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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.

Mark Delic
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    Equation $59$ here https://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html ... otherwise looks pretty routine ? – Donald Splutterwit Jun 15 '20 at 19:58
  • Should I use quotient rule on the first function? – Mark Delic Jun 15 '20 at 20:08
  • That's right. Differentiate using the quotient rule ... then use $59$ to deal with the derivatives of the Bessel functions. ... I have not actually done it, but it looks quite easy ? – Donald Splutterwit Jun 15 '20 at 20:23
  • After applying quotient rule I get two derivative terms $\frac{d}{dz}[J_{n+1}(z)]$ and $\frac{d}{dz}[zJ_{n}(z)]$ and I don't know what to do next. I've used 59 before, but when n is defined. – Mark Delic Jun 15 '20 at 20:47
  • The other formula you will need is \begin{eqnarray} z(J_{n+1}(z)+J_{n-1}(z))=2nJ_n(z). \end{eqnarray} – Donald Splutterwit Jun 15 '20 at 21:01

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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.

Jack D'Aurizio
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