Legendre polynomials and Bessel function of the first kind

AI Thread Summary
The discussion focuses on proving the identity involving Legendre polynomials and the Bessel function of the first kind, specifically that the sum of Legendre polynomials multiplied by an exponential function equals a product of the exponential function and the Bessel function. The attempt to solve this involves expanding the exponential and Bessel function into series and manipulating the resulting sums. Suggestions include using Rodrigues's formula and considering the physical interpretation of the problem in terms of solutions to Laplace's equation in different coordinate systems. The discussion emphasizes the need for careful algebraic manipulation and transformation between coordinate systems to establish the proof.
Rulonegger
Messages
14
Reaction score
0

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!
 
Physics news on Phys.org
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.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top