Let $$f(x) = \sum_{n=1}^{\infty} \frac{x^n}{2^n} \cos{nx}$$ Proof that $f$ is differentiable on $(-2,2)$

my approach

let $ m := \frac{x}{2} $ so $m<1$ $$ \left| \frac{x^n}{2^n} \cos{nx} \right| \le m^n \rightarrow 0 $$ So by comparison we have point convergence. But I don't know how to deal with diffentiable..

  • $\begingroup$ It's not enough that $m^n \to 0$, you need convergence of $\sum\limits_{n=1}^\infty m^n$ to use comparison. $\endgroup$ – mihaild May 21 at 15:19

Series of derivatives, $\sum\limits_{n=1}^\infty \frac{n x^{n - 1} \cos nx - n x^n \sin nx}{2^n}$ converges uniformly on any subsegment of $(-2, 2)$ (per comparison again) and series itself converges. Then the series converges to differentiable function (and it's derivative is sum of series of derivatives) - see, for example, theorem 7.17 of Rudin, "Principles of Mathematical Analysis".

To prove the series is uniformly convergent on $(-2 + \alpha, 2 - \alpha)$, we can note that $|n x^{n - 1} \cos nx - nx^n \sin nx| \leqslant 2 \cdot n \cdot (2 - \alpha)^n$, and it is less then $(2 - \frac{\alpha}{2})^n$ for large enough $n$. And the series $\sum_{n=1}^\infty \frac{(2 - \frac{\alpha}{2})^n}{2^n} = \sum\limits_{n=1}^\infty \left(1 - \frac{\alpha}{4}\right)^n$ converges.

  • $\begingroup$ how you got $\sum\limits_{n=1}^\infty -\frac{n^2 x^{n - 1}}{2^n}\sin nx$? $\endgroup$ – trolley May 21 at 15:34
  • $\begingroup$ By differentiating each term. $\endgroup$ – mihaild May 21 at 15:35
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    $\begingroup$ By differentiating each term we get $2^{-n} n x^{n-1} \cos (n x)-2^{-n} n x^n \sin (n x) \neq \frac{n^2 x^{n - 1}}{2^n}\sin nx $ $\endgroup$ – trolley May 21 at 15:37
  • $\begingroup$ It was a stupid mistake from me indeed. However correct series of derivatives converges uniformly too. $\endgroup$ – mihaild May 21 at 15:43
  • $\begingroup$ how can I prove uniform convergence of that $\sum \frac{n x^{n - 1} \cos nx - n x^n \sin nx}{2^n}$ ? $ \frac{n x^{n - 1} \cos nx - n x^n \sin nx}{2^n} = \frac{nx^{n-1}}{2^n}(\cos{nx} - x \sin {nx}) < \frac{3n}{2^n}\cdot 2^{n-1} = \frac{3}{2}n$ $\endgroup$ – trolley May 21 at 15:57

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