From 797fc438f811aa78b4a254734d13d89ff964a224 Mon Sep 17 00:00:00 2001
From: Nao Pross <np@0hm.ch>
Date: Thu, 28 Apr 2022 02:22:24 +0200
Subject: Start copying proofs

---
 notes/FourierOnS2.tex | 248 +++++++++++++++++++++++++++++++++++++++++++++++++-
 1 file changed, 245 insertions(+), 3 deletions(-)

(limited to 'notes/FourierOnS2.tex')

diff --git a/notes/FourierOnS2.tex b/notes/FourierOnS2.tex
index 2dc30c1..de15f01 100644
--- a/notes/FourierOnS2.tex
+++ b/notes/FourierOnS2.tex
@@ -15,6 +15,7 @@
 \newtheorem{lemma}{Lemma}
 \newtheorem{definition}{Definition}
 \newtheorem{remark}{Remark}
+\newtheorem{proposition}{Proposition}
 
 %
 % Metadata
@@ -219,14 +220,255 @@ Though, unfortunately this time the separation process is more involved, and the
 
 Equation \eqref{eqn:separation-phi} is easy, the solutions are complex exponentials \(e^{im\varphi}\), while \eqref{eqn:separation-theta} is known as the \emph{associated Legendre equation}. Though, normally the equation is written in term of \(x\) and \(y(x)\), so \eqref{eqn:separation-theta} is brought to a more familiar form by using the substitution \(x = \cos\vartheta\) and \(y = \Theta\):
 \[
-  \left(1 - x^2 \right) \frac{d^2 y}{dx} - 2x \frac{dy}{dx} + \left[
-    n(n+1) - \frac{m^2}{1 - x^2}
-  \right] y(x) = 0.
+  \left( 1 - x^2 \right) \frac{d^2 y}{dx^2} - 2x \frac{dy}{dx}
+    + \left[ n(n+1) - \frac{m^2}{1 - x^2} \right] y(x) = 0.
 \]
 Finding the solutions to this equation is so involved, that it deserves its own section.
 
 \subsection{The associated Legendre polynomials}
 
+In this section we would like to find the solutions to the \emph{associated} Legendre equation, which is actually a generalization of Legendre equation:
+\begin{equation} \label{eqn:legendre}
+  \left( 1 - x^2 \right) \frac{d^2 y}{dx^2}
+    - 2x \frac{dy}{dx} + n(n + 1) y(x) = 0.
+\end{equation}
+Thus we first need examine the solutions to this equation before constructing the more general solution.
+
+\begin{proposition}
+  The polynomials
+  \begin{equation} \label{eqn:legendre-poly}
+    P_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} 
+    \frac{(-1)^k (2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k},
+  \end{equation}
+  known as Legendre's polynomials are solutions to Legendre's equation \eqref{eqn:legendre}.
+\end{proposition}
+
+\begin{lemma} The expression
+\begin{equation} \label{eqn:legendre-rodrigues}
+  P_n(x) = \frac{1}{n!2^n}\frac{d^n}{dx^n}(x^2-1)^n.
+\end{equation}
+is equivalent to \eqref{eqn:legendre-poly}.
+\end{lemma}
+
+\begin{proof}
+  We start expanding the term \((x^2-1)^n\); According to the binomial theorem
+  \begin{equation*}
+    (x^2-1)^n=\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2(n-k)}.
+  \end{equation*}
+  Substituting the above, \eqref{eqn:legendre-rodrigues} becomes
+  \begin{align*}
+    \frac{1}{n!2^n} \frac{d^n}{dx^n} (x^2 - 1)^n 
+      &= \frac{1}{n!2^n} \sum_{k=0}^n(-1)^k \binom{n}{k} \frac{d^n}{dx^n}x^{2(n-k)} \\
+      &= \frac{1}{n!2^n} \sum_{k=0}^{\lfloor n / 2 \rfloor}(-1)^k
+         \binom{n}{k}\frac{d^n}{dx^n}x^{2(n-k)}.
+  \end{align*}
+  Recall that
+  \[
+    \frac{d^n}{dx^n}x^\alpha
+      % = \alpha(\alpha-1)(\alpha-2)\hdots(\alpha-n+1)x^{\alpha-n}
+      = \frac{\alpha!}{(\alpha-n)!}x^{\alpha-n},
+  \]
+  thus
+  \begin{align*}
+    \frac{1}{n!2^n} &\sum_{k=0}^{\lfloor n / 2 \rfloor}(-1)^k
+       \binom{n}{k}\frac{d^n}{dx^n}x^{2(n-k)} \\
+    &= \frac{1}{n!2^n} \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{k}
+        \frac{(2n-2k)!}{(n-2k)!}x^{n-2k} \\
+    &= \frac{1}{n!2^n} \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \frac{n!}{k!(n-k)!}
+        \frac{(2n-2k)!}{(n-2k)!}x^{n-2k} \\
+    &= \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{(-1)^k (2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k}.
+    \qedhere
+  \end{align*}
+\end{proof}
+
 \subsection{Spherical harmonics}
 
+\clearpage
+\appendix
+\section{Proofs}
+\subsection{Legendre Polynomials}
+
+\begin{lemma}
+  The polynomial
+  \[
+    P_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k
+      \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k},
+  \]
+  is a solution to Legendre's equation
+  \[
+    \left( 1 - x^2 \right) \frac{d^2 y}{dx^2}
+      - 2x \frac{dy}{dx} + n(n + 1) y(x) = 0
+  \]
+  for \(n > 0\).
+\end{lemma}
+\begin{proof}
+To solve \eqref{eqn:legendre} we use the power series ansatz
+\begin{equation} \label{eqn:legendre-power-ansatz}
+  y(x) = \sum_{k=0}^\infty a_k x^k,
+\end{equation}
+from which follows that
+\[
+  y' = \sum_{k = 0}^\infty k a_k x^{k-1}, \text{ and }
+  y'' = \sum_{k = 0}^\infty k (k-1) a_k x^{k-2}.
+\]
+By substituting the above and \eqref{eqn:legendre-power-ansatz} into \eqref{eqn:legendre} we get the that first term
+\begin{align*}
+  \Big( 1 &- x^2 \Big) y''
+    = \left( 1 - x^2 \right) \sum_{k = 0}^\infty k (k-1) a_k x^{k - 2} \\
+    &= \sum_{k = 0}^\infty k (k-1) a_k x^{k-2} + k (k-1) a_k x^{k} \\
+    &= \sum_{k = 0}^\infty \left[
+         (k + 1) (k + 2) a_{k + 2} + k (k-1) a_k 
+       \right] x^{k},
+\end{align*}
+where in the last step to factor out \(x^k\) we shifted the index in the coefficients by 2, i.e.
+\[
+  \sum_{k = 0}^\infty k (k - 1)a_k x^{k - 2}
+  = \sum_{k = 0}^\infty (k + 2) (k + 1) a_{k + 2} x^k.
+\]
+Similarly, the second term:
+\[
+  -2xy' = -2x \sum_{k = 0}^\infty k a_k x^{k - 1}
+        = \sum_{k = 0}^\infty - 2k a_k x^k.
+\]
+Finally, combining the above the complete substitution yields
+\begin{gather*}
+  \Big( 1 - x^2 \Big) y'' - 2xy' + n(n + 1) y  = 0 \\
+    \implies \sum_{k = 0}^\infty \big[
+         (k + 1) (k + 2) a_{k + 2} + k (k-1) a_k \\
+         \qquad \qquad
+         - 2k a_k + n(n + 1) a_k
+       \big] x^k = 0,
+\end{gather*}
+from which we can extract the recurrence relation
+\begin{gather*}
+  \begin{aligned}
+    (k + 1) (k + 2) a_{k + 2} &+ k (k-1) a_k \\
+    &- 2k a_k + n(n + 1) a_k = 0
+  \end{aligned} \\
+  \iff a_{k + 2} = \frac{(k-n)(k+n+1)}{(k+2)(k+1)} a_k.
+\end{gather*}
+
+% TODO: finish reviewing proof
+\if 0
+
+We can derive a recursion formula for $a_{k+2}$ from Eq.\eqref{eq:condition_2}, which can be expressed as
+\begin{equation}\label{eq:recursion}
+a_{k+2}= \frac{k (k-1) - 2 k + n(n+1)}{(k+1)(k+2)}a_k = \frac{(k-n)(k+n+1)}{(k+2)(k+1)}a_k.
+\end{equation}
+All coefficients can be calculated using the latter. 
+
+Following Eq.\eqref{eq:recursion}, if we want to compute $a_6$ we would have
+\begin{align*}
+a_{6}= -\frac{(n-4)(n+5)}{6\cdot 5}a_4 &= -\frac{(n-4)(5+n)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3}  a_2 \\
+&= -\frac{(n-4)(n+5)}{6 \cdot 5} -\frac{(n-2)(n+3)}{4 \cdot 3} -\frac{n(n+1)}{2 \cdot 1}  a_0 \\
+&= -\frac{(n+5)(n+3)(n+1)n(n-2)(n-4)}{6!} a_0.
+\end{align*}
+One can generalize this relation for the $i^\text{th}$ even coefficient as
+\begin{equation*}
+a_{2k} = (-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots (n-(2k-2)+2)(n-(2k-2))}{(2k)!}a_0
+\end{equation*}
+where $i=2k$.
+
+A similar expression can be written for the odd coefficients $a_{2k-1}$. In this case, the equation starts from $a_1$ and to find the pattern we can write the recursion for an odd coefficient, $a_7$ for example
+\begin{align*}
+a_{7}= -\frac{(n-5)(n+6)}{7\cdot 6}a_5 &= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4}  a_3 \\
+&= - \frac{(n-5)(n+6)}{7\cdot 6} -\frac{(n-3)(n+4)}{5 \cdot 4}  -\frac{(n-1)(n+2)}{3 \cdot 2}  a_1 \\
+&= -\frac{(n+6)(n+4)(n+2)(n-1)(n-3)(n-5)}{7!} a_1.
+\end{align*}
+As before, we can generalize this equation for the $i^\text{th}$ odd coefficient
+\begin{equation*}
+a_{2k+1} = (-1)^k \frac{(n + 2k)(n+2k-2)\hdots(n-(2k-1)+2)(n-(2k-1))}{(2k+1)!}a_1
+\end{equation*}
+where $i=2k+1$.
+
+Let be 
+\begin{align*}
+y_\text{e}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n+(2k-1))(n+(2k-1)-2)\hdots \color{red}(n-(2k-2)+2)(n-(2k-2))}{(2k)!} x^{2k}, \\
+y_\text{o}^K(x) &:= \sum_{k=0}^K(-1)^k \frac{(n + 2k)(n+2k-2)\hdots \color{blue} (n-(2k-1)+2)(n-(2k-1))}{(2k+1)!} x^{2k+1}.
+\end{align*}
+The solution to the Eq.\eqref{eq:legendre} can be written as
+\begin{equation}\label{eq:solution}
+y(x) = \lim_{K \to \infty} \left[ a_0 y_\text{e}^K(x) + a_1 y_\text{o}^K(x) \right].
+\end{equation}
+
+The colored parts can be analyzed separately:
+\begin{itemize}
+  \item[\textcolor{red}{\textbullet}] Suppose that $n=n_0$ is an even number. Then the red part, for a specific value of $k=k_0$, will follow the following relation:
+\begin{equation*}
+n_0-(2k_0-2)=0. 
+\end{equation*}
+From that point on, given the recursive nature of Eq.\eqref{eq:recursion}, all the subsequent coefficients will also be 0, making the sum finite.
+\begin{equation*}
+a_{2k}=0 \iff y_{\text{o}}^{2k}(x)=y_{\text{o}}^{2k_0}(x), \quad \forall k>k_0
+\end{equation*} 
+  \item[\textcolor{blue}{\textbullet}] Suppose that $n=n_0$ is an odd number. Then the blue part, for a specific value of $k=k_0$, will follow the following relation 
+\begin{equation*}
+n_0-(2k_0-1)=0.  
+\end{equation*}
+From that point on, for the same reason as before, all the subsequent coefficients will also be 0, making the sum finite.
+\begin{equation*}
+a_{2k+1}=0 \iff y_{\text{o}}^{2k+1}(x)=y_{\text{o}}^{2k_0+1}(x), \quad \forall k>k_0
+\end{equation*}
+\end{itemize} 
+
+There is the possibility of expressing the solution in Eq.\eqref{eq:solution} in a more compact form, combining the two solutions $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$. They are both a polynomial of maximum degree $n$, assuming $n \in \mathbb{N}$. In the case where $n$ is even, the polynomial solution
+\begin{equation*}
+\lim_{K\to \infty} y_\text{e}^K(x)
+\end{equation*}
+will be a finite sum. If instead $n$ is odd, will be 
+\begin{equation*}
+\lim_{K\to \infty} y_\text{o}^K(x)
+\end{equation*}
+to be a finite sum. 
+
+Depending on the coefficient we start with, $a_1$ or $a_0$, we will obtain the odd or even polynomial respectively. Starting with the last coefficient $a_n$ and, recursively, calculating all the others in descending order, we can express the two parts $y_\text{o}^K(x)$ and $y_\text{e}^K(x)$ with a single sum. Hence, because we start with the last coefficient, the choice concerning $a_1$ and $a_0$ will be at the end of the sum, and not at the beginning. To compact Eq.\eqref{eq:solution}, Eq.\eqref{eq:recursion} can be reconsidered to calculate the coefficient $a_{k-2}$, using $a_k$
+\begin{equation*}
+a_{k-2} = -\frac{(k+2)(k+1)}{(k-n)(k+n+1)}a_k
+\end{equation*}
+Now the game is to find a pattern, as before. Remember that $n$ is a fixed parameter of Eq.\eqref{eq:legendre}.  
+\begin{align*}
+a_{n-2} &= -\frac{n(n-1)}{2(2n-1)}a_n, \\
+a_{n-4} &= -\frac{(n-2)(n-3)}{4(2n-3)}a_{n-2} \\
+&= -\frac{(n-2)(n-3)}{4(2n-3)}-\frac{n(n-1)}{2(2n-1)}a_n.
+\end{align*}
+In general 
+\begin{equation}\label{eq:general_recursion}
+a_{n-2k} = (-1)^k \frac{n(n-1)(n-2)(n-3) \hdots (n-2k+1)}{2\cdot4\hdots 2k(2n-1)(2n-3)\hdots(2n-2k+1)}a_n
+\end{equation}
+The whole solution can now be written as
+\begin{align}
+y(x) &= a_n x^n + a_{n-2} x^{n-2} + a_{n-4} x^{n-4} + a_{n-6} x^{n-6} + \hdots + \begin{cases} 
+a_1 x, \quad &\text{if } n \text{ odd} \\ 
+a_0, \quad  &\text{if } n \text{ even} 
+\end{cases} \nonumber \\
+&= \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} a_{n-2k}x^{n-2k} \label{eq:solution_2}
+\end{align}
+By considering
+\begin{align}
+(2n-1)(2n-3)\hdots (2n-2k+1)&=\frac{2n(2n-1)(2n-2)(2n-3)\hdots(2n-2k+1)}
+{2n(2n-2)(2n-4)(2n-6)\hdots(2n-2k+2)} \nonumber \\ 
+&=\frac{\frac{(2n)!}{(2n-2k)!}}
+{2^kn(n-1)(n-2)(n-3)\hdots(n-k+1)} \nonumber \\
+&=\frac{\frac{(2n)!}{(2n-2k)!}}
+{2^k\frac{n!}{(n-k)!}}=\frac{(n-k)!(2n)!}{n!(2n-2k)!2^k} \label{eq:1_sub_recursion}, \\
+2 \cdot 4  \hdots 2k &= 2^r 1\cdot2 \hdots r = 2^r r!\label{eq:2_sub_recursion}, \\
+n(n-1)(n-2)(n-3) \hdots (n-2k+1) &= \frac{n!}{(n-2k)!}\label{eq:3_sub_recursion}.
+\end{align}
+Eq.\eqref{eq:solution_2} can be rewritten as
+\begin{equation}\label{eq:solution_3}
+y(x)=a_n \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{n!^2(2n-2k)!}{k!(n-2k)!(n-k)!(2n)!}  x^{n-2k}.
+\end{equation}
+Eq.\eqref{eq:solution_3} is defined for any $a_n$. By letting $a_n$ be declared as
+\begin{equation*}
+a_{n} := \frac{(2n)!}{2^n n!^2},
+\end{equation*}
+the so called \emph{Legendre polynomial} emerges
+\begin{equation}\label{eq:leg_poly}
+P_n(x):=\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^k \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k} 
+\end{equation}
+\fi
+
+\end{proof}
+
+
 \end{document}
-- 
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