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The following is an identity concerning the Bessel functions of the first kind $J_n(x)$ for integers $n$ and $m$:

$$\sum_{n=-\infty}^\infty J_n(x) J_{n+m}(x) = \delta(m)$$ where $\delta(x)$ is the Kronecker delta function.

This can be derived from the Jacobi-Anger identity, but is there a simpler way to derive it, for instance using well-known recurrence relations of the Bessel functions?

Community
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Tobin Fricke
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    a side comment: Jacobi-Anger identity is IMHO more natural than recurrence relations. It's a splitting of the wave $\psi(x,y)=\exp(ix)=$ (or of $\exp(ikx)$) to functions with given angular momentum, $\exp(ix)=\sum i^nJ_n(r)\exp(in\theta)$. As $\Delta\psi=-\psi$, also $\Delta(J_n(r)\exp(in\theta))=-J_n(r)\exp(in\theta)$, which is the Bessel equation for $J_n$. – user8268 Mar 24 '11 at 20:15

2 Answers2

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If $m = 2k+1$, this sum always contains the two different terms $J_n(x)J_{n+m}(x)$ and $J_{-(n+m)}(x)J_{-n}(x)$. I believe we can rewrite this sum as $$\sum_{n=-k}^\infty (J_n(x)J_{n+m}(x) + J_{-(n+m)}(x)J_{-n}(x))$$ and then by using the identity $J_{-r}(x) = (-1)^{r}J_r(x)$ we can rewrite this as $$\sum_{n=-k}^\infty (J_n(x)J_{n+m}(x) + (-1)^{2n+m} J_{n+m}(x)J_n(x) = \sum_{n=-k}^\infty (J_n(x)J_{n+m}(x) - J_{n+m}(x)J_n(x)) = 0.$$ So that does it for odd $m$, assuming that reordering the terms is legal in this case.

For even $m = 2k \neq 0$ I haven't a clue. You can try this same trick but it doesn't work, because all the $-1$s cancel out completely rather than leaving one behind to cancel out the whole sum. Plus there's a single awkward $J_{-k}(x)J_{k}(x)$ term in the middle. I'm betting the trick is to use the recurrence $${2r \over x}J_r(x) = J_{r-1}(x) + J_{r+1}(x)$$ to unzip the whole thing starting from the awkward term into two similar sums with $m' = m \pm 1$ (which obviously won't work if $m = 0$, as expected), and then use the above to show both parts are zero. But I have no concrete reason to believe that this will actually work, just a gut instinct.

Paul Z
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  • Yes, it's exactly the m=even case that prompted the question. That recurrence looks tempting, but also a bit messy since it brings out a factor of "x", and the identity should hold for all x. I'll give it a try and see what happens. – Tobin Fricke Mar 16 '11 at 02:48
  • I'm not worried about the x, it falls completely out of the whole problem since we're not changing it. The problem is the 2r, which I have no idea what happens with that as it's summed over. – Paul Z Mar 16 '11 at 02:52
  • Ah, good point. I think all the Jn(x) are non-negative (for n≥0) before the first zero of Jn. Which means the 'awkward term' is the only negative one (for certain values of x) and thus the only possibility for cancellation. – Tobin Fricke Mar 16 '11 at 03:44
  • It looks like this is a special case of the Neumann Addition Formula: http://dlmf.nist.gov/10.23#E4 with $n=1$. Unfortunately, that doesn't lend much insight. Maybe I'll have to settle for "magical property of the Bessel functions". – Tobin Fricke Mar 16 '11 at 04:20
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Here's a hand-wavey argument using the generating function $e^{\frac{x}{2}(t-1/t)} = \sum_{m=-\infty}^\infty t^m J_m(z)$:

$$\begin{eqnarray} e^{\frac{x}{2}(t-1/t)+\frac{x}{2}(u-1/u)} &=& e^{\frac{x}{2}(t-1/t)}e^{\frac{x}{2}(u-1/u)}\\\ &=& \left(\sum_{m=-\infty}^\infty t^m J_m(x)\right)\left(\sum_{n=-\infty}^\infty u^n J_n(x)\right)\\\ &=&\sum_{m=-\infty}^\infty\;\sum_{n=-\infty}^\infty t^{m}u^{n}J_m(x)J_n(x)\\\ &=&\sum_{m=-\infty}^\infty\;\sum_{k=-\infty}^\infty t^{m}u^{m+k}J_m(x)J_{m+k}(x)\\\ &=&\sum_{k=-\infty}^\infty u^k\sum_{m=-\infty}^\infty (tu)^{m}J_m(x)J_{m+k}(x)\;. \end{eqnarray}$$

Now let $t=1/u$ and we get

$$1 = \sum_{k=-\infty}^\infty u^k \left(\sum_{m=-\infty}^\infty J_m(x)J_{m+k}(x)\right)\;,$$

which gives the desired result (since the parenthesized term is independent of $u$).

[Thanks to Joriki for cleaning up the LaTeX.]

Apollo
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