Graduate Bessel function, Generating function

Click For Summary
The generating function for Bessel functions is defined as G(x,t)=e^{\frac{x}{2}(t-\frac{1}{t})} and can be expressed as a Laurent series around t=0, indicating singularity for x ≠ 0. This function leads to the eigenvalue equation (\nabla^2 + 1) f(r,φ) = 0 when analyzed in polar coordinates. The Laplace operator applied to e^{r(e^{iφ}- e^{-iφ})/2} yields e^{r sin(φ)}, confirming its role in solving the eigenvalue problem. The resulting expression shows that the condition can only be satisfied if it aligns with Bessel's differential equation. Thus, the discussion highlights the connection between generating functions, eigenvalue equations, and Bessel's differential equation.
LagrangeEuler
Messages
711
Reaction score
22
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?
 
Physics news on Phys.org
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.
 
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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K