Relation for Bessel functions

Question

I have function $$Q_{n}(z)=\frac{J_{n+1}(z)}{zJ_{n}(z)},$$where $J_{n+1}(z)$ and $J_{n}(z)$ are Bessel functions of the first kind. I need to prove $$\frac{dQ_{n}}{dz}=\frac{1}{z}-\frac{2(n+1)}{z}Q_{n}(z)+zQ_{n}^{2}(z)$$ but I don't know where to start.

Answer 1

The shown identity is more or less straightforward according to your definition of $J_n(z)$.
Assuming the series definition $$ J_n(x) = \sum_{m\geq 0}\frac{x^{2m}(-1)^m}{2^n 4^m n!(n+m)!} $$ we clearly have that $J_n$ is an entire function and $2\,J_n'(z)=J_{n-1}(z)-J_{n+1}(z)$ (which is even more evident from the integral definition $J_n(z)=\frac{1}{\pi}\int_{0}^{\pi}\cos(z\sin\theta - n\theta)\,d\theta$ and the cosine addition formulas), so

$$ Q_n'(z) = -\frac{J_{n+1}(z)}{z^2 J_n(z)}+\frac{J_{n+1}'(z)}{z J_n(z)}-\frac{J_{n+1}(z)J_n'(z)}{z J_n(z)^2} $$ equals $$ -\frac{J_{n+1}(z)}{z^2 J_n(z)}+\frac{J_n(z)-J_{n+2}(z)}{2z J_n(z)}-\frac{J_{n+1}(z)(J_{n-1}(z)-J_{n+1}(z))}{2z J_n(z)^2} $$ while $$ \frac{1}{z}-\frac{2(n+1)}{z}Q_n(z)+zQ_n^2(z)=\frac{1}{z}-\frac{2(n+1)J_{n+1}(z)}{z^2 J_n(z)}+\frac{J_{n+1}(z)^2}{z J_n(z)^2} $$ so the given identity is equivalent to $$ -2J_{n+1}(z)J_n(z)+zJ_n(z)(J_n(z)-J_{n+2}(z))-z J_{n+1}(z)(J_{n-1}(z)-J_{n+1}(z))\\= 2z J_n(z)^2-4(n+1)J_n(z) J_{n+1}(z)+2z J_{n+1}(z)^2$$ which is a consequence of $\frac{2n}{x}J_n(x) = J_{n-1}(x)+J_{n+1}(x)$.

This is usually done "in the opposite way", i.e. by proving that the solutions of certain differential equations of the Riccati type are given by ratios of adjacent Bessel functions.