- #1
Rulonegger
- 14
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Homework Statement
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.
Homework 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}}
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!
