# Legendre polynomials and Bessel function of the first kind

1. May 8, 2013

### Rulonegger

1. The problem statement, all variables and given/known data
Prove that $$\sum_{n=0}^{\infty}{\frac{r^n}{n!}P_{n}(\cos{\theta})}=e^{r\cos{\theta}}J_{0}(r\sin{\theta})$$ where $P_{n}(x)$ is the n-th legendre polynomial and $J_{0}(x)$ is the first kind Bessel function of order zero.

2. Relevant equations
$$P_{n}(\cos{\theta})=\frac{1}{2^{2n}}\sum_{k=0}^{n}{\frac{(2n-2k)!(2k)!}{(k!)^2[(n-k)!]^2}e^{i(2k-n)\theta}}$$
$$J_{\nu}(x)=\sum_{s=0}^{\infty}{\frac{(-1)^s}{s!(\nu+s)!}\left(\frac{x}{2}\right)^{\nu+2s}}$$
$$\left(\sum_{n=0}^{\infty}{a_n}\right)\cdot\left(\sum_{n=0}^{\infty}{b_n}\right)=\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{a_{n-k}b_k}}$$

3. The attempt at a solution
Using that $$e^{r\cos{\theta}}=\sum_{n=0}^{\infty}{\frac{(r\cos{\theta})^{n}}{n!}}$$
I get $$e^{r\cos{\theta}}J_{0}(r\sin{\theta})=\sum_{n=0}^{\infty}{\frac{(r\cos{\theta})^{n}}{n!}}\cdot\sum_{s=0}^{\infty}{\frac{(-1)^s}{(s!)^2}\left(\frac{r\sin{\theta}}{2}\right)^{2s}}=\sum_{n=0}^{∞}{\sum_{k=0}^{n}{\frac{(r\cos{\theta})^{n-2k}}{(n-2k)!}}\frac{(-1)^k}{(k!)^2}\left(\frac{r\sin{\theta}}{2}\right)^{2k}}$$
$$\implies e^{r\cos{\theta}}J_{0}(r\sin{\theta})=\sum_{n=0}^{\infty}{\frac{r^n}{n!}\left(\sum_{k=0}^{n}{\frac{n!(-1)^k}{(n-2k)!(k!)^2}\frac{\cos^{n-2k}{\theta}\sin^{2k}{\theta}}{2^{2k}}}\right)}$$
So I just need to prove that
$$\frac{1}{2^{2n}}\sum_{k=0}^{n}{\frac{(2n-2k)!(2k)!}{(k!)^2[(n-k)!]^2}e^{i(2k-n)\theta}}=\sum_{k=0}^{n}{\frac{n!(-1)^k}{(n-2k)!(k!)^2}\frac{\cos^{n-2k}{\theta}\sin^{2k}{\theta}}{2^{2k}}}$$
I've tried to expand the exponential as an infinite sum, or write the sine and cosine functions as exponentials, but I don't get anything, and I seriously doubt the former steps were made without any mistake. Any help would be greatly appreciated!

2. May 15, 2013

### Thaakisfox

Try using some physical arguments. The left hand side is the solution to Laplace's equation considering azimuthal symmetry. So you can assume it is the solution with a given potential on the surface of a sphere. So the left hand side should be also a solution to Laplace's equation, but in cylindrical coordinates.
So these two represent a solution of Laplace's equation in different coordinate systems. So just transform the left to cylindrical and do the math.
Also try using Rodrigues's formula and the integral representation of the Bessel function.