Merge pull request #58 from NaoPross/master

kugel: Eigenvalue problem and Legendre functions

27 files changed, 1339 insertions, 10 deletions

diff --git a/buch/papers/kugel/Makefile b/buch/papers/kugel/Makefile
index f798a55..995206b 100644
--- a/buch/papers/kugel/Makefile
+++ b/buch/papers/kugel/Makefile
@@ -5,5 +5,6 @@
 #
 
 images:
-	@echo "no images to be created in kugel"
+	$(MAKE) -C ./figures/povray/
+	$(MAKE) -C ./figures/tikz/
 
diff --git a/buch/papers/kugel/figures/flux.pdf b/buch/papers/kugel/figures/flux.pdf
new file mode 100644
index 0000000..6a87288
--- /dev/null
+++ b/buch/papers/kugel/figures/flux.pdf
diff --git a/buch/papers/kugel/images/Makefile b/buch/papers/kugel/figures/povray/Makefile
index 4226dab..4226dab 100644
--- a/buch/papers/kugel/images/Makefile
+++ b/buch/papers/kugel/figures/povray/Makefile
diff --git a/buch/papers/kugel/figures/povray/curvature.jpg b/buch/papers/kugel/figures/povray/curvature.jpg
new file mode 100644
index 0000000..6448966
--- /dev/null
+++ b/buch/papers/kugel/figures/povray/curvature.jpg
diff --git a/buch/papers/kugel/images/curvature.maxima b/buch/papers/kugel/figures/povray/curvature.maxima
index 6313642..6313642 100644
--- a/buch/papers/kugel/images/curvature.maxima
+++ b/buch/papers/kugel/figures/povray/curvature.maxima
diff --git a/buch/papers/kugel/figures/povray/curvature.png b/buch/papers/kugel/figures/povray/curvature.png
new file mode 100644
index 0000000..20268f2
--- /dev/null
+++ b/buch/papers/kugel/figures/povray/curvature.png
diff --git a/buch/papers/kugel/images/curvature.pov b/buch/papers/kugel/figures/povray/curvature.pov
index 3b15d77..3b15d77 100644
--- a/buch/papers/kugel/images/curvature.pov
+++ b/buch/papers/kugel/figures/povray/curvature.pov
diff --git a/buch/papers/kugel/images/curvgraph.m b/buch/papers/kugel/figures/povray/curvgraph.m
index 75effd6..75effd6 100644
--- a/buch/papers/kugel/images/curvgraph.m
+++ b/buch/papers/kugel/figures/povray/curvgraph.m
diff --git a/buch/papers/kugel/images/spherecurve.cpp b/buch/papers/kugel/figures/povray/spherecurve.cpp
index 8ddf5e5..8ddf5e5 100644
--- a/buch/papers/kugel/images/spherecurve.cpp
+++ b/buch/papers/kugel/figures/povray/spherecurve.cpp
diff --git a/buch/papers/kugel/figures/povray/spherecurve.jpg b/buch/papers/kugel/figures/povray/spherecurve.jpg
new file mode 100644
index 0000000..cd2e7c8
--- /dev/null
+++ b/buch/papers/kugel/figures/povray/spherecurve.jpg
diff --git a/buch/papers/kugel/images/spherecurve.m b/buch/papers/kugel/figures/povray/spherecurve.m
index 99d5c9a..99d5c9a 100644
--- a/buch/papers/kugel/images/spherecurve.m
+++ b/buch/papers/kugel/figures/povray/spherecurve.m
diff --git a/buch/papers/kugel/images/spherecurve.maxima b/buch/papers/kugel/figures/povray/spherecurve.maxima
index 1e9077c..1e9077c 100644
--- a/buch/papers/kugel/images/spherecurve.maxima
+++ b/buch/papers/kugel/figures/povray/spherecurve.maxima
diff --git a/buch/papers/kugel/figures/povray/spherecurve.png b/buch/papers/kugel/figures/povray/spherecurve.png
new file mode 100644
index 0000000..ff24371
--- /dev/null
+++ b/buch/papers/kugel/figures/povray/spherecurve.png
diff --git a/buch/papers/kugel/images/spherecurve.pov b/buch/papers/kugel/figures/povray/spherecurve.pov
index b1bf4b8..b1bf4b8 100644
--- a/buch/papers/kugel/images/spherecurve.pov
+++ b/buch/papers/kugel/figures/povray/spherecurve.pov
diff --git a/buch/papers/kugel/figures/tikz/Makefile b/buch/papers/kugel/figures/tikz/Makefile
new file mode 100644
index 0000000..4ec4e5a
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/Makefile
@@ -0,0 +1,12 @@
+FIGURES := spherical-coordinates.pdf curvature-1d.pdf
+
+all: $(FIGURES)
+
+%.pdf: %.tex
+	pdflatex $<
+
+curvature-1d.pdf: curvature-1d.tex curvature-1d.dat
+	pdflatex curvature-1d.tex
+
+curvature-1d.dat: curvature-1d.py
+	python3 $<
diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.dat b/buch/papers/kugel/figures/tikz/curvature-1d.dat
new file mode 100644
index 0000000..6622398
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/curvature-1d.dat
@@ -0,0 +1,500 @@
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diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.pdf b/buch/papers/kugel/figures/tikz/curvature-1d.pdf
new file mode 100644
index 0000000..6425af6
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/curvature-1d.pdf
diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.py b/buch/papers/kugel/figures/tikz/curvature-1d.py
new file mode 100644
index 0000000..4710fc8
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/curvature-1d.py
@@ -0,0 +1,32 @@
+import numpy as np
+import matplotlib.pyplot as plt
+
+
+@np.vectorize
+def fn(x):
+    return (x ** 2) * 2 / 100 + (1 + x / 4) + np.sin(x)
+
+@np.vectorize
+def ddfn(x):
+    return 2 * 5 / 100 - np.sin(x)
+
+x = np.linspace(0, 10, 500)
+y = fn(x)
+ddy = ddfn(x)
+
+cmap = ddy - np.min(ddy)
+cmap = cmap * 1000 / np.max(cmap)
+
+plt.plot(x, y)
+plt.plot(x, ddy)
+# plt.plot(x, cmap)
+
+plt.show()
+
+fname = "curvature-1d.dat"
+np.savetxt(fname, np.array([x, y, cmap]).T, delimiter="  ")
+
+# with open(fname, "w") as f:
+#     # f.write("x   y   cmap\n")
+#     for xv, yv, cv in zip(x, y, cmap):
+#         f.write(f"{xv}   {yv}   {cv}\n")
diff --git a/buch/papers/kugel/figures/tikz/curvature-1d.tex b/buch/papers/kugel/figures/tikz/curvature-1d.tex
new file mode 100644
index 0000000..6983fb0
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/curvature-1d.tex
@@ -0,0 +1,21 @@
+% vim:ts=2 sw=2 et:
+\documentclass[tikz, border=5mm]{standalone}
+\usepackage{pgfplots}
+
+\begin{document}
+\begin{tikzpicture}
+  \begin{axis}[
+      clip = false,
+      width = 8cm, height = 6cm,
+      xtick = \empty, ytick = \empty,
+      colormap name = viridis,
+      axis lines = middle,
+      axis line style = {ultra thick, -latex}
+    ]
+    \addplot+[
+      smooth, mark=none, line width = 3pt, mesh,
+      point meta=explicit,
+    ] file {curvature-1d.dat};
+  \end{axis}
+\end{tikzpicture}
+\end{document}
diff --git a/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf b/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf
new file mode 100644
index 0000000..1bff016
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/spherical-coordinates.pdf
diff --git a/buch/papers/kugel/figures/tikz/spherical-coordinates.tex b/buch/papers/kugel/figures/tikz/spherical-coordinates.tex
new file mode 100644
index 0000000..3a45385
--- /dev/null
+++ b/buch/papers/kugel/figures/tikz/spherical-coordinates.tex
@@ -0,0 +1,99 @@
+\documentclass[tikz]{standalone}
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{bm}
+\usepackage{lmodern}
+\usepackage{tikz-3dplot}
+
+\usetikzlibrary{arrows}
+\usetikzlibrary{intersections}
+\usetikzlibrary{math}
+\usetikzlibrary{positioning}
+\usetikzlibrary{arrows.meta}
+\usetikzlibrary{shapes.misc}
+\usetikzlibrary{calc}
+
+\begin{document}
+
+\tdplotsetmaincoords{60}{130}
+\pgfmathsetmacro{\l}{2}
+
+\begin{tikzpicture}[
+    >=latex,
+    tdplot_main_coords,
+    dot/.style = {
+      black, fill = black, circle,
+      outer sep = 0, inner sep = 0,
+      minimum size = .8mm 
+    },
+    round/.style = {
+      draw = orange, thick, circle,
+      minimum size = 1mm,
+      inner sep = 0pt, outer sep = 0pt,
+    },
+    cross/.style = {
+      cross out, draw = magenta, thick,
+      minimum size = 1mm, 
+      inner sep = 0pt, outer sep = 0pt
+    },
+  ]
+
+  % origin
+  \coordinate (O) at (0,0,0);
+
+  % poles
+  \coordinate (NP) at (0,0,\l);
+  \coordinate (SP) at (0,0,-\l);
+  
+  % \draw (SP) node[dot, gray] {};
+  % \draw (NP) node[dot, gray] {};
+
+  % gray unit circle
+  \tdplotdrawarc[gray]{(O)}{\l}{0}{360}{}{};
+  \draw[gray, dashed] (-\l, 0, 0) to (\l, 0, 0);
+  \draw[gray, dashed] (0, -\l, 0) to (0, \l, 0);
+
+  % axis
+  \draw[->] (O) -- ++(1.25*\l,0,0) node[left] {\(\mathbf{\hat{x}}\)};
+  \draw[->] (O) -- ++(0,1.25*\l,0) node[right] {\(\mathbf{\hat{y}}\)};
+  \draw[->] (O) -- ++(0,0,1.25*\l) node[above] {\(\mathbf{\hat{z}}\)};
+
+  % meridians
+  \foreach \phi in {0, 30, 60, ..., 150}{
+    \tdplotsetrotatedcoords{\phi}{90}{0};
+    \tdplotdrawarc[lightgray, densely dotted, tdplot_rotated_coords]{(O)}{\l}{0}{360}{}{};
+  }
+
+  % dot above and its projection
+  \pgfmathsetmacro{\phi}{120}
+  \pgfmathsetmacro{\theta}{40}
+
+  \pgfmathsetmacro{\px}{cos(\phi)*sin(\theta)*\l}
+  \pgfmathsetmacro{\py}{sin(\phi)*sin(\theta)*\l}
+  \pgfmathsetmacro{\pz}{cos(\theta)*\l})
+  
+  % point A
+  \coordinate (A) at (\px,\py,\pz);
+  \coordinate (Ap) at (\px,\py, 0);
+  
+  % lines
+  \draw[red!80!black, ->] (O) -- (A);
+	\draw[red!80!black, densely dashed] (O) -- (Ap) -- (A)
+		node[above right] {\(\mathbf{\hat{r}}\)};
+  
+  % arcs
+  \tdplotdrawarc[blue!80!black, ->]{(O)}{.8\l}{0}{\phi}{}{};
+  \node[below right, blue!80!black] at (.8\l,0,0) {\(\bm{\hat{\varphi}}\)};
+  
+  \tdplotsetrotatedcoords{\phi-90}{-90}{0};
+  \tdplotdrawarc[blue!80!black, ->, tdplot_rotated_coords]{(O)}{.95\l}{0}{\theta}{}{};
+  \node[above right = 1mm, blue!80!black] at (0,0,.8\l) {\(\bm{\hat{\vartheta}}\)};
+
+
+  % dots
+  \draw (O) node[dot] {};
+  \draw (A) node[dot, fill = red!80!black] {};
+
+\end{tikzpicture}
+\end{document}
+% vim:ts=2 sw=2 et:
diff --git a/buch/papers/kugel/main.tex b/buch/papers/kugel/main.tex
index 98d9cb2..d063f87 100644
--- a/buch/papers/kugel/main.tex
+++ b/buch/papers/kugel/main.tex
@@ -14,6 +14,7 @@
 \input{papers/kugel/preliminaries}
 \input{papers/kugel/spherical-harmonics}
 \input{papers/kugel/applications}
+\input{papers/kugel/proofs}
 
 \printbibliography[heading=subbibliography]
 \end{refsection}
diff --git a/buch/papers/kugel/packages.tex b/buch/papers/kugel/packages.tex
index 61f91ad..ead7653 100644
--- a/buch/papers/kugel/packages.tex
+++ b/buch/papers/kugel/packages.tex
@@ -1,3 +1,4 @@
+% vim:ts=2 sw=2 et:
 %
 % packages.tex -- packages required by the paper kugel
 %
@@ -7,4 +8,13 @@
 % if your paper needs special packages, add package commands as in the
 % following example
 %\usepackage{packagename}
+\usepackage{cases}
 
+\newcommand{\kugeltodo}[1]{\textcolor{red!70!black}{\texttt{[TODO: #1]}}}
+\newcommand{\kugelplaceholderfig}[2]{ \begin{tikzpicture}%
+    \fill[lightgray!20] (0, 0) rectangle (#1, #2);%
+    \node[gray, anchor = center] at ({#1 / 2}, {#2 / 2}) {\Huge \ttfamily \bfseries TODO};
+  \end{tikzpicture}}
+
+\DeclareMathOperator{\sphlaplacian}{\nabla^2_{\mathit{S}}}
+\DeclareMathOperator{\surflaplacian}{\nabla^2_{\partial \mathit{S}}}
diff --git a/buch/papers/kugel/preliminaries.tex b/buch/papers/kugel/preliminaries.tex
index 03cd421..e48abe4 100644
--- a/buch/papers/kugel/preliminaries.tex
+++ b/buch/papers/kugel/preliminaries.tex
@@ -44,23 +44,23 @@ numbers \(\mathbb{R}\).
   \)
 \end{definition}
 
-\texttt{TODO: Text here.}
+\kugeltodo{Text here.}
 
 \begin{definition}[Span]
 \end{definition}
 
-\texttt{TODO: Text here.}
+\kugeltodo{Text here.}
 
 \begin{definition}[Linear independence]
 \end{definition}
 
 
-\texttt{TODO: Text here.}
+\kugeltodo{Text here.}
 
 \begin{definition}[Basis]
 \end{definition}
 
-\texttt{TODO: Text here.}
+\kugeltodo{Text here.}
 
 \begin{definition}[Inner product]
   \label{kugel:def:inner-product} \nocite{axler_linear_2014}
diff --git a/buch/papers/kugel/proofs.tex b/buch/papers/kugel/proofs.tex
new file mode 100644
index 0000000..143caa8
--- /dev/null
+++ b/buch/papers/kugel/proofs.tex
@@ -0,0 +1,245 @@
+% vim:ts=2 sw=2 et spell tw=80:
+\section{Proofs}
+
+\subsection{Legendre Functions} \label{kugel:sec:proofs:legendre}
+
+\kugeltodo{Fix theorem numbers to match, review text.}
+
+\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}
+
+
+\begin{lemma}
+  If $Z_n(z)$ is a solution of the Legendre equation \eqref{kugel:eqn:legendre},
+  then
+  \begin{equation*}
+    P^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}.
+\end{lemma}
+% \begin{proof} [TODO: modificare la $m$ (è già usata come costante di separazione) o forse è giusta (?)]
+\begin{proof}
+    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}
diff --git a/buch/papers/kugel/references.bib b/buch/papers/kugel/references.bib
index b74c5cd..e5d6452 100644
--- a/buch/papers/kugel/references.bib
+++ b/buch/papers/kugel/references.bib
@@ -192,4 +192,15 @@ Created by Henry Reich},
 	urldate = {2022-08-01},
 	date = {2022},
 	file = {Metric Spaces\: Completeness:/Users/npross/Zotero/storage/5JYEE8NF/completeness.html:text/html},
+}
+
+@book{bell_special_2004,
+	location = {Mineola, {NY}},
+	title = {Special functions for scientists and engineers},
+	isbn = {978-0-486-43521-3},
+	series = {Dover books on mathematics},
+	pagetotal = {247},
+	publisher = {Dover Publ},
+	author = {Bell, William Wallace},
+	date = {2004},
 }
\ No newline at end of file
diff --git a/buch/papers/kugel/spherical-harmonics.tex b/buch/papers/kugel/spherical-harmonics.tex
index 6b23ce5..2ded50b 100644
--- a/buch/papers/kugel/spherical-harmonics.tex
+++ b/buch/papers/kugel/spherical-harmonics.tex
@@ -1,13 +1,410 @@
-% 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}
+
+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.
+
+\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{kuvel: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) \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(
+      \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 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}
+  - 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}
+  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} \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}
+  See section \ref{kugel:sec:proofs:legendre}.
+\end{proof}
+
+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]
+  \label{kugel:def:ferrers-functions}
+  The functions
+  \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}
+
+\kugeltodo{Discuss $|m| \leq n$.}
+
+\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*}
+    \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.
+\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}
+    \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 $P_{m,n}(x)$
+\fi
 
 \subsection{Properties}
 
 \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 $S^2$}