0
$\begingroup$

I want to show that series $$f(x)=\sum_{k=1}^{\infty}2^k\sin (3^{-k}x)$$ is continuously differentiable.

We have that each $f_k=2^k\sin (3^{-k}x)$ is differentiable and its derivative is equal to $f_k'(x)=2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)$.

It holds that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}$$

To show that the series is continuously differentiable do we have to show that it converges uniformly?

$\endgroup$
6
  • 1
    $\begingroup$ Observe that $2 <3$ $\endgroup$ Apr 7 '18 at 19:37
  • $\begingroup$ Related math.stackexchange.com/questions/2714034/… $\endgroup$ Apr 7 '18 at 21:42
  • $\begingroup$ So we have that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$$ right? Do we use the geometric series? @Salahamam_Fatima $\endgroup$
    – Mary Star
    Apr 7 '18 at 22:27
  • $\begingroup$ @MaryStar Yes {}{}{[{[{[{ $\endgroup$ Apr 7 '18 at 22:30
  • $\begingroup$ We have that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$$ and the series $\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k$ converges, since it is a geometric series. By the Comparison Test we get that the series $\sum_{k=1}^{\infty}\left |\frac{2^k}{3^k}\cos(x/3^k)\right |$ converges. So the series $\sum_{k=1}^{\infty}\frac{2^k}{3^k}\cos(x/3^k)$ is absolutely convergent. Does it hold that every absolutely convergent is also continuously differentiable? Or how do we continue from here? @Salahamam_Fatima $\endgroup$
    – Mary Star
    Apr 7 '18 at 22:36
2
$\begingroup$

$$\sum_{k\geq 1}\frac{2^k}{3^k}\cos(x/3^k) $$ is an absolutely convergent series of continuous functions, hence a continuous function which can be termwise-integrated, leading to a continuously differentiable function, $f(x)$.

$\endgroup$
3
  • $\begingroup$ We have that $$|2^k\cdot 3^{-k}\cdot \cos (3^{-k}x)|\leq 2^k\cdot 3^{-k}=\left (\frac{2}{3}\right )^k$$ and the series $\sum_{k=1}^{\infty}\left (\frac{2}{3}\right )^k$ converges, since it is a geometric series. By the Comparison Test we get that the series $\sum_{k\geq 1}\left |\frac{2^k}{3^k}\cos(x/3^k)\right |$ converges. So the series $\sum_{k\geq 1}\frac{2^k}{3^k}\cos(x/3^k)$ is absolutely convergent. How could we continue from here? Could you explain it further to me? $\endgroup$
    – Mary Star
    Apr 7 '18 at 22:34
  • $\begingroup$ We have that $f(x)$ and $f'(x)$ converge absolutely, right? Does this imply that the series is continuously differentiable? $\endgroup$
    – Mary Star
    Apr 8 '18 at 2:58
  • $\begingroup$ @MaryStar: by the $3\varepsilon$-theorem, an absolutely convergent series of continuous functions is a continuous function. $\endgroup$ Apr 8 '18 at 3:28

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.