# Bessel function, Generating function

LagrangeEuler
Generating function for Bessel function is defined by

$$G(x,t)=e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=-\infty}J_n(x)t^n$$
Why here we have Laurent series, even in case when functions are of real variables?

Staff Emeritus
Homework Helper
Gold Member
It is only a Laurent series if you look at it as an expansion around ##t = 0##, at which the function is clearly singular for all ##x \neq 0##. The more relevant point is to look at ##f(r,\varphi) = G(r,e^{i\varphi})## (i.e., ##|t| = 1##), which solves the eigenvalue equation ##(\nabla^2 + 1) f(r,\varphi) = 0##, where ##\nabla^2## is the Laplace operator written in polar coordinates. You will find that
$$\nabla^2 e^{r(e^{i\varphi}- e^{-i\varphi})/2} = \nabla^2 e^{r\sin(\varphi)} = e^{r\sin(\varphi)}.$$
Correspondingly, you will find that
$$(\nabla^2 + 1) f(r,\varphi) = \sum_{n=-\infty}^\infty e^{in\varphi} \left( \frac{1}{r} \partial_r r \partial_r J_n(r) + \left(1 - \frac{n^2}{r^2}\right) J_n(r)\right) = 0.$$
Since each ##e^{in\varphi}## is linearly independent, this can only be satisfied if
$$\frac{1}{r} \partial_r r \partial_r J_n(r) + \left(1 - \frac{n^2}{r^2}\right) J_n(r) = 0,$$
which is Bessel's differential equation.

Staff Emeritus
$$\nabla^2 e^{r(e^{i\varphi}- e^{-i\varphi})/2} = \nabla^2 e^{r\sin(\varphi)} = e^{r\sin(\varphi)}.$$
$$(\nabla^2 + 1) f(r,\varphi) = \sum_{n=-\infty}^\infty e^{in\varphi} \left( \frac{1}{r} \partial_r r \partial_r J_n(r) + \left(1 - \frac{n^2}{r^2}\right) J_n(r)\right) = 0.$$
$$\frac{1}{r} \partial_r r \partial_r J_n(r) + \left(1 - \frac{n^2}{r^2}\right) J_n(r) = 0,$$