Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [legendre-polynomials]

For questions about Legendre polynomials, which are solutions to a particular differential equation that frequently arises in physics.

The Legendre polynomials $P_n$ are defined to be solutions of Legendre's differential equation

$$\frac{d}{dx} \left[(1 - x^2) \frac{d}{dx} P_n(x)\right] + n(n + 1) P_n(x) = 0$$

Alternatively, by Rodrigues' formula,

$$P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \left[(x^2 - 1)^n\right]$$

The Legendre polynomials occur frequently in physics, and in particular in solving Laplace's equation in spherical coordinates.

641 questions
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Prove Bonnet's recursion formula for Legendre polynomials

I'm struggling with the following problem: In order to calculate $\int_{a}^{b}f(x)dx$ we use the Gaussian quadrature formula $\int_{a}^{b}f(x)dx\approx\sum_{i=0}^{n}A_if(x_i)$, where $A_i$ are the weights and $x_i$ are the roots of the polynomial…
Filip Stankovski
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Prove: $\int_{-1}^1 x^n P_n(x) dx = \frac{2^{n+1}n!^2}{(2n+1)!}$

I'm trying to prove: $$\int_{-1}^1 x^n P_n(x) dx = \frac{2^{n+1}n!^2}{(2n+1)!}$$ My attempt consisted in applying the Rodrigues formula: $$\int_{-1}^1 x^n P_n(x)dx = \dfrac{1}{2^n n!} \int_{-1}^1 x^n \dfrac{d^n}{dx^n}(x^2-1)^ndx$$ and integrating by…
Carlos Eduardo Staudt
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First Derivative of Legendre Polynomials

I have been tasked with calculating the first derivative of the k-th Legendre polynomial $$P^{'}_k(1)$$ I was given the hint to use the generalized product rule $$\frac{d^{n}}{ds^{n}}[F(s)G(s)]=\sum_{j=0}^n {n \choose j}F^{(n-j)}(s)G^{(j)}(s)$$ but…
Anonymous
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Applying properties of Legendre functions

In addition to a former question. I wish to show that: $$\int_{-1}^1 x^n P_n (x) dx = \frac{n}{2n+1}\int_{-1}^1 x^{n-1} P_{n-1} (x) dx$$ Assuming the result of: $$P'_{n+1}-P'_{n-1} = (2n+1)P_{n}$$ Multiplying both sides with $x^n$. We obtain:…
mathpieuler
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Legendre polynomial with argument zero $P_n ( 0 )$

I want to find the expression for $P_n(x)$ with $x = 0$, ie $P_n(0)$ for any $n$. The first few non-zero legendre polynomials with $x=0$ are $P_0(0) = 1$, $P_2(0) = -\frac{1}{2}$, $P_4(0) = \frac{3}{8}$, $P_6(0) = -\frac{5}{16}$, $P_8(0) =…
user3264392
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Associated Legendre Special Properties

Prove the relationship of the equation that : $$ P_n^{-m}(x)=(-1)^m\frac{(n-m)!}{(n+m)!}P_n^{m}(x) -------> equation (1)$$ This is what I know: $$ P_n^{m}(x)=\frac{1}{2^nn!}(1-x^2)^{\frac{m}{2}}\frac{d^{n+m}}{dx^{n+m}}(x^2-1)^n$$ Leibnitz's…
Aschoolar
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Prove this relation for the legendre polynomials

I'm asked to prove that$$P_{2n}(0)=(-1)^n\frac{1\cdot3\cdots(2n-1)}{2\cdot 4\cdots (2n)}$$ given that $$\frac{1}{\sqrt{1-2xu+u^2}}=\sum_{n=0}^\infty P_n(x)u^n$$ I tried this: Let $x=0$ and use $$(u^2+1)^{-1/2}=\sum_{m=0}^\infty \begin{pmatrix} -1/2…
Cristian Rodríguez
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Legendre polynomial property

How do i show from the Legendre's Polynomial equation $$P_n(x)=\sum_{k=0}^N \dfrac{(-1)^k(2n-2k)!}{2^nk!(n-k)!(n-2k)!}x^{n-2k}$$ where $N=n/2$ for even $n$ and $ N=(n-1)/2$ for odd $n$. Using just this information how to show $P_n(1)=1$? Can…
Upstart
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Integral form of legendre polynomial of second kind

I found a definite integral form of Legendre polynomial of second kind, $$ Q_{n}(z)=\frac{1}{2}\int^{+1}_{-1}\frac{P_{n}(t)}{z-t}dt $$ when n is an integer. I wonder how to evaluate this integral. I tried to use contour integral by changing $t$ to…
tard
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Fourier-Legendre

Question: Let $f:[-1,1] \rightarrow \mathbb{R}$ be defined by $$ f(x) = \sqrt{\frac{1-x}{2}}. $$ By multiplying both sides of the equation $$ \frac{1}{\sqrt{1-2xt+t^2}} = \sum_{n=1}^{\infty} P_n(x)t^n $$ by $f(x)$ and integrating both sides with…
Joe
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Prove that the even (odd) degree Legendre polynomials are even > (odd) functions of $t$.

a.) Prove that the even (odd) degree Legendre polynomials are even (odd) functions of $t$. b.) Prove that if $p(t) = p(-t)$ is an even polynomial, then all the odd order coefficents $c_{2j+1} = 0$ in its Legendre expansion $p(t) = c_0q_0(t) +…
diimension
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Maximum of Legendre polynomial

I wrote down some proofs long time ago and I stated the following. The (shifted) Legendre polynomial $p_i(x)$ with degree $i$ satisfies $\max\{|p_i(x)|:0\leq x\leq 1\} = \sqrt{2i+1}$ $\max\{|p_i'(x)|:0\leq x\leq 1\} = i(i+1)\sqrt{2i+1}$…
simko
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How to prove the Legendre Polynomials are unique up to a scalar multiple?

I've gotten this question in a book I am reading. My plan is consistent with the typical methods for uniqueness. I want to assume that there exist some other orthogonal basis of polynomials such that $ =0$ when $n \neq m$, and $R_n$ is an…
Rellek
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evaluate associated legendre polynomial

Show that: $$P_{2n}^1(0)=0$$ $$P_{2n}^1(0)=(-1)^n\frac{(2n+1)!}{(2^nn!)^2} $$ by each of three methods: a) use of recurrence relations $$(2n+1)(1-x^2)^{\frac{1}{2}}P_{n}^m=(n+m)(n+m-1)P_{n-1}^{m-1}-(n-m+1)(n-m-2)P_{n+1}^{m-1}$$ b) expansion of…
Aschoolar
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Proving Rodrigues's formula

Recall that the Legendre polynomials $\{P_n(t)\}, n = 0, 1,\dotsc$ are defined by applying the Gram–Schmidt process to the monomials $\{1, t, t^2,\dotsc \}$ in $L^2[−1, 1]$ and by rescaling the resulting orthonormal vectors ${x_n(t)}$ as…
Bolys
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