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## Answered question

2021-09-04

Using generating function for Legendre polynomials, prove that
$\sum _{0}^{l}\left(2l+1\right){P}_{l}\left(x\right){t}^{l}=\frac{1-{t}^{2}}{{\left(1-2xt+{t}^{2}\right)}^{\frac{3}{2}}}$

### Answer & Explanation

jlo2niT

Skilled2021-09-05Added 96 answers

Introduction
The legendre polynomial denoted by ${P}_{n}\left(x\right)$.
The solution of physical problems is the class of function called legendre polynomials.
Explanation
To prove,
$\sum _{0}^{l}{P}_{l}\left(x\right){t}^{l}\left(l+l+1\right)=\frac{xt-{t}^{2}+xt-{t}^{2}}{{\left(1-2xt+{t}^{2}\right)}^{\frac{3}{2}}}+\frac{1}{{\left(1-2xt+{t}^{2}\right)}^{\frac{1}{2}}}$
Use generating function for legendre polynomial,
$\sum _{x=0}^{\mathrm{\infty }}{P}_{n}\left(x\right){t}^{n}={\left(1-2xt+{t}^{2}\right)}^{-\frac{1}{2}}$
Consider the limit 0 to l
$\sum _{n=0}^{l}{P}_{l}\left(x\right){t}^{l}={\left(1-2xt+{t}^{2}\right)}^{-\frac{1}{2}}$
Differentiating with respect to t,
$\sum _{0}^{l}{P}_{l}\left(x\right){<}^{l-1}=\left(-\frac{1}{2}\right){\left(1-2xt+{t}^{2}\right)}^{-\frac{3}{2}}\left(-2x+2t\right)$
$\sum _{0}^{l}{P}_{l}\left(x\right){<}^{l-1}=\frac{\left(x-t\right)}{{\left(1-2xt+{t}^{2}\right)}^{\frac{3}{2}}}$
Multiplying t on both sides we get,
$\sum _{0}^{l}{P}_{l}\left(x\right){<}^{l-1}t=\frac{\left(x-t\right)t}{{\left(1-2xt+{t}^{2}\right)}^{\frac{3}{2}}}$
$\sum _{0}^{l}{P}_{l}\left(x\right){<}^{l}=\frac{xt-{t}^{2}}{{\left(1-2xt+{t}^{2}\right)}^{\frac{3}{2}}}$
Let us consider,

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