2
$\begingroup$

For the following series representation of the Bessel function:

$w = J_n = \sum_{k=0}^{\infty} \frac{(-1)^k z^{n+2k}}{k!(n+k)!2^{n+2k}}.$

I want to show that w is indeed the Bessel function, such that $w'' + \frac{1}{z}w' + (1-\frac{n^2}{z^2})w = 0$, with the series definition.

I got the following first and second derivatives:

$w' = \sum_{k=1}^{\infty} \frac{(-1)^k (n+2k) z^{n+2k-1}}{k!(n+k)!2^{n+2k}},$

and

$w'' = \sum_{k=1}^{\infty} \frac{(-1)^k (n+2k) (n+2k-1) z^{n+2k-2}}{k!(n+k)!2^{n+2k}}.$

I tried summing the equations by brute force, which got me as far as

$w'' + \frac{1}{z}w' + (1-\frac{n^2}{z^2})w = \sum_{k=1}^{\infty} \frac{(-1)^k (4nk+4k^2) z^{n+2k-2}+(-1)^k z^{n+2k}}{k!(n+k)!2^{n+2k}} + (1-\frac{n^2}{z^2})(\frac{z^n}{n!2^n}),$ but that doesn't seem to equal zero.

Any ideas much appreciated.

1 Answers 1

3

Collecting all terms containing $z^{n+2j-2}$, we have a contribution

$ \frac{(-1)^j(n+2j)(n+2j-1)}{j!(n+j)!2^{n+2j}} $

from $w''$, a contribution

$ \frac{(-1)^j(n+2j)}{j!(n+j)!2^{n+2j}} $

from $w'$, a contribution

$ \frac{(-1)^{j-1}}{(j-1)!(n+j-1)!2^{n+2(j-1)}} $

from $w$ and a contribution

$ -n^2\frac{(-1)^j}{j!(n+j)!2^{n+2j}} $

from $-wn^2/z^2$. Adding these up and multiplying through by $(-1)^jj!(n+j)!2^{n+2j}$ yields

$ (n+2j)(n+2j-1)+(n+2j)-4j(n+j)-n^2=0. $

  • 0
    @dhz: You're welcome!2012-11-26