Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

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


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

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.

share|cite|improve this question
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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.