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,

$\sum _{0}^{l}{P}_{l}\left(x\right){t}^{l}(l+l+1)=\frac{xt-{t}^{2}+xt-{t}^{2}}{{(1-2xt+{t}^{2})}^{\frac{3}{2}}}+\frac{1}{{(1-2xt+{t}^{2})}^{\frac{1}{2}}}$

Use generating function for legendre polynomial,

$\sum _{x=0}^{\mathrm{\infty}}{P}_{n}\left(x\right){t}^{n}={(1-2xt+{t}^{2})}^{-\frac{1}{2}}$

Consider the limit 0 to l

$\sum _{n=0}^{l}{P}_{l}\left(x\right){t}^{l}={(1-2xt+{t}^{2})}^{-\frac{1}{2}}$

Differentiating with respect to t,

$\sum _{0}^{l}{P}_{l}\left(x\right){<}^{l-1}=(-\frac{1}{2}){(1-2xt+{t}^{2})}^{-\frac{3}{2}}(-2x+2t)$

$\sum _{0}^{l}{P}_{l}\left(x\right){<}^{l-1}=\frac{(x-t)}{{(1-2xt+{t}^{2})}^{\frac{3}{2}}}$

Multiplying t on both sides we get,

$\sum _{0}^{l}{P}_{l}\left(x\right){<}^{l-1}t=\frac{(x-t)t}{{(1-2xt+{t}^{2})}^{\frac{3}{2}}}$

$\sum _{0}^{l}{P}_{l}\left(x\right){<}^{l}=\frac{xt-{t}^{2}}{{(1-2xt+{t}^{2})}^{\frac{3}{2}}}$

Let us consider,