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I've been struggling with this problem for a couple hours and finally decided I need some guidance.

Statement of Problem

Use the product of generating functions $G(x,t)G(-x,t)=1$ to derive the identity $J_0(x)^2+2\sum_{n=1}^{\infty}J_n(x)^2=1$.

$J_n$ is a Bessel function of the first kind, and note that the generating function being used means that $n$ must be an integer (I'm pretty sure).

Relevant Info / Attempt

I know the following:

$$G(x,t)=\sum_{n=-\infty}^{\infty} t^n J_n(x)$$

$$J_{n}(x)=\sum_{m=0}^\infty\frac{(-1)^m}{m!\,(m+n)!}\left(\frac{x}{2}\right)^{2m+n}$$

$$J_n(-x) = (-1)^n J_n(x)$$

$$G(x,t) = J_0(x)+\sum_{n=1}^\infty J_n(x)\bigl(t^n+(-1)^nt^{-n}\bigr)$$

$$G(-x,t) = J_0(x)+\sum_{n=1}^\infty J_n(x)(-1)^n\bigl(t^n+(-1)^n t^{-n}\bigr)$$

My most successful attempt was to use the last two of the above equations and multiply the right hand sides. That gave me:

$$1 = J_0(x)^2 + 2\sum_{n=1}^{\infty}J_n(x)^2 + \text{three more terms that I can't get to vanish}$$

That seems so close it feels like I must be doing something right. I have to run right now, if no one has answered by the time I get back I'll update with a little more detail. Thanks!

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  • $\begingroup$ Acouple of hours is not much time! $\endgroup$ – Mhenni Benghorbal Nov 9 '13 at 4:56
  • $\begingroup$ By a couple of hours I mean like 6. $\endgroup$ – Yelneerg Nov 9 '13 at 6:30
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Let's try it directly : \begin{align} \tag{1}G(x,t)G(-x,t)&=\sum_{m=-\infty}^{\infty} t^m J_m(x)\sum_{n=-\infty}^{\infty} t^n J_n(-x)\\ \tag{2}1&=\sum_{m=-\infty}^{\infty} t^m J_m(x)\sum_{n=-\infty}^{\infty} (-t)^n J_n(x)\\ \end{align} Since $\;J_n(-x) = (-1)^n J_n(x)\;$ and $\;\displaystyle e^{\frac x2\left(t-\frac 1t\right)}e^{-\frac x2\left(t-\frac 1t\right)}=1$.

The left of $(2)$ is simply $1$ while the Laurent expansion in $t$ of the product at the right must be unique. The only terms contributing to the constant part will have to verify $n+m=0$ and we will get : \begin{align} 1&=\sum_{n=-\infty}^{\infty} t^{-n}J_{-n}(x)(-t)^n J_n(x)\\ 1&=\sum_{n=-\infty}^{\infty} (-1)^nJ_{-n}(x)J_n(x)\\ \tag{3}1&=\sum_{n=-\infty}^{\infty} J_n(x)^2\\ \end{align} since $J_{-n}(x) = (-1)^n \,J_n(x)\,$.

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  • $\begingroup$ I follow you to that point. Although that isn't quite what I'm supposed to show. Can I say that the sum from -\infty to \infty of J_n(x)^2 is just the same thing as J_0(x)^2 + 2*\sum from 1 to \infty J_n(x)^2? That would have to assume that all the negative J's summed equals all the positive J's summed. That feels kinda handwavy... $\endgroup$ – Yelneerg Nov 9 '13 at 22:36
  • $\begingroup$ @Yelneerg: Yes, you'll obtain the wished relation using $\,J_{-n}(x) = (-1)^n \,J_n(x)\,$ again (for the negative values of $n$ only) getting $\,J_{-n}(x)^2 = J_n(x)^2$. $\endgroup$ – Raymond Manzoni Nov 9 '13 at 23:40
  • $\begingroup$ Btw we have too $\;\displaystyle G(-x,t)=G(x,-t)= e^{-\frac x2\left(t-\frac 1t\right)}\;$ and $\;J_{-n}(x) = J_n(-x)$. $\endgroup$ – Raymond Manzoni Nov 9 '13 at 23:47

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