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authorNao Pross <np@0hm.ch>2022-04-28 15:48:30 +0200
committerNao Pross <np@0hm.ch>2022-04-28 15:48:30 +0200
commit1ff8774bb6669b11e8599641f0acd7066913da57 (patch)
treedc9952fc30919c9574765791335a58b7e370ebbe /notes
parentAdd definition with hypergeometric function 2F1 (diff)
downloadFourierOnS2-1ff8774bb6669b11e8599641f0acd7066913da57.tar.gz
FourierOnS2-1ff8774bb6669b11e8599641f0acd7066913da57.zip
Add draftwatermark and associated legendre polynomials
Diffstat (limited to 'notes')
-rw-r--r--notes/FourierOnS2.tex26
-rw-r--r--notes/build/FourierOnS2.pdfbin56389 -> 61780 bytes
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diff --git a/notes/FourierOnS2.tex b/notes/FourierOnS2.tex
index edd4813..ced6b8d 100644
--- a/notes/FourierOnS2.tex
+++ b/notes/FourierOnS2.tex
@@ -7,6 +7,7 @@
%% styling for this document
\usepackage{tex/docstyle}
+\usepackage[firstpageonly, color={[gray]{0.9}}]{draftwatermark}
%% Theorems, TODO: move to docstyle
\usepackage{amsthm}
@@ -101,6 +102,7 @@ The answer has to do with solving other problems. That is because originally Fou
+ \frac{\partial^2 f}{\partial \nu^2} = 0.
\end{equation}
+\texttt{[FIXME: Should be an eigenvalue problem \(\nabla_2^2 f = \lambda f\)]}
This PDE is known as Laplace's equation, and can be solved by separation using the ansatz
\[
f(\mu, \nu) = M(\mu)N(\nu),
@@ -242,6 +244,7 @@ Thus we first need examine the solutions to this equation before constructing th
\end{equation}
are solutions to Legendre's equation \eqref{eqn:legendre} when \(n > 0\).
\end{proposition}
+\begin{proof} See appendix. \end{proof}
The proof for this proposition is quite algebraically involved and is thus left in the appendix. Since this is a power series \eqref{eqn:legendre-poly} can also be rewritten using Gauss' Hypergeometric function.
@@ -306,6 +309,28 @@ In some applications, such as in quantum mechanics, it is more common to see it
\end{align*}
\end{proof}
+Now, using the solutions to the Legendre equation we can construct the solution to the more general problem:
+\begin{align}
+ \left( 1 - x^2 \right) \frac{d^2 y}{dx^2} &- 2x \frac{dy}{dx} \nonumber \\
+ & + \left[ n(n+1) - \frac{m^2}{1 - x^2} \right] y(x) = 0.
+ \label{eqn:legendre-as}
+\end{align}
+
+This equation is considerably more difficult, and again, we will just analyze the solution.
+
+\begin{definition}[Associated Legendre Polynomials]
+ Let \(m \in \mathbb{N}_0\). The polynomials
+ \begin{equation} \label{eqn:legendre-poly-as}
+ P_{m, n} (x) = \left( 1 - x^2 \right)^{m/2} \frac{d^m}{dx^m} P_n (x),
+ \end{equation}
+ are called the associated Legendre polynomials.
+\end{definition}
+
+\begin{lemma}
+ The associated Legendre polynomials \eqref{eqn:legendre-poly-as} are solutions to the associated Legendre differential equation \eqref{eqn:legendre-as}.
+\end{lemma}
+\begin{proof} See appendix. \end{proof}
+
\subsection{Spherical harmonics}
\clearpage
@@ -374,6 +399,7 @@ from which we can extract the recurrence relation
\end{gather*}
% TODO: finish reviewing proof
+ \texttt{[TODO: finish copying proof]}
\if 0
We can derive a recursion formula for $a_{k+2}$ from Eq.\eqref{eq:condition_2}, which can be expressed as
diff --git a/notes/build/FourierOnS2.pdf b/notes/build/FourierOnS2.pdf
index d40262c..d60d9bf 100644
--- a/notes/build/FourierOnS2.pdf
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