0
$\begingroup$

On a recent test I had the question was asked:

Is the series $\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^3}$ termwise differentiable on an interval $I\subseteq \mathbb{R}$? Explain why it is or why it isn't and specify the cases if there are any.

What i knew about termwise differentiability is from the power series theorem that says:

Let $\sum_{n=1}^{\infty} a_nx^n$ be a power series and $R$ it's radius of convergence. Then for every segment $[a,b]\subseteq (-R,R)$ the series can be differentiated and integrated term by term and the resulting series will have the same region of convergence.

The consequence being that this theorem can be applied arbitrary number of times on a single series.

This series is a functional series and not a power series. All the theorems I knew applied to the power series. All i could figure out was that:

$(\forall x \in \mathbb{R})$, $\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^3}\leq \sum_{n=1}^{\infty} \frac{1}{n^3}$. Now I know that the series converges for all values of $x$. Since it converges for all value of $x$ it's region of convergence is $(-\infty, +\infty)$. If i take $d/dx$ of the series and notice that $(\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^3})'\leq \sum_{n=1}^{\infty} \frac{1}{n^2}$ it remains that it's region of convergence is the same as the original series. If the main term can be differentiated it follows that it's terwise differentiable as well. This however does not hold for the second differentiation of the series, so it's not like in the original theorem.

That is all i could think of for this question. I'm not sure if it's correct so i asked before i got the results to check what the correct answer would be if i missed it.

What would the correct answer to this question be?

$\endgroup$
1
$\begingroup$

The standard theorem is this:

If $\sum_{n=1}^{\infty}f_n(x)$ converges at least at one point, and the series of derivatives $\sum_{n=1}^{\infty}f_n^{'}(x)$ converges uniformly in some interval $I$, then in that interval the series can be differentiated term by term, meaning that the sum of the derivatives converges to the derivative of the sum.

In your case it is easy to see that $\sum_{n=1}^{\infty}\frac{n\cos(nx)}{n^3}$ converges uniformly on $\mathbb{R}$, and so the correct answer is that the given series is termwise differentiable in any interval $I\subset\mathbb{R}$.

$\endgroup$
6
  • $\begingroup$ It even converges normally. $\endgroup$ – Bernard Jul 8 '18 at 20:07
  • $\begingroup$ What do you mean by "normally"? $\endgroup$ – uniquesolution Jul 8 '18 at 20:08
  • $\begingroup$ Convergence in norm. It implies uniform convergence. $\endgroup$ – Bernard Jul 8 '18 at 20:12
  • $\begingroup$ What norm are you referring to? $\endgroup$ – uniquesolution Jul 8 '18 at 20:14
  • $\begingroup$ Why, the $\sup$ norm. $\endgroup$ – Bernard Jul 8 '18 at 20:15

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.