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-% vim:ts=2 sw=2 et spell:
+% vim:ts=2 sw=2 et spell tw=80:
-\section{Spherical Harmonics}
+\section{Construction of the Spherical Harmonics}
-\subsection{Eigenvalue Problem in Spherical Coordinates}
+\kugeltodo{Review text, or rewrite if preliminaries becomes an addendum}
-\subsection{Properties}
+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.
-\subsection{Recurrence Relations}
+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.
-\section{Series Expansions in \(C(S^2)\)}
+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.
+
+\subsection{Eigenvalue Problem}
+\label{kugel:sec:construction:eigenvalue}
+
+\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*}
+ \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*}
+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{kugel:eqn:eigen}
+ \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)\footnote{
+ Considering the fact that we are dealing with a PDE,
+ you may be wondering what are the boundary conditions. Well, since this eigenvalue problem is been developed on
+ the spherical surface (boundary of a sphere), the boundary in this case are empty, i.e no boundary condition has to be considered.}.
+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}
+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}
+ f(\vartheta, \varphi) = \Theta(\vartheta)\Phi(\varphi).
+\end{equation}
+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*}
+ \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 by $\Theta(\vartheta)\Phi(\varphi)$ and introducing an auxiliary
+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^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^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^2}{\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 the
+coordinate systems requires that $\Phi(\varphi + 2\pi) = \Phi(\varphi)$.
+Unfortunately, solving \eqref{kugel:eqn:ode-theta} is not 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}
+
+\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
+\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-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 $Z(z) = \Theta(\cos^{-1}z)$
+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^2}
+ - 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^2}
+ - 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:thm: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}
+ This results is derived in section \ref{kugel:sec:proofs:legendre}.
+\end{proof}
+
+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 n!} \frac{d^n}{dz^n} (z^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 to patch the solutions
+such that they also become solutions of the associated Legendre equation
+\eqref{kugel:eqn:associated-legendre}.
+
+\begin{lemma} \label{kugel:thm: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}
+ See section \ref{kugel:sec:proofs:legendre}.
+\end{proof}
+
+What is happening in lemma \ref{kugel:thm: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]
+ \label{kugel:def:ferrers-functions}
+ The functions
+ \begin{equation}
+ P^m_n (z) = (1-z^2)^{\frac{m}{2}}\frac{d^{m}}{dz^{m}} P_n(z)
+ = \frac{1}{2^n n!}(1-z^2)^{\frac{m}{2}}\frac{d^{m+n}}{dz^{m+n}}(1-z^2)^n, \quad |m|<n
+ \end{equation}
+ are known as Ferrers or associated Legendre functions.
+\end{definition}
+The constraint $|m|<n$, can be justified by considering eq.\eqref{kugel:eq:associated_leg_func}, where we differentiate $m+n$ times. We all know that a differentiation, to be well defined, must have an order that is greater than zero \kugeltodo{is that always true?}. 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, that would be a trivial solution. This is the power of zero: It is almost always a (boring) solution.
+
+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*}
+
+\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 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}, \quad |m|<n.
+\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}, \quad |m|<n
+ \end{equation*}
+ where $m, n \in \mathbb{Z}$ are called (unnormalized) spherical
+ harmonics.
+\end{definition}
+
+\begin{figure}
+ \centering
+ \kugelplaceholderfig{\textwidth}{.8\paperheight}
+ \caption{
+ \kugeltodo{Big picture with the first few spherical harmonics.}
+ \label{kugel:fig:spherical-harmonics}
+ }
+\end{figure}
+
+\kugeltodo{Describe how they look like with fig.
+\ref{kugel:fig:spherical-harmonics}}
+
+\subsection{Orthogonality of $P_n$, $P^m_n$ and $Y^m_n$}
+
+We shall now discuss an important property of the spherical harmonics: they form
+an orthogonal system. And since the spherical harmonics contain the Ferrers or
+associated Legendre functions, we need to discuss their orthogonality first.
+But the Ferrers functions themselves depend on the Legendre polynomials, so that
+will be our starting point.
+
+\begin{lemma} For the Legendre polynomials $P_n(z)$ and $P_k(z)$ it holds that
+ \label{kugel:thm:legendre-poly-ortho}
+ \begin{equation*}
+ \int_{-1}^1 P_n(z) P_k(z) \, dz
+ = \frac{2}{2n + 1} \delta_{nk}
+ = \begin{cases}
+ \frac{2}{2n + 1} & \text{if } n = k, \\
+ 0 & \text{otherwise}.
+ \end{cases}
+ \end{equation*}
+\end{lemma}
+\begin{proof}
+ To start, consider the fact that the Legendre equation
+ \eqref{kugel:eqn:legendre}, of which two distinct Legendre polynomials
+ $P_n(z)$ and $P_k(z)$ are a solution ($n \neq k$), can be rewritten in the
+ following form:
+ \begin{equation}
+ \frac{d}{dz} \left[
+ \left( 1 - z^2 \right) \frac{dZ}{dz}
+ \right] + n(n+1) Z(z) = 0.
+ \end{equation}
+ So we rewrite the Legendre equations for $P_n(z)$ and $P_k(z)$:
+ \begin{align*}
+ \frac{d}{dz} \left[
+ \left( 1 - z^2 \right) \frac{dP_n}{dz}
+ \right] + n(n+1) P_n(z) &= 0,
+ &
+ \frac{d}{dz} \left[
+ \left( 1 - z^2 \right) \frac{dP_k}{dz}
+ \right] + k(k+1) P_k(z) &= 0,
+ \end{align*}
+ then we multiply the former by $P_k(z)$ and the latter by $P_n(z)$ and
+ subtract the two to get
+ \begin{equation*}
+ \frac{d}{dz} \left[
+ \left( 1 - z^2 \right) \frac{dP_n}{dz}
+ \right] P_k(z) + n(n+1) P_n(z) P_k(z)
+ -
+ \frac{d}{dz} \left[
+ \left( 1 - z^2 \right) \frac{dP_k}{dz}
+ \right] P_n(z) - k(k+1) P_k(z) P_n(z) = 0.
+ \end{equation*}
+ By grouping terms, making order and integrating with respect to $z$ from $-1$
+ to 1 we obtain
+ \begin{gather}
+ \int_{-1}^1 \left\{
+ \frac{d}{dz} \left[
+ \left( 1 - z^2 \right) \frac{dP_n}{dz}
+ \right] P_k(z)
+ -
+ \frac{d}{dz} \left[
+ \left( 1 - z^2 \right) \frac{dP_k}{dz}
+ \right] P_n(z) - k(k+1) P_k(z) P_n(z)
+ \right\} \,dz \nonumber \\
+ + \left[ n(n+1) - k(k+1) \right] \int_{-1}^1 P_k(z) P_n(z) \, dz = 0.
+ \label{kugel:thm:legendre-poly-ortho:proof:1}
+ \end{gather}
+ Since by the product rule
+ \begin{equation*}
+ \frac{d}{dz} \left[ (1 - z^2) \frac{dP_k}{dz} P_n(z) \right]
+ =
+ \frac{d}{dz} \left[ (1 - z^2) \frac{dP_n}{dz} \right] P_k(z)
+ + (1 - z^2) \frac{dP_n}{dz} \frac{dP_k}{dz},
+ \end{equation*}
+ we can simplify the first term in
+ \eqref{kugel:thm:legendre-poly-ortho:proof:1} to get
+ \begin{gather*}
+ \int_{-1}^1 \left\{
+ \frac{d}{dz} \left[ (1 - z^2) \frac{dP_k}{dz} P_n(z) \right]
+ - \cancel{(1 - z^2) \frac{dP_n}{dz} \frac{dP_k}{dz}}
+ - \frac{d}{dz} \left[ (1 - z^2) \frac{dP_n}{dz} P_k(z) \right]
+ + \cancel{(1 - z^2) \frac{dP_k}{dz} \frac{dP_n}{dz}}
+ \right\} \, dz \\
+ = \int_{-1}^1 \frac{d}{dz} \left\{ (1 - z^2) \left[
+ \frac{dP_k}{dz} P_n(z) - \frac{dP_n}{dz} P_k(z)
+ \right] \right\} \, dz
+ = (1 - z^2) \left[
+ \frac{dP_k}{dz} P_n(z) - \frac{dP_n}{dz} P_k(z)
+ \right] \Bigg|_{-1}^1,
+ \end{gather*}
+ which always equals 0 because the product contains $1 - z^2$ and the bounds
+ are at $\pm 1$. Thus, of \eqref{kugel:thm:legendre-poly-ortho:proof:1} only
+ the second term remains and the equation becomes
+ \begin{equation*}
+ \left[ n(n+1) - k(k+1) \right] \int_{-1}^1 P_k(z) P_n(z) \, dz = 0.
+ \end{equation*}
+ By dividing by the constant in front of the integral we have our first result.
+ Now we need to show that when $n = k$ the integral equals $2 / (2n + 1)$.
+ % \begin{equation*}
+ % \end{equation*}
+ \kugeltodo{Finish proof. Can we do it without the generating function of
+ $P_n$?}
+\end{proof}
+
+In a similarly algebraically tedious fashion, we can also continue to check for
+orthogonality for the Ferrers functions $P^m_n(z)$, since they are related to
+$P_n(z)$ by a $m$-th derivative, and obtain the following result.
+
+\begin{lemma} For the associated Legendre functions
+ \label{kugel:thm:associated-legendre-ortho}
+ \begin{equation*}
+ \int_{-1}^1 P^m_n(z) P^{m}_{n'}(z) \, dz
+ = \frac{2(m + n)!}{(2n + 1)(n - m)!} \delta_{nn'}
+ = \begin{cases}
+ \frac{2(m + n)!}{(2n + 1)(n - m)!}
+ & \text{if } n = n', \\
+ 0 & \text{otherwise}.
+ \end{cases}
+ \end{equation*}
+\end{lemma}
+\begin{proof}
+ To show that the expression equals zero when $n \neq n'$ we can perform
+ exactly the same steps as in the proof of lemma
+ \ref{kugel:thm:legendre-poly-ortho}, so we will not repeat them here and prove
+ instead only the case when $n = n'$.
+ \kugeltodo{Finish proof, or not? I have to look and decide if it is
+ interesting enough.}
+\end{proof}
+
+By having the orthogonality relations of the Legendre functions we can finally
+show that spherical harmonics are also orthogonal under the following inner
+product:
+
+\begin{definition}[Inner product in $S^2$]
+ \label{kugel:def:inner-product-s2}
+ For two complex valued functions $f(\vartheta, \varphi)$ and $g(\vartheta,
+ \varphi)$ on the surface of the sphere the inner product is defined to be
+ \begin{equation*}
+ \langle f, g \rangle
+ = \int_{0}^\pi \int_0^{2\pi}
+ f(\vartheta, \varphi) \overline{g(\vartheta, \varphi)}
+ \sin \vartheta \, d\varphi \, d\vartheta.
+ \end{equation*}
+\end{definition}
+
+
+\begin{theorem} For the (unnormalized) spherical harmonics
+ \label{kugel:thm:spherical-harmonics-ortho}
+ \begin{align}
+ \langle Y^m_n, Y^{m'}_{n'} \rangle
+ &= \int_{0}^\pi \int_0^{2\pi}
+ Y^m_n(\vartheta, \varphi) \overline{Y^{m'}_{n'}(\vartheta, \varphi)}
+ \sin \vartheta \, d\varphi \, d\vartheta
+ \label{kugel:eq:spherical-harmonics-inner-prod} \\
+ &= \frac{4\pi}{2n + 1} \frac{(m + n)!}{(n - m)!} \delta_{nn'} \delta_{mm'}
+ = \begin{cases}
+ \frac{4\pi}{2n + 1} \frac{(m + n)!}{(n - m)!}
+ & \text{if } n = n' \text{ and } m = m', \nonumber \\
+ 0 & \text{otherwise}.
+ \end{cases}
+ \end{align}
+\end{theorem}
+\begin{proof}
+ We will begin by doing a bit of algebraic maipulaiton:
+ \begin{align*}
+ \int_{0}^\pi \int_0^{2\pi}
+ Y^m_n(\vartheta, \varphi) \overline{Y^{m'}_{n'}(\vartheta, \varphi)}
+ \sin \vartheta \, d\varphi \, d\vartheta
+ &= \int_{0}^\pi \int_0^{2\pi}
+ e^{im\varphi} P^m_n(\cos \vartheta)
+ e^{-im'\varphi} P^{m'}_{n'}(\cos \vartheta)
+ \, d\varphi \sin \vartheta \, d\vartheta
+ \\
+ &= \int_{0}^\pi
+ P^m_n(\cos \vartheta) P^{m'}_{n'}(\cos \vartheta) \sin \vartheta \, d\vartheta
+ \int_0^{2\pi} e^{i(m - m')\varphi}
+ \, d\varphi.
+ \end{align*}
+ First, notice that the associated Legendre polynomials are assumed to be real,
+ and are thus unaffected by the complex conjugation. Then, we can see that when
+ $m = m'$ the inner integral simplifies to $\int_0^{2\pi} 1 \, d\varphi$ which
+ equals $2\pi$, so in this case the expression becomes
+ \begin{equation*}
+ 2\pi \int_{0}^\pi
+ P^m_n(\cos \vartheta) P^{m'}_{n'}(\cos \vartheta)
+ \sin \vartheta \, d\vartheta
+ = -2\pi \int_{1}^{-1} P^m_n(z) P^{m'}_{n'}(z) \, dz
+ = \frac{4\pi(m + n)!}{(2n + 1)(n - m)!} \delta_{nn'},
+ \end{equation*}
+ where in the second step we performed the substitution $z = \cos\vartheta$;
+ $d\vartheta = \frac{d\vartheta}{dz} dz= - dz / \sin \vartheta$, and then we
+ used lemma \ref{kugel:thm:associated-legendre-ortho}.
+ We are allowed to use
+ the lemma because $m = m'$. After the just mentioned substitution we can write eq.\eqref{kugel:eq:spherical-harmonics-inner-prod} in this form
+ \begin{equation*}
+ \langle Y^m_n, Y^{m'}_{n'} \rangle_{\partial S} = \langle P^m_n, P^{m'}_{n'} \rangle_z \; \langle e^{im\varphi}, e^{-im'\varphi} \rangle_\varphi.
+ \end{equation*}
+ Now we just need look at the case when $m \neq m'$. Fortunately this is
+ easier: the inner integral is $\int_0^{2\pi} e^{i(m - m')\varphi} d\varphi$,
+ or in other words we are integrating a complex exponential over the entire
+ period, which always results in zero. Thus, we do not need to do anything and
+ the proof is complete.
+\end{proof}
+
+These proofs for the various orthogonality relations were quite long and
+algebraically tedious, mainly because they are ``low level'', by which we mean
+that they (arguably) do not rely on very abstract theory. However, if we allow
+ourselves to use the more abstract Sturm Liouville theory discussed in chapters
+\ref{buch:integrale:subsection:sturm-liouville-problem} and \kugeltodo{reference
+to chapter 17 of haddouche and Löffler} the proofs can become ridiculously
+short. Let's do for example lemma \ref{kugel:thm:associated-legendre-ortho}.
+
+\begin{proof}[
+ Shorter proof of lemma \ref{kugel:thm:associated-legendre-ortho}
+ ]
+ The associated Legendre polynomials, of which we would like to prove an
+ orthogonality relation, are the solution to the associated Legendre equation,
+ which we can write as $LZ(z) = 0$, where
+ \begin{equation*}
+ L = \frac{d}{dz} (1 - z^2) \frac{d}{dz}
+ + n(n+1) - \frac{m^2}{1 - z^2}.
+ \end{equation*}
+ Notice that $L$ is in fact a Sturm-Liouville operator of the form
+ \begin{equation*}
+ L = \frac{1}{w(z)} \left[
+ \frac{d}{dz} p(z) \frac{d}{dz} - \lambda + q(z)
+ \right],
+ \end{equation*}
+ if we let $w(z) = 1$, $p(z) = (1 - z^2 )$, $q(z) = -m^2 / (1 - z^2)$, and
+ $\lambda = -n(n+1)$. By the theory of Sturm-Liouville operators, we know that
+ the each solution of the problem $LZ(z) = 0$, namely $P^m_n(z)$, is orthogonal
+ to every other solution that has a different $\lambda$. In our case $\lambda$
+ varies with $n$, so $P^m_n(z)$ with different $n$'s are orthogonal to each
+ other.
+\end{proof}
+
+But that was still rather informative and had a bit of explanation, which is
+terrible. Real snobs, such as Wikipedia contributors, some authors and
+regrettably sometimes even ourselves, would write instead:
+
+\begin{proof}[
+ Infuriatingly short proof of lemma \ref{kugel:thm:associated-legendre-ortho}
+ ]
+ The associated Legendre polynomials are solutions of the associated Legendre
+ equation which is a Sturm-Liouville problem and are thus orthogonal to each
+ other. The factor in front Kronecker delta is left as an exercise to the
+ reader.
+\end{proof}
+
+Lemma \ref{kugel:thm:legendre-poly-ortho} has a very similar proof, while the theorem \ref{kugel:thm:spherical-harmonics-ortho} for the spherical harmonics is proved by the following argument.
+The spherical harmonics are the solutions to the eigenvalue problem $\surflaplacian f = -\lambda f$,
+which as discussed in the previous section is solved using the separation Ansatz. So to
+prove their orthogonality using the Sturm-Liouville theory we argue that
+\begin{equation*}
+ \surflaplacian = L_\vartheta L_\varphi \iff
+ \surflaplacian f(\vartheta, \varphi)
+ = L_\vartheta \Theta(\vartheta) L_\varphi \Phi(\varphi),
+\end{equation*}
+then we show that both $L_\vartheta$ and $L_\varphi$ are both Sturm-Liouville
+operators (we just did the former in the shorter proof above). Since both are
+Sturm-Liouville operators their combination, the surface spherical Laplacian, is
+also a Sturm-Liouville operator, which then implies orthogonality.
+
+\subsection{Normalization and the Phase Factor}
+
+At this point we have shown that the spherical harmonics form an orthogonal
+system, but in many applications we usually also want a normalization of some
+kind. For example the most obvious desirable property could be for the spherical
+harmonics to be ortho\emph{normal}, by which we mean that $\langle Y^m_n,
+Y^{m'}_{n'} \rangle = \delta_{nn'}$. To obtain orthonormality, we simply add an
+ugly normalization factor in front of the previous definition
+\ref{kugel:def:spherical-harmonics} as follows.
+
+\begin{definition}[Orthonormal spherical harmonics]
+ \label{kugel:def:spherical-harmonics-orthonormal}
+ The functions
+ \begin{equation*}
+ Y^m_n(\vartheta, \varphi)
+ = \sqrt{\frac{2n + 1}{4\pi} \frac{(n-m)!}{(m+n)!}}
+ P^m_n(\cos \vartheta) e^{im\varphi}
+ \end{equation*}
+ where $m, n \in \mathbb{Z}$ and $|m| < n$ are the orthonormal spherical
+ harmonics.
+\end{definition}
+
+Orthornomality is very useful, but it is not the only common normalization that
+is found in the literature. In physics, geomagnetism to be more specific, it is
+common to use the so called Schmidt semi-normalization (or sometimes also called
+quasi-normalization).
+
+\begin{definition}[Schmidt semi-normalized spherical harmonics]
+ \label{kugel:def:spherical-harmonics-schmidt}
+ The Schmidt semi-normalized spherical harmonics are
+ \begin{equation*}
+ Y^m_n(\vartheta, \varphi)
+ = \sqrt{2 \frac{(n - m)!}{(n + m)!}}
+ P^m_n(\cos \vartheta) e^{im\varphi}
+ \end{equation*}
+ where $m, n \in \mathbb{Z}$ and $|m| < n$.
+\end{definition}
+
+Additionally, there is another quirk in the literature that should be mentioned.
+In some other branches of physics such as seismology and quantum mechanics there
+is a so called Condon-Shortley phase factor $(-1)^m$ in front of the square root
+in the definition of the normalized spherical harmonics. It is yet another
+normalization that is added for physical reasons that are not very relevant to
+our discussion, but we mention this potential source of confusion since many
+numerical packages (such as \texttt{SHTOOLS} \kugeltodo{Reference}) offer an
+option to add or remove it from the computation.
+
+Though, for our purposes we will mostly only need the orthonormal spherical
+harmonics, so from now on, unless specified otherwise when we say spherical
+harmonics or write $Y^m_n$, we mean the orthonormal spherical harmonics of
+definition \ref{kugel:def:spherical-harmonics-orthonormal}.
+
+\subsection{Recurrence Relations}\kugeltodo{replace x with z}
+The idea of this subsection is to introduce first some recursive relations regarding the Associated Legendre Functions, defined in eq.\eqref{kugel:def:ferrers-functions}. Subsequently we will extend them, in order to derive recurrence formulas for the case of Spherical Harmonic functions as well.
+\subsubsection{Associated Legendre Functions}
+To start this journey, we can first write the following equations, which relate the Associated Legendre functions of different indeces $m$ and $n$ recursively:
+\begin{subequations}
+ \begin{align}
+ P^m_n(z) &= \dfrac{1}{(2n+1)x} \left[ (m+n) P^m_{n-1}(z) + (n-m+1) P^m_{n+1}(z) \right] \label{kugel:eq:rec-leg-1} \\
+ P^m_n(z) &= \dfrac{\sqrt{1-z^2}}{2mz} \left[ P^{m+1}_n(z) + [n(n+1)-m(m-1)] P^{m-1}_n(z) \right] \label{kugel:eq:rec-leg-2} \\
+ P^m_n(z) &= \dfrac{1}{(2n+1)\sqrt{1-z^2}} \left[ P^{m+1}_{n+1}(z) - P^{m+1}_{n-1}(z) \right] \label{kugel:eq:rec-leg-3} \\
+ P^m_n(z) &= \dfrac{1}{(2n+1)\sqrt{1-z^2}} \left[ (n+m)(n+m-1)P^{m-1}_{n-1}(z) - (n-m+1)(n-m+2)P^{m-1}_{n+1}(z) \right] \label{kugel:eq:rec-leg-4}
+ \end{align}
+\end{subequations}
+Much of the effort will be proving this bunch of equalities. Then, in the second part, where we will derive the recursion equations for $Y^m_n(\vartheta,\varphi)$, we will basically reuse the ones presented above.
+
+Maybe it is worth mentioning at least one use case for these relations: In some software implementations (that include lighting computations in computer graphics, antenna modelling softwares, 3-D modelling in medical applications, etc.)
+they are widely used, as they lead to better numerical accuracy and computational cost lower by a factor of six\cite{usecase_recursion_paper}.
+\begin{enumerate}[(i)]
+ \item
+ \begin{proof}
+ This is the relation that links the associated Legendre functions with the same $m$ index but different $n$. Using \ref{} \kugeltodo{search the general equation of recursion for orthogonal polynomials (is somewhere in the book)}, we have
+ \begin{equation*}
+ (n+1)P_{n+1}(z)-(2n+1)xP_n(z)+nP_{n-1}(z)=0,
+ \end{equation*}
+ that can be differentiated $m$ times, obtaining
+ \begin{equation}\label{kugel:eq:rec_1}
+ (n+1)\frac{d^mP_{n+1}}{dz^m}-(2n+1) \left[z \frac{d^m P_n}{dz^m}+ m\frac{d^{m-1}P_{n-1}}{dz^{m-1}} \right] + n\frac{d^m P_{n-1}}{dz^m}=0.
+ \end{equation}
+ To continue this derivation, we need the following relation:
+ \begin{equation}\label{kugel:eq:rec_2}
+ \frac{dP_{n+1}}{dz} - \frac{dP_{n-1}}{dz} = (2n+1)P_n.
+ \end{equation}
+ The latter will not be derived, because it suffices to use the definition of the Legendre Polynomials $P_n(x)$ to check it.
+
+ We can now differentiate the just presented eq.\eqref{kugel:eq:rec_2} $m-1$ times, that will become
+ \begin{equation}\label{kugel:eq:rec_3}
+ \frac{d^mP_{n+1}}{dx^m} - \frac{d^mP_{n-1}}{dx^m} = (2n+1)\frac{d^{m-1}P_n}{dx^{m-1}}.
+ \end{equation}
+ Then, using eq.\eqref{kugel:eq:rec_3} in eq.\eqref{kugel:eq:rec_1}, we will have
+ \begin{equation}\label{kugel:eq:rec_4}
+ (n+1)\frac{d^mP_{n+1}}{dx^m}- (2n+1)\frac{d^mP_{n+1}}{dx^m} -m\left[\frac{d^m P_{n+1}}{dx^m}+ \frac{d^{m}P_{n-1}}{dx^m}\right] + n\frac{d^m P_{n-1}}{dx^m}=0.
+ \end{equation}
+ Finally, multiplying both sides by $(1-x^2)^{\frac{m}{2}}$ and simplifying the expression, we can rewrite eq.\eqref{kugel:eq:rec_4} in terms of $P^m_n(x)$, namely
+ \begin{equation*}
+ (n+1-m)P^m_{n+1}(x)-(2n+1)xP^m_n(x)+(m+n)P^m_{n-1}(x)=0,
+ \end{equation*}
+ that rearranged, will be
+ \begin{equation*}
+ (2n+1) x P^m_n(x)= (m+n) P^m_{n-1}(x) + (n-m+1) P^m_{n+1}(x).
+ \end{equation*}
+ \end{proof}
+
+ \item
+ \begin{proof}
+ This relation, unlike the previous one, link three expression with the same $n$ index but different $m$.
+
+ In the proof of Lemma \ref{kugel:lemma:sol_associated_leg_eq}, at some point we ran into this expression.
+ \begin{equation*}
+ (1-x^2)\frac{d^{m+2}P_n}{dx^{m+2}} - 2(m+1)x \frac{d^{m+1}P_n}{dx^{m+1}} + [n(n+1)-m(m+1)]\frac{d^mP_n}{dx^m} = 0,
+ \end{equation*}
+ that, if multiplied by $(1-x^2)^{\frac{m}{2}}$, will be
+ \begin{equation*}
+ (1-x^2)^{\frac{m}{2}+1}\frac{d^{m+2}P_n}{dx^{m+2}} - 2(m+1)x (1-x^2)^{\frac{m}{2}}\frac{d^{m+1}P_n}{dx^{m+1}} + [n(n+1)-m(m+1)](1-x^2)^{\frac{m}{2}}\frac{d^mP_n}{dx^m} = 0.
+ \end{equation*}
+ Therefore, as before, expressing it in terms of $P^m_n(x)$:
+ \begin{equation*}
+ P^{m+2}_n(x) - \frac{2(m+1)x}{\sqrt{1-x^2}}P^{m+1}_n(x) + [n(n+1)-m(m+1)]P^m_n(x)=0.
+ \end{equation*}
+ Further, we can adjust the indeces and terms, obtaining
+ \begin{equation*}
+ \frac{2mx}{\sqrt{(1-x^2)}} P^m_n(x) = P^{m+1}_n(x) + [n(n+1)-m(m-1)] P^{m-1}_n(x).
+ \end{equation*}
+
+ \end{proof}
+
+ \item
+ \begin{proof}
+ To derive this expression, we can multiply eq.\eqref{kugel:eq:rec_3} by $(1-x^2)^{\frac{m}{2}}$ and, as always, we could express it in terms of $P^m_n(x)$:
+ \begin{equation*}
+ P^m_{n+1}(x) - P^m_{n-1}(x) = (2n+1)\sqrt{1-x^2}P^{m-1}_n(x).
+ \end{equation*}
+ Afer that we can divide by $2n+1$ resulting in
+ \begin{equation}\label{kugel:eq:helper}
+ \frac{1}{2n+1}[P^m_{n+1}(x) - P^m_{n-1}(x)] = \sqrt{1-x^2}P^{m-1}_n(x).
+ \end{equation}
+ To conclude, we arrange the indeces differently:
+ \begin{equation*}
+ \sqrt{1-x^2}P^{m}_n(x)=\frac{1}{2n+1}[P^{m+1}_{n+1}(x) - P^{m+1}_{n-1}(x)].
+ \end{equation*}
+ \end{proof}
+
+ \item
+ \begin{proof}
+ For this proof we can rely on eq.\eqref{kugel:eq:rec-leg-1}, and therefore rewrite eq.\eqref{kugel:eq:rec-leg-2} as
+ \begin{equation*}
+ \frac{2m}{(2n+1)\sqrt{1-x^2}} \left[ (m+n)P^m_{n-1}(x) + (n-m+1)P^m_{n+1}(x) \right] = P^{m+1}_n(x) + [ n(n+1)-m(m-1) ]P^{m-1}_n(x).
+ \end{equation*}
+ Rewriting then $P^{m-1}_n(x)$ using eq.\eqref{kugel:eq:helper}, we will have
+ \begin{align*}
+ \frac{2m}{(2n+1)\sqrt{1-x^2}} &\left[ (m+n)P^m_{n-1}(x) + (n-m+1)P^m_{n+1}(x) \right] = P^{m+1}_n(x) \\
+ &+ \frac{n(n+1)-m(m-1)}{(2n+1)\sqrt{1-x^2}} \left[ P^m_{n+1}(x)-P^m_{n-1}(x) \right].
+ \end{align*}
+ The last equation, after some algebric rearrangements, it is easy to show that it is equivalent to
+ \begin{equation*}
+ \sqrt{1-x^2} P^m_n(x) = \dfrac{1}{2n+1} \left[ (n+m)(n+m-1)P^{m-1}_{n-1}(x) - (n-m+1)(n-m+2)P^{m-1}_{n+1}(x) \right]
+ \end{equation*}
+ \end{proof}
+
+\end{enumerate}
+
+\subsubsection{Spherical Harmonics}
+The goal of this subsection's part is to apply the recurrence relations of the $P^m_n(z)$ functions to the Spherical Harmonics.
+With some little adjustments we will be able to have recursion equations for them too. As previously written the most of the work is already done. Now it is only a matter of minor mathematical operations/rearrangements.
+
+We can start by listing all of them:
+\begin{subequations}
+ \begin{align}
+ Y^m_n(\vartheta, \varphi) &= \dfrac{1}{(2n+1)\cos \vartheta} \left[ (m+n)Y^m_{n-1}(\vartheta, \varphi) + (m-n+1)Y^m_{n+1}(\vartheta, \varphi) \right] \label{kugel:eq:rec-sph_harm-1} \\
+ Y^m_n(\vartheta, \varphi) &= \dfrac{\tan \vartheta}{2m}\left[ Y^{m+1}_n(\vartheta, \varphi)e^{-i\varphi} + [n(n+1)-m(m-1)]Y^{m-1}_n(\vartheta, \varphi)e^{i\varphi} \right] \label{kugel:eq:rec-sph_harm-2} \\
+ Y^m_n(\vartheta, \varphi) &= \dfrac{e^{-i\varphi}}{ (2n+1)\sin \vartheta } \left[ Y^{m+1}_{n+1}(\vartheta, \varphi) - Y^{m+1}_{n-1}(\vartheta, \varphi) \right] \label{kugel:eq:rec-sph_harm-3} \\
+ Y^m_n(\vartheta, \varphi) &= \dfrac{e^{i\varphi}}{(2n+1)\sin \vartheta} \left[ (n+m)(n+m-1)Y^{m-1}_{n-1}(\vartheta, \varphi) - (n-m+1)(n-m+2)Y^{m-1}_{n+1}(\vartheta, \varphi) \right] \label{kugel:eq:rec-sph_harm-4}
+ \end{align}
+\end{subequations}
+
+\begin{enumerate}[(i)]
+ \item
+ \begin{proof}
+ We can multiply both sides of equality in eq.\eqref{kugel:eq:rec-leg-1} by $e^{im \varphi}$ and perform the substitution $z=\cos \vartheta$. After a few simple algebraic steps, we will obtain the relation we are looking for
+ \end{proof}
+ \item
+ \begin{proof}
+ In this proof, as before, we can perform the substitution $z=\cos \vartheta$, and notice that $\sqrt{1-z^2}=\sin \vartheta$, hence, the relation in eq.\eqref{kugel:eq:rec-leg-2} will be
+ \begin{equation*}
+ \frac{2m \cos \vartheta}{\sin \vartheta} P^m_n(\cos \vartheta) = P^{m+1}_n(\cos \vartheta) + [n(n+1)-m(m-1)]P^{m-1}_n P^m_n(\cos \vartheta).
+ \end{equation*}
+ The latter, multiplied by $e^{im\varphi}$, becomes
+ \begin{align*}
+ \frac{2m \cos \vartheta}{\sin \vartheta} P^m_n(\cos \vartheta)e^{im\varphi} &= P^{m+1}_n(\cos \vartheta)e^{im\varphi} + [n(n+1)-m(m-1)]P^{m-1}_n P^m_n(\cos \vartheta)e^{im\varphi} \\
+ &= P^{m+1}_n(\cos \vartheta)e^{i(m+1)\varphi}e^{-i\varphi} + [n(n+1)-m(m-1)]P^{m-1}_n (\cos \vartheta)e^{i(m-1)\varphi}e^{i\varphi} \\
+ &= Y^{m+1}_n(\vartheta, \varphi)e^{-i\varphi} + [n(n+1)-m(m-1)]Y^{m-1}_n(\vartheta, \varphi)e^{i\varphi}.
+ \end{align*}
+ Finally, after some ``cleaning''
+ \begin{equation*}
+ Y^m_n(\vartheta, \varphi) = \frac{\tan \vartheta}{2m} \left[ Y^{m+1}_n(\vartheta, \varphi)e^{-i\varphi} + [n(n+1)-m(m-1)]Y^{m-1}_n(\vartheta, \varphi)e^{i\varphi} \right]
+ \end{equation*}
+ \end{proof}
+ \item
+ \begin{proof}
+ Now we can consider eq.\eqref{kugel:eq:rec-leg-3}, and multiply it by $e^{im\varphi}$. After the usual substitution $z=\cos \vartheta$, we have
+ \begin{align*}
+ \sin \vartheta P^m_n(\cos \vartheta)e^{im\varphi} &= \dfrac{e^{im\varphi}}{2n+1}\left[ P^{m+1}_{n+1}(\cos \vartheta) - P^{m+1}_{n-1}(\cos \vartheta)\right] \\
+ &= \dfrac{e^{-i\varphi}}{2n+1}\left[ P^{m+1}_{n+1}(\cos \vartheta)e^{i(m+1)\varphi} - P^{m+1}_{n-1}(\cos \vartheta)e^{i(m+1)\varphi}\right].
+ \end{align*}
+ A few manipulations later, we will obtain
+ \begin{equation*}
+ Y^m_n(\vartheta, \varphi) = \frac{e^{-i\varphi}}{(2n+1)\sin \vartheta} \left[ Y^{m+1}_{n+1}(\vartheta, \varphi)-Y^{m+1}_{n-1}(\vartheta, \varphi) \right].
+ \end{equation*}
+ \end{proof}
+ \item
+ \begin{proof}
+ This proof is very similar to the previous one. We just have to perform the substitution $z = \cos \vartheta$, as always. Secondly we can multiply the right side by $e^{im\varphi}$ and the left one too but in a different form, namely $e^{im\varphi}=e^{i(m-1)\varphi}e^{i\varphi}$. Then it is only a question of recalling the definition of $Y^m_n(\vartheta, \varphi)$.
+ \end{proof}
+\end{enumerate}
+
+\section{Series Expansions in $L^2(S^2)$}
+We have now reach a point where we have all the tools that are necessary to build something truly amazing: a general series expansion formula for
+function on the surface of the sphere.
+Before starting we want to recall the definition of the inner product on the spherical surface of definition \ref{kugel:def:inner-product-s2}
+\begin{equation*}
+ \langle f, g \rangle
+ = \int_{0}^\pi \int_0^{2\pi}
+ f(\vartheta, \varphi) \overline{g(\vartheta, \varphi)}
+ \sin \vartheta \, d\varphi \, d\vartheta.
+\end{equation*}
+To be a bit technical we can say that the set of spherical harmonic functions, toghether with the inner product just showed,
+form something that we call Hilbert Space\footnote{For more details about Hilber space you can take a look in section \ref{kugel:sec:preliminaries}}.
+This function space is defined over the space of ``well-behaved'' \footnote{The definitions of ``well-behaved'' is pretty ambigous, even for mathematicians.
+It depends basically on the context.
+You can sumarize it by saying: functions for which the theory we are considering (Fourier theorem) is always true. In our case we can say that well-behaved functions
+are functions that follow some convergence contraints (pointwise, uniform, absolute, ...) that we don't want to consider further anyway.} functions.
+We can say that the theory we are about to show can be applied on all twice differentiable complex valued functions,
+to be more concise: complex valued $L^2$ functions $S^2 \to \mathbb{C}$.
+
+All these jargons are not really necessary for the practical applications of us mere mortals, namely physicists and engineers.
+From now on we will therefore assume that the functions we will dealing with fulfill these ``minor'' conditions.
+
+The insiders could turn up their nose, but we don't want to dwell too much on the concept of Hilbert space, convergence, metric, well-behaved functions etc.
+We simply think that this rigorousness could be at the expense of the possibility to appreciate the beauty and elegance of this theory.
+Furthermore, the risk of writing 300+ pages to prove that $1+1=2$\cite{principia-mathematica} is just around the corner (we apologize in advance to Mr. Whitehead and Mr. Russel for using their effort with a negative connotation).
+
+Despite all, if you desire having definitions a bit more rigorous (as rigorous as two engineers can be), you could take a look at the chapter \ref{}.
+
+\subsection{Spherical Harmonics Series}
+
+To talk about a \emph{series expansion} we first need a series, so we shall
+build one using the spherical harmonics.
+
+\begin{definition}[Spherical harmonic series]
+ \label{kugel:definition:spherical-harmonics-series}
+ \begin{equation}
+ f(\vartheta, \varphi)
+ = \sum_{n=0}^\infty \sum_{m =-n}^n
+ c_{m,n} Y^m_n(\vartheta, \varphi). \label{kugel:definition:spherical-harmonics-series}
+ \end{equation}
+\end{definition}
+
+With this definition we are basically saying that any function defined on the spherical surface can be represented as a linear combination of spherical harmonics.
+Does eq.\eqref{kugel:definition:spherical-harmonics-series} sound familiar? Well that is prefectly normal, since this is analog to the classical Fourier theory.
+In the latter is stated that ``any'' $T$-periodic function $f(x)$, on any interval $[x_0-T/2,x_0+T/2]$, can be represented as a linear combination of complex exponentials. More compactly:
+\begin{equation*}
+ f(x) = \sum_{n \in \mathbb{Z}} c_n e^{i \omega_0 x}, \quad \omega_0=\frac{2\pi}{T}
+\end{equation*}
+In the case of definition \ref{kugel:definition:spherical-harmonics-series} the kernels, instead of $e^{i\omega_0x}$, have become $Y^m_n$. In addition, the sum is now over the two indices $m$ and $n$.
+
+\begin{lemma}[Spherical harmonic coefficients]
+ \label{kugel:lemma:spherical-harmonic-coefficient}
+ \begin{align*}
+ c_{m,n}
+ &= \langle f, Y^m_n \rangle_{\partial S} \\
+ &= \int_0^\pi \int_0^{2\pi} f(\vartheta,\varphi) \overline{Y^m_n(\vartheta,\varphi)} \sin\vartheta \,d\varphi\,d\vartheta
+ \end{align*}
+\end{lemma}
+\begin{proof}
+ To develop this proof we will take advantage of the orthogonality property of the Spherical Harmonics. We can start and finish by applying the inner product on both sides of eq.\eqref{kugel:definition:spherical-harmonics-series}:
+ \begin{align*}
+ \langle f, Y^{m}_{n} \rangle_{\partial S}
+ &= \left\langle \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'}
+ c_{m',n'} Y^{m'}_{n'}(\vartheta, \varphi) \right\rangle_{\partial S} \\
+ &= \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'}
+ \langle c_{m',n'} Y^{m'}_{n'}, Y^{m}_{n} \rangle_{\partial S} \\
+ &= \sum_{n'=0}^\infty \sum_{m' =-n'}^{n'} c_{m',n'} \langle Y^{m'}_{n'}, Y^{m}_{n} \rangle_{\partial S} = c_{m,n}
+ \end{align*}
+ We omitted the $\vartheta, \varphi$ dependency to avoid overloading the notation.
+\end{proof}
+Thanks to Lemma \ref{kugel:lemma:spherical-harmonic-coefficient} we can now calculate the series expansion defined in \ref{kugel:definition:spherical-harmonics-series}.
+
+It can be shown that, for the famous ``well-behaved functions'' $f(\vartheta, \varphi)$ mentioned before, theorem \ref{fourier-theorem-spherical-surface} is true
+\begin{theorem}[Fourier Theorem on $\partial S$]
+ \label{fourier-theorem-spherical-surface}
+ \begin{equation*}
+ \lim_{N \to \infty}
+ \int_0^\pi \int_0^{2\pi} \left\| f(\vartheta,\varphi) - \sum_{n=0}^N\sum_{m=-n}^n c_{m,n} Y^m_n(\vartheta,\varphi)
+ \right\|_2 \sin\vartheta \,d\varphi\,d\vartheta = 0
+ \end{equation*}
+\end{theorem}
+The connection to Theorem \ref{fourier-theorem-1D} is pretty obvious.
+
+\subsection{Spectrum}
+
+\begin{figure}
+ \centering
+ \kugelplaceholderfig{.8\textwidth}{5cm}
+ \caption{\kugeltodo{Rectangular signal and his spectrum.}}
+ \label{kugel:fig:1d-fourier}
+\end{figure}
+
+In the case of the classical one-dimensional Fourier theory, we call \emph{Spectrum} the relation between the fourier coefficients $c_n$ and the multiple
+of the fundamental frequency $2\pi/T$, namely $n 2\pi/T$. In the most general case $c_n$ are complex numbers, so we divide the concept of spectrum in
+\emph{Amplitude Spectrum} and \emph{Phase Spectrum}. In fig.\ref{kugel:fig:1d-fourier} a function $f(x)$ is presented along with the amplitude spectrum.
+
+\begin{figure}
+ \centering
+ \kugelplaceholderfig{.8\textwidth}{7cm}
+ \caption{\kugeltodo{Confront between image reconstructed only with phase and one only with amplitued}}
+ \label{kugel:fig:phase&amplitude-2d-fourier}
+\end{figure}
+
+The thing that is easiest for us humans to visualize and understand is often the Amplitude Spectrum.
+This is a huge limitation, since for example in Image Processing can be showed in a nice way that much more information is contained in the phase part (see fig.\ref{kugel:fig:phase-2d-fourier}).
+
+\begin{figure}
+ \centering
+ \kugelplaceholderfig{.8\textwidth}{9cm}
+ \caption{\kugeltodo{fig that show fourier style reconstruction on sphere (with increasing index)}}
+ \label{kugel:fig:fourier-on-sphere-increasing-index}
+\end{figure}
+
+The same logic can be extended to the spherical harmonic coefficients $c_{m,n}$. In fig.\ref{kugel:fig:fourier-on-sphere-increasing-index} you can see the same concept as in fig.\ref{kugel:fig:1d-fourier}
+but with a spherical function $f(\vartheta, \varphi)$.
+
+\subsection{Energy of a function $f(\vartheta, \varpi)$}
+
+\begin{lemma}[Energy of a spherical function (\emph{Parseval's theorem})]
+ \begin{equation*}
+ \int_0^{2\pi}\int_0^\pi |f(\vartheta, \varphi)|^2 \sin\vartheta \, d\varphi \, d\varphi = \sum_{n=0}^\infty \frac{1}{2n+1} \sum_{m=-n}^n |c_{m,n}|^2.
+ \end{equation*}
+\end{lemma}
+\begin{proof}
+\end{proof}
+
+\subsection{Visualization} \ No newline at end of file
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