How to derive $J_v(x)$

Question

I've seen many sources say that $\frac{\text{d}}{\text{d}x}J_v(x) = J_{v-1}(x) - \frac{v}{x}J_v(x)$, but every time I try to derive it, I get the conjugate, $\frac{v}{x}J_v(x) - J_{v-1}(x)$. Could someone walk me through the steps to arrive at this derivative, so I may see where I went wrong?

$$ J_v(x) = \sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v}}\\ \begin{align} \frac{\text{d}}{\text{d}x}J_v(x) & = \sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\frac{\text{d}}{\text{d}x}\left(\frac{x}{2}\right)^{2n+v}} \\ & = \sum_{n=0}^\infty{\frac{(-1)^n(2n+v)}{2\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v-1}} \\ & = \sum_{n=0}^\infty{\frac{(-1)^nn}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v-1}} + \frac{v}{2}\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v-1}} \\ & = \left(\frac{x}{2}\right)^{-1}\left[\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)(n-1)!}\left(\frac{x}{2}\right)^{2n+v}} + \frac{v}{2}\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v}}\right] \\ & = \left(\frac{2}{x}\right)\left[\sum_{n=1}^\infty{\frac{(-1)^{n+1}}{\Pi(v+n-1)n!}\left(\frac{x}{2}\right)^{2(n-1)+v}} + \frac{v}{2}\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v}}\right] \\ & = -\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi((v-1)+n)n!}\left(\frac{x}{2}\right)^{2n+(v-1)}} + \frac{v}{x}\sum_{n=0}^\infty{\frac{(-1)^n}{\Pi(v+n)n!}\left(\frac{x}{2}\right)^{2n+v}} \\ & = \frac{v}{x}J_v(x) - J_{v-1}(x) \end{align} $$

Answer 1

We have

$$\frac{dJ_{\nu}}{dx} = \sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu+1)} (2m +\nu)\frac{x^{2m+\nu-1}}{2^{2m+\nu}} = \sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu)} \frac{2m +\nu}{2m+2\nu}\left({x\over 2}\right)^{2m+\nu-1}$$

where we have used $\Gamma(m+\nu+1)=(m+\nu)\Gamma(m+\nu)$. Now use $$\frac{2m+\nu}{2m+2\nu} = 1 -\frac{\nu}{2(m+\nu)}$$

The sum splits into two sums. The first sum is

$$\sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu)}\cdot 1 \cdot\left({x\over 2}\right)^{2m+\nu-1} \equiv J_{\nu -1}$$

and the second sum is

$$~~~~~~\sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu)}\cdot \left(-\frac{\nu}{2(m+\nu)} \right)\cdot\left({x\over 2}\right)^{2m+\nu-1} \\= -\frac{\nu}{x}\sum_{m=0} \frac{(-1)^m}{m!\Gamma(m+\nu+1)} \cdot\left({x\over 2}\right)^{2m+\nu} \equiv -\frac{\nu}{x}J_\nu$$

where we have used $\Gamma(m+\nu)(m+\nu) = \Gamma(m+\nu+1)$.

Answer 2

I think it is a bit faster to use the relation: $$ J_\nu(x) = \frac{1}{\pi}\int_{0}^{\pi}\cos(\nu t-x\sin t)\,dt \tag{1}$$ from which: $$\frac{d}{dx} J_\nu(x) = \frac{1}{\pi}\int_{0}^{\pi}\sin t\,\sin(\nu t-x\sin t)\,dt = \frac{1}{2}\left(J_{\nu-1}(x)-J_{\nu+1}(x)\right)$$ follows by the Briggs formulas. Hence we have to prove: $$J_{\nu-1}(x)-\frac{\nu}{x}J_{\nu}(x)=\frac{1}{2}\left(J_{\nu-1}(x)-J_{\nu+1}(x)\right)$$ or: $$J_{\nu-1}(x)-\frac{2\nu}{x}J_{\nu}(x)+J_{\nu+1}(x)=0\tag{2}$$ that follows from: $$J_{\nu+1}(x)+J_{\nu-1}(x)=\frac{2}{\pi}\int_{0}^{\pi}\cos(x\sin t)\cos(\nu t-x\sin t)\,dt =\frac{2\nu}{x}J_{\nu}(x).$$