j)$,
+ we use the inverse transforms. This transform allows us to rewrite equation \eq{eq:Cheb-sum} in the form \eq{eq:w-sum}.
+ Since the inverse transform differs between the Gauss-Chebyshev-Lobotto and Gauss-Chebyshev-Radau cases, we treat each 
+ case separately below. 
+ 
+Weights on Gauss-Chebyshev-Lobotto points
+===============================================
+  We use the inverse transformation 
+ ```math
+ \begin{equation} a_{n} = \frac{\q_{n}}{2N}\sum^{2N-1}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N}\right], \label{eq:inverse-transform-GCL}\end{equation}
+ ```
+ where 
+ ```math
+ \begin{equation} \q_{n} = \left\{\begin{tabular}{l} $2 {\rm ~if~} n\neq0,N $ \\ $1 {\rm ~if~} n=0,N$  \end{tabular}\right.,\end{equation}
+ ```
+and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
+```math
+\begin{equation} \hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j}){\rm~for~} N+1 \leq j \leq 2N-1. \end{equation}
+```
+ With this inverse tranformation, we can write 
+```math
+\begin{equation} \begin{split}\sum_{n=0}^N \J_{n} a_{n} & =  \sum^{2N-1}_{n=0} \frac{a_{n}\J_{n}}{\q_{n}} \\
+ & = \sum^{2N-1}_{j=0}\sum^{2N-1}_{n=0} \frac{\hat{f}_j \J_{n}}{2N} \exp\left[-i \frac{2\pi n j}{2N}\right] \\ 
+ & = \sum^{2N-1}_{j=0} \hat{f}_j \v_{j} = \sum^{N}_{j=0} \hat{f}_j \q_{j}\v_{j},\end{split}\label{eq:weights-working}\end{equation}
+```
+ where in the first step we have extended the sum from $N$ to $2N-1$ and used FFT-order definitions of $\J_{n}$ and $a_{n}$
+```math
+\begin{equation} \J_{j} = \J_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1,\end{equation}
+```
+```math
+\begin{equation} a_{j} = a_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1.\end{equation}
+```
+In the second step we use the definition of the inverse transform \eq{eq:inverse-transform-GCR}, and 
+in the third step we define 
+```math
+\begin{equation} \v_{j} = \sum_{n=0}^{2N-1}\frac{\J_{n}}{2N}\exp\left[-i \frac{2\pi n j}{2N}\right].\end{equation}
+```
+Finally, we can compare equations \eq{eq:w-sum} and \eq{eq:weights-working} and deduce that 
+```math
+\begin{equation} w_{j} = \q_{j}\v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
+```
+We can write $\v_{j}$ in terms of a discrete cosine transform, i.e.,
+```math
+\begin{equation} \v_{j} = \frac{1}{2N}\left(\J_{0} + (-1)^j\J_{N} + 2\sum_{n=1}^{N-1}\J_{n}\cos\left(\frac{\pi n j}{N}\right)\right).\end{equation}
+```
+ 
+Weights on Gauss-Chebyshev-Radau points
+===============================================
+We use the inverse transformation 
+```math
+\begin{equation} a_{n} = \frac{\q_{n}}{2N+1}\sum^{2N}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N+1}\right], \label{eq:inverse-transform-GCR}\end{equation}
+```
+where 
+```math
+\begin{equation} \q_{n} = \left\{\begin{tabular}{l} $2 {\rm ~if~} n > 0 $ \\ $1 {\rm ~if~} n=0$  \end{tabular}\right.,\end{equation}
+```
+and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
+```math
+\begin{equation} \hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j+1}){\rm~for~} N+1 \leq j \leq 2N. \end{equation}
+```
+Note that the details of what is the appropriate FFT order depends on the order in which the points $x_j$ are stored.
+The key detail in the Chebyshev-Radau scheme is that (in the notation above)
+$x_0 = 1$ is not a repeated point, and must occupy $\hat{f}_0$. 
+With this inverse tranformation, we can write 
+```math
+\begin{equation} \begin{split}\sum_{n=0}^N \J_{n} a_{n} & =  \sum^{2N}_{n=0} \frac{a_{n}\J_{n}}{\q_{n}} \\
+& = \sum^{2N}_{j=0}\sum^{2N}_{n=0} \frac{\hat{f}_j \J_{n}}{2N+1} \exp\left[-i \frac{2\pi n j}{2N+1}\right] \\ 
+& = \sum^{2N}_{j=0} \hat{f}_j \v_{j} = \sum^{N}_{j=0} \hat{f}_j \q_{j}\v_{j},\end{split}\label{eq:weights-working}\end{equation}
+```
+where in the first step we have extended the sum from $N$ to $2N$ and used FFT-order definitions of $\J_{n}$ and $a_{n}$
+```math
+\begin{equation} \J_{j} = \J_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N,\end{equation}
+```
+```math
+\begin{equation} a_{j} = a_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N.\end{equation}
+```
+In the second step we use the definition of the inverse transform \eq{eq:inverse-transform-GCR}, and 
+in the third step we define 
+```math
+\begin{equation} \v_{j} = \sum_{n=0}^{2N}\frac{\J_{n}}{2N+1}\exp\left[-i \frac{2\pi n j}{2N+1}\right].\end{equation}
+```
+Finally, we can compare equations \eq{eq:w-sum} and \eq{eq:weights-working} and deduce that 
+```math
+\begin{equation} w_{j} = \q_{j}\v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
+```
\ No newline at end of file

From 957264f426fa92a6828e78bbbf11bcab2c0e2de1 Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:21:15 +0100
Subject: [PATCH 03/15] Debug chebyshev.md

---
 docs/src/chebyshev.md | 54 +++++++++++++++++++++----------------------
 1 file changed, 26 insertions(+), 28 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index e5f3b17f2..e70c5e339 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -28,7 +28,7 @@ We pick points
 ```
 Then 
 ```math
-\begin{equation} T_n{x_j} = \cos \frac{n j \pi}{N}.\end{equation}
+\begin{equation} T_n(x_j) = \cos \frac{n j \pi}{N}.\end{equation}
 ```
 Assuming that $M = 2N$, with $N$ an integer, and $b_{k} = b_{M-k}$ for $k>0$, we have that 
 ```math
@@ -124,9 +124,7 @@ we obtain the (unqiuely-determined) relations
 ```math
 \begin{equation} \begin{split} &d_{N-1} = 2Na_{N},\quad d_{N-2} = 2(N-1)a_{N-1}, \\ 
 & d_{k} = 2(k+1) a_{k+1} + d_{k+2}, \quad d_{0} = \frac{d_{2}}{2} + a_{1}.\end{split} \label{eq:dn-result-U}\end{equation}
-```
-Note the lack of a second relation for $d_{0}$, but that otherwise the expressions \eq{eq:dn-result-U}
-agree with those of the last section.       
+```       
 
 Clenshaw-Curtis integration weights
 ===============================================
@@ -156,12 +154,12 @@ to write
  \begin{equation} \frac{\cos(n \pi) +1}{1 - n^2} = \left\{\begin{tabular}{l} $0 {\rm ~if~} n = 2 r + 1, ~r \in \mathbb{Z} $ \\ $2/(1 - n^2) {\rm ~if~} n=2r,~r. \in \mathbb{Z}$  \end{tabular}\right. \end{equation}
  ```
  We define ```math
- \begin{equation} \J_{n} = \frac{\cos(n \pi) +1}{1 - n^2}. \end{equation}
+ \begin{equation} J_{n} = \frac{\cos(n \pi) +1}{1 - n^2}. \end{equation}
  ```
  Using this definition, we can write the integral of $f(x)$ can be written 
  in terms of a sum over of the Chebyshev coefficients:
  ```math
- \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{n=0}^N \J_{n} a_{n}. \label{eq:Cheb-sum}\end{equation}
+ \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{n=0}^N J_{n} a_{n}. \label{eq:Cheb-sum}\end{equation}
  ```
  
  To avoid computing the set of coefficients $\{a_{n}\}$ every time we wish to integrate $f(x_j)$,
@@ -173,25 +171,25 @@ Weights on Gauss-Chebyshev-Lobotto points
 ===============================================
   We use the inverse transformation 
  ```math
- \begin{equation} a_{n} = \frac{\q_{n}}{2N}\sum^{2N-1}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N}\right], \label{eq:inverse-transform-GCL}\end{equation}
+ \begin{equation} a_{n} = \frac{q_{n}}{2N}\sum^{2N-1}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N}\right], \label{eq:inverse-transform-GCL}\end{equation}
  ```
  where 
  ```math
- \begin{equation} \q_{n} = \left\{\begin{tabular}{l} $2 {\rm ~if~} n\neq0,N $ \\ $1 {\rm ~if~} n=0,N$  \end{tabular}\right.,\end{equation}
+ \begin{equation} q_{n} = \left\{\begin{tabular}{l} $2 {\rm ~if~} n\neq0,N $ \\ $1 {\rm ~if~} n=0,N$  \end{tabular}\right.,\end{equation}
  ```
 and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math
-\begin{equation} \hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j}){\rm~for~} N+1 \leq j \leq 2N-1. \end{equation}
+\begin{equation} \hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,quad \hat{f}_j = f(x_{2N-j}){\rm~for~} N+1 \leq j \leq 2N-1. \end{equation}
 ```
  With this inverse tranformation, we can write 
 ```math
-\begin{equation} \begin{split}\sum_{n=0}^N \J_{n} a_{n} & =  \sum^{2N-1}_{n=0} \frac{a_{n}\J_{n}}{\q_{n}} \\
- & = \sum^{2N-1}_{j=0}\sum^{2N-1}_{n=0} \frac{\hat{f}_j \J_{n}}{2N} \exp\left[-i \frac{2\pi n j}{2N}\right] \\ 
- & = \sum^{2N-1}_{j=0} \hat{f}_j \v_{j} = \sum^{N}_{j=0} \hat{f}_j \q_{j}\v_{j},\end{split}\label{eq:weights-working}\end{equation}
+\begin{equation} \begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N-1}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
+ & = \sum^{2N-1}_{j=0}\sum^{2N-1}_{n=0} \frac{\hat{f}_j J_{n}}{2N} \exp\left[-i \frac{2\pi n j}{2N}\right] \\ 
+ & = \sum^{2N-1}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working}\end{equation}
 ```
- where in the first step we have extended the sum from $N$ to $2N-1$ and used FFT-order definitions of $\J_{n}$ and $a_{n}$
+ where in the first step we have extended the sum from $N$ to $2N-1$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
 ```math
-\begin{equation} \J_{j} = \J_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1,\end{equation}
+\begin{equation} J_{j} = J_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1,\end{equation}
 ```
 ```math
 \begin{equation} a_{j} = a_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1.\end{equation}
@@ -199,26 +197,26 @@ and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 In the second step we use the definition of the inverse transform \eq{eq:inverse-transform-GCR}, and 
 in the third step we define 
 ```math
-\begin{equation} \v_{j} = \sum_{n=0}^{2N-1}\frac{\J_{n}}{2N}\exp\left[-i \frac{2\pi n j}{2N}\right].\end{equation}
+\begin{equation} v_{j} = \sum_{n=0}^{2N-1}\frac{J_{n}}{2N}\exp\left[-i \frac{2\pi n j}{2N}\right].\end{equation}
 ```
 Finally, we can compare equations \eq{eq:w-sum} and \eq{eq:weights-working} and deduce that 
 ```math
-\begin{equation} w_{j} = \q_{j}\v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
+\begin{equation} w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
 ```
-We can write $\v_{j}$ in terms of a discrete cosine transform, i.e.,
+We can write $v_{j}$ in terms of a discrete cosine transform, i.e.,
 ```math
-\begin{equation} \v_{j} = \frac{1}{2N}\left(\J_{0} + (-1)^j\J_{N} + 2\sum_{n=1}^{N-1}\J_{n}\cos\left(\frac{\pi n j}{N}\right)\right).\end{equation}
+\begin{equation} v_{j} = \frac{1}{2N}\left(J_{0} + (-1)^jJ_{N} + 2\sum_{n=1}^{N-1}J_{n}\cos\left(\frac{\pi n j}{N}\right)\right).\end{equation}
 ```
  
 Weights on Gauss-Chebyshev-Radau points
 ===============================================
 We use the inverse transformation 
 ```math
-\begin{equation} a_{n} = \frac{\q_{n}}{2N+1}\sum^{2N}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N+1}\right], \label{eq:inverse-transform-GCR}\end{equation}
+\begin{equation} a_{n} = \frac{q_{n}}{2N+1}\sum^{2N}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N+1}\right], \label{eq:inverse-transform-GCR}\end{equation}
 ```
 where 
 ```math
-\begin{equation} \q_{n} = \left\{\begin{tabular}{l} $2 {\rm ~if~} n > 0 $ \\ $1 {\rm ~if~} n=0$  \end{tabular}\right.,\end{equation}
+\begin{equation} q_{n} = \left\{\begin{tabular}{l} $2 {\rm ~if~} n > 0 $ \\ $1 {\rm ~if~} n=0$  \end{tabular}\right.,\end{equation}
 ```
 and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math
@@ -229,13 +227,13 @@ The key detail in the Chebyshev-Radau scheme is that (in the notation above)
 $x_0 = 1$ is not a repeated point, and must occupy $\hat{f}_0$. 
 With this inverse tranformation, we can write 
 ```math
-\begin{equation} \begin{split}\sum_{n=0}^N \J_{n} a_{n} & =  \sum^{2N}_{n=0} \frac{a_{n}\J_{n}}{\q_{n}} \\
-& = \sum^{2N}_{j=0}\sum^{2N}_{n=0} \frac{\hat{f}_j \J_{n}}{2N+1} \exp\left[-i \frac{2\pi n j}{2N+1}\right] \\ 
-& = \sum^{2N}_{j=0} \hat{f}_j \v_{j} = \sum^{N}_{j=0} \hat{f}_j \q_{j}\v_{j},\end{split}\label{eq:weights-working}\end{equation}
+\begin{equation} \begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
+& = \sum^{2N}_{j=0}\sum^{2N}_{n=0} \frac{\hat{f}_j J_{n}}{2N+1} \exp\left[-i \frac{2\pi n j}{2N+1}\right] \\ 
+& = \sum^{2N}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working}\end{equation}
 ```
-where in the first step we have extended the sum from $N$ to $2N$ and used FFT-order definitions of $\J_{n}$ and $a_{n}$
+where in the first step we have extended the sum from $N$ to $2N$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
 ```math
-\begin{equation} \J_{j} = \J_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N,\end{equation}
+\begin{equation} J_{j} = J_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N,\end{equation}
 ```
 ```math
 \begin{equation} a_{j} = a_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N.\end{equation}
@@ -243,9 +241,9 @@ where in the first step we have extended the sum from $N$ to $2N$ and used FFT-o
 In the second step we use the definition of the inverse transform \eq{eq:inverse-transform-GCR}, and 
 in the third step we define 
 ```math
-\begin{equation} \v_{j} = \sum_{n=0}^{2N}\frac{\J_{n}}{2N+1}\exp\left[-i \frac{2\pi n j}{2N+1}\right].\end{equation}
+\begin{equation} v_{j} = \sum_{n=0}^{2N}\frac{J_{n}}{2N+1}\exp\left[-i \frac{2\pi n j}{2N+1}\right].\end{equation}
 ```
 Finally, we can compare equations \eq{eq:w-sum} and \eq{eq:weights-working} and deduce that 
 ```math
-\begin{equation} w_{j} = \q_{j}\v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
-```
\ No newline at end of file
+\begin{equation} w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
+```

From 5e69ba5b66ffcaec6c1711d30499a48c85fdedf6 Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:27:22 +0100
Subject: [PATCH 04/15] Debug chebyshev.md

---
 docs/src/chebyshev.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index e70c5e339..61285204d 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -80,8 +80,8 @@ Chebyshev coefficients of derivatives of a function
 
 Starting from the expression of $f$ as a sum Chebyshev polynomials, equation \eq{eq:cheb-expansion},
 we can obtain an expression for the derivative
-```math \begin{equation}
-\frac{d f}{d x} = \sum^N_{n=0} a_{n}\frac{d T_{n}}{d x}. \label{eq:derivative-def}\end{equation}
+```math
+\begin{equation} \frac{d f}{d x} = \sum^N_{n=0} a_{n}\frac{d T_{n}}{d x}. \label{eq:derivative-def}\end{equation}
 ```
 We note that we must be able to express ${d f}/{d x}$ as a sum 
 of Chebyshev polynomials of up to order $N-1$, i.e.,

From 2cf21670ab257e457e4da052264c1f64a3af7073 Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:29:32 +0100
Subject: [PATCH 05/15] Debug chebyshev.md

---
 docs/src/chebyshev.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index 61285204d..54c19aea0 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -229,7 +229,7 @@ With this inverse tranformation, we can write
 ```math
 \begin{equation} \begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
 & = \sum^{2N}_{j=0}\sum^{2N}_{n=0} \frac{\hat{f}_j J_{n}}{2N+1} \exp\left[-i \frac{2\pi n j}{2N+1}\right] \\ 
-& = \sum^{2N}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working}\end{equation}
+& = \sum^{2N}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working-radau}\end{equation}
 ```
 where in the first step we have extended the sum from $N$ to $2N$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
 ```math
@@ -243,7 +243,7 @@ in the third step we define
 ```math
 \begin{equation} v_{j} = \sum_{n=0}^{2N}\frac{J_{n}}{2N+1}\exp\left[-i \frac{2\pi n j}{2N+1}\right].\end{equation}
 ```
-Finally, we can compare equations \eq{eq:w-sum} and \eq{eq:weights-working} and deduce that 
+Finally, we can compare equations \eq{eq:w-sum} and \eq{eq:weights-working-radau} and deduce that 
 ```math
 \begin{equation} w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
 ```

From f12082c7e79a636694fe12f4637245836035b9f6 Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:33:06 +0100
Subject: [PATCH 06/15] Debug chebyshev.md

---
 docs/src/chebyshev.md | 7 ++++---
 1 file changed, 4 insertions(+), 3 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index 54c19aea0..d2f5caf66 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -150,10 +150,11 @@ to write
 \begin{equation} \int^\pi_0 \cos(n \theta) \sin \theta \; d \theta = \frac{\cos(n \pi) +1}{1 - n^2} {\rm~for~} n \geq 0.\end{equation}
 ```
  Also note that 
- ```math
- \begin{equation} \frac{\cos(n \pi) +1}{1 - n^2} = \left\{\begin{tabular}{l} $0 {\rm ~if~} n = 2 r + 1, ~r \in \mathbb{Z} $ \\ $2/(1 - n^2) {\rm ~if~} n=2r,~r. \in \mathbb{Z}$  \end{tabular}\right. \end{equation}
+```math
+\begin{equation} \frac{\cos(n \pi) +1}{1 - n^2} = \left\{\begin{tabular}{l} $0 {\rm ~if~} n = 2 r + 1, ~r \in \mathbb{Z} $ \\ $2/(1 - n^2) {\rm ~if~} n=2r,~r. \in \mathbb{Z}$  \end{tabular}\right. \end{equation}
  ```
- We define ```math
+ We define 
+ ```math
  \begin{equation} J_{n} = \frac{\cos(n \pi) +1}{1 - n^2}. \end{equation}
  ```
  Using this definition, we can write the integral of $f(x)$ can be written 

From 616be7a39f0fbcec46d66dc4c2f5b82ae399da1a Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:33:58 +0100
Subject: [PATCH 07/15] Debug chebyshev.md

---
 docs/src/chebyshev.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index d2f5caf66..f14e013b1 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -180,7 +180,7 @@ Weights on Gauss-Chebyshev-Lobotto points
  ```
 and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math
-\begin{equation} \hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,quad \hat{f}_j = f(x_{2N-j}){\rm~for~} N+1 \leq j \leq 2N-1. \end{equation}
+\begin{equation} \hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j}){\rm~for~} N+1 \leq j \leq 2N-1. \end{equation}
 ```
  With this inverse tranformation, we can write 
 ```math

From d379912cb9e627b9a1684be9d2138be7470addf8 Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:46:42 +0100
Subject: [PATCH 08/15] Debug chebyshev.md

---
 docs/src/chebyshev.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index f14e013b1..721960a89 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -118,7 +118,7 @@ we find that equation  \eq{eq:dn-def} becomes
 Using the orthogonality relation 
 ```math
 \begin{equation} \int^1_{-1} U_{m}(x)U_{n}(x)\sqrt{1-x^2} \; d x = 
-\left\{\begin{tabular}{l} $0 {\rm ~if~} n\neq m $ \\ $\pi/2 {\rm ~if~} n=m$  \end{tabular}\right.,\end{equation}
+\left\{\begin{array}{l} 0 {\rm ~if~} n\neq m  \\ \pi/2 {\rm ~if~} n=m \\ \end{array} \right.,\end{equation}
 ```
 we obtain the (unqiuely-determined) relations 
 ```math

From 1893d93ed437bd7fe2392e0e25756973846c1f11 Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:48:43 +0100
Subject: [PATCH 09/15] Debug chebyshev.md

---
 docs/src/chebyshev.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index 721960a89..76c76cbf1 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -118,7 +118,7 @@ we find that equation  \eq{eq:dn-def} becomes
 Using the orthogonality relation 
 ```math
 \begin{equation} \int^1_{-1} U_{m}(x)U_{n}(x)\sqrt{1-x^2} \; d x = 
-\left\{\begin{array}{l} 0 {\rm ~if~} n\neq m  \\ \pi/2 {\rm ~if~} n=m \\ \end{array} \right.,\end{equation}
+\left\{\begin{array}{l} 0 {\rm ~if~} n\neq m  \\ \pi/2 {\rm ~if~} n=m \\ \end{array} \right.\end{equation}
 ```
 we obtain the (unqiuely-determined) relations 
 ```math
@@ -151,7 +151,7 @@ to write
 ```
  Also note that 
 ```math
-\begin{equation} \frac{\cos(n \pi) +1}{1 - n^2} = \left\{\begin{tabular}{l} $0 {\rm ~if~} n = 2 r + 1, ~r \in \mathbb{Z} $ \\ $2/(1 - n^2) {\rm ~if~} n=2r,~r. \in \mathbb{Z}$  \end{tabular}\right. \end{equation}
+\begin{equation} \frac{\cos(n \pi) +1}{1 - n^2} = \left\{\begin{array}{l} 0 {\rm ~if~} n = 2 r + 1, ~r \in \mathbb{Z}  \\ 2/(1 - n^2) {\rm ~if~} n=2r,~r. \in \mathbb{Z}  \end{array}\right. \end{equation}
  ```
  We define 
  ```math
@@ -176,7 +176,7 @@ Weights on Gauss-Chebyshev-Lobotto points
  ```
  where 
  ```math
- \begin{equation} q_{n} = \left\{\begin{tabular}{l} $2 {\rm ~if~} n\neq0,N $ \\ $1 {\rm ~if~} n=0,N$  \end{tabular}\right.,\end{equation}
+ \begin{equation} q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n\neq0,N  \\ 1 {\rm ~if~} n=0,N  \end{array}\right.,\end{equation}
  ```
 and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math
@@ -217,7 +217,7 @@ We use the inverse transformation
 ```
 where 
 ```math
-\begin{equation} q_{n} = \left\{\begin{tabular}{l} $2 {\rm ~if~} n > 0 $ \\ $1 {\rm ~if~} n=0$  \end{tabular}\right.,\end{equation}
+\begin{equation} q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n > 0  \\ 1 {\rm ~if~} n=0  \end{array}\right.,\end{equation}
 ```
 and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math

From 9a0a65172db48da996ef7ce4c1256a9c1dbeb1eb Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:49:54 +0100
Subject: [PATCH 10/15] Debug chebyshev.md

---
 docs/src/chebyshev.md | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index 76c76cbf1..8ff0a5de0 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -176,7 +176,7 @@ Weights on Gauss-Chebyshev-Lobotto points
  ```
  where 
  ```math
- \begin{equation} q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n\neq0,N  \\ 1 {\rm ~if~} n=0,N  \end{array}\right.,\end{equation}
+ \begin{equation} q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n\neq0,N  \\ 1 {\rm ~if~} n=0,N  \end{array}\right.\end{equation}
  ```
 and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math
@@ -217,7 +217,7 @@ We use the inverse transformation
 ```
 where 
 ```math
-\begin{equation} q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n > 0  \\ 1 {\rm ~if~} n=0  \end{array}\right.,\end{equation}
+\begin{equation} q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n > 0  \\ 1 {\rm ~if~} n=0  \end{array}\right.\end{equation}
 ```
 and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math

From c4c17636a19258a345b1eb2b154275283638b2ee Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 10:59:22 +0100
Subject: [PATCH 11/15] Update chebyshev.md to have a manual label.

---
 docs/src/chebyshev.md | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index 8ff0a5de0..20f772448 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -7,7 +7,7 @@ Chebyshev tranform via Fourier transform
 ```
 We express a function $f$ as a sum of Chebyshev polynomials
 ```math 
-\begin{equation} f(x) = \sum^N_{n=0} a_{n}T_n(x)\label{eq:cheb-expansion}\end{equation}
+\begin{equation} f(x) = \sum^N_{n=0} a_{n}T_n(x)\label{eq:cheb-expansion} \tag{1}\end{equation}
 ```
 The Chebyshev polynomials are defined by 
 ```math

From 37e49bdca3bd97c4e990a2bfb15b158468946306 Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 11:04:44 +0100
Subject: [PATCH 12/15] Update chebyshev.md to have manual labels.

---
 docs/src/chebyshev.md | 28 ++++++++++++++--------------
 1 file changed, 14 insertions(+), 14 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index 20f772448..8d9f42173 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -16,7 +16,7 @@ The Chebyshev polynomials are defined by
 We can see how to find $\{a_{n}\}$ given $\{f(x_j)\}$ via Fourier transform. 
 The Fourier series representation of $f$ on a uniform grid indexed by $j$ is defined by 
 ```math
-\begin{equation} f_j = \sum_{k=0}^{M-1} b_{k}\exp\left[i \frac{2\pi k j}{M}\right].\label{eq:fourier-series}\end{equation}
+\begin{equation} f_j = \sum_{k=0}^{M-1} b_{k}\exp\left[i \frac{2\pi k j}{M}\right].\label{eq:fourier-series}\tag{2}\end{equation}
 ```
 
 Gauss-Chebyshev-Lobotto points
@@ -61,7 +61,7 @@ these cases we choose the points
 Writing out the Chebyshev series \eq{eq:cheb-expansion}, 
 we have that 
 ```math
-\begin{equation} \begin{split} f(x_j) = & \sum^N_{n=0} a_{n} \cos \frac{2 n j \pi}{2 N + 1} \\ & = a_{0} + \sum^N_{n=1} \frac{a_{n}}{2}\left(\exp\left[i \frac{2\pi n j}{2N +1}\right] + \exp\left[-i \frac{2\pi n j}{2N +1}\right]\right).\end{split} \label{eq:cheb-expansion-radau-points}\end{equation}
+\begin{equation} \begin{split} f(x_j) = & \sum^N_{n=0} a_{n} \cos \frac{2 n j \pi}{2 N + 1} \\ & = a_{0} + \sum^N_{n=1} \frac{a_{n}}{2}\left(\exp\left[i \frac{2\pi n j}{2N +1}\right] + \exp\left[-i \frac{2\pi n j}{2N +1}\right]\right).\end{split} \label{eq:cheb-expansion-radau-points}\tag{3}\end{equation}
 ```
 The form of the series \eq{eq:cheb-expansion-radau-points} is identical to the form of 
 a Fourier series on an odd number of points, i.e., taking $M = 2 N + 1$ in equation \eq{eq:fourier-series},
@@ -81,7 +81,7 @@ Chebyshev coefficients of derivatives of a function
 Starting from the expression of $f$ as a sum Chebyshev polynomials, equation \eq{eq:cheb-expansion},
 we can obtain an expression for the derivative
 ```math
-\begin{equation} \frac{d f}{d x} = \sum^N_{n=0} a_{n}\frac{d T_{n}}{d x}. \label{eq:derivative-def}\end{equation}
+\begin{equation} \frac{d f}{d x} = \sum^N_{n=0} a_{n}\frac{d T_{n}}{d x}. \label{eq:derivative-def}\tag{4}\end{equation}
 ```
 We note that we must be able to express ${d f}/{d x}$ as a sum 
 of Chebyshev polynomials of up to order $N-1$, i.e.,
@@ -91,7 +91,7 @@ of Chebyshev polynomials of up to order $N-1$, i.e.,
 We must determine the set $\{d_{n}\}$ in terms of the set $\{a_{n}\}$.
 First, we equate the two expressions to find that 
 ```math
-\begin{equation} \sum^N_{k=0} a_{k}\frac{d T_{k}}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}. \label{eq:dn-def}\end{equation}
+\begin{equation} \sum^N_{k=0} a_{k}\frac{d T_{k}}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}. \label{eq:dn-def}\tag{5}\end{equation}
 ```
 We use the Chebyshev polynomials of the second kind $U_n{x}$ to aid us in the calculation of the set $\{d_{n}\}$. 
 These polynomials are defined by 
@@ -113,7 +113,7 @@ we find that equation  \eq{eq:dn-def} becomes
 ```math
 \begin{equation} \begin{split}\sum^N_{n=1} a_{n} n U_{n-1}(x) =& \frac{d_{N-1}}{2}U_{N-1}+\frac{d_{N-2}}{2}U_{N-2} 
 \\ & + \sum^{N-3}_{k=1} \frac{d_{k}-d_{k+2}}{2}U_{k} + \left(d_{0} - \frac{d_{2}}{2}\right)U_{0}. \end{split}
-\label{eq:dn-def-U}\end{equation}
+\label{eq:dn-def-U}\tag{6}\end{equation}
 ```
 Using the orthogonality relation 
 ```math
@@ -123,7 +123,7 @@ Using the orthogonality relation
 we obtain the (unqiuely-determined) relations 
 ```math
 \begin{equation} \begin{split} &d_{N-1} = 2Na_{N},\quad d_{N-2} = 2(N-1)a_{N-1}, \\ 
-& d_{k} = 2(k+1) a_{k+1} + d_{k+2}, \quad d_{0} = \frac{d_{2}}{2} + a_{1}.\end{split} \label{eq:dn-result-U}\end{equation}
+& d_{k} = 2(k+1) a_{k+1} + d_{k+2}, \quad d_{0} = \frac{d_{2}}{2} + a_{1}.\end{split} \label{eq:dn-result-U}\tag{7}\end{equation}
 ```       
 
 Clenshaw-Curtis integration weights
@@ -132,18 +132,18 @@ Clenshaw-Curtis integration weights
 We require the integration weights for the set of points $\{x_j\}$ chosen 
 in our numerical scheme. The weights $w_{j}$ are defined implicitly by 
 ```math
-\begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{j=0}^N f(x_j) w_{j}. \label{eq:w-sum}\end{equation}
+\begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{j=0}^N f(x_j) w_{j}. \label{eq:w-sum}\tag{8}\end{equation}
 ```
 In the Chebyshev scheme we use the change of variables $x = \cos \theta$
 to write 
 ```math
-\begin{equation} \int^{1}_{-1} f(x) \; d x = \int^\pi_0 f(\cos\theta) \sin \theta \; d \theta . \label{eq:change-of-variables-integral} \end{equation}
+\begin{equation} \int^{1}_{-1} f(x) \; d x = \int^\pi_0 f(\cos\theta) \sin \theta \; d \theta . \label{eq:change-of-variables-integral} \tag{9}\end{equation}
 ```
  Using the series expansion \eq{eq:cheb-expansion} in equation \eq{eq:change-of-variables-integral}
  we find that 
  ```math
  \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum^N_{n=0} a_{n}\int^\pi_0 \cos (n \theta) \sin \theta \; d \theta
- . \label{eq:series-integral} \end{equation}
+ . \label{eq:series-integral}\tag{10} \end{equation}
  ```
  Note the integral identity
  ```math
@@ -160,7 +160,7 @@ to write
  Using this definition, we can write the integral of $f(x)$ can be written 
  in terms of a sum over of the Chebyshev coefficients:
  ```math
- \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{n=0}^N J_{n} a_{n}. \label{eq:Cheb-sum}\end{equation}
+ \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{n=0}^N J_{n} a_{n}. \label{eq:Cheb-sum}\tag{11}\end{equation}
  ```
  
  To avoid computing the set of coefficients $\{a_{n}\}$ every time we wish to integrate $f(x_j)$,
@@ -172,7 +172,7 @@ Weights on Gauss-Chebyshev-Lobotto points
 ===============================================
   We use the inverse transformation 
  ```math
- \begin{equation} a_{n} = \frac{q_{n}}{2N}\sum^{2N-1}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N}\right], \label{eq:inverse-transform-GCL}\end{equation}
+ \begin{equation} a_{n} = \frac{q_{n}}{2N}\sum^{2N-1}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N}\right], \label{eq:inverse-transform-GCL}\tag{12}\end{equation}
  ```
  where 
  ```math
@@ -186,7 +186,7 @@ and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math
 \begin{equation} \begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N-1}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
  & = \sum^{2N-1}_{j=0}\sum^{2N-1}_{n=0} \frac{\hat{f}_j J_{n}}{2N} \exp\left[-i \frac{2\pi n j}{2N}\right] \\ 
- & = \sum^{2N-1}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working}\end{equation}
+ & = \sum^{2N-1}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working}\tag{13}\end{equation}
 ```
  where in the first step we have extended the sum from $N$ to $2N-1$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
 ```math
@@ -213,7 +213,7 @@ Weights on Gauss-Chebyshev-Radau points
 ===============================================
 We use the inverse transformation 
 ```math
-\begin{equation} a_{n} = \frac{q_{n}}{2N+1}\sum^{2N}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N+1}\right], \label{eq:inverse-transform-GCR}\end{equation}
+\begin{equation} a_{n} = \frac{q_{n}}{2N+1}\sum^{2N}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N+1}\right], \label{eq:inverse-transform-GCR}\tag{14}\end{equation}
 ```
 where 
 ```math
@@ -230,7 +230,7 @@ With this inverse tranformation, we can write
 ```math
 \begin{equation} \begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
 & = \sum^{2N}_{j=0}\sum^{2N}_{n=0} \frac{\hat{f}_j J_{n}}{2N+1} \exp\left[-i \frac{2\pi n j}{2N+1}\right] \\ 
-& = \sum^{2N}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working-radau}\end{equation}
+& = \sum^{2N}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working-radau}\tag{15}\end{equation}
 ```
 where in the first step we have extended the sum from $N$ to $2N$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
 ```math

From a721ba7e7df33c5306dd336c2577c754915c602b Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 11:13:05 +0100
Subject: [PATCH 13/15] Update chebyshev.md to reference manual labels.

---
 docs/src/chebyshev.md | 22 +++++++++++-----------
 1 file changed, 11 insertions(+), 11 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index 8d9f42173..d7da3da0a 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -58,13 +58,13 @@ these cases we choose the points
 ```math
 \begin{equation} x_j = \cos \theta_j, \quad \theta_j = \frac{2 j \pi}{2 N + 1} \quad 0 \leq j \leq N.\end{equation}
 ```
-Writing out the Chebyshev series \eq{eq:cheb-expansion}, 
+Writing out the Chebyshev series (1), <!-- \eq{eq:cheb-expansion}, -->
 we have that 
 ```math
 \begin{equation} \begin{split} f(x_j) = & \sum^N_{n=0} a_{n} \cos \frac{2 n j \pi}{2 N + 1} \\ & = a_{0} + \sum^N_{n=1} \frac{a_{n}}{2}\left(\exp\left[i \frac{2\pi n j}{2N +1}\right] + \exp\left[-i \frac{2\pi n j}{2N +1}\right]\right).\end{split} \label{eq:cheb-expansion-radau-points}\tag{3}\end{equation}
 ```
-The form of the series \eq{eq:cheb-expansion-radau-points} is identical to the form of 
-a Fourier series on an odd number of points, i.e., taking $M = 2 N + 1$ in equation \eq{eq:fourier-series},
+The form of the series (3) <!--\eq{eq:cheb-expansion-radau-points}--> is identical to the form of 
+a Fourier series on an odd number of points, i.e., taking $M = 2 N + 1$ in equation (2) <--\eq{eq:fourier-series}-->,
 and assuming $b_{k} = b_{M -k}$ for $k>1$,
 we have that 
 ```math
@@ -78,7 +78,7 @@ We can thus take a Chebyshev transform using a Fourier transform on Gauss-Chebys
 Chebyshev coefficients of derivatives of a function
 ===============================================
 
-Starting from the expression of $f$ as a sum Chebyshev polynomials, equation \eq{eq:cheb-expansion},
+Starting from the expression of $f$ as a sum Chebyshev polynomials, equation (1) <!--\eq{eq:cheb-expansion}-->,
 we can obtain an expression for the derivative
 ```math
 \begin{equation} \frac{d f}{d x} = \sum^N_{n=0} a_{n}\frac{d T_{n}}{d x}. \label{eq:derivative-def}\tag{4}\end{equation}
@@ -109,7 +109,7 @@ Using these identities, which may be obtained from the trigonometric definition
 ```math
 \begin{equation}  U_{n}(\cos \theta) \sin \theta = \sin \left((n+1)\theta\right),\end{equation}
 ```
-we find that equation  \eq{eq:dn-def} becomes 
+we find that equation (5) <!--\eq{eq:dn-def}--> becomes 
 ```math
 \begin{equation} \begin{split}\sum^N_{n=1} a_{n} n U_{n-1}(x) =& \frac{d_{N-1}}{2}U_{N-1}+\frac{d_{N-2}}{2}U_{N-2} 
 \\ & + \sum^{N-3}_{k=1} \frac{d_{k}-d_{k+2}}{2}U_{k} + \left(d_{0} - \frac{d_{2}}{2}\right)U_{0}. \end{split}
@@ -139,7 +139,7 @@ to write
 ```math
 \begin{equation} \int^{1}_{-1} f(x) \; d x = \int^\pi_0 f(\cos\theta) \sin \theta \; d \theta . \label{eq:change-of-variables-integral} \tag{9}\end{equation}
 ```
- Using the series expansion \eq{eq:cheb-expansion} in equation \eq{eq:change-of-variables-integral}
+ Using the series expansion (1) <!--\eq{eq:cheb-expansion}--> in equation (9) <!--\eq{eq:change-of-variables-integral}-->
  we find that 
  ```math
  \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum^N_{n=0} a_{n}\int^\pi_0 \cos (n \theta) \sin \theta \; d \theta
@@ -164,7 +164,7 @@ to write
  ```
  
  To avoid computing the set of coefficients $\{a_{n}\}$ every time we wish to integrate $f(x_j)$,
- we use the inverse transforms. This transform allows us to rewrite equation \eq{eq:Cheb-sum} in the form \eq{eq:w-sum}.
+ we use the inverse transforms. This transform allows us to rewrite equation (11) <!--\eq{eq:Cheb-sum}--> in the form (8) <!--\eq{eq:w-sum}-->.
  Since the inverse transform differs between the Gauss-Chebyshev-Lobotto and Gauss-Chebyshev-Radau cases, we treat each 
  case separately below. 
  
@@ -195,12 +195,12 @@ and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math
 \begin{equation} a_{j} = a_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1.\end{equation}
 ```
-In the second step we use the definition of the inverse transform \eq{eq:inverse-transform-GCR}, and 
+In the second step we use the definition of the inverse transform (14) <!--\eq{eq:inverse-transform-GCR}-->, and 
 in the third step we define 
 ```math
 \begin{equation} v_{j} = \sum_{n=0}^{2N-1}\frac{J_{n}}{2N}\exp\left[-i \frac{2\pi n j}{2N}\right].\end{equation}
 ```
-Finally, we can compare equations \eq{eq:w-sum} and \eq{eq:weights-working} and deduce that 
+Finally, we can compare equations (8) <!--\eq{eq:w-sum}--> and (13) <!--\eq{eq:weights-working}--> and deduce that 
 ```math
 \begin{equation} w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
 ```
@@ -239,12 +239,12 @@ where in the first step we have extended the sum from $N$ to $2N$ and used FFT-o
 ```math
 \begin{equation} a_{j} = a_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N.\end{equation}
 ```
-In the second step we use the definition of the inverse transform \eq{eq:inverse-transform-GCR}, and 
+In the second step we use the definition of the inverse transform (14) <!--\eq{eq:inverse-transform-GCR}-->, and 
 in the third step we define 
 ```math
 \begin{equation} v_{j} = \sum_{n=0}^{2N}\frac{J_{n}}{2N+1}\exp\left[-i \frac{2\pi n j}{2N+1}\right].\end{equation}
 ```
-Finally, we can compare equations \eq{eq:w-sum} and \eq{eq:weights-working-radau} and deduce that 
+Finally, we can compare equations (8) <!--\eq{eq:w-sum}--> and (15) <!--\eq{eq:weights-working-radau}--> and deduce that 
 ```math
 \begin{equation} w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
 ```

From b2ccf62fc579a1b01108e8c1d056a0f74189e1d6 Mon Sep 17 00:00:00 2001
From: mrhardman <29800382+mrhardman@users.noreply.github.com>
Date: Mon, 22 Jul 2024 11:15:24 +0100
Subject: [PATCH 14/15] Update chebyshev.md

---
 docs/src/chebyshev.md | 3 +--
 1 file changed, 1 insertion(+), 2 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index d7da3da0a..908daaf09 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -35,8 +35,7 @@ Assuming that $M = 2N$, with $N$ an integer, and $b_{k} = b_{M-k}$ for $k>0$, we
 \begin{equation} f_j = b_{0} + b_{N}(-1)^j + \sum_{n=1}^{N-1}
 b_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right).\end{equation}
 ```
-Comparing this to the expression for $f(x_j)$ in the Chebyshev representation, 
-using the form of $T_n(x_j)$, 
+Comparing this to the expression for $f(x_j)$ in the Chebyshev representation,
 ```math
 \begin{equation} f_j = a_{0} + a_{N}(-1)^j + \frac{1}{2}\sum_{n=1}^{N-1}
 a_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right),\end{equation}

From aa365d55f3155b5e21c358b7cf26f2ed26f4c913 Mon Sep 17 00:00:00 2001
From: Michael Hardman <29800382+mrhardman@users.noreply.github.com>
Date: Sat, 26 Oct 2024 20:12:35 +0100
Subject: [PATCH 15/15] Correct formatting of chebyshev.md so that
 Documenter.jl can correctly render the document.

---
 docs/src/chebyshev.md | 285 +++++++++++++++++++++++++++---------------
 1 file changed, 186 insertions(+), 99 deletions(-)

diff --git a/docs/src/chebyshev.md b/docs/src/chebyshev.md
index 908daaf09..c4d38f8ae 100644
--- a/docs/src/chebyshev.md
+++ b/docs/src/chebyshev.md
@@ -1,22 +1,24 @@
 Chebyshev tranform via Fourier transform
 ===============================================
 
-```math
-\begin{equation}
-\end{equation}
-```
 We express a function $f$ as a sum of Chebyshev polynomials
 ```math 
-\begin{equation} f(x) = \sum^N_{n=0} a_{n}T_n(x)\label{eq:cheb-expansion} \tag{1}\end{equation}
+\begin{equation}
+f(x) = \sum^N_{n=0} a_{n}T_n(x) \tag{1}
+\end{equation}
 ```
 The Chebyshev polynomials are defined by 
 ```math
-\begin{equation} T_n(\cos \theta) = \cos n \theta, {\rm~with~}x = \cos \theta. \end{equation}
+\begin{equation}
+T_n(\cos \theta) = \cos n \theta, {\rm~with~}x = \cos \theta.
+\end{equation}
 ```
 We can see how to find $\{a_{n}\}$ given $\{f(x_j)\}$ via Fourier transform. 
 The Fourier series representation of $f$ on a uniform grid indexed by $j$ is defined by 
 ```math
-\begin{equation} f_j = \sum_{k=0}^{M-1} b_{k}\exp\left[i \frac{2\pi k j}{M}\right].\label{eq:fourier-series}\tag{2}\end{equation}
+\begin{equation}
+f_j = \sum_{k=0}^{M-1} b_{k}\exp\left[i \frac{2\pi k j}{M}\right].\tag{2}
+\end{equation}
 ```
 
 Gauss-Chebyshev-Lobotto points
@@ -24,26 +26,36 @@ Gauss-Chebyshev-Lobotto points
 
 We pick points 
 ```math
-\begin{equation} x_j = \cos \theta_j, \quad \theta_j = \frac{j \pi}{N} \quad 0 \leq j \leq N.\end{equation}
+\begin{equation}
+x_j = \cos \theta_j, \quad \theta_j = \frac{j \pi}{N} \quad 0 \leq j \leq N.
+\end{equation}
 ```
 Then 
 ```math
-\begin{equation} T_n(x_j) = \cos \frac{n j \pi}{N}.\end{equation}
+\begin{equation}
+T_n(x_j) = \cos \frac{n j \pi}{N}.
+\end{equation}
 ```
 Assuming that $M = 2N$, with $N$ an integer, and $b_{k} = b_{M-k}$ for $k>0$, we have that 
 ```math
-\begin{equation} f_j = b_{0} + b_{N}(-1)^j + \sum_{n=1}^{N-1}
-b_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right).\end{equation}
+\begin{equation}
+f_j = b_{0} + b_{N}(-1)^j + \sum_{n=1}^{N-1}
+b_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right).
+\end{equation}
 ```
 Comparing this to the expression for $f(x_j)$ in the Chebyshev representation,
 ```math
-\begin{equation} f_j = a_{0} + a_{N}(-1)^j + \frac{1}{2}\sum_{n=1}^{N-1}
-a_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right),\end{equation}
+\begin{equation}
+f_j = a_{0} + a_{N}(-1)^j + \frac{1}{2}\sum_{n=1}^{N-1}
+a_{n}\left(\exp\left[i \frac{\pi n j}{N}\right]+\exp\left[-i \frac{\pi n j}{N}\right]\right),
+\end{equation}
 ```
 we find that the Chebyshev representation on the Chebyshev points is equivalent 
 to the Fourier representation on the uniform grid points, if we identify
 ```math
-\begin{equation} b_{0} = a_{0}, \quad  b_{N} = a_{N}, \quad b_{j} = \frac{a_{j}}{2} {\rm~for~} 1 \leq j \leq N-1. \end{equation}
+\begin{equation}
+b_{0} = a_{0}, \quad  b_{N} = a_{N}, \quad b_{j} = \frac{a_{j}}{2} {\rm~for~} 1 \leq j \leq N-1.
+\end{equation}
 ```
 This fact allows us to carry out the Chebyshev tranform by Fourier transforming the $\{f_j\}$ data
 and carrying out the correct normalisation of the resulting coefficients. 
@@ -55,74 +67,103 @@ The last subsection dealt with grids which contain both endpoints on the $[-1,1]
 Certain problems require domains which contain a single endpoint, i.e., $x \in (-1,1]$. For 
 these cases we choose the points 
 ```math
-\begin{equation} x_j = \cos \theta_j, \quad \theta_j = \frac{2 j \pi}{2 N + 1} \quad 0 \leq j \leq N.\end{equation}
+\begin{equation}
+x_j = \cos \theta_j, \quad \theta_j = \frac{2 j \pi}{2 N + 1} \quad 0 \leq j \leq N.
+\end{equation}
 ```
-Writing out the Chebyshev series (1), <!-- \eq{eq:cheb-expansion}, -->
+Writing out the Chebyshev series (1),
 we have that 
 ```math
-\begin{equation} \begin{split} f(x_j) = & \sum^N_{n=0} a_{n} \cos \frac{2 n j \pi}{2 N + 1} \\ & = a_{0} + \sum^N_{n=1} \frac{a_{n}}{2}\left(\exp\left[i \frac{2\pi n j}{2N +1}\right] + \exp\left[-i \frac{2\pi n j}{2N +1}\right]\right).\end{split} \label{eq:cheb-expansion-radau-points}\tag{3}\end{equation}
+\begin{equation}
+\begin{split}f(x_j) = & \sum^N_{n=0} a_{n} \cos \frac{2 n j \pi}{2 N + 1} \\ &
+= a_{0} + \sum^N_{n=1} \frac{a_{n}}{2}\left(\exp\left[i \frac{2\pi n j}{2N +1}\right] + \exp\left[-i \frac{2\pi n j}{2N +1}\right]\right).\end{split}
+\tag{3}
+\end{equation}
 ```
-The form of the series (3) <!--\eq{eq:cheb-expansion-radau-points}--> is identical to the form of 
-a Fourier series on an odd number of points, i.e., taking $M = 2 N + 1$ in equation (2) <--\eq{eq:fourier-series}-->,
+The form of the series (3) is identical to the form of 
+a Fourier series on an odd number of points, i.e., taking $M = 2 N + 1$ in equation (2),
 and assuming $b_{k} = b_{M -k}$ for $k>1$,
 we have that 
 ```math
-\begin{equation} f_j = b_{0} + \sum_{k=1}^{N} b_{k}\left(\exp\left[i \frac{2\pi k j}{2N+1}\right] + \exp\left[-i \frac{2\pi k j}{2N+1} \right]\right). \end{equation}
+\begin{equation}
+f_j = b_{0} + \sum_{k=1}^{N} b_{k}\left(\exp\left[i \frac{2\pi k j}{2N+1}\right] + \exp\left[-i \frac{2\pi k j}{2N+1} \right]\right).
+\end{equation}
 ```
 We can thus take a Chebyshev transform using a Fourier transform on Gauss-Chebyshev-Radau points if we identify 
 ```math
-\begin{equation} b_{0} = a_{0}, \quad b_{j} = \frac{a_{j}}{2} {\rm~for~} 1 \leq j \leq N. \end{equation}
+\begin{equation}
+b_{0} = a_{0}, \quad b_{j} = \frac{a_{j}}{2} {\rm~for~} 1 \leq j \leq N.
+\end{equation}
 ```
 
 Chebyshev coefficients of derivatives of a function
 ===============================================
 
-Starting from the expression of $f$ as a sum Chebyshev polynomials, equation (1) <!--\eq{eq:cheb-expansion}-->,
+Starting from the expression of $f$ as a sum Chebyshev polynomials, equation (1),
 we can obtain an expression for the derivative
 ```math
-\begin{equation} \frac{d f}{d x} = \sum^N_{n=0} a_{n}\frac{d T_{n}}{d x}. \label{eq:derivative-def}\tag{4}\end{equation}
+\begin{equation}
+\frac{d f}{d x} = \sum^N_{n=0} a_{n}\frac{d T_{n}}{d x}.\tag{4}
+\end{equation}
 ```
 We note that we must be able to express ${d f}/{d x}$ as a sum 
 of Chebyshev polynomials of up to order $N-1$, i.e.,
 ```math
-\begin{equation} \frac{d f}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}. \end{equation}
+\begin{equation}
+\frac{d f}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}.
+\end{equation}
 ```
 We must determine the set $\{d_{n}\}$ in terms of the set $\{a_{n}\}$.
 First, we equate the two expressions to find that 
 ```math
-\begin{equation} \sum^N_{k=0} a_{k}\frac{d T_{k}}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}. \label{eq:dn-def}\tag{5}\end{equation}
+\begin{equation}
+\sum^N_{k=0} a_{k}\frac{d T_{k}}{d x} = \sum^{N-1}_{n=0} d_{n}T_{n}.\tag{5}
+\end{equation}
 ```
 We use the Chebyshev polynomials of the second kind $U_n{x}$ to aid us in the calculation of the set $\{d_{n}\}$. 
 These polynomials are defined by 
 ```math
-\begin{equation} U_{0}(x) = 1, \quad U_{1}(x) = 2x, \quad U_{n+1} = 2 x U_{n}(x) - U_{n-1}(x).\end{equation}
+\begin{equation}
+U_{0}(x) = 1, \quad U_{1}(x) = 2x, \quad U_{n+1} = 2 x U_{n}(x) - U_{n-1}(x).
+\end{equation}
 ```
 Note the useful relations 
 ```math
-\begin{equation} \frac{d T_{n}}{d x} = n U_{n-1}, {\rm~for~}n\geq 1, \quad \frac{d T_{0}}{d x} = 0,\end{equation}
+\begin{equation}
+\frac{d T_{n}}{d x} = n U_{n-1}, {\rm~for~}n\geq 1, \quad \frac{d T_{0}}{d x} = 0,
+\end{equation}
 ```
 ```math
-\begin{equation} T_{n} = \frac{1}{2}\left(U_{n} - U_{n -2}\right), T_{0} = U_{0}\quad, {\rm ~and~} \quad 2 T_{1} = U_{1}. \end{equation}
+\begin{equation}
+T_{n} = \frac{1}{2}\left(U_{n} - U_{n -2}\right), T_{0} = U_{0}\quad, {\rm ~and~} \quad 2 T_{1} = U_{1}.
+\end{equation}
 ```
 Using these identities, which may be obtained from the trigonometric definition of $U_{n}(\cos \theta)$
 ```math
-\begin{equation}  U_{n}(\cos \theta) \sin \theta = \sin \left((n+1)\theta\right),\end{equation}
+\begin{equation}
+U_{n}(\cos \theta) \sin \theta = \sin \left((n+1)\theta\right),
+\end{equation}
 ```
-we find that equation (5) <!--\eq{eq:dn-def}--> becomes 
+we find that equation (5) becomes 
 ```math
-\begin{equation} \begin{split}\sum^N_{n=1} a_{n} n U_{n-1}(x) =& \frac{d_{N-1}}{2}U_{N-1}+\frac{d_{N-2}}{2}U_{N-2} 
-\\ & + \sum^{N-3}_{k=1} \frac{d_{k}-d_{k+2}}{2}U_{k} + \left(d_{0} - \frac{d_{2}}{2}\right)U_{0}. \end{split}
-\label{eq:dn-def-U}\tag{6}\end{equation}
+\begin{equation}
+\begin{split}\sum^N_{n=1} a_{n} n U_{n-1}(x) =& \frac{d_{N-1}}{2}U_{N-1}+\frac{d_{N-2}}{2}U_{N-2} 
+\\ & + \sum^{N-3}_{k=1} \frac{d_{k}-d_{k+2}}{2}U_{k} + \left(d_{0} - \frac{d_{2}}{2}\right)U_{0}. \end{split}\tag{6}
+\end{equation}
 ```
 Using the orthogonality relation 
 ```math
-\begin{equation} \int^1_{-1} U_{m}(x)U_{n}(x)\sqrt{1-x^2} \; d x = 
-\left\{\begin{array}{l} 0 {\rm ~if~} n\neq m  \\ \pi/2 {\rm ~if~} n=m \\ \end{array} \right.\end{equation}
+\begin{equation}
+\int^1_{-1} U_{m}(x)U_{n}(x)\sqrt{1-x^2} \; d x = 
+\left\{\begin{array}{l} 0 {\rm ~if~} n\neq m  \\ \pi/2 {\rm ~if~} n=m \\ \end{array} \right.
+\end{equation}
 ```
 we obtain the (unqiuely-determined) relations 
 ```math
-\begin{equation} \begin{split} &d_{N-1} = 2Na_{N},\quad d_{N-2} = 2(N-1)a_{N-1}, \\ 
-& d_{k} = 2(k+1) a_{k+1} + d_{k+2}, \quad d_{0} = \frac{d_{2}}{2} + a_{1}.\end{split} \label{eq:dn-result-U}\tag{7}\end{equation}
+\begin{equation}
+\begin{split} &d_{N-1} = 2Na_{N},\quad d_{N-2} = 2(N-1)a_{N-1}, \\ 
+& d_{k} = 2(k+1) a_{k+1} + d_{k+2}, \quad d_{0} = \frac{d_{2}}{2} + a_{1}.\end{split}\tag{7}
+\end{equation}
 ```       
 
 Clenshaw-Curtis integration weights
@@ -131,119 +172,165 @@ Clenshaw-Curtis integration weights
 We require the integration weights for the set of points $\{x_j\}$ chosen 
 in our numerical scheme. The weights $w_{j}$ are defined implicitly by 
 ```math
-\begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{j=0}^N f(x_j) w_{j}. \label{eq:w-sum}\tag{8}\end{equation}
+\begin{equation}
+\int^{1}_{-1} f(x) \; d x = \sum_{j=0}^N f(x_j) w_{j}.\tag{8}
+\end{equation}
 ```
 In the Chebyshev scheme we use the change of variables $x = \cos \theta$
 to write 
 ```math
-\begin{equation} \int^{1}_{-1} f(x) \; d x = \int^\pi_0 f(\cos\theta) \sin \theta \; d \theta . \label{eq:change-of-variables-integral} \tag{9}\end{equation}
+\begin{equation}
+\int^{1}_{-1} f(x) \; d x = \int^\pi_0 f(\cos\theta) \sin \theta \; d \theta .\tag{9}
+\end{equation}
 ```
- Using the series expansion (1) <!--\eq{eq:cheb-expansion}--> in equation (9) <!--\eq{eq:change-of-variables-integral}-->
+ Using the series expansion (1) in equation (9)
  we find that 
- ```math
- \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum^N_{n=0} a_{n}\int^\pi_0 \cos (n \theta) \sin \theta \; d \theta
- . \label{eq:series-integral}\tag{10} \end{equation}
- ```
- Note the integral identity
- ```math
-\begin{equation} \int^\pi_0 \cos(n \theta) \sin \theta \; d \theta = \frac{\cos(n \pi) +1}{1 - n^2} {\rm~for~} n \geq 0.\end{equation}
-```
- Also note that 
-```math
-\begin{equation} \frac{\cos(n \pi) +1}{1 - n^2} = \left\{\begin{array}{l} 0 {\rm ~if~} n = 2 r + 1, ~r \in \mathbb{Z}  \\ 2/(1 - n^2) {\rm ~if~} n=2r,~r. \in \mathbb{Z}  \end{array}\right. \end{equation}
- ```
- We define 
- ```math
- \begin{equation} J_{n} = \frac{\cos(n \pi) +1}{1 - n^2}. \end{equation}
- ```
- Using this definition, we can write the integral of $f(x)$ can be written 
- in terms of a sum over of the Chebyshev coefficients:
- ```math
- \begin{equation} \int^{1}_{-1} f(x) \; d x = \sum_{n=0}^N J_{n} a_{n}. \label{eq:Cheb-sum}\tag{11}\end{equation}
- ```
- 
- To avoid computing the set of coefficients $\{a_{n}\}$ every time we wish to integrate $f(x_j)$,
- we use the inverse transforms. This transform allows us to rewrite equation (11) <!--\eq{eq:Cheb-sum}--> in the form (8) <!--\eq{eq:w-sum}-->.
- Since the inverse transform differs between the Gauss-Chebyshev-Lobotto and Gauss-Chebyshev-Radau cases, we treat each 
- case separately below. 
- 
+```math
+\begin{equation}
+\int^{1}_{-1} f(x) \; d x = \sum^N_{n=0} a_{n}\int^\pi_0 \cos (n \theta) \sin \theta \; d \theta
+.\tag{10}
+\end{equation}
+```
+Note the integral identity
+```math
+\begin{equation}
+\int^\pi_0 \cos(n \theta) \sin \theta \; d \theta = \frac{\cos(n \pi) +1}{1 - n^2} {\rm~for~} n \geq 0.
+\end{equation}
+```
+Also note that 
+```math
+\begin{equation}
+\frac{\cos(n \pi) +1}{1 - n^2} = \left\{\begin{array}{l} 0 {\rm ~if~} n = 2 r + 1, ~r \in \mathbb{Z}  \\ 2/(1 - n^2) {\rm ~if~} n=2r,~r. \in \mathbb{Z}  \end{array}\right.
+\end{equation}
+```
+We define 
+```math
+\begin{equation}
+J_{n} = \frac{\cos(n \pi) +1}{1 - n^2}.
+\end{equation}
+```
+Using this definition, we can write the integral of $f(x)$ can be written 
+in terms of a sum over of the Chebyshev coefficients:
+```math
+\begin{equation}
+\int^{1}_{-1} f(x) \; d x = \sum_{n=0}^N J_{n} a_{n}.\tag{11}
+\end{equation}
+```
+To avoid computing the set of coefficients $\{a_{n}\}$ every time we wish to integrate $f(x_j)$,
+we use the inverse transforms. This transform allows us to rewrite equation (11) in the form (8).
+Since the inverse transform differs between the Gauss-Chebyshev-Lobotto and Gauss-Chebyshev-Radau cases, we treat each 
+case separately below. 
+
 Weights on Gauss-Chebyshev-Lobotto points
 ===============================================
-  We use the inverse transformation 
- ```math
- \begin{equation} a_{n} = \frac{q_{n}}{2N}\sum^{2N-1}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N}\right], \label{eq:inverse-transform-GCL}\tag{12}\end{equation}
- ```
- where 
- ```math
- \begin{equation} q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n\neq0,N  \\ 1 {\rm ~if~} n=0,N  \end{array}\right.\end{equation}
- ```
+We use the inverse transformation 
+```math
+\begin{equation}
+a_{n} = \frac{q_{n}}{2N}\sum^{2N-1}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N}\right], \tag{12}
+\end{equation}
+```
+where 
+```math
+\begin{equation}
+q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n\neq0,N  \\ 1 {\rm ~if~} n=0,N  \end{array}\right.
+\end{equation}
+```
 and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math
-\begin{equation} \hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j}){\rm~for~} N+1 \leq j \leq 2N-1. \end{equation}
+\begin{equation}
+\hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j}){\rm~for~} N+1 \leq j \leq 2N-1.
+\end{equation}
 ```
- With this inverse tranformation, we can write 
+With this inverse tranformation, we can write 
 ```math
-\begin{equation} \begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N-1}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
- & = \sum^{2N-1}_{j=0}\sum^{2N-1}_{n=0} \frac{\hat{f}_j J_{n}}{2N} \exp\left[-i \frac{2\pi n j}{2N}\right] \\ 
- & = \sum^{2N-1}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working}\tag{13}\end{equation}
+\begin{equation}
+\begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N-1}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
+& = \sum^{2N-1}_{j=0}\sum^{2N-1}_{n=0} \frac{\hat{f}_j J_{n}}{2N} \exp\left[-i \frac{2\pi n j}{2N}\right] \\ 
+& = \sum^{2N-1}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\tag{13}
+\end{equation}
 ```
- where in the first step we have extended the sum from $N$ to $2N-1$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
+where in the first step we have extended the sum from $N$ to $2N-1$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
 ```math
-\begin{equation} J_{j} = J_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1,\end{equation}
+\begin{equation}
+J_{j} = J_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1,
+\end{equation}
 ```
 ```math
-\begin{equation} a_{j} = a_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1.\end{equation}
+\begin{equation}
+a_{j} = a_{2N-j}, {\rm~for~} N+1 \leq j \leq 2N-1.
+\end{equation}
 ```
-In the second step we use the definition of the inverse transform (14) <!--\eq{eq:inverse-transform-GCR}-->, and 
+In the second step we use the definition of the inverse transform (14) , and 
 in the third step we define 
 ```math
-\begin{equation} v_{j} = \sum_{n=0}^{2N-1}\frac{J_{n}}{2N}\exp\left[-i \frac{2\pi n j}{2N}\right].\end{equation}
+\begin{equation}
+v_{j} = \sum_{n=0}^{2N-1}\frac{J_{n}}{2N}\exp\left[-i \frac{2\pi n j}{2N}\right].
+\end{equation}
 ```
-Finally, we can compare equations (8) <!--\eq{eq:w-sum}--> and (13) <!--\eq{eq:weights-working}--> and deduce that 
+Finally, we can compare equations (8) and (13) and deduce that 
 ```math
-\begin{equation} w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
+\begin{equation}
+w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.
+\end{equation}
 ```
 We can write $v_{j}$ in terms of a discrete cosine transform, i.e.,
 ```math
-\begin{equation} v_{j} = \frac{1}{2N}\left(J_{0} + (-1)^jJ_{N} + 2\sum_{n=1}^{N-1}J_{n}\cos\left(\frac{\pi n j}{N}\right)\right).\end{equation}
+\begin{equation}
+v_{j} = \frac{1}{2N}\left(J_{0} + (-1)^jJ_{N} + 2\sum_{n=1}^{N-1}J_{n}\cos\left(\frac{\pi n j}{N}\right)\right).
+\end{equation}
 ```
  
 Weights on Gauss-Chebyshev-Radau points
 ===============================================
 We use the inverse transformation 
 ```math
-\begin{equation} a_{n} = \frac{q_{n}}{2N+1}\sum^{2N}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N+1}\right], \label{eq:inverse-transform-GCR}\tag{14}\end{equation}
+\begin{equation}
+a_{n} = \frac{q_{n}}{2N+1}\sum^{2N}_{j=0} \hat{f}_j \exp\left[- i \frac{2\pi n j}{2N+1}\right],\tag{14}
+\end{equation}
 ```
 where 
 ```math
-\begin{equation} q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n > 0  \\ 1 {\rm ~if~} n=0  \end{array}\right.\end{equation}
+\begin{equation}
+q_{n} = \left\{\begin{array}{l} 2 {\rm ~if~} n > 0  \\ 1 {\rm ~if~} n=0  \end{array}\right.
+\end{equation}
 ```
 and $\hat{f}_j$ is $f(x_j)$ on the extended domain in FFT order, i.e.,
 ```math
-\begin{equation} \hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j+1}){\rm~for~} N+1 \leq j \leq 2N. \end{equation}
+\begin{equation}
+\hat{f}_j = f(x_{j}) {\rm~for~} 0 \leq j \leq N ,\quad \hat{f}_j = f(x_{2N-j+1}){\rm~for~} N+1 \leq j \leq 2N.
+\end{equation}
 ```
 Note that the details of what is the appropriate FFT order depends on the order in which the points $x_j$ are stored.
 The key detail in the Chebyshev-Radau scheme is that (in the notation above)
 $x_0 = 1$ is not a repeated point, and must occupy $\hat{f}_0$. 
 With this inverse tranformation, we can write 
 ```math
-\begin{equation} \begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
+\begin{equation}
+\begin{split}\sum_{n=0}^N J_{n} a_{n} & =  \sum^{2N}_{n=0} \frac{a_{n}J_{n}}{q_{n}} \\
 & = \sum^{2N}_{j=0}\sum^{2N}_{n=0} \frac{\hat{f}_j J_{n}}{2N+1} \exp\left[-i \frac{2\pi n j}{2N+1}\right] \\ 
-& = \sum^{2N}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\label{eq:weights-working-radau}\tag{15}\end{equation}
+& = \sum^{2N}_{j=0} \hat{f}_j v_{j} = \sum^{N}_{j=0} \hat{f}_j q_{j}v_{j},\end{split}\tag{15}\end{equation}
 ```
 where in the first step we have extended the sum from $N$ to $2N$ and used FFT-order definitions of $J_{n}$ and $a_{n}$
 ```math
-\begin{equation} J_{j} = J_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N,\end{equation}
+\begin{equation}
+J_{j} = J_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N,
+\end{equation}
 ```
 ```math
-\begin{equation} a_{j} = a_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N.\end{equation}
+\begin{equation}
+a_{j} = a_{2N+1-j}, {\rm~for~} N+1 \leq j \leq 2N.
+\end{equation}
 ```
-In the second step we use the definition of the inverse transform (14) <!--\eq{eq:inverse-transform-GCR}-->, and 
+In the second step we use the definition of the inverse transform (14), and 
 in the third step we define 
 ```math
-\begin{equation} v_{j} = \sum_{n=0}^{2N}\frac{J_{n}}{2N+1}\exp\left[-i \frac{2\pi n j}{2N+1}\right].\end{equation}
+\begin{equation}
+v_{j} = \sum_{n=0}^{2N}\frac{J_{n}}{2N+1}\exp\left[-i \frac{2\pi n j}{2N+1}\right].
+\end{equation}
 ```
-Finally, we can compare equations (8) <!--\eq{eq:w-sum}--> and (15) <!--\eq{eq:weights-working-radau}--> and deduce that 
+Finally, we can compare equations (8) and (15) and deduce that 
 ```math
-\begin{equation} w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.  \end{equation}
+\begin{equation}
+w_{j} = q_{j}v_{j} {\rm~for~} 0 \leq j \leq N.
+\end{equation}
 ```