# 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 $\closedint {-1} 1$.

## First few Legendre polynomials

The first five Legendre polynomials are:

$P_0 \left({x}\right) = 1$
$P_1 \left({x}\right) = x$
$P_2 \left({x}\right) = \dfrac 1 2 \left({3 x^2 - 1}\right)$
$P_3 \left({x}\right) = \dfrac 1 2 \left({5 x^3 - 3 x}\right)$
$P_4 \left({x}\right) = \dfrac 1 8 \left({35 x^4 - 30 x^2 + 3}\right)$

## Bonnet's Recursion Formula

Legendre polynomials can be found using Bonnet's Recursion Formula.

$\paren {n + 1} \map {P_{n + 1} } x = \paren {2 n + 1} x \map {P_n} x - n \map {P_{n - 1} } x$

## Length of Legendre Polynomial

$\displaystyle \norm {\map {P_n} x} = \sqrt {\int_{-1}^1 \paren {\map {P_n} X}^2 \rd x} = \sqrt {\frac 2 {2 n + 1} }$

## Proof of Length

Applying Bonnet's Recursion Formula for $n - 1$:

$n \map {P_n} x = \paren {2 n - 1} x \map {P_{n - 1} } x - \paren {n - 1} \map {P_{n - 2} } x$

so:

$\map {P_n} x = \dfrac {2 n - 1} n x \map {P_{n - 1} } x - \dfrac {n - 1} n \map {P_{n - 2} } x$

Substituting for $\map {P_n} x$:

 $\displaystyle \norm {\map {P_n} x}^2$ $=$ $\displaystyle \int_{-1}^1 \map {P_n} x \, \map {P_n} x \rd x$ $\displaystyle$ $=$ $\displaystyle \int_{-1}^1 \map {P_n} x \paren {\frac {2 n - 1} n x \map {P_{n - 1} } x - \frac {n - 1} n \map {P_{n - 2} } x} \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {2 n - 1} n \int_{-1}^1 x \map {P_n} x \, \map {P_{n - 1} } x \rd x - \frac {n - 1} n \int_{-1}^1 \map {P_n} x \, \map {P_{n - 2} } x \rd x$ Linear Combination of Integrals
$\displaystyle \int_{-1}^1 \map {P_n} x \, \map {P_m} x \rd x = 0 \iff n \ne m$

so:

$\displaystyle (1): \quad \norm {\map {P_n} x}^2 = \frac {2 n - 1} n \int_{-1}^1 x \map {P_n} x \, \map {P_{n - 1} } x \rd x$
$\displaystyle x \map {P_n} x = \frac {n + 1} {2 n + 1} \map {P_{n + 1} } x + \frac n {2 n + 1} \map {P_{n - 1} } x$

Substituting for $x \map {P_n} x$ in $(1)$:

 $\displaystyle \norm {\map {P_n} x}^2$ $=$ $\displaystyle \frac {2 n - 1} n \frac {n + 1} {2 n + 1} \int_{-1}^1 \map {P_{n + 1} } x \, \map {P_{n - 1} } x \rd x + \frac {2 n - 1} {2 n + 1} \int_{-1}^1 \map {P_{n - 1} } x \, \map {P_{n - 1} } x \rd x$ $\displaystyle$ $=$ $\displaystyle \frac{2 n - 1} {2 n + 1} \int_{-1}^1 \map {P_{n - 1} } x \, \map {P_{n - 1} } x \rd x$ $\displaystyle$ $=$ $\displaystyle \frac {2 n - 1} {2 n + 1} \norm {\map {P_{n - 1} } x}^2$

Thus:

 $\displaystyle \norm {\map {P_n} x}^2$ $=$ $\displaystyle \frac {2 n - 1} {2 n + 1} \norm {\map {P_{n - 1} } x}^2$ $\displaystyle$ $=$ $\displaystyle \frac {2 n - 1} {2 n + 1} \frac {2 n - 3} {2 n - 1} \norm {\map {P_{n - 2} } x}^2$ $\displaystyle$ $=$ $\displaystyle \cdots$ $\displaystyle$ $=$ $\displaystyle \frac {2 n - 1} {2 n + 1} \frac {2 n - 3} {2 n - 1} \frac {2 n - 5} {2 n - 3} \dotsm \frac 3 5 \frac 1 3 \norm {\map {P_0} x}^2$

Most of this cancels out, leaving:

$\norm {\map {P_n} x}^2 = \dfrac {\norm {\map {P_0} x}^2} {2 n + 1}$

It remains to compute the length of the first Legendre polynomial:

 $\displaystyle \norm {\map {P_0} x}^2$ $=$ $\displaystyle \int_{-1}^1 1 \rd x$ $\displaystyle$ $=$ $\displaystyle \bigintlimits x {-1} 1$ Primitive of Constant $\displaystyle$ $=$ $\displaystyle 2$

Thus:

$\norm {\map {P_n} x}^2 = \dfrac 2 {2 n + 1}$

and so taking the square root:

$\norm {\map {P_n} x} = \sqrt {\dfrac 2 {2 n + 1} }$

$\blacksquare$

## Source of Name

This entry was named for Adrien-Marie Legendre.