Recurrence relation of Bessel Function

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aryajur

Joined: 23 Oct 2004
Posts: 689
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Location: Sunnyvale, USA

04 Oct 2005 8:16 recurrence relation for bessel equation

How can I prove the recurrance relation of the Bessel Function of the 1st kind? i.e.

J(p-1, x) + J(p+1, x) = 2p/x J(p, x)

Any hints would be appreciated.

Last edited by aryajur on 06 Oct 2005 5:25; edited 1 time in total

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me2please

Joined: 07 Aug 2004
Posts: 362
Helped: 64

04 Oct 2005 8:35 bessel function recurrence relation

Use generating function
$3$g(x,t) = e^{\frac{x}{2}( t-\frac{1}{t})}$
From
$3$ e^{\frac{x}{2}( t-\frac{1}{t})} = \sum_{=-\infty}^{\infty} J_n (x) t^n$

$3$ \frac{\partial}{\partial t} g(x,t) = \frac{1}{2} x (1+ \frac{1}{t^2})e^{\frac{x}{2}( t-\frac{1}{t})}$
$3$=\sum_{n= - \infty}^{\infty} n J_n(x) t^{n-1}$

Plug back generating function:
$3$\frac{x}{2} ( \sum_{n= - \infty}^{\infty} J_n(x) t^n + \sum_{n= - \infty}^{\infty} J_n(x) t^{n-2} ) = \sum_{n= - \infty}^{\infty} n J_n(x) t^{n-1}$

compare the coef. of the polynomial with the same power