Bessel function, Generating function

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SUMMARY

The generating function for the Bessel function is defined as G(x,t)=e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=-\infty}J_n(x)t^n. This function exhibits a Laurent series expansion around t = 0, indicating singular behavior for all x ≠ 0. The function f(r,φ) = G(r,e^{iφ}) solves the eigenvalue equation (∇² + 1) f(r,φ) = 0, where ∇² is the Laplace operator in polar coordinates. The resulting expression leads to Bessel's differential equation, confirming the relationship between the generating function and Bessel functions.

PREREQUISITES
  • Understanding of Bessel functions and their properties
  • Familiarity with generating functions in mathematical analysis
  • Knowledge of differential equations, specifically Bessel's differential equation
  • Basic concepts of complex analysis, particularly Laurent series
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  • Study the properties and applications of Bessel functions in mathematical physics
  • Learn about the derivation and applications of generating functions
  • Explore the solutions to Bessel's differential equation and their significance
  • Investigate the use of the Laplace operator in polar coordinates in various mathematical contexts
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Mathematicians, physicists, and engineers interested in applied mathematics, particularly those working with Bessel functions and differential equations in polar coordinates.

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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?
 
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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.
 

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