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diff --git a/notes/FourierOnS2.tex b/notes/FourierOnS2.tex index 2dc30c1..de15f01 100644 --- a/notes/FourierOnS2.tex +++ b/notes/FourierOnS2.tex @@ -15,6 +15,7 @@ \newtheorem{lemma}{Lemma} \newtheorem{definition}{Definition} \newtheorem{remark}{Remark} +\newtheorem{proposition}{Proposition} % % Metadata @@ -219,14 +220,255 @@ Though, unfortunately this time the separation process is more involved, and the Equation \eqref{eqn:separation-phi} is easy, the solutions are complex exponentials \(e^{im\varphi}\), while \eqref{eqn:separation-theta} is known as the \emph{associated Legendre equation}. Though, normally the equation is written in term of \(x\) and \(y(x)\), so \eqref{eqn:separation-theta} is brought to a more familiar form by using the substitution \(x = \cos\vartheta\) and \(y = \Theta\): \[ - \left(1 - x^2 \right) \frac{d^2 y}{dx} - 2x \frac{dy}{dx} + \left[ - n(n+1) - \frac{m^2}{1 - x^2} - \right] y(x) = 0. + \left( 1 - x^2 \right) \frac{d^2 y}{dx^2} - 2x \frac{dy}{dx} + + \left[ n(n+1) - \frac{m^2}{1 - x^2} \right] y(x) = 0. \] Finding the solutions to this equation is so involved, that it deserves its own section. \subsection{The associated Legendre polynomials} +In this section we would like to find the solutions to the \emph{associated} Legendre equation, which is actually a generalization of Legendre equation: +\begin{equation} \label{eqn:legendre} + \left( 1 - x^2 \right) \frac{d^2 y}{dx^2} + - 2x \frac{dy}{dx} + n(n + 1) y(x) = 0. +\end{equation} +Thus we first need examine the solutions to this equation before constructing the more general solution. + +\begin{proposition} + The polynomials + \begin{equation} \label{eqn:legendre-poly} + P_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} + \frac{(-1)^k (2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k}, + \end{equation} + known as Legendre's polynomials are solutions to Legendre's equation \eqref{eqn:legendre}. +\end{proposition} + +\begin{lemma} The expression +\begin{equation} \label{eqn:legendre-rodrigues} + P_n(x) = \frac{1}{n!2^n}\frac{d^n}{dx^n}(x^2-1)^n. +\end{equation} +is equivalent to \eqref{eqn:legendre-poly}. +\end{lemma} + +\begin{proof} + We start expanding the term \((x^2-1)^n\); According to the binomial theorem + \begin{equation*} + (x^2-1)^n=\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2(n-k)}. + \end{equation*} + Substituting the above, \eqref{eqn:legendre-rodrigues} becomes + \begin{align*} + \frac{1}{n!2^n} \frac{d^n}{dx^n} (x^2 - 1)^n + &= \frac{1}{n!2^n} \sum_{k=0}^n(-1)^k \binom{n}{k} \frac{d^n}{dx^n}x^{2(n-k)} \\ + &= \frac{1}{n!2^n} \sum_{k=0}^{\lfloor n / 2 \rfloor}(-1)^k + \binom{n}{k}\frac{d^n}{dx^n}x^{2(n-k)}. + \end{align*} + Recall that + \[ + \frac{d^n}{dx^n}x^\alpha + % = \alpha(\alpha-1)(\alpha-2)\hdots(\alpha-n+1)x^{\alpha-n} + = \frac{\alpha!}{(\alpha-n)!}x^{\alpha-n}, + \] + thus + \begin{align*} + \frac{1}{n!2^n} &\sum_{k=0}^{\lfloor n / 2 \rfloor}(-1)^k + \binom{n}{k}\frac{d^n}{dx^n}x^{2(n-k)} \\ + &= \frac{1}{n!2^n} \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \binom{n}{k} + \frac{(2n-2k)!}{(n-2k)!}x^{n-2k} \\ + &= \frac{1}{n!2^n} \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k \frac{n!}{k!(n-k)!} + \frac{(2n-2k)!}{(n-2k)!}x^{n-2k} \\ + &= \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{(-1)^k (2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k}. + \qedhere + \end{align*} +\end{proof} + \subsection{Spherical harmonics} +\clearpage +\appendix +\section{Proofs} +\subsection{Legendre Polynomials} + +\begin{lemma} + The polynomial + \[ + P_n(x) = \sum_{k=0}^{\lfloor n/2 \rfloor} (-1)^k + \frac{(2n-2k)!}{2^n k! (n-k)!(n-2k)!} x^{n-2k}, + \] + is a solution to Legendre's equation + \[ + \left( 1 - x^2 \right) \frac{d^2 y}{dx^2} + - 2x \frac{dy}{dx} + n(n + 1) y(x) = 0 + \] + for \(n > 0\). +\end{lemma} +\begin{proof} +To solve \eqref{eqn:legendre} we use the power series ansatz +\begin{equation} \label{eqn:legendre-power-ansatz} + y(x) = \sum_{k=0}^\infty a_k x^k, +\end{equation} +from which follows that +\[ + y' = \sum_{k = 0}^\infty k a_k x^{k-1}, \text{ and } + y'' = \sum_{k = 0}^\infty k (k-1) a_k x^{k-2}. +\] +By substituting the above and \eqref{eqn:legendre-power-ansatz} into \eqref{eqn:legendre} we get the that first term +\begin{align*} + \Big( 1 &- x^2 \Big) y'' + = \left( 1 - x^2 \right) \sum_{k = 0}^\infty k (k-1) a_k x^{k - 2} \\ + &= \sum_{k = 0}^\infty k (k-1) a_k x^{k-2} + k (k-1) a_k x^{k} \\ + &= \sum_{k = 0}^\infty \left[ + (k + 1) (k + 2) a_{k + 2} + k (k-1) a_k + \right] x^{k}, +\end{align*} +where in the last step to factor out \(x^k\) we shifted the index in the coefficients by 2, i.e. +\[ + \sum_{k = 0}^\infty k (k - 1)a_k x^{k - 2} + = \sum_{k = 0}^\infty (k + 2) (k + 1) a_{k + 2} x^k. +\] +Similarly, the second term: +\[ + -2xy' = -2x \sum_{k = 0}^\infty k a_k x^{k - 1} + = \sum_{k = 0}^\infty - 2k a_k x^k. +\] +Finally, combining the above the complete substitution yields +\begin{gather*} + \Big( 1 - x^2 \Big) y'' - 2xy' + n(n + 1) y = 0 \\ + \implies \sum_{k = 0}^\infty \big[ + (k + 1) (k + 2) a_{k + 2} + k (k-1) a_k \\ + \qquad \qquad + - 2k a_k + n(n + 1) a_k + \big] x^k = 0, +\end{gather*} +from which we can extract the recurrence relation +\begin{gather*} + \begin{aligned} + (k + 1) (k + 2) a_{k + 2} &+ k (k-1) a_k \\ + &- 2k a_k + n(n + 1) a_k = 0 + \end{aligned} \\ + \iff a_{k + 2} = \frac{(k-n)(k+n+1)}{(k+2)(k+1)} a_k. +\end{gather*} + +% TODO: finish reviewing proof +\if 0 + +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} +\fi + +\end{proof} + + \end{document} diff --git a/notes/build/FourierOnS2.pdf b/notes/build/FourierOnS2.pdf Binary files differindex 0201255..c327e27 100644 --- a/notes/build/FourierOnS2.pdf +++ b/notes/build/FourierOnS2.pdf |