I was reading a book and found a question "analyze the applicability of term by term differentiation for $\sum_{n \geq 1} \arctan \frac {x} {n^2}$", What does that means? How to solve this?
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I don't understand the question – Ethan Splaver Apr 14 '13 at 04:31
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@Ethan: I think what is asked is to justify $\frac{d}{dx}(\sum\tan^{-1}\frac{x}{n^2})=\sum\frac{d}{dx}(\tan^{-1}\frac{x}{n^2})$ – Apr 14 '13 at 04:52
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@Shahab I'm not sure that's relevant. I believe you are saying that you can differentiate a power series by differentiating its terms one at a time. But OP's function is not a power series, and not clearly analytic. (Or is it?) And if OP's function is analytic, these terms are not the terms of its power series. – 2'5 9'2 Apr 14 '13 at 05:03
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You are being asked to investigate whether or not $$\frac{d}{dx}\sum_{n\ge1}\arctan\left(\frac{x}{n^2}\right)=\sum_{n\ge1}\frac{d}{dx}\arctan\left(\frac{x}{n^2}\right)$$ I believe that answers your first question. Your second question is how to accomplish this. But now that the first is more clear, you might have some ideas.
2'5 9'2
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Just note that, the series
$$ \sum_{n=1}^{\infty}\frac{n^2}{n^4+x^2}, $$
which is the derivative of our original series, converges uniformly $\forall x\in \mathbb{R}$, since
$$ \sum_{n=1}^{\infty}\frac{n^2}{n^4+x^2} \leq \sum_{n=1}^{\infty}\frac{1}{n^2} <\infty .$$
That implies we can differentiate the series term by term. There are other verifications you should take in consideration. See theorem 2, page 1 .
Mhenni Benghorbal
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