0
$\begingroup$

I want to show that the function $f(x) = \sum_{n = 1}^{\infty} \frac{\sin(nx)}{2^{n}}$ is infinitely differentiable on $\mathbb{R}$. I have no idea how to do this. I think it's pretty obvious that the function converges since $2^{n}$ increases really quickly. I tried doing something with analyticity which implies infinitely differentiable; however, I got nowhere.

$\endgroup$
4
  • 2
    $\begingroup$ This series is uniformly convergent on $\Bbb R$, and the series of the derivatives converges uniformly on $\Bbb R$. This is enough to conclude that the derivative of $f(x)$ is simply the series of the derivatives. $\endgroup$ – Crostul Dec 14 '18 at 16:38
  • $\begingroup$ how do we know the series is uniformly convergent? $\endgroup$ – user614735 Dec 14 '18 at 17:08
  • $\begingroup$ Do you know Weierstrass' M-test? It's so simple to use here. $\endgroup$ – Crostul Dec 14 '18 at 17:14
  • $\begingroup$ i am allowed to use weierstrass m test, yes $\endgroup$ – user614735 Dec 14 '18 at 17:17
0
$\begingroup$

Write $\sin(nx)={1\over 2i}\bigl(e^{inx}-e^{-inx}\bigr)$, so that your $f$ appears as $$f(x)={1\over 2i}\left(\sum_{n\geq1} p^n-\sum_{n\geq1} q^n\right)\ .$$ Simplifying in the end will give you a simple expression for $f$, namely $$f(x)={2\sin x\over5-4\cos x}\ .$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy