Normalisation of Legendre polynomials

Question

Legendre Polynomials have been normalized such that P(1)=1. But how is this justified? If P(1) is anything other than 0, I can divide the polynomial by that constant and move on. But how can we be sure that P(1) is never zero where P is any Legendre Polynomial.

Answer 1

As discussed on the Wikipedia page, these polynomials obey the "three term recurrence relation..."

$$(n+1)P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x)$$

Using induction with $P_0(1)=1$, suppose that $\forall n\le k,P_n(1)=P_k(1)=1$. Then we have

$$(k+1)P_{k+1}(1)=(2k+1)P_k(1)-kP_{k-1}(1)\\ (k+1)P_{k+1}(1)=2k+1-k\\ P_{k+1}(1)=1$$

and by reapplying the same process, we can get $P_{k+2}(1)=1,\dots$ and for all $n, P_n(1)=1$.

Answer 2

The generating function of the Legendre polynomials is $$ \frac{1}{\sqrt{1-2xt+t^2}} = \sum_{n=0}^{\infty} P_n(x) t^n. $$ Now stick $x=1$, and you obtain (since $t$ is as small as we like) $$ \sum_{n=0}^{\infty} P_n(1) t^n = \frac{1}{\sqrt{1-2t+t^2}} = \frac{1}{1-t} = \sum_{n=0}^{\infty} 1t^n, $$ so $P_n(1)=1$ for all $n$.