I'm trying to prove a property of the Bessel function, specifically that $$J'_p = \frac{1}{2}(J_{p-1} - J_{p+1})$$ I've worked on this problem for quite awhile with little to show. I've compared terms and they never seem to agree. Any help would be appreciated.
Note: We only consider $p\in \mathbb{Z}^+$, so the Bessel function can be written as $$J_p(x) = \sum_{n=0}^\infty \frac{(-1)^n}{n!(n+p)!} \Big(\frac{x}{2}\Big)^{2n+p}$$
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Too many formulas for a comment. If $$ J_p(x) = \sum_{n=0}^\infty \frac{(-1)^n}{n!(n+p)!} \Big(\frac{x}{2}\Big)^{2n+p} $$ then $$ J_{p-1}(x) = \sum_{n=0}^\infty\frac{(-1)^n}{n!(n+p-1)!}\Big(\frac{x}{2}\Big)^{2n+p-1}, $$ $$ J_{p+1}(x) = \sum_{n=0}^\infty\frac{(-1)^n}{n!(n+p+1)!}\Big(\frac{x}{2}\Big)^{2n+p+1} = \sum_{n=1}^\infty\frac{(-1)^{n-1}}{(n-1)!(n+p)!}\Big(\frac{x}{2}\Big)^{2n+p-1}, $$ $$ J_p'(x) = \sum_{n=1}^\infty\frac{(-1)^n(2n+p)}{n!(n+p)!2}\Big(\frac{x}{2}\Big)^{2n+p-1}. $$ And now, compare coefficients
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In addition to the series-based proofs already given (and since we only need to consider integer values of $p$), this follows almost automatically from the generating function of the Bessel functions. We have $$ e^{z (t - 1/t)/2} = \sum_{t = -\infty}^{\infty} t^p J_p(z) $$ Differentiating both sides with respect to $z$, we get \begin{align*} \frac{1}{2} (t - t^{-1}) \sum_{t = -\infty}^{\infty} t^p J_p(z) &= \sum_{t = -\infty}^{\infty} t^p J'_p(z) \\ \sum_{t = -\infty}^{\infty} \frac{1}{2} t^{p+1} J_p(z) - \sum_{t = -\infty}^{\infty} \frac{1}{2} t^{p-1} J_p(z) &= \sum_{t = -\infty}^{\infty} t^p J'_p(z) \\ \sum_{t = -\infty}^{\infty} \frac{1}{2} t^{p} J_{p-1}(z) - \sum_{t = -\infty}^{\infty} \frac{1}{2} t^{p} J_{p+1}(z) &= \sum_{t = -\infty}^{\infty} t^p J'_p(z) \end{align*} The identity then follows by comparing coefficients of $t^p$ on the left- and right-hand sides.
\begin{eqnarray*} J_p(x) =\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+p)!} \left( \frac{x}{2} \right)^{2n+p} . \end{eqnarray*} \begin{eqnarray*} \frac{ \partial}{\partial x} J_p(x) &=& \frac{1}{2} \sum_{n=0}^{\infty} \frac{\overbrace{(2n+p)}^{n+p+n }(-1)^n}{n!(n+p)!} \left( \frac{x}{2} \right)^{2n+p-1} \\ &=& \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+p-1)!} \left( \frac{x}{2} \right)^{2n+p-1} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{(-1)^n}{(n-1)!(n+p)!} \left( \frac{x}{2} \right)^{2n+p-1} \\ &=& \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+p-1)!} \left( \frac{x}{2} \right)^{2n+p-1} + \frac{-1}{2} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+p+1)!} \left( \frac{x}{2} \right)^{2n+p+1} \\ &=& \frac{1}{2}(J_{p-1}(x)-J_{p+1}(x)). \end{eqnarray*}