Definition:Legendre Polynomial

Definition
The Legendre polynomials are the solutions to Legendre's Differential Equation.

These solutions form a polynomial sequence of orthogonal polynomials on the interval $\left[{-1 \,.\,.\, 1}\right]$.

First few Legendre polynomials
The first five Legendre polynomials are:

$P_0 (x) = 1$

$P_1 (x) = x$

$P_2 (x) = \frac{1}{2}(3x^2 - 1)$

$P_3 (x) = \frac{1}{2}(5x^3 - 3x)$

$P_4 (x) = \frac{1}{8}(35x^4 - 30x^2 + 3)$

Legendre polynomials can be found using the Bonnet's recursion formula.

Bonnet's Recursion Formula
$(n+1)P_{n+1}(x) = (2n + 1) x P_n (x) - n P_{n-1} (x)$

Length of Legendre polynomial
$||P_n (x)|| = \sqrt{\displaystyle \int_{-1}^{1} \left(P_n (x)\right)^2\,\mathrm{d}x} = \sqrt{\displaystyle \frac{2}{2n+1}}$

Proof of Length
This proof is based on Bonnet's recursion formula. We want to proof the following statement.

$||P_n (x)|| = \sqrt{\displaystyle \frac{2}{2n+1}}$

To proof this statement we need to shift Bonnet's formula from $n$ to $n-1$, after shifting we recieve modified Bonnet's formula:

$n P_n (x) = (2n - 1) x P_{n-1} (x) - (n-1) P_{n-2} (x)$

so $P_n (x) = \frac{2n-1}{n} x P_{n-1} (x) - \frac{n-1}{n} P_{n-2} (x)$

Now we plug this into following: $\displaystyle || P_n (x) ||^2 = \int_{-1}^{1} P(x) P(x)\,\mathrm{d}x = \int_{-1}^{1} P(x) \left(\frac{2n-1}{n} x P_{n-1} (x) - \frac{n-1}{n} P_{n-2} (x)\right)\,\mathrm{d}x$

$\displaystyle = \frac{2n-1}{n} \int_{-1}^{1} x P_n (x) P_{n-1} (x)\,\mathrm{d}x - \frac{n-1}{n} \int_{-1}^1 P_n (x) P_{n-2} (x)\,\mathrm{d}x$

Two different Legendre polynomials are orthogonal on the interval $[-1,1]$ which means, that:

$\displaystyle \int_{-1}^1 P_n (x) P_m (x)\,\mathrm{d}x = 0$ if and only if $n \not= m$, so we now know that:

$\displaystyle || P_n (x) ||^2 = \frac{2n-1}{n} \int_{-1}^{1} x P_n (x) P_{n-1} (x)\,\mathrm{d}x$

Now from the original Bonnet's formula we see that:

$\displaystyle x P_n(x) = \frac{n+1}{2n+1}P_{n+1}(x) + \frac{n}{2n+1}P_{n-1}(x)$

Now we replace the $x P_n(x)$ in the integral and we get:

$\displaystyle ||P_n(x)||^2 = \frac{2n-1}{n}\frac{n+1}{2n+1}\int_{-1}^1 P_{n+1}(x)P_{n-1}(x)\,\mathrm{d}x + \frac{2n-1}{2n+1}\int_{-1}^1 P_{n-1}(x)P_{n-1}(x)\,\mathrm{d}x = \frac{2n-1}{2n+1}\int_{-1}^1 P_{n-1}(x)P_{n-1}(x)\mathrm{d}x = \frac{2n-1}{2n+1} ||P_{n-1}(x)||^2$

So we see that:

$\displaystyle ||P_n(x)||^2 = \frac{2n-1}{2n+1}||P_{n-1}(x)||^2 = \dots = \frac{2n-1}{2n+1} \frac{2n-3}{2n-1} \frac{2n-5}{2n-3} \dots \frac{3}{5} \frac{1}{3} ||P_0 (x)||^2$

We see that almost everything nicely cancel out and we get:

$\displaystyle ||P_n(x)||^2 = \frac{||P_0(x)||^2}{2n+1}$

All what is left is to compute the length of first Legendre polynomial: $\displaystyle ||P_0(x)||^2 = \int_{-1}^1 1\,\mathrm{d}x = 2$

So we get: $\displaystyle ||P_n(x)||^2 = \frac{2}{2n+1}$ and the length is the square root of this: $\displaystyle ||P_n(x)|| = \sqrt{\frac{2}{2n+1}}$.