Prove that: $j_0(\sqrt{x^2-2xt})=\sum_{n=0}^{\infty}\dfrac{t^n}{n!}j_n(x)$ (Bessel Functions)

Question

Prove that:

$j_0(\sqrt{x^2-2xt})=\sum_{n=0}^{\infty}\dfrac{t^n}{n!}j_n(x)$

the result can be obtained by applying the Taylor expansion for a conveniently chosen function, with appropriate changes of variable, and the help from

$j_n(x)=(-1)^nx^n\left(\dfrac{1}{x}\dfrac{d}{dx} \right)^n\left( \dfrac{\sin(x)}{x} \right)$

I think it can be obtained from the generating function for spherical Bessel functions, but I can't find a way to relate it.

Answer 1

The (exponential) generating function of the spherical Bessel functions of the first kind is $$ \frac{\cos\left(\sqrt{x^2-2xt}\right)}{x} = \sum \limits_{m=0}^\infty \frac{t^m}{m!} \operatorname{j}_{m-1}(x) \, . $$ Taking the derivative with respect to $t$ we obtain \begin{align} \sum \limits_{n=0}^\infty \frac{t^n}{n!} \operatorname{j}_n(x) &= \sum \limits_{m=1}^\infty \frac{t^{m-1}}{(m-1)!} \operatorname{j}_{m-1}(x) = \frac{\partial}{\partial t} \sum \limits_{m=0}^\infty \frac{t^m}{m!} \operatorname{j}_{m-1}(x) = \frac{\partial}{\partial t} \frac{\cos\left(\sqrt{x^2-2xt}\right)}{x} \\ &= \frac{\sin\left(\sqrt{x^2-2xt}\right)}{\sqrt{x^2-2xt}} = \operatorname{j}_0 \left(\sqrt{x^2-2xt}\right) \, . \end{align}