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Show that $\sum_{n=1}^{\infty}\frac{sin(x^{n})}{n!}$ converges uniformly for $x \in \mathbb R$ to a $C^{1}$ function $f:\mathbb R \rightarrow \mathbb R$, compute an expression for the derivative.

My attempt: For uniform convergence, It is clear that $|\sum_{n=1}^{\infty}\frac{sin(x^{n})}{n!}|<\sum_{n=1}^{\infty}\frac{1}{n!}$.

By comparison test, we know that $\sum_{n=1}^{\infty}\frac{1}{n!}$ is convergent. So by WM test $\sum_{n=1}^{\infty}\frac{sin(x^{n})}{n!}$ converges uniformly for $x \in \mathbb R$.

Can anyone suggest me about the second part? Is the question about term by term differentiation?

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    $\begingroup$ Can you do term-by-term differentiation? $\endgroup$
    – GEdgar
    Dec 13, 2019 at 1:14
  • $\begingroup$ @User124356 Recall that $ e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} \ldots $ so then your infinite sum $ \sum_{n=1}^{\infty}\frac{1}{n!} = e-1$ . I don't know if this will help I just saw that and thought I would add this. $\endgroup$
    – Dclrk
    Dec 13, 2019 at 1:23
  • $\begingroup$ Your proof of uniform convergence looks weird. Instead of bounding the sum, you should bound each term. If WM stands for Weirsrass M test, then you indeed need to bound each term. $\endgroup$
    – Michael
    Dec 13, 2019 at 1:36
  • $\begingroup$ @Michael do you mean by this $\sum_{n=1}^{\infty}\frac{sin(x^{n})}{n!}=\frac{sin(x)}{1!}+\frac{sin(x^{2})}{2!}+\frac{sin(x^{3})}{3!}+.........+\frac{sin(x^{n})}{n!}$. Then I take mod to bound the terms? $\endgroup$
    – User124356
    Dec 13, 2019 at 1:44
  • $\begingroup$ Can you set $y=sin(x^n)$ so that it converges to $e^y-1$ and use the chain rule? $\endgroup$ Dec 13, 2019 at 2:27

1 Answer 1

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Since $\sum_{n=1}^\infty \frac{\sin(x^n)}{n!}$ converges uniformly for $x\in\mathbb R$, we may differentiate term-by-term:

\begin{align} \frac{\mathsf d}{\mathsf dx} \sum_{n=1}^\infty \frac{\sin(x^n)}{n!} &= \sum_{n=1}^\infty \frac{\mathsf d}{\mathsf dx}\frac{\sin(x^n)}{n!} \\ &= \sum_{n=1}^\infty \frac{n x^{n-1}\cos(x^n)}{n!}\\ &= \sum_{n=1}^\infty \frac{x^{n-1}\cos(x^n)}{(n-1)!}\\ &= \sum_{n=0}^\infty \frac{x^n\cos(x^{n-1})}{n!}\\ \end{align} I do not know how to further simplify this, and will defer that to someone with more expertise.

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  • $\begingroup$ I think the assertion is incomplete, for example: Define $f_n(x) =\frac{\sin\left(x^{(n^4)}\right)}{n^2}$ for $n \in \{1, 2, 3, …\}$. Then $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly for $x \in \mathbb{R}$, but we cannot differentiate term-by-term. $\endgroup$
    – Michael
    Dec 14, 2019 at 19:11
  • $\begingroup$ It converges absolutely as well. What else is needed for term-by-term differentiation? $\endgroup$
    – Math1000
    Dec 14, 2019 at 20:49

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