# Find a recurrence relation and the Fourier-Legendre Series

Rodrique's Formula for the $n$th Legendre Polynomial is $$P_n\left(x\right)=\dfrac{1}{2^nn!}\dfrac{d^n}{dx^n}\left(\left(x^2-1\right)^n\right)$$

The Fourier-Legendre series of a function f is written $$f\left(x\right)\sum^{\infty}_{n=0}a_nP_n\left(x\right)$$ where the coefficients of the series are given by $$a_n=\dfrac{2n+1}{2}\int^1_{-1}f\left(x\right)P_n\left(x\right)dx$$

For the case $f\left(x\right)=\exp\left(-x\right)$, find a recurrence relation for $a_n$ and the Fourier-Legendre series.

I have begun substituting for $P_n$ in $a_n$ using Rodriques formula:

$$a_n=\dfrac{2n+1}{2^{n+1}n!}\int^1_{-1}\exp\left(-x\right)\dfrac{d^n}{dx^n}\left(\left(x^2-1\right)^n\right)dx$$

I then assume we integrate by parts, but I'm not sure where to go from there.

• Try writing out what one integration-by-parts on $a_{n}$ gives you. – Semiclassical May 24 '16 at 13:42
• I'm confused as to how we integrate this by parts @Semiclassical – Sophie Filer May 24 '16 at 19:24

You may use Bonnet's recursion formula $$(n+1)P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1} (x) \tag{1}$$ multiply both sides by $e^{-x}$ and integrate over $[-1,1]$, getting:
$$\frac{2n+2}{2n+3} a_{n+1} = -\frac{2n}{2n-1} a_{n-1} + (2n+1)\int_{-1}^{1} x e^{-x} P_n(x)\,dx \tag{2}$$ Now it is enough to use integration by parts and express $\frac{d}{dx}\left(x\cdot P_{n}(x)\right)$ as a linear combination of $P_{n}(x),P_{n-1}(x),\ldots$ and plug it into $(2)$ to get the wanted recursion.