<s>[INST] Bessel functions of the first kind How would I show that
$$J_1(x)+J_3(x)=\frac 4x J_2(x)$$
Using  the series definition of the Bessel Function, which is
$$J_p(x)=\sum ^\infty _{n=0} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac x2\right)^{2n+p}$$
I know that it can be achieved using some reindexing, but I am not sure how. I would greatly appreciate any help. Thank you.
 [/INST] Yes, there is an index shift.  Rewrite
$$J_1(x) = \frac{x}{2} + \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n+1)!} \left ( \frac{x}{2} \right ) ^{2 n+1}$$
$$J_3(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n-1)!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+1}$$
Then
$$\begin{align}J_1(x)+J_3(x) &= \frac{x}{2} + \sum_{n=1}^{\infty}  (-1)^n \left [ \frac{1}{n! (n+1)!} -\frac{1}{(n-1)! (n+2)!}  \right ]\left ( \frac{x}{2} \right ) ^{2 n+1}\\ &= \frac{x}{2} + 2 \sum_{n=1}^{\infty}  \frac{(-1)^n}{n!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+1} \\ &= \frac{4}{x} \left[\frac{1}{2} \left ( \frac{x}{2} \right )^2 + \sum_{n=1}^{\infty}  \frac{(-1)^n}{n!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+2} \right ]\\ &= \frac{4}{x}\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+2} \\\therefore J_1(x)+J_3(x) &= \frac{4}{x} J_2(x)  \end{align}$$
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