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-rw-r--r--buch/papers/kugel/proofs.tex321
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diff --git a/buch/papers/kugel/proofs.tex b/buch/papers/kugel/proofs.tex
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--- a/buch/papers/kugel/proofs.tex
+++ b/buch/papers/kugel/proofs.tex
@@ -1,147 +1,184 @@
% vim:ts=2 sw=2 et spell tw=80:
-\section{(long) Proofs}
+\section{Long Proofs}
-\subsection{Legendre Functions} \label{kugel:sec:proofs:legendre}
+Here, we will give the long and tedious proofs we skipped earlier.
-\kugeltodo{Fix theorem numbers to match, review text.}
+\subsection{Legendre Polynomials} \label{kugel:sec:proofs:legendre}
-\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*}
+\begin{proof}[Proof of lemma \ref{kugel:thm:legendre-poly}]
+ It was stated that the polynomial function
+ \begin{equation*}
+ 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}
+ \end{equation*}
+ is the only finite solution of the Legendre equation
+ \begin{equation}
+ \label{kugel:eqn:legendre-bis}
+ (1 - z^2)\frac{d^2 Z}{dz^2}
+ - 2z\frac{d Z}{dz}
+ + n(n + 1) Z(z) = 0,
+ \end{equation}
+ when $n \in \mathbb{Z}$ and $z \in [-1; 1]$. In order to prove this fact, we
+ begin with the power series \emph{Ansatz}
+ \begin{equation*}
+ Z(x) = \sum_{k=0}^\infty a_k z^k,
+ \quad\text{from which follows that}\quad
+ \frac{dZ}{dz} = \sum_{k=0}^\infty k a_k z^{k-1}, \qquad
+ \frac{d^2 Z}{dz^2} = \sum_{k=0}^\infty k (k-1) a_k z^{k-2}.
+ \end{equation*}
+ Since the power series method converges only up to the nearest singularity,
+ which is at $z=1$ (and $z=-1$), we shall remark that we will find a solution
+ only for $|z|<1$. The Legendre equation \eqref{kugel:eqn:legendre-bis} can be
+ therefore rewritten as
+ \begin{align}
+ 0 &= (1-z^2) \sum_{k=0}^\infty k (k-1) a_k z^{k-2}
+ - 2z\sum_{k=0}^\infty k a_k z^{k-1}
+ + n(n+1)\sum_{k=0}^\infty a_k z^k \nonumber \\
+ &= \sum_{k=0}^\infty k (k-1) a_k z^{k-2}
+ - \sum_{k=0}^\infty k (k-1) a_k z^{k}
+ - 2z\sum_{k=0}^\infty k a_k z^{k-1}
+ + n(n+1)\sum_{k=0}^\infty a_k z^k. \label{kugel:eqn:legendre-ansatz}
+ \end{align}
+ Considers that by shifting the index $k$ in the sum of first term
+ \begin{equation*}
+ \sum_{k=0}^\infty k (k-1) a_k z^{k-2}
+ = \sum_{k=-2}^\infty (k+2)(k+1) a_{k+2} z^k
+ = \sum_{k=0}^\infty (k+2)(k+1) a_k z^k,
+ \end{equation*}
+ since when $k = -1$ or $-2$ the summand is zero. This means that
+ \eqref{kugel:eqn:legendre-ansatz} becomes
+ \begin{align*}
+ \sum_{k=0}^\infty &(k+1)(k+2) a_{k+2} z^{k}
+ - \sum_{k=0}^\infty k (k-1) a_k z^{k}
+ - 2\sum_{k=0}^\infty k a_k z^k
+ + n(n+1)\sum_{k=0}^\infty a_k z^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] z^k \stackrel{!}{=} 0,
+ \end{align*}
+ which is equivalent to saying that
+ \begin{equation*}
+ (k+1)(k+2) a_{k+2} - k (k-1) a_k - 2 k a_k + n(n+1) a_k = 0,
+ \end{equation*}
+ so we can derive a recurrence relation for $a_{k+2}$:
+ \begin{equation}
+ \label{kugel:eqn:coeff-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}
+ Following the relation \eqref{kugel:eqn:coeff-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
+ &= \left( -\frac{(n-4)(5+n)}{6 \cdot 5} \right)
+ \left( -\frac{(n-2)(n+3)}{4 \cdot 3} \right) a_2 \\
+ &= \left( -\frac{(n-4)(n+5)}{6 \cdot 5} \right)
+ \left( -\frac{(n-2)(n+3)}{4 \cdot 3} \right)
+ \left( -\frac{n(n+1)}{2 \cdot 1} \right) 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 ($i = 2k$)
+ coefficient and obtain
+ \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*}
+ and a similar expression can also be written for the odd coefficients
+ $a_{2k-1}$. In the latter case, the equation starts from $a_1$ and to find the
+ pattern we can write the recursion for an odd coefficient, for example for
+ $a_7$:
+ \begin{align*}
+ a_{7} = -\frac{(n-5)(n+6)}{7\cdot 6} a_5
+ &= \left( -\frac{(n-5)(n+6)}{7 \cdot 6} \right)
+ \left( -\frac{(n-3)(n+4)}{5 \cdot 4} \right) a_3 \\
+ &= \left( -\frac{(n-5)(n+6)}{7 \cdot 6} \right)
+ \left( -\frac{(n-3)(n+4)}{5 \cdot 4} \right)
+ \left( -\frac{(n-1)(n+2)}{3 \cdot 2} \right) 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 ($i =
+ 2k+1$) coefficient and get
+ \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*}
+ Now, if we let
+ \begin{align*}
+ Z_\text{e}^K(z) &:=
+ \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)!} z^{2k}, \\
+ Z_\text{o}^K(z) &:=
+ \sum_{k=0}^K(-1)^k \frac{
+ (n + 2k)(n+2k-2)\hdots \color{blue} (n-(2k-1)+2)(n-(2k-1))
+ }{(2k+1)!} z^{2k+1},
+ \end{align*}
+ the solution to the Legendre equation \eqref{kugel:eqn:legendre-bis} can be
+ written as
+ \begin{equation}\label{eq:solution}
+ Z(z) = \lim_{K \to \infty} \left[
+ a_0 Z_\text{e}^K(z) + a_1 Z_\text{o}^K(z)
+ \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*}
+ \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*}
+ \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*}
+ \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}
+ \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}
+ \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}
+ 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}
+ \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)!}}
@@ -150,21 +187,23 @@
{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}
+ \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*}
+ \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}
+ \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{equation}
\end{proof}
+\subsection{Associated Legendre Equation}
+\label{kugel:sec:proofs:associated-legendre}
\begin{lemma}\label{kugel:lemma:sol_associated_leg_eq}
If $Z_n(z)$ is a solution of the Legendre equation \eqref{kugel:eqn:legendre},
@@ -243,3 +282,5 @@
\end{align*}
implies $\hat{y}_m(x)$ being a solution of Eq.\eqref{kugel:eq:associated_leg_eq}
\end{proof}
+
+\subsection{Orthogonality}