From b31c40a22e885205373ff1a3284935d544c99889 Mon Sep 17 00:00:00 2001
From: canuel <cattaneo.manuel@hotmail.com>
Date: Tue, 16 Aug 2022 14:32:49 +0200
Subject: started the chapter about spherical harmonics (more specific about
 the derivation of them)

---
 buch/papers/kugel/spherical-harmonics.tex | 387 +++++++++++++++++++++++++++++-
 1 file changed, 386 insertions(+), 1 deletion(-)

(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 6b23ce5..c76e757 100644
--- a/buch/papers/kugel/spherical-harmonics.tex
+++ b/buch/papers/kugel/spherical-harmonics.tex
@@ -1,9 +1,394 @@
 % vim:ts=2 sw=2 et spell:
 
 \section{Spherical Harmonics}
+We finally arrived at the main section, which gives our chapter its name. The idea is to discuss spherical harmonics, their mathematical derivation and some of their properties and applications.\newline
+The subsection \ref{} will be devoted to the Eigenvalue problem of the Laplace operator. Through the latter, we will derive the set of Eigenfunctions that obey the equation presented in \ref{}[TODO: reference to eigenvalue equation], which will be defined as \emph{Spherical Harmonics}. In fact, this subsection will present their mathematical derivation.\newline
+In the subsection \ref{}, on the other hand, some interesting properties related to them will be discussed. Some of these will come back to help us understand in more detail why they are useful in various real-world applications, which will be presented in the section \ref{}.\newline
+One specific property will be studied in more detail in the subsection \ref{}, namely the recursive property.
+The last subsection is devoted to one of the most beautiful applications (In our humble opinion), namely the derivation of a Fourier-style series expansion but defined on the sphere instead of a plane.\newline 
+More importantly, this subsection will allow us to connect all the dots we have created with the previous sections, concluding that Fourier is just a specific case of the application of the concept of orthogonality.\newline
+Our hope is that after reading this section you will appreciate the beauty and power of generalization that mathematics offers us.
 
-\subsection{Eigenvalue Problem in Spherical Coordinates}
+\subsection{Eigenvalue Problem on the Spherical surface}
+\subsubsection{Unormalized Spherical Harmonics}
+From the chapter \ref{}, we know that the spherical Laplacian is defined as. \begin{equation*}
+    \nabla^2_S := \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r} \right) + \frac{1}{r^2} 
+    \left[
+        \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial}{\partial\vartheta} \right)
+                + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}
+    \right] 
+\end{equation*}
+But we do not want to consider this algebraic monster entirely, since this includes the whole set $\mathbb{R}^3$; rather, we want to focus only on the spherical surface (as the title suggests). We can then further concretise our calculations by selecting any number for the variable $r$, so that we have a sphere and, more importantly, a spherical surface on which we can ``play''.\newline
+Surely you have already heard of the unit circle, a geometric entity used extensively in many mathematical contexts. The most famous and basic among them is surely trigonometry.\newline
+Extending this concept into three dimensions, we will talk about the unit sphere. This is a very famous sphere, as is the unit circle. So since we need a sphere why not use the most famous one? Thus imposing $r=1$.\newline
+Now, since the variable $r$ became a constant, we can leave out all derivatives with respect to $r$, setting them to zero. Then substituting the value of $r$ for 1, we will obtain the operator we will refer to as \emph{Spherical Surface Operator}:
+\begin{equation*}
+    \nabla^2_{\partial S} := \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial}{\partial\vartheta} \right)
+                + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}.
+\end{equation*}
+As can be seen, for this definition, the subscript ``$\partial S$'' was used to emphasize the fact that we are on the spherical surface, which can be understood as a boundary of the sphere.\newline
+Now that we have defined an operator, we can go on to calculate its eigenfunctions. As mentioned earlier, we can translate this problem at first abstract into a much more concrete problem, which has to do with the field of \emph{Partial Differential Equaitons} (PDEs). The functions we want to find are simply functions that respect the following expression:
+\begin{equation}\label{kugel:eq:sph_srfc_laplace}
+    \nabla^2_{\partial S} f = \lambda f 
+\end{equation}
+Which is traditionally written as follows:
+\begin{equation*}
+    \nabla^2_{\partial S} f = -\lambda f 
+\end{equation*}
+Perhaps the fact that we are dealing with a PDE may not be obvious at first glance, but if we extend the operator $\nabla^2_{\partial S}$ according to Eq.(\ref{kugel:eq:sph_srfc_laplace}), we will get:
+\begin{equation}\label{kugel:eq:PDE_sph}
+    \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial f}{\partial\vartheta} \right)
+                + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 f}{\partial\varphi^2} + \lambda f = 0, 
+\end{equation}
+making it emerge.\newline
+All functions satisfying Eq.(\ref{kugel:eq:PDE_sph}), are called eigenfunctions. Our new goal is therefore to solve this PDE. The task seems very difficult but we can simplify it with a well-known technique, namely the \emph{separation Ansatz}. The latter consists in assuming that the function $f(\vartheta, \varphi)$ we are looking for can be factorized in the following form 
+\begin{equation}\label{kugel:eq:sep_ansatz_0}
+    f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi). 
+\end{equation}
+In short, we are saying that the effect of the two independent variables can be described using the multiplication of two functions that describe their effect separately. If we include this assumption in Eq.(\ref{kugel:eq:PDE_sph}), we have:
+\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.  \label{kugel:eq:sep_ansatz_1}
+\end{equation}
+Dividing Eq.(\ref{kugel:eq:sep_ansatz_1}) by $\Theta(\vartheta)\Phi(\varphi)$ and inserting an auxiliary variable $m$, which we will call the separating constant, we will have:
+\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,
+\end{equation*}
+which is equivalent to the following system of two \emph{Ordinary Differential Equations} (ODEs)
+\begin{align}
+    \frac{d^2\Phi(\varphi)}{d\varphi^2} &= -m \Phi(\varphi) \label{kugel:eq:ODE_1} \\ 
+    \sin \vartheta \frac{d}{d \vartheta} \left( \sin \vartheta \frac{d \Theta}{d \vartheta} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \label{kugel:eq:ODE_2} 
+\end{align}
+The solution of Eq.(\ref{kugel:eq:ODE_1}) is quite trivial. The complex exponential is obviously the function we are looking for, so we can write
+\begin{equation*}
+    \Phi_m(\varphi) = e^{j m \varphi}, \quad m \in \mathbb{Z}.
+\end{equation*}
+The restriction for the separation constant $m$ arises from the fact that we require the following periodic constraint $\Phi_m(\varphi + 2\pi) = \Phi_m(\varphi)$.\newline
+As for Eq.(\ref{kugel:eq:ODE_2}), the resolution will not be so straightforward. 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.
+\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 
+\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})
+\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}
+\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}
+\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}
+\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}
+\end{definition}
+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),
+\end{equation*}
+obtaining the much sought function $\Theta(\vartheta)$. \newline
+So we finally reached the end of this tortuous path. Now we just need to put together all the information we have to construct $f(\vartheta, \varphi)$ in the following way:
+\begin{equation}\label{kugel:eq:sph_harm_0}
+    f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi) = P_{m,n}(\cos \vartheta)e^{jm\varphi}, \quad |m|\leq n.
+\end{equation}
+The constraint $|m|<n$, can be justified by considering Eq.\eqref{kugel:eq:associated_leg_func}, in which the derivative of degree $m+n$ is present. A derivative to be well defined must have an order that is greater than zero. Furthermore, it can be seen that this derivative is applied on a polynomial of degree $2n$. As is known from Calculus 1, if you derive a polynomial of degree $2n$ more than $2n$ times, you get zero, which is a trivial solution in which we are not interested.\newline
+We can thus summarize these two conditions by writing:
+\begin{equation*}
+    \begin{rcases}
+        m+n \leq 2n &\implies m \leq n \\
+        m+n \geq 0  &\implies  m \geq -n
+    \end{rcases} |m| \leq n.
+\end{equation*}
+The set of functions in Eq.\eqref{kugel:eq:sph_harm_0} is named \emph{Spherical Harmonics}, which are the eigenfunctions of the Laplace operator on the \emph{spherical surface domain}, which is exactly what we were looking for at the beginning of this section.
+\begin{definition}{Spherical Harmonics}
+    \begin{equation}\label{kugel:eq:sph_harm_1}
+    \tilde{Y}_{m,n}(\vartheta, \varphi) := P_{m,n}(\cos \vartheta)e^{jm\varphi}, \quad |m|\leq n.
+    \end{equation}
+\end{definition}
 
+\subsection{Normalization}
+As explained in the chapter \ref{}, the concept of orthogonality is very important and at the practical level it is very useful, because it allows us to develop very powerful techniques at the mathematical level.\newline 
+Throughout this book we have been confronted with the Sturm-Liouville theory (see chapter \ref{}). The latter, among other things, carries with it the concept of orthogonality. Indeed, if we consider the solutions of the Sturm-Liouville equation, which can be expressed in this form
+\begin{equation}\label{kugel:eq:sturm_liouville}
+    \mathcal{S}f := \frac{d}{dx}\left[p(x)\frac{df}{dx}\right]+q(x)f(x)
+\end{equation}
+possiamo dire che formano una base ortogonale.\newline
+Adesso possiamo dare un occhiata alle due equazioni che abbiamo ottenuto tramite la Separation Ansatz (Eqs.\eqref{kugel:eq:associated_leg_eq}\eqref{kugel:eq:ODE_1}), le quali possono essere riscritte come:
+\begin{align*}
+    \frac{d}{dx} \left[ (1-x^2) \cdot \frac{dP_{m,n}}{dx} \right] &+ \left(n(n+1)-\frac{m}{1-x^2} \right) \cdot P_{m,n}(x) = 0, \\
+    \frac{d}{d\varphi} \left[ 1 \cdot \frac{ d\Phi }{d\varphi} \right] &+ 1 \cdot \Phi(\varphi) = 0. 
+\end{align*}
+Si può concludere in modo diretto che sono due casi dell'equazione di Sturm-Liouville. Questo significa che le loro soluzioni sono ortogonali sotto l'inner product con weight function $w(x)=1$, dunque:
+\begin{align}
+\int_{0}^{2\pi} \Phi_m(\varphi)\Phi_m'(\varphi) d\varphi &= \delta_{m'm}, \nonumber \\
+\int_{-1}^1 P_{m,m'}(x)P_{n,n'}(x) dx &= \delta_{m'm}\delta_{n'n}. \label{kugel:eq:orthogonality_associated_func}
+\end{align}
+Inoltre, possiamo provare l'ortogonalità di $\Theta(\vartheta)$ utilizzando \eqref{kugel:eq:orthogonality_associated_func}:
+\begin{align}
+    x
+\end{align}
+Ora, visto che la soluzione dell'eigenfunction problem è formata dalla moltiplicazione di $\Phi_m(\varphi)$ e $P_{m,n}(x)$
+\begin{lemma}
+
+\end{lemma}
 \subsection{Properties}
 
 \subsection{Recurrence Relations}
-- 
cgit v1.2.1


From c4cf68ac67f7fbadaacae64597ae713a6879f944 Mon Sep 17 00:00:00 2001
From: Nao Pross <np@0hm.ch>
Date: Tue, 16 Aug 2022 23:39:59 +0200
Subject: kugel: Comment out preliminaries, review manu's work until legendre

---
 buch/papers/kugel/spherical-harmonics.tex | 229 +++++++++++++++++++++++-------
 1 file changed, 174 insertions(+), 55 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 c76e757..70657c9 100644
--- a/buch/papers/kugel/spherical-harmonics.tex
+++ b/buch/papers/kugel/spherical-harmonics.tex
@@ -1,70 +1,189 @@
-% vim:ts=2 sw=2 et spell:
+% vim:ts=2 sw=2 et spell tw=80:
 
 \section{Spherical Harmonics}
-We finally arrived at the main section, which gives our chapter its name. The idea is to discuss spherical harmonics, their mathematical derivation and some of their properties and applications.\newline
-The subsection \ref{} will be devoted to the Eigenvalue problem of the Laplace operator. Through the latter, we will derive the set of Eigenfunctions that obey the equation presented in \ref{}[TODO: reference to eigenvalue equation], which will be defined as \emph{Spherical Harmonics}. In fact, this subsection will present their mathematical derivation.\newline
-In the subsection \ref{}, on the other hand, some interesting properties related to them will be discussed. Some of these will come back to help us understand in more detail why they are useful in various real-world applications, which will be presented in the section \ref{}.\newline
-One specific property will be studied in more detail in the subsection \ref{}, namely the recursive property.
-The last subsection is devoted to one of the most beautiful applications (In our humble opinion), namely the derivation of a Fourier-style series expansion but defined on the sphere instead of a plane.\newline 
-More importantly, this subsection will allow us to connect all the dots we have created with the previous sections, concluding that Fourier is just a specific case of the application of the concept of orthogonality.\newline
-Our hope is that after reading this section you will appreciate the beauty and power of generalization that mathematics offers us.
 
-\subsection{Eigenvalue Problem on the Spherical surface}
-\subsubsection{Unormalized Spherical Harmonics}
-From the chapter \ref{}, we know that the spherical Laplacian is defined as. \begin{equation*}
-    \nabla^2_S := \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r} \right) + \frac{1}{r^2} 
-    \left[
-        \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial}{\partial\vartheta} \right)
-                + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}
-    \right] 
-\end{equation*}
-But we do not want to consider this algebraic monster entirely, since this includes the whole set $\mathbb{R}^3$; rather, we want to focus only on the spherical surface (as the title suggests). We can then further concretise our calculations by selecting any number for the variable $r$, so that we have a sphere and, more importantly, a spherical surface on which we can ``play''.\newline
-Surely you have already heard of the unit circle, a geometric entity used extensively in many mathematical contexts. The most famous and basic among them is surely trigonometry.\newline
-Extending this concept into three dimensions, we will talk about the unit sphere. This is a very famous sphere, as is the unit circle. So since we need a sphere why not use the most famous one? Thus imposing $r=1$.\newline
-Now, since the variable $r$ became a constant, we can leave out all derivatives with respect to $r$, setting them to zero. Then substituting the value of $r$ for 1, we will obtain the operator we will refer to as \emph{Spherical Surface Operator}:
+\if 0
+\kugeltodo{Rewrite this section if the preliminaries become an addendum}
+We finally arrived at the main section, which gives our chapter its name. The
+idea is to discuss spherical harmonics, their mathematical derivation and some
+of their properties and applications.
+
+The subsection \ref{} \kugeltodo{Fix references} will be devoted to the
+Eigenvalue problem of the Laplace operator. Through the latter we will derive
+the set of Eigenfunctions that obey the equation presented in \ref{}
+\kugeltodo{reference to eigenvalue equation}, which will be defined as
+\emph{Spherical Harmonics}. In fact, this subsection will present their
+mathematical derivation.
+
+In the subsection \ref{}, on the other hand, some interesting properties
+related to them will be discussed. Some of these will come back to help us
+understand in more detail why they are useful in various real-world
+applications, which will be presented in the section \ref{}.
+
+One specific property will be studied in more detail in the subsection \ref{},
+namely the recursive property.  The last subsection is devoted to one of the
+most beautiful applications (In our humble opinion), namely the derivation of a
+Fourier-style series expansion but defined on the sphere instead of a plane.
+More importantly, this subsection will allow us to connect all the dots we have
+created with the previous sections, concluding that Fourier is just a specific
+case of the application of the concept of orthogonality. Our hope is that after
+reading this section you will appreciate the beauty and power of generalization
+that mathematics offers us.
+\fi
+
+\subsection{Eigenvalue Problem}
+
+\begin{figure}
+  \centering
+  \includegraphics{papers/kugel/figures/tikz/spherical-coordinates}
+  \caption{
+    Spherical coordinate system. Space is described with the free variables $r
+    \in \mathbb{R}_0^+$, $\vartheta \in [0; \pi]$ and $\varphi \in [0; 2\pi)$.
+    \label{kugel:fig:spherical-coordinates}
+  }
+\end{figure}
+
+From Section \ref{buch:pde:section:kugel}, we know that the spherical Laplacian
+in the spherical coordinate system (shown in Figure
+\ref{kugel:fig:spherical-coordinates}) is is defined as
 \begin{equation*}
-    \nabla^2_{\partial S} := \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial}{\partial\vartheta} \right)
-                + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}.
+    \sphlaplacian :=
+      \frac{1}{r^2} \frac{\partial}{\partial r} \left(
+        r^2 \frac{\partial}{\partial r}
+      \right)
+      + \frac{1}{r^2} \left[
+          \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left(
+            \sin\vartheta \frac{\partial}{\partial\vartheta}
+          \right)
+        + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2}
+      \right].
 \end{equation*}
-As can be seen, for this definition, the subscript ``$\partial S$'' was used to emphasize the fact that we are on the spherical surface, which can be understood as a boundary of the sphere.\newline
-Now that we have defined an operator, we can go on to calculate its eigenfunctions. As mentioned earlier, we can translate this problem at first abstract into a much more concrete problem, which has to do with the field of \emph{Partial Differential Equaitons} (PDEs). The functions we want to find are simply functions that respect the following expression:
-\begin{equation}\label{kugel:eq:sph_srfc_laplace}
-    \nabla^2_{\partial S} f = \lambda f 
+But we will not consider this algebraic monstrosity in its entirety. As the
+title suggests, we will only care about the \emph{surface} of the sphere.  This
+is for many reasons, but mainly to simplify reduce the already broad scope of
+this text. Concretely, we will always work on the unit sphere, which just means
+that we set $r = 1$ and keep only $\vartheta$ and $\varphi$ as free variables.
+Now, since the variable $r$ became a constant, we can leave out all derivatives
+with respect to $r$ and substitute all $r$'s with 1's to obtain a new operator
+that deserves its own name.
+
+\begin{definition}[Surface spherical Laplacian]
+  \label{kugel:def:surface-laplacian}
+  The operator
+  \begin{equation*}
+      \surflaplacian :=
+        \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left(
+          \sin\vartheta \frac{\partial}{\partial\vartheta}
+        \right)
+        + \frac{1}{\sin^2 \vartheta} \frac{\partial^2}{\partial\varphi^2},
+  \end{equation*}
+  is called the surface spherical Laplacian.
+\end{definition}
+
+In the definition, the subscript ``$\partial S$'' was used to emphasize the
+fact that we are on the spherical surface, which can be understood as being the
+boundary of the sphere. But what does it actually do? To get an intuition,
+first of all, notice the fact that $\surflaplacian$ have second derivatives,
+which means that this a measure of \emph{curvature}; But curvature of what? To
+get an even stronger intuition we will go into geometry, were curvature can be
+grasped very well visually. Consider figure \ref{kugel:fig:curvature} where the
+curvature is shown using colors. First we have the curvature of a curve in 1D,
+then the curvature of a surface (2D), and finally the curvature of a function on
+the surface of the unit sphere.
+
+\begin{figure}
+  \centering
+  \includegraphics[width=.3\linewidth]{papers/kugel/figures/tikz/curvature-1d}
+  \hskip 5mm
+  \includegraphics[width=.3\linewidth]{papers/kugel/figures/povray/curvature}
+  \hskip 5mm
+  \includegraphics[width=.3\linewidth]{papers/kugel/figures/povray/spherecurve}
+  \caption{
+    \kugeltodo{Fix alignment / size, add caption. Would be nice to match colors.}
+    \label{kugel:fig:curvature}
+  }
+\end{figure}
+
+Now that we have defined an operator, we can go and study its eigenfunctions,
+which means that we would like to find the functions $f(\vartheta, \varphi)$
+that satisfy the equation
+\begin{equation} \label{kuvel:eqn:eigen}
+    \surflaplacian f = -\lambda f.
 \end{equation}
-Which is traditionally written as follows:
-\begin{equation*}
-    \nabla^2_{\partial S} f = -\lambda f 
-\end{equation*}
-Perhaps the fact that we are dealing with a PDE may not be obvious at first glance, but if we extend the operator $\nabla^2_{\partial S}$ according to Eq.(\ref{kugel:eq:sph_srfc_laplace}), we will get:
-\begin{equation}\label{kugel:eq:PDE_sph}
-    \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left( \sin\vartheta \frac{\partial f}{\partial\vartheta} \right)
-                + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 f}{\partial\varphi^2} + \lambda f = 0, 
+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
+$\nabla^2_{\partial S}$ according to definition
+\ref{kugel:def:surface-laplacian}, we get:
+\begin{equation} \label{kugel:eqn:eigen-pde}
+    \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left(
+      \sin\vartheta \frac{\partial f}{\partial\vartheta}
+    \right)
+    + \frac{1}{\sin^2 \vartheta} \frac{\partial^2 f}{\partial\varphi^2}
+    + \lambda f = 0.
 \end{equation}
-making it emerge.\newline
-All functions satisfying Eq.(\ref{kugel:eq:PDE_sph}), are called eigenfunctions. Our new goal is therefore to solve this PDE. The task seems very difficult but we can simplify it with a well-known technique, namely the \emph{separation Ansatz}. The latter consists in assuming that the function $f(\vartheta, \varphi)$ we are looking for can be factorized in the following form 
-\begin{equation}\label{kugel:eq:sep_ansatz_0}
+Since all functions satisfying \eqref{kugel:eqn:eigen-pde} are the
+\emph{eigenfunctions} of $\surflaplacian$, our new goal is to solve this PDE.
+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}
     f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi). 
 \end{equation}
-In short, we are saying that the effect of the two independent variables can be described using the multiplication of two functions that describe their effect separately. If we include this assumption in Eq.(\ref{kugel:eq:PDE_sph}), we have:
-\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.  \label{kugel:eq:sep_ansatz_1}
+In other words, we are saying that the effect of the two independent variables
+can be described using the multiplication of two functions that describe their
+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}
+    \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}
-Dividing Eq.(\ref{kugel:eq:sep_ansatz_1}) by $\Theta(\vartheta)\Phi(\varphi)$ and inserting an auxiliary variable $m$, which we will call the separating constant, we will have:
-\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,
-\end{equation*}
-which is equivalent to the following system of two \emph{Ordinary Differential Equations} (ODEs)
-\begin{align}
-    \frac{d^2\Phi(\varphi)}{d\varphi^2} &= -m \Phi(\varphi) \label{kugel:eq:ODE_1} \\ 
-    \sin \vartheta \frac{d}{d \vartheta} \left( \sin \vartheta \frac{d \Theta}{d \vartheta} \right) + \left( \lambda - \frac{m}{\sin^2 \vartheta} \right)\Theta(\vartheta) &= 0 \label{kugel:eq:ODE_2} 
-\end{align}
-The solution of Eq.(\ref{kugel:eq:ODE_1}) is quite trivial. The complex exponential is obviously the function we are looking for, so we can write
+Dividing by $\Theta(\vartheta)\Phi(\varphi)$ and introducing an auxiliary
+variable $m$, the separation constant, yields:
 \begin{equation*}
-    \Phi_m(\varphi) = e^{j m \varphi}, \quad m \in \mathbb{Z}.
+  \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,
 \end{equation*}
-The restriction for the separation constant $m$ arises from the fact that we require the following periodic constraint $\Phi_m(\varphi + 2\pi) = \Phi_m(\varphi)$.\newline
-As for Eq.(\ref{kugel:eq:ODE_2}), the resolution will not be so straightforward. We can begin by considering the substitution $x = \cos \vartheta$. The operator $\frac{d}{d \vartheta}$ will be:
+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),
+      \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)
+      \Theta(\vartheta) = 0
+      \label{kugel:eqn:ode-theta}.
+  \end{gather}
+\end{subequations}
+The solution of \eqref{kugel:eqn:ode-phi} is easy to find: The complex
+exponential is obviously the function we are looking for. So we can directly
+write the solutions
+\begin{equation} \label{kugel:eqn:ode-phi-sol}
+    \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.
+
+\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}.
-- 
cgit v1.2.1


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 $S^2$}
-- 
cgit v1.2.1


From 494636b6d00b0697bda4c5840a3666b0867f22e8 Mon Sep 17 00:00:00 2001
From: Nao Pross <np@0hm.ch>
Date: Wed, 17 Aug 2022 17:18:13 +0200
Subject: kugel: Minor changes

---
 buch/papers/kugel/spherical-harmonics.tex | 87 ++++++++++++++++++++++---------
 1 file changed, 63 insertions(+), 24 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 5645941..2ded50b 100644
--- a/buch/papers/kugel/spherical-harmonics.tex
+++ b/buch/papers/kugel/spherical-harmonics.tex
@@ -2,8 +2,8 @@
 
 \section{Construction of the Spherical Harmonics}
 
-\if 0
-\kugeltodo{Rewrite this section if the preliminaries become an addendum}
+\kugeltodo{Review text, or rewrite if preliminaries becomes an addendum}
+
 We finally arrived at the main section, which gives our chapter its name. The
 idea is to discuss spherical harmonics, their mathematical derivation and some
 of their properties and applications.
@@ -29,9 +29,9 @@ created with the previous sections, concluding that Fourier is just a specific
 case of the application of the concept of orthogonality. Our hope is that after
 reading this section you will appreciate the beauty and power of generalization
 that mathematics offers us.
-\fi
 
 \subsection{Eigenvalue Problem}
+\label{kugel:sec:construction:eigenvalue}
 
 \begin{figure}
   \centering
@@ -111,8 +111,9 @@ 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) \kugeltodo{Boundary conditions?}. If we unpack the notation of the operator
-$\nabla^2_{\partial S}$ according to definition
+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}
     \frac{1}{\sin\vartheta} \frac{\partial}{\partial \vartheta} \left(
@@ -139,7 +140,8 @@ convenience. If we substitute this assumption in
     \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}
+    + \frac{1}{\sin^2 \vartheta}
+      \frac{\partial^2 \Phi(\varphi)}{\partial\varphi^2}
       \Theta(\vartheta)
     + \lambda \Theta(\vartheta)\Phi(\varphi) = 0.
 \end{equation*}
@@ -182,6 +184,14 @@ require a dedicated section of its own.
 
 \subsection{Legendre Functions}
 
+\begin{figure}
+  \centering
+  \kugelplaceholderfig{.8\textwidth}{5cm}
+  \caption{
+    \kugeltodo{Why $z = \cos \vartheta$.}
+  }
+\end{figure}
+
 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
@@ -298,26 +308,19 @@ Legendre equation, which is not possible only using power series
 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]
+\begin{definition}[Ferrers or associated Legendre functions]
+  \label{kugel:def:ferrers-functions}
   The functions
-  \begin{equation}\label{kugel:eq:associated_leg_func}
+  \begin{equation}
     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}
+\kugeltodo{Discuss $|m| \leq n$.}
 
-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),
-\end{equation*}
-obtaining the much sought function $\Theta(\vartheta)$. \newline
-So we finally reached the end of this tortuous path. Now we just need to put together all the information we have to construct $f(\vartheta, \varphi)$ in the following way:
-\begin{equation}\label{kugel:eq:sph_harm_0}
-    f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi) = P_{m,n}(\cos \vartheta)e^{jm\varphi}, \quad |m|\leq n.
-\end{equation}
+\if 0
 The constraint $|m|<n$, can be justified by considering Eq.\eqref{kugel:eq:associated_leg_func}, in which the derivative of degree $m+n$ is present. A derivative to be well defined must have an order that is greater than zero. Furthermore, it can be seen that this derivative is applied on a polynomial of degree $2n$. As is known from Calculus 1, if you derive a polynomial of degree $2n$ more than $2n$ times, you get zero, which is a trivial solution in which we are not interested.\newline
 We can thus summarize these two conditions by writing:
 \begin{equation*}
@@ -327,13 +330,50 @@ We can thus summarize these two conditions by writing:
     \end{rcases} |m| \leq n.
 \end{equation*}
 The set of functions in Eq.\eqref{kugel:eq:sph_harm_0} is named \emph{Spherical Harmonics}, which are the eigenfunctions of the Laplace operator on the \emph{spherical surface domain}, which is exactly what we were looking for at the beginning of this section.
-\begin{definition}{Spherical Harmonics}
-    \begin{equation}\label{kugel:eq:sph_harm_1}
-    \tilde{Y}_{m,n}(\vartheta, \varphi) := P_{m,n}(\cos \vartheta)e^{jm\varphi}, \quad |m|\leq n.
-    \end{equation}
+\fi
+
+\subsection{Spherical Harmonics}
+
+Finally, we can go back to solving our boundary value problem we started in
+section \ref{kugel:sec:construction:eigenvalue}. We had left off in the middle
+of the separation, were we had used the Ansatz $f(\vartheta, \varphi) =
+\Theta(\vartheta) \Phi(\varphi)$ to find that $\Phi(\varphi) = e^{im\varphi}$,
+and we were solving for $\Theta(\vartheta)$.  As you may recall, previously we
+performed the substitution $z = \cos \vartheta$. Now we can finally to bring back the
+solution to the associated Legendre equation $P^m_n(z)$ into the $\vartheta$
+domain and combine it with $\Phi(\varphi)$ to get the full result:
+\begin{equation*}
+    f(\vartheta, \varphi)
+      = \Theta(\vartheta)\Phi(\varphi)
+      = P^m_n (\cos \vartheta) e^{im\varphi}.
+\end{equation*}
+This family of functions, which recall are the solutions of the eigenvalue
+problem of the surface spherical Laplacian, are the long anticipated
+\emph{complex spherical harmonics}, and they are usually denoted with
+$Y^m_n(\vartheta, \varphi)$.
+
+\begin{definition}[Spherical harmonics]
+  \label{kugel:def:spherical-harmonics}
+  The functions
+  \begin{equation*}
+    Y_{m,n}(\vartheta, \varphi) = P^m_n(\cos \vartheta) e^{im\varphi},
+  \end{equation*}
+  where $m, n \in \mathbb{Z}$ and $|m| < n$ are called spherical harmonics.
 \end{definition}
 
+\begin{figure}
+  \centering
+  \kugelplaceholderfig{\textwidth}{.8\paperheight}
+  \caption{
+    \kugeltodo{Big picture with the first few spherical harmonics.}
+  }
+\end{figure}
+
 \subsection{Normalization}
+
+\kugeltodo{Discuss various normalizations.}
+
+\if 0
 As explained in the chapter \ref{}, the concept of orthogonality is very important and at the practical level it is very useful, because it allows us to develop very powerful techniques at the mathematical level.\newline 
 Throughout this book we have been confronted with the Sturm-Liouville theory (see chapter \ref{}). The latter, among other things, carries with it the concept of orthogonality. Indeed, if we consider the solutions of the Sturm-Liouville equation, which can be expressed in this form
 \begin{equation}\label{kugel:eq:sturm_liouville}
@@ -355,9 +395,8 @@ Inoltre, possiamo provare l'ortogonalità di $\Theta(\vartheta)$ utilizzando \eq
     x
 \end{align}
 Ora, visto che la soluzione dell'eigenfunction problem è formata dalla moltiplicazione di $\Phi_m(\varphi)$ e $P_{m,n}(x)$
-\begin{lemma}
+\fi
 
-\end{lemma}
 \subsection{Properties}
 
 \subsection{Recurrence Relations}
-- 
cgit v1.2.1

