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This derivative just showed up in a past paper as part of a question, i don't know what to do with it because of the summation etc?? Please help

$$\frac{\partial}{\partial h} \sum_{n=-\infty}^{\infty} h^n J_n(x)$$

J is just any function of x i think

naslundx
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Amy
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  • Is the sum from -infinity to infinity or 0 to infinity? – Cbjork Apr 03 '14 at 19:36
  • Hint: $\frac{\partial}{\partial h}J_n(x)=0$. – Ian Coley Apr 03 '14 at 19:36
  • If it all works well, this should be $\sum_{\substack{n=-\infty}\{n\ne0}}^\infty nh^{n-1}J_n(x)$, but whether this is indeed the case, depends on what the $J_n$ are, what interval this is computed over, etc. – Andrés E. Caicedo Apr 03 '14 at 19:37
  • I would expect $J_n$ to be a Bessel function. They have many sum relations. Your partial becomes $nh^{n-1}$ – Ross Millikan Apr 03 '14 at 19:40
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    "Master, what is the difference between a humanistic, monastic system of belief in which wisdom is sought by means of an apparently nonsensical system of questions and answers, and a lot of mystic gibberish made up on the spur of the moment?" Wen considered this for some time, and at last said, "A fish!" And Clodpole went away, satisfied. – Will Jagy Apr 03 '14 at 19:44

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If $J_n(x)$ does not depend on $h$, then it's simply: \begin{equation} \frac{\partial}{\partial h} \sum_{n=-\infty}^\infty h^n J_n(x)= \sum_{n=-\infty}^\infty \left(\frac{\partial}{\partial h} h^n \right) J_n(x)= \sum_{n=-\infty}^\infty nh^{n-1}J_n(x) \end{equation}

Danijel
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