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Let $f(x) = \sum_{n=1}^{\infty} \frac{x^{3/2}\cos(nx)}{n^{5/2}}$ on $(0, \infty)$.

(i) Is $f(x)$ differentiable on $(0, \infty)$?

(ii) Does the series uniformly converge to $f$ on $(0, \infty)$?

Any help or hints will be appreciated.

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Edit: What I know about the differentiability of a sequence of functions is that

If $(f_n(x))$ is defined on $I = [a,b]$ and

(i) Each $f_n$ is differentiable on $I$.

(ii) For some $x_0 \in I$, $(f_n(x_0))$ converges.

(iii) $(f'_n)$ converges uniformly.

Then $(f_n)$ converges uniformly, $f$ is differentiable on $I$ and

$ f'(x) = \lim_{n \rightarrow \infty} f'_n(x)$ holds.

The above case seems not to satisfy the conditions of the above theorem.

At this point, I don't know how to proceed.

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2 Answers 2

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Hint for uniform convergence: If the series converged uniformly on $(0,\infty),$ then

$$\sup_{0<x<\infty} \left |\frac{x^{3/2}\cos (nx)}{n^{5/2}}\right | \to 0$$

as $n\to \infty.$ Is this true?

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  • $\begingroup$ I see. if some series $\sum_{n=1}^{\infty} f_n (x)$ converges uniformly on $I$, then $sup_{x \in I} f_n(x) \rightarrow 0$, right? $\endgroup$
    – glimpser
    Jul 31, 2020 at 4:27
  • $\begingroup$ Yes. If the series converges uniformly, then the general terms tend to $0$ uniformly. $\endgroup$
    – Mark Viola
    Jul 31, 2020 at 4:29
  • $\begingroup$ @MarkViola Thanks. $\endgroup$
    – glimpser
    Jul 31, 2020 at 4:33
  • $\begingroup$ It was my pleasure. $\endgroup$
    – Mark Viola
    Jul 31, 2020 at 4:34
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HINT:

For any positive number $L$, we see that the series obtained by differentiation of the general terms, namely $\displaystyle \sum_{n=1}^\infty \left(\frac32 \frac{x^{1/2}\cos(nx)}{n^{5/2}}-\frac{x^{3/2}\sin(nx)}{n^{3/2}} \right)$, converges uniformly for $x\in (0,L]$.

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  • $\begingroup$ Ah! I have kept calculating the term with $sin(nx)$ wrongly, that's why I thought that the above series seems not to satisfy the hypotheses of the above theorem...Thanks. $\endgroup$
    – glimpser
    Jul 31, 2020 at 4:30
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Aug 21, 2020 at 16:05

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