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diff --git a/buch/papers/kugel/proofs.tex b/buch/papers/kugel/proofs.tex index 93b3857..c3d4f01 100644 --- 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} |