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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.

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1 Answer 1

up vote 3 down vote accepted

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

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

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Just a quick clarification - how did you get the contribution from w to not contain z? Thanks in advance! –  dhz Nov 26 '12 at 10:42
    
@dhz: I'm afraid you'll have to first explain why you would expect it to contain $z$ :-) –  joriki Nov 26 '12 at 10:44
    
Ohh I see you changed the sum for w a little bit to get the $z^{n+2j−2}$ contribution. Got it. Thanks so much!! –  dhz Nov 26 '12 at 11:36
    
@dhz: You're welcome! –  joriki Nov 26 '12 at 11:37

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