A question (Problem $7.4$) in my textbook (Mathematical Methods in the Physical Sciences – 3rd Edition by Mary L. Boas P578) asks me to Use $$\int_{x=-1}^{1}(P_L(x)\cdot\text{any polynomial of degree < L})\,\mathrm{d}x=0\tag{A}$$ to prove that $$\displaystyle\int_{x=-1}^{1}P_L(x)P_{L-1}\acute (x)\,\mathrm{d}x=0\tag{1}$$and gives the hint: $\color{#180}{\fbox{What is the degree of $P_{L−1}(x)$}}$? Also, show that $$\displaystyle\int_{x=-1}^{1}P_L\acute(x)P_{L+1} (x)\,\mathrm{d}x=0\tag{2}$$ Where $P_L(x),P_{L-1}(x)$ represent any general […]

This comes up in relation to Legendre functions. The claim is made that for $n =0,1,2,3,\cdots$ and $m=0,1,2,3,\cdots,n$, there is a constant $C_{n,m}$ such that $$ \frac{d^{n-m}}{dx^{n-m}}(1-x^2)^n=C_{n,m}(1-x^2)^{m}\frac{d^{n+m}}{dx^{n+m}}(1-x^2)^n $$ A typical suggested proof is writing out all of the polynomial coefficients. Does anyone have a clever way to see this must be true? I’ve tried many […]

If a polynomial is given by $$y=\color{red}{a_0\left[1-\frac{l(l+1)}{2!}x^2+\frac{l(l+1)(l-2)(l+3)}{4!}x^4-\cdots\right]}+\color{blue}{a_1\left[x-\frac{(l-1)(l+2)}{3!}x^3+\frac{(l-1)(l+2)(l-3)(l+4)}{5!}x^5-\cdots\right]}\tag{1}$$ where $l$ is a constant and $a_0,a_1$ are coefficients. The recurrence relation is given by $$a_{n+2}=-\frac{(l-n)(l+n+1)}{(n+2)(n+1)}a_n\tag{2}$$ The objective is to find the first few Legendre polynomials $P_l(x)$ such that $P_l(1)=1$ without using Rodrigues’ formula: $$\fbox{$P_l(x)=\frac{1}{2^{l}l!}\frac{\mathrm{d}^l}{\mathrm{d}x^l}{\left(x^2-1\right)}^l$}$$ The method given in my textbook states that: If the value of $a_0$ […]

From the Generating function for Legendre Polynomials: $$\Phi(x,h)=(1-2xh+h^2)^{-1/2}\quad\text{for}\quad \mid{h}\,\mid\,\lt 1$$ My text states that: For $x=1$ $$\Phi(1,h)=\color{red}{(1-2h+h^2)^{-1/2}=\frac{1}{1-h}}=1+h+h^2+\cdots$$ My question is about the justification of the equality marked $\color{red}{\mathrm{red}}$. Since although $$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((1-h)(1-h)\Big)^{-1/2}$$$$=\Big((1-h)^2\Big)^{-1/2}=\color{#180}{\frac{1}{1-h}}=1+h+h^2+\cdots\tag{1}$$ as required. I could also write $$\Phi(1,h)=(1-2h+h^2)^{-1/2}=\Big((h-1)(h-1)\Big)^{-1/2}$$$$=\Big((h-1)^2\Big)^{-1/2}=\color{blue}{\frac{1}{h-1}}=\frac{1}{h}\left(1-\frac{1}{h}\right)^{-1}\ne 1+h+h^2+\cdots\tag{2}$$ Why is it that $\Phi(1,h)$ is equal to $(1)$ but not equal to $(2)$? I think […]

The Generating function for Legendre Polynomials is: $$\Phi(x,h)=(1-2xh+h^2)^{-1/2} \quad\text{for}\quad |h|\lt 1\tag{5.1}$$ or $$\Phi(x,h)=\sum_{l=0}^\infty h^l P_l(x)\quad\text{for}\quad |h|\lt 1\tag{5.2}$$ In order for the question I have to make any sense whatsoever I must include some background information as written in my textbook: Now, suppose that I have a potential $V_q$ which is given by $$V_q=\frac{q}{\sqrt{R^2-2aR\cos\theta+a^2}}$$ and I […]

The Generating function for Legendre Polynomials is: $$\Phi(x,h)=(1-2xh+h^2)^{-1/2}\quad\text{for}\quad \mid{h}\,\mid\,\lt 1\tag{1}$$ and Legendres’ Differential equation is $$\begin{align*} (1-x^2)y^{\prime\prime}-2xy^{\prime}+l(l+1)y=0\tag{2} \end{align*}$$ The question in my text asks me to: Verify the identity $$\color{blue}{\left(1-x^2\right)\frac{\partial^2 \Phi}{\partial x^2}-2x\frac{\partial\Phi}{\partial x}+h\frac{\partial^2}{\partial h^2}\left(h\Phi \right)=0}\tag{A}$$ by straightforward differentiation of $(1)$ and some algebra. End of question. My attempt: I started by taking partial derivatives of […]

Let $f : (-1,1)\to \mathbb{R}$ be given by $$f(x) = \begin{cases}0, & x\in (-1,0) \\ 1, & x \in [0,1)\end{cases}$$ and suppose we want to compute the Fourier-Legendre series of $f$, that is we want to write $$f = \sum_{n}c_n P_n$$ with $$c_n = \dfrac{\langle f,c_n\rangle }{|P_n|^2} = \dfrac{2}{2n+1}\int_{-1}^{1}f(x)P_n(x)dx.$$ This integral, though is quite complicated. […]

The question that follows is the final stage of the previous $3$ stages found here: Stage 1, Stage 2 and Stage 3 which are needed as part of a derivation of the Associated Legendre Functions Normalization Formula: $$\color{blue}{\displaystyle\int_{x=-1}^{1}[{P_{L}}^m(x)]^2\,\mathrm{d}x=\left(\frac{2}{2L+1}\right)\frac{(L+m)!}{(L-m)!}}\tag{1}$$ where for each $m$, the functions $${P_L}^m(x)=\frac{1}{2^LL!}\left(1-x^2\right)^{m/2}\frac{\mathrm{d}^{L+m}}{\mathrm{d}x^{L+m}}\left(x^2-1\right)^L\tag{2}$$ are a set of Associated Legendre functions on $[−1, 1]$. […]

I am trying to show that, \begin{align} I = \int_{-1}^1 x^nP_n(x)\,\mathrm{d}x = \frac{2^{n+1}n!n!}{(2n+1)!} \end{align} So far I have done the following. Rodrigues formula is as follows: \begin{align*} P_n(x) = \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} x^{n-2k} \end{align*} where, \begin{align*} \begin{aligned} N&=n/2, && \text{if} \quad n=\text{even} \\ N&=(n-1)/2, && \text{if} \quad n=\text{odd} \end{aligned} \end{align*} Substitute Rodrigues formula, \begin{align*} I […]

In a recent question, I proved that the Fourier-Legendre expansion of the function $f(x)=\text{sign}(x)$ over $(-1,1)$ is given by: $$2\sum_{m\geq 0}\frac{4m+3}{4m+4}\cdot\frac{(-1)^m}{4^m}\binom{2m}{m}\cdot P_{2m+1}(x)\tag{1}$$ with $P_n(x)$ being the $n$-th Legendre polynomial. On the other hand, it is well known that the Fourier series of $f(x)$ over $(-1,1)$ is given by: $$ \frac{4}{\pi}\sum_{m\geq 0}\frac{\sin((2m+1)\pi x)}{(2m+1)}.\tag{2}$$ When considering truncated […]

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