Given a series:

$$f(x) = \sum_{n=1}^\infty (a_n \sin(nx)+b_n \cos(nx))$$

Provide a sufficient condition on the sequences {$a_n$} and {$b_n$} to be able to differentiate $f(x)$ term by term with

$$f'(x) = \sum_{n=1}^\infty (na_n \cos(nx)-nb_n \sin(nx))$$

I know we can interchange sums with differentiation if

(i)$\sum_{n=1}^\infty f_n(x_0)$ converges for some $x\in[a,b]$

(ii)$\sum_{n=1}^\infty f_n'$ converges uniformly to a differentiable function.

I can't seem to figure out how I go about it. My intuition is that $na_n$ and $nb_n$ should converge.

I also tried looking at it from Fourier series perspective but it didn't kead me to any conclusions. I would appreciate any help!

  • $\begingroup$ If $g(x) = \sum_n c_n in e^{inx}$ converges uniformly on $x\in [A,B]$ then it is easy and for $x \in [A,B]$ : $\int_A^x g(t)dt = cst+ \sum_n c_n e^{inx}$. Otherwise if you can't find an interval where the Fourier series of the derivative converges uniformly, it might be complicated. $\endgroup$ – reuns Nov 2 '16 at 16:21
  • $\begingroup$ @user1952009 If $[A,B]$ contains no multiple of $\pi/2$, then all we need is, I believe, for $na_n$ and $nb_n$ to monotonically to $0$. $\endgroup$ – Mark Viola Nov 2 '16 at 17:25
  • $\begingroup$ @Dr.MV I never said it is not true, I said that in general you don't have $a_n \ge 0$, so you can't apply the Dirichlet test to $n a_n$. And if the Dirichlet test for $\sum_n c_n in e^{inx},x \ \in [A,B]$ works, then you also have the uniform convergence on $[A,B]$ $\endgroup$ – reuns Nov 2 '16 at 17:42
  • $\begingroup$ @user1952009 Perhaps we are saying the same things in slightly different ways. Yes, the case for which $a_n$ is not non-negative doesn't conform to the criteria of the Dirichlet Test (DT). However, if the signs alternate, then we simply "bunch" the $(-1)^n$ term with the sine (or cosine) function and continue with the DT. And we can use a similar approach if the sign of $a_n$ vary in other ways. I am unsure as to a general approach here, and that therein, is perhaps what the OP is asking. Let me know your thoughts. As always, I enjoy our discussions. -Mark $\endgroup$ – Mark Viola Nov 2 '16 at 21:14
  • $\begingroup$ Thank you both for your comments! I think I have to keep the $a_n$ and $b_n$ and not switch to the complex exponential form, but I do get the concerns about the difficulties. One of my main concerns was that some friends tried to use the Weierstrass M-test that they used checking for $f(x)$ and noting that $\sum_n(|a_n|+|b_n|) $should be finite. However I was confused as it seems they leave out the (ii) condition for term by term differentiation and apply rules for power series. $\endgroup$ – Kat.m Nov 3 '16 at 1:33

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