From 7e51ae842c61ba338aec179d71fab2d041ebe8c5 Mon Sep 17 00:00:00 2001
From: Nao Pross <np@0hm.ch>
Date: Wed, 17 Aug 2022 16:29:41 +0200
Subject: kugel: Review manu's text, improve legendre functions

---
 buch/papers/kugel/spherical-harmonics.tex | 424 ++++++++++--------------------
 1 file changed, 139 insertions(+), 285 deletions(-)

(limited to 'buch/papers/kugel/spherical-harmonics.tex')

diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex
index 70657c9..5645941 100644
--- a/buch/papers/kugel/spherical-harmonics.tex
+++ b/buch/papers/kugel/spherical-harmonics.tex
@@ -1,6 +1,6 @@
 % vim:ts=2 sw=2 et spell tw=80:
 
-\section{Spherical Harmonics}
+\section{Construction of the Spherical Harmonics}
 
 \if 0
 \kugeltodo{Rewrite this section if the preliminaries become an addendum}
@@ -111,7 +111,7 @@ that satisfy the equation
     \surflaplacian f = -\lambda f.
 \end{equation}
 Perhaps it may not be obvious at first glance, but we are in fact dealing with a
-partial differential equation (PDE). If we unpack the notation of the operator
+partial differential equation (PDE) \kugeltodo{Boundary conditions?}. If we unpack the notation of the operator
 $\nabla^2_{\partial S}$ according to definition
 \ref{kugel:def:surface-laplacian}, we get:
 \begin{equation} \label{kugel:eqn:eigen-pde}
@@ -126,7 +126,7 @@ Since all functions satisfying \eqref{kugel:eqn:eigen-pde} are the
 The task may seem very difficult but we can simplify it with a well-known
 technique: \emph{the separation Ansatz}. It consists in assuming that the
 function $f(\vartheta, \varphi)$ can be factorized in the following form:
-\begin{equation} \label{kugel:eqn:sep-ansatz:0}
+\begin{equation}
     f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi). 
 \end{equation}
 In other words, we are saying that the effect of the two independent variables
@@ -135,34 +135,34 @@ effect separately. This separation process was already presented in section
 \ref{buch:pde:section:kugel}, but we will briefly rehearse it here for
 convenience. If we substitute this assumption in
 \eqref{kugel:eqn:eigen-pde}, we have:
-\begin{equation} \label{kugel:eqn:sep-ansatz:1}
+\begin{equation*}
     \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left(
       \sin\vartheta \frac{\partial  \Theta(\vartheta)}{\partial\vartheta}
     \right) \Phi(\varphi)
     + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2}
       \Theta(\vartheta)
     + \lambda \Theta(\vartheta)\Phi(\varphi) = 0.
-\end{equation}
+\end{equation*}
 Dividing by $\Theta(\vartheta)\Phi(\varphi)$ and introducing an auxiliary
-variable $m$, the separation constant, yields:
+variable $m^2$, the separation constant, yields:
 \begin{equation*}
   \frac{1}{\Theta(\vartheta)}\sin \vartheta \frac{d}{d \vartheta} \left(
     \sin \vartheta \frac{d \Theta}{d \vartheta}
   \right)
   + \lambda \sin^2 \vartheta
   = -\frac{1}{\Phi(\varphi)} \frac{d^2\Phi(\varphi)}{d\varphi^2}
-  = m,
+  = m^2,
 \end{equation*}
 which is equivalent to the following system of 2 first order differential
 equations (ODEs):
 \begin{subequations}
   \begin{gather}
-    \frac{d^2\Phi(\varphi)}{d\varphi^2} = -m \Phi(\varphi),
+    \frac{d^2\Phi(\varphi)}{d\varphi^2} = -m^2 \Phi(\varphi),
       \label{kugel:eqn:ode-phi} \\ 
     \sin \vartheta \frac{d}{d \vartheta} \left(
       \sin \vartheta \frac{d \Theta}{d \vartheta}
     \right)
-    + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)
+    + \left( \lambda - \frac{m^2}{\sin^2 \vartheta} \right)
       \Theta(\vartheta) = 0
       \label{kugel:eqn:ode-theta}.
   \end{gather}
@@ -174,291 +174,141 @@ write the solutions
     \Phi(\varphi) = e^{i m \varphi}, \quad m \in \mathbb{Z}.
 \end{equation}
 The restriction that the separation constant $m$ needs to be an integer arises
-from the fact that we require a $2\pi$-periodicity in $\varphi$ since
-$\Phi(\varphi + 2\pi) = \Phi(\varphi)$. Unfortunately, solving
-\eqref{kugel:eqn:ode-theta} is not so straightforward. Actually it is quite
-difficult, and the process is so involved that it will require a dedicated
-section of its own.
+from the fact that we require a $2\pi$-periodicity in $\varphi$ since the
+coordinate systems requires that $\Phi(\varphi + 2\pi) = \Phi(\varphi)$.
+Unfortunately, solving \eqref{kugel:eqn:ode-theta} is as straightforward,
+actually, it is quite difficult, and the process is so involved that it will
+require a dedicated section of its own.
 
 \subsection{Legendre Functions}
 
-To solve \eqref{kugel:eqn:ode-theta}
-We can begin by considering the substitution $x = \cos \vartheta$. The operator $\frac{d}{d \vartheta}$ will be:
-\begin{align*}
-    \frac{d}{d \vartheta} = \frac{dx}{d \vartheta}\frac{d}{dx} &= -\sin \vartheta \frac{d}{dx} \\
-    &= -\sqrt{1-x^2} \frac{d}{dx}.
-\end{align*} 
-Eq.(\ref{kugel:eq:ODE_2}) will then become.
+To solve \eqref{kugel:eqn:ode-theta} we start with the substitution $z = \cos
+\vartheta$ \kugeltodo{Explain geometric origin with picture}. The operator
+$\frac{d}{d \vartheta}$ becomes
+\begin{equation*}
+    \frac{d}{d \vartheta}
+    = \frac{dz}{d \vartheta}\frac{d}{dz}
+    = -\sin \vartheta \frac{d}{dz}
+    = -\sqrt{1-z^2} \frac{d}{dz},
+\end{equation*} 
+since $\sin \vartheta = \sqrt{1 - \cos^2 \vartheta} = \sqrt{1 - z^2}$, and
+then \eqref{kugel:eqn:ode-theta} becomes 
 \begin{align*}
-   \frac{-\sqrt{1-x^2}}{\sqrt{1-x^2}} \frac{d}{dx} \left( \left(\sqrt{1-x^2}\right) \left(-\sqrt{1-x^2}\right) \frac{d \Theta}{dx} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \\
-   \frac{d}{dx} \left( (1-x^2) \frac{d \Theta}{dx} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \\
-   (1-x^2)\frac{d^2 \Theta}{dx} - 2x\frac{d \Theta}{dx} + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \\
-   (1-x^2)\frac{d^2 \Theta}{dx} - 2x\frac{d \Theta}{dx} + \left( \lambda - \frac{m}{1-x^2} \right)\Theta(\vartheta) &= 0 
+  \frac{-\sqrt{1-z^2}}{\sqrt{1-z^2}} \frac{d}{dz} \left[
+    \left(\sqrt{1-z^2}\right) \left(-\sqrt{1-z^2}\right) \frac{d \Theta}{dz}
+  \right]
+  + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0,
+  \\
+  \frac{d}{dz} \left[ (1-z^2) \frac{d \Theta}{dz} \right]
+  + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0,
+  \\
+  (1-z^2)\frac{d^2 \Theta}{dz} - 2z\frac{d \Theta}{dz}
+  + \left( \lambda - \frac{m^2}{1 - z^2} \right)\Theta(\vartheta) &= 0.
 \end{align*}
-By making two final cosmetic substitutions, namely $\Theta(\vartheta)=\Theta(\cos^{-1}x):=y(x)$ and $\lambda=n(n+1)$, we will be able to define the \emph{Associated Legendre Equation} in its standard and most familiar form
-\begin{definition}{Associated Legendre Equation}
-    \begin{equation}\label{kugel:eq:associated_leg_eq}
-        (1-x^2)\frac{d^2 y}{dx} - 2x\frac{d y}{dx} + \left( n(n+1) - \frac{m}{1-x^2} \right)y(x) = 0. 
-    \end{equation}
-\end{definition}
-Our new goal then became solving Eq.(\ref{kugel:eq:asssociated_leg_eq}). After that we can fit the solution into Eq.(\ref{kugel:eq:sep_ansatz_0}), obtaining $f(\vartheta, \varphi)$, the solution of the eigenvalue problem. \newline
-We simplified the problem somewhat but the task still remains very difficult. We can rely on a lemma to continue but first we need to define an additional equation, namely the \emph{Legendre Equation}
-\begin{definition}{Legendre equation}\newline
-    Setting $m=0$ in Eq.(\ref{kugel:eq:asssociated_leg_eq}), we get
-    \begin{equation}\label{kugel:eq:leg_eq}
-        (1-x^2)\frac{d^2 y}{dx} - 2x\frac{d y}{dx} + n(n+1)y(x) = 0,
-    \end{equation}
-    also known as \emph{Legendre Equation}.
-\end{definition}
-Now we can continue with the lemma
-\begin{lemma}\label{kugel:lemma_1}
-    If $y_n(x)$ is a solution of Eq.(\ref{kugel:eq:leg_eq}), then the function
-    \begin{equation*}
-        y_{m,n}(x) = (1-x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}y_n(x)
-    \end{equation*}
-    satisfies Eq.(\ref{kugel:eq:associated_leg_eq})
+By making two final cosmetic substitutions, namely $Z(z) = \Theta(\cos^{-1}z)$
+and $\lambda = n(n+1)$, we obtain what is known in the literature as the
+\emph{associated Legendre equation of order $m$}:
+\nocite{olver_introduction_2013}
+\begin{equation} \label{kugel:eqn:associated-legendre}
+  (1 - z^2)\frac{d^2 Z}{dz}
+  - 2z\frac{d Z}{dz}
+  + \left( n(n + 1) - \frac{m^2}{1 - z^2} \right) Z(z) = 0,
+  \quad
+  z \in [-1; 1], m \in \mathbb{Z}.
+\end{equation}
+
+Our new goal has therefore become to solve
+\eqref{kugel:eqn:associated-legendre}, since if we find a solution for $Z(z)$ we
+can perform the substitution backwards and get back to our eigenvalue problem.
+However, the associated Legendre equation is not any easier, so to attack the
+problem we will look for the solutions in the easier special case when $m = 0$.
+This reduces the problem because it removes the double pole, which is always
+tricky to deal with. In fact, the reduced problem when $m = 0$ is known as the
+\emph{Legendre equation}:
+\begin{equation} \label{kugel:eqn:legendre}
+  (1 - z^2)\frac{d^2 Z}{dz}
+  - 2z\frac{d Z}{dz}
+  + n(n + 1) Z(z) = 0,
+  \quad
+  z \in [-1; 1].
+\end{equation}
+
+The Legendre equation is a second order differential equation, and therefore it
+has 2 independent solutions, which are known as \emph{Legendre functions} of the
+first and second kind. For the scope of this text we will only derive a special
+case of the former that is known known as the \emph{Legendre polynomials}, since
+we only need a solution between $-1$ and $1$.
+
+\begin{lemma}[Legendre polynomials]
+  \label{kugel:lem:legendre-poly}
+  The polynomial function
+  \[
+    P_n(z) = \sum^{\lfloor n/2 \rfloor}_{k=0}
+      \frac{(-1)^k}{2^n s^k!} \frac{(2n - 2k)!}{(n - k)! (n-2k)!} z^{n - 2k}
+  \]
+  is the only finite solution of the Legendre equation
+  \eqref{kugel:eqn:legendre} when $n \in \mathbb{Z}$ and $z \in [-1; 1]$.
 \end{lemma}
-\begin{proof} [TODO: modificare la $m$ (è già usata come costante di separazione) o forse è giusta (?)]
-    To begin, we can start by differentiating $m$ times Eq.\eqref{kugel:eq:leg_eq} (which is staisfied by $y(x)$), obtaining
-    \begin{equation}\label{eq:lagrange_mderiv}
-    \frac{d^m}{dx^m}\left[ (1-x^2)\frac{d^2y}{dx^2} \right] -2 \frac{d^m}{dx^m}\left[ x\frac{dy}{dx} \right] + n(n+1)\frac{d^m}{dx^m}y=0.
-    \end{equation}
-    \emph{Leibniz's theorem} says, that if we want to differentiate $m$ times a multiplication of two functions, we can use the binomial coefficients to build up a sum. This allows us to be more compact, obtaining 
-    \begin{equation}\label{eq:leibniz}
-    \frac{d^m}{dx^m}[u(x)v(x)] = \sum_{i=0}^m \binom{n}{i} \frac{d^{m-i}u}{dx^{m-1}} \frac{d^{i}v}{dx^i}.
-    \end{equation}
-    Using Eq.\eqref{eq:leibniz} in Eq.\eqref{eq:lagrange_mderiv}, we have
-    \begin{align}
-    (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} &+ m \frac{d}{dx}(1-x^2)\frac{d^{m+1}y}{dx^{m+1}} + \frac{m(m-1)}{2}\frac{d^{2}}{dx^{2}}(1-x^2)\frac{d^{m}y}{dx^{m}} + n(n+1)\frac{d^m{}y}{dx^{m}} \nonumber \\
-    &-2\left(x\frac{d^{m+1}y}{dx^{m+1}} + m\frac{d}{dx}x\frac{d^{m}y}{dx^{m}} \right) \nonumber \\
-    &= (1-x^2)\frac{d^{m+2}y}{dx^{m+2}} -2x(m+1)\frac{d^{m+1}y}{dx^{m+1}}+(n(n+1)-m(m-1)-2m)\frac{d^{m}y}{dx^{m}}=0. \label{eq:aux_3}
-    \end{align}
-    To make the notation easier to follow, a new function can be defined
-    \begin{equation*}
-    \frac{d^{m}y}{dx^{m}} := y_m.
-    \end{equation*}
-    Eq.\eqref{eq:aux_3} now becomes
-    \begin{equation}\label{eq:1st_subs}
-    (1-x^2)\frac{d^{2}y_m}{dx^{2}} -2x(m+1)\frac{dy_m}{dx}+(n(n+1)-m(m+1))y_m=0
-    \end{equation}
-    A second function can be further defined as
-    \begin{equation*}
-    (1-x^2)^{\frac{m}{2}}\frac{d^{m}y}{dx^{m}} = (1-x^2)^{\frac{m}{2}}y_m := \hat{y}_m,
-    \end{equation*}
-    allowing to write Eq.\eqref{eq:1st_subs} as
-    \begin{equation}\label{eq:2st_subs}
-    (1-x^2)\frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] -2(m+1)x\frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] + (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0.
-    \end{equation}
-    The goal now is to compute the two terms 
-    \begin{align*}
-    \frac{d^2}{dx^2}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &=  \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\
-    &+ m\left( \frac{d\hat{y}_m}{dx} x (1-x^2)^{-\frac{m}{2}-1} + \hat{y}_m (1-x^2)^{-\frac{m}{2}-1} - \hat{y}_m x (-\frac{m}{2}-1)(1-x^2)^{-\frac{m}{2}} 2x\right) \\
-    &= \frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} + \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1}\\
-    &+ m\hat{y}_m  (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2}
-    \end{align*}
-    and
-    \begin{align*}
-    \frac{d}{dx}[\hat{y}_m(1-x^2)^{-\frac{m}{2}}] &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_m\frac{m}{2}(1-x^2)^{-\frac{m}{2}-1}2x \\
-    &= \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x,
-    \end{align*}
-    to use them in Eq.\eqref{eq:2st_subs}, obtaining
-    \begin{align*}
-    (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} (1-x^2)^{-\frac{m}{2}} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-\frac{m}{2}-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-\frac{m}{2}-1} \\ 
-    &+ m\hat{y}_m  (1-x^2)^{-\frac{m}{2}-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-\frac{m}{2}-2}\biggr] \\
-    &-2(m+1)x\left[  \frac{d\hat{y}_m}{dx}(1-x^2)^{-\frac{m}{2}} + \hat{y}_mm(1-x^2)^{-\frac{m}{2}-1}x \right] \\
-    &+ (n(n+1)-m(m+1))\hat{y}_m(1-x^2)^{-\frac{m}{2}}=0.\\
-    \end{align*}
-    We can now divide by $(1-x^2)^{-\frac{m}{2}}$, obtaining
-    \begin{align*}
-    (1-x^2)\biggl[\frac{d^2\hat{y}_m}{dx^2} &+ \frac{d\hat{y}_m}{dx}mx (1-x^2)^{-1} + m\frac{d\hat{y}_m}{dx}x (1-x^2)^{-1} + m\hat{y}_m  (1-x^2)^{-1} + m\hat{y}_m x^2(m+2)(1-x^2)^{-2}\biggr] \\
-    &-2(m+1)x\left[  \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\
-    &= \frac{d^2\hat{y}_m}{dx^2} + \frac{d\hat{y}_m}{dx}mx + m\frac{d\hat{y}_m}{dx}x + m\hat{y}_m + m\hat{y}_m x^2(m+2)(1-x^2)^{-1} \\
-    &-2(m+1)x\left[  \frac{d\hat{y}_m}{dx} + \hat{y}_mm(1-x^2)^{-1}x \right] + (n(n+1)-m(m+1))\hat{y}_m\\
-    \end{align*}
-    and collecting some terms
-    \begin{equation*}
-    (1-x^2)\frac{d^2\hat{y}_m}{dx^2} - 2x\frac{d\hat{y}_m}{dx} + \left( -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1)\right)\hat{y}_m=0.
-    \end{equation*}
-    Showing that 
-    \begin{align*}
-    -x^2 \frac{m^2}{1-x^2} + m+n(n+1)-m(m+1) &= n(n+1)- m^2 -x^2 \frac{m^2}{1-x^2} \\
-    &= n(n+1)- \frac{m}{1-x^2}
-    \end{align*}
-    implies $\hat{y}_m(x)$ being a solution of Eq.\eqref{kugel:eq:associated_leg_eq}
+\begin{proof}
+  This results is derived in section \ref{kugel:sec:proofs:legendre}.
 \end{proof}
-In simpler words, if we find a solution to Eq.\eqref{kugel:eq:leg_eq}, we can extend the latter according to the Lemma \ref{kugel:lemma_1} obtaining the solution of Eq.\eqref{kugel:eq:associated_leg_eq}.\newline
-We can say that we are going in the right direction, as the problem to be solved is decreasing in difficulty. We moved from having to find a solution to Eq.\eqref{kugel:eq:associated_leg_eq} to finding a solution to Eq.\eqref{kugel:eq:leg_eq}, which is much more approachable as a problem. Luckily for us, the lemma we will present below will help us extensively, which is something of an euphemism, since it will give us the solution directly.
-\begin{lemma}
-    The polynomial function
-    \begin{align*}
-        y_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}\\
-        &= \frac{1}{n!2^n}\frac{d^n}{dx^n}(1-x^2)^n =: P_n(x),
-    \end{align*}
-    is a solution to the second order differential equation
-    \begin{equation}\label{kugel:eq:sol_leg}
-        (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx} + n(n+1)y=0, \quad \forall n>0.
-    \end{equation}
+
+Since the Legendre \emph{polynomials} are indeed polynomials, they can also be
+expressed using the hypergeometric functions described in section
+\ref{buch:rekursion:section:hypergeometrische-funktion}, so in fact
+\begin{equation}
+  P_n(z) = {}_2F_1 \left( \begin{matrix}
+    n + 1, & -n \\ \multicolumn{2}{c}{1}
+  \end{matrix} ; \frac{1 - z}{2} \right).
+\end{equation}
+Further, there are a few more interesting but not very relevant forms to write
+$P_n(z)$ such as \emph{Rodrigues' formula} and \emph{Laplace's integral
+representation} which are
+\begin{equation*}
+  P_n(z) = \frac{1}{2^n} \frac{d^n}{dz^n} (x^2 - 1)^n,
+  \qquad \text{and} \qquad
+  P_n(z) = \frac{1}{\pi} \int_0^\pi \left(
+    z + \cos\vartheta \sqrt{z^2 - 1}
+  \right) \, d\vartheta
+\end{equation*}
+respectively, both of which we will not prove (see chapter 3 of
+\cite{bell_special_2004} for a proof). Now that we have a solution for the
+Legendre equation, we can make use of the following lemma patch the solutions
+such that they also become solutions of the associated Legendre equation
+\eqref{kugel:eqn:associated-legendre}.
+
+\begin{lemma} \label{kugel:lem:extend-legendre}
+  If $Z_n(z)$ is a solution of the Legendre equation \eqref{kugel:eqn:legendre},
+  then
+  \begin{equation*}
+    Z^m_n(z) = (1 - z^2)^{m/2} \frac{d^m}{dz^m}Z_n(z)
+  \end{equation*}
+  solves the associated Legendre equation \eqref{kugel:eqn:associated-legendre}.
+  \nocite{bell_special_2004}
 \end{lemma}
 \begin{proof}
-    In order to find a solution to Eq.\eqref{eq:legendre}, the following Ansatz can be performed:
-    \begin{equation}\label{eq:ansatz}
-    y(x) = \sum_{k=0}^\infty a_k x^k.
-    \end{equation}
-    Given Eq.\eqref{eq:ansatz}, then
-    \begin{align*}
-    \frac{dy}{dx} &= \sum_{k=0}^\infty k a_k x^{k-1}, \\
-    \frac{d^2y}{dx^2} &= \sum_{k=0}^\infty k (k-1) a_k x^{k-2}.
-    \end{align*}
-    Eq.\eqref{eq:legendre} can be therefore written as
-    \begin{align}
-    &(1-x^2)\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0 \label{eq:ansatz_in_legendre} \\
-    &=\sum_{k=0}^\infty k (k-1) a_k x^{k-2} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2x\sum_{k=0}^\infty k a_k x^{k-1} + n(n+1)\sum_{k=0}^\infty a_k x^k=0. \nonumber
-    \end{align}
-    If one consider the term
-    \begin{equation}\label{eq:term}
-    \sum_{k=0}^\infty k (k-1) a_k x^{k-2},
-    \end{equation}
-    the substitution $\tilde{k}=k-2$ yields Eq.\eqref{eq:term} to
-    \begin{equation*}
-    \sum_{\tilde{k}=-2}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}+2} x^{\tilde{k}}=\sum_{\tilde{k}=0}^\infty (\tilde{k}+2) (\tilde{k}+1) a_{\tilde{k}} x^{\tilde{k}}.
-    \end{equation*}
-    This means that Eq.\eqref{eq:ansatz_in_legendre} becomes
-    \begin{align}
-    &\sum_{k=0}^\infty (k+1)(k+2) a_{k+2} x^{k} - \sum_{k=0}^\infty k (k-1) a_k x^{k} - 2\sum_{k=0}^\infty k a_k x^k + n(n+1)\sum_{k=0}^\infty a_k x^k \nonumber \\
-    = &\sum_{k=0}^\infty \big[ (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k \big] x^k \stackrel{!}{=} 0. \label{eq:condition}
-    \end{align}
-    The condition in Eq.\eqref{eq:condition} is equivalent to 
-    \begin{equation}\label{eq:condition_2}
-    (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k = 0.
-    \end{equation}
-    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}
+  See section \ref{kugel:sec:proofs:legendre}.
 \end{proof}
-As can be seen, the solution is a $n$-dependent power series, traditionally denoted as $P_n(x)$. This set of polynomials are called \emph{Legendre Polynomials}, because precisely they are polynomials satisfying the Legendre equation.\newline
-Now that we have a solution to Eq.\eqref{kugel:eq:leg_eq}, we can then extend Eq.\eqref{kugel:eq:sol_leg}, as stated in Lemma \ref{kugel:lemma_1}. We will then have
-\begin{align*}
-y_{m,n}(x) &= (1-x^2)^{\frac{m}{2}}\frac{d^m}{dx^m}P_n(x) \\
-&= \frac{1}{n!2^n}(1-x^2)^{\frac{m}{2}}\frac{d^{m+n}}{dx^{m+n}}(1-x^2)^n 
-\end{align*}
-This set of functions are defined as \emph{Associated Legendre functions}, because similarly to before, they solve the Associated Legendre equation, defined in Eq.\eqref{kugel:eq:eq_leg}.
-\begin{definition}{Associated Legendre Functions}
-\begin{equation}\label{kugel:eq:associated_leg_func}
-P_{m,n}(x) := \frac{1}{n!2^n}(1-x^2)^{\frac{m}{2}}\frac{d^{m+n}}{dx^{m+n}}(1-x^2)^n 
-\end{equation}
+
+What is happening in lemma \ref{kugel:lem:extend-legendre}, is that we are
+essentially inserting a square root function in the solution in order to be able
+to reach the parts of the domain near the poles at $\pm 1$ of the associated
+Legendre equation, which is not possible only using power series
+\kugeltodo{Reference book theory on extended power series method.}. Now, since
+we have a solution in our domain, namely $P_n(z)$, we can insert it in the lemma 
+obtain the \emph{associated Legendre functions}.
+
+\begin{definition}[Ferrers or Associated Legendre functions]
+  The functions
+  \begin{equation}\label{kugel:eq:associated_leg_func}
+    P^m_n (z) = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m}}{dz^{m}} P_n(z)
+      = \frac{1}{n!2^n}(1-z^2)^{\frac{m}{2}}\frac{d^{m+n}}{dz^{m+n}}(1-z^2)^n 
+  \end{equation}
+  are known as Ferrers or associated Legendre functions.
 \end{definition}
+
+\subsection{Spherical Harmonics}
+
 As you may recall, previously we performed the substitution $x=\cos \vartheta$. Now we need to return to the old domain, which can be done straightforwardly:
 \begin{equation*}
     \Theta(\vartheta) = P_{m,n}(\cos \vartheta),
@@ -512,6 +362,10 @@ Ora, visto che la soluzione dell'eigenfunction problem è formata dalla moltipli
 
 \subsection{Recurrence Relations}
 
-\section{Series Expansions in \(C(S^2)\)}
+\section{Series Expansions in $C(S^2)$}
 
-\nocite{olver_introduction_2013}
+\subsection{Orthogonality of $P_n$, $P^m_n$ and $Y^m_n$}
+
+\subsection{Series Expansion}
+
+\subsection{Fourier on $S^2$}
-- 
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