Deduce $\frac{d J_{0}}{dx}=-J_{1}$ from identities

Question

For Bessel function the following identities hold

$$J_{\nu-1}+J_{\nu+1}=\frac{2 \nu}{x} J_{\nu}$$

$$J_{\nu-1}-J_{\nu+1}=2\frac{d J_{\nu}}{dx}$$

How can I prove the following?

$$\frac{d J_{0}}{dx}=-J_{1}$$ I'm trying to use the previous relations but I can't understand how it is deduced. If I use the second identity a factor $2$ should appear, but it is not there and I do not see why. What am I missing?

Answer 1

We have

$$J_{\nu-1}+J_{\nu+1}=\frac{2 \nu}{x} J_{\nu}\implies J_{-1}+J_{1}=0,$$ and

$$J_{\nu-1}-J_{\nu+1}=2\frac{d J_{\nu}}{dx} \implies J_{-1}-J_{1}=2\frac{d J_{0}}{dx}.$$ Add both equalities and you are done.

Answer 2

We have $$ J_\nu(x)=\frac{1}{\pi}\int_{0}^{\pi}\cos(x\sin\theta-\nu\theta)\,d\theta \tag{1}$$ hence $$J_{\nu-1}(x)+J_{\nu+1}(x)=\frac{1}{\pi}\int_{0}^{\pi}2\cos\theta\cos(x\sin\theta-\nu\theta)=\frac{2\nu}{x}J_\nu(x)\tag{2}$$ simply follows from integration by parts. In a similar fashion $$ \frac{d}{dx}J_0(x)=-\frac{1}{\pi}\int_{0}^{\pi}\sin\theta\sin(x\sin\theta)\,d\theta=-\frac{J_{-1}(x)+J_1(x)}{2}=-J_1(x).\tag{3} $$