Question on the infinite sum of Bessel functions multiplied by $\frac{x^n}{n!}$

Question

How would I show that

$$\sum_{n=0}^{\infty} \frac{x^n}{n!} J_n(a) = J_{0}(\sqrt{a^2-2ax})$$

I have tried using the infinite series representation of $J_0$ but didn't get anywhere.

Thanks.

Answer 1

A possible approach is to use the generating function $$e^{(z/2)(t-1/t)}=\sum_{n\in\mathbb{Z}}J_n(z)t^n\qquad(z,t\in\mathbb{C},\ t\neq 0)$$ or rather its consequence, the integral representation $$J_n(z)=\frac1{2\pi i}\oint t^{-n-1}e^{(z/2)(t-1/t)}\,dt\qquad(n\in\mathbb{Z})$$ (with integration along a simple contour encircling $t=0$). Then $$J_0(2\sqrt{ab})=\frac1{2\pi i}\oint\exp\left(az-\frac bz\right)\frac{dz}z\qquad(a,b\in\mathbb{C})$$ (reduces to the above after $z=t\sqrt{b/a}$), and we have enough ingredients: $$\sum_{n=0}^\infty\frac{x^n}{n!}J_n(a)=\frac1{2\pi i}\sum_{n=0}^\infty\frac{x^n}{n!}\oint z^{-n-1}e^{(a/2)(z-1/z)}\,dz\\=\frac1{2\pi i}\oint\exp\left(\frac{az}2-\frac{a-2x}{2z}\right)\frac{dz}z=J_0(\sqrt{a^2-2ax}).$$