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The question is to prove that $$\frac{d}{dx}[xJ_n(x)J_{n+1}(x)]=x[J_n^{2}(x)-J_{n+1}^{2}(x)]$$

I have attempted using the product rule and the definition of Bessel but I just end up with a mess and no end in sight. Any help would be greatly appreciated.

projectilemotion
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Aksel'sRose
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1 Answers1

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I think you didn't try very hard using the chain rule.

Our answer is supposed to be in terms of $J_n(x)$ and $J_{n+1}(x)$, so write everything in terms of them (use the recurrence to write $J_{n+2}$ in terms of $J_{n+1}$ and $J_n$).
$$ J_n'(x) = \frac{n}{x} J_n(x) - J_{n+1}(x) \tag{1} $$

$$ J_{n+1}'(x) = J_n(x) - \frac{n+1}{x} J_{n+1}(x) \tag{2} $$

Chain rule: $$ \frac{d}{dx}\big[xJ_n(x)J_{n+1}(x)\big] = J_n(x)J_{n+1}(x) +xJ_n'(x)J_{n+1}(x) +xJ_n(x)J_{n+1}'(x) $$ Plug in $(1)$ and $(2)$. Expand.

GEdgar
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  • I have all of what you have written, which makes me happy, but I must have messed up somewhere or just given up too soon. I shall continue in this strain. Thank you. – Aksel'sRose Aug 04 '16 at 01:19
  • I am stuck with 3 extra pieces after expanding and attempting to simplify. Any advice on how to complete this? $$ J_{n+1}(x)[J_n(x)+nJ_n(x)-(n+1)]-xJ_{n+1}^{2}(x)+xJ_n^{2}$$ – Aksel'sRose Aug 04 '16 at 20:33
  • You made a mistake. Every term must have two J's in it. Your $(n+1)$ is therefore obviously wrong. Perhaps it should have been $(n+1)J_n(x)$? – GEdgar Aug 04 '16 at 23:36
  • Ah, yep, thanks! – Aksel'sRose Aug 05 '16 at 01:33