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Convergence of trigonometric(Fourier) series

  1. Oct 23, 2015 #1
    1. The problem statement, all variables and given/known data
    Which of the following series is point-wise convergent, absolutely convergent? Which ones are ##L^2(-\pi,\pi)##-convergent.
    A) ##\sum_1^\infty \frac{\cos n \theta}{n+1}##
    B) ##\sum_1^\infty \frac{(-1)^n\cos n \theta}{n+1}##
    2. Relevant equations
    Abel's test:

    Suppose ##\sum a_n## is a convergent series , ##\{ b_n \}## is a monotone sequence and bounded then ##\sum a_nb_n## is also convergent.
    Dirichlet's test:
    If ##\{a_n\}## is a sequence of real numbers and ##\{ b_n \}## a sequence of complex numbers satisfying
    ##a_n \ge a_{n+1}##
    ##\lim_{n\to \infty} a_n = 0##
    ##\left| \sum_1^N b_n \right| \le M## where ##M## is some constant then
    ##\sum_1^\infty a_nb_n## converges.


    3. The attempt at a solution
    To prove A is convergent by applying the Dirichlet test I need to show that every partial sum of ##\cos n \theta## is bounded. (excluding ##\theta=0## since then the series obviously diverges.) Writing ##\cos n\theta= \frac{e^{in\theta}+ e^{-in\theta}}{2}## we have two geometric series that are bounded, hence ##\sum \cos n \theta## is bounded.
    Since ##\frac{1}{n+1}## is decreasing and and approaches ##0## the series A is convergent. I suspect the series ain't absolutely convergent but I'm not sure how to prove that, If the series was convergent I guess I could use weierstrass M test too prove that but I suspect it's divergent.

    The next series (B) is convergent since ##\sum \frac{(-1)^n}{1+n}## is convergent by the alternating series test and it then follows from Abel's test that the series converge since ##\sum \cos n\theta## is bounded (for ##\theta \ne 0##). Is the above correct? For absolute convergence I got the same problem as for (A) which I don't know how to prove.

    For the part about ##L^2##-convergence I'm not sure what is meant by the question. It isn't supposed to be ##l^2## convergence? Or do I have a integral around that like ##\int \sum |f_n(\theta)| < \infty##?

     
  2. jcsd
  3. Oct 23, 2015 #2

    Samy_A

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    I think they mean (taking the first series as an example):

    set ##f(\theta)=\sum_1^\infty \frac{\cos n \theta}{n+1}## for ##\theta \in (-\pi,\pi)## (assuming this is well defined), is the function ##f \in L^2(-\pi,\pi)##?
     
  4. Oct 23, 2015 #3
    This could be it. But then I'm still not entirely sure. If Fourier series does indeed converge to the function then it also has to converge in norm i.e. in L^2 I think.
    Does this have something to do with that a Fourier series that is absolutely convergent is continuous but may not be be continuous if it's only point wise convergent (is this true?). Don't understand any if this too well I'm afraid.
     
  5. Oct 23, 2015 #4

    Samy_A

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    Is that so?

    Consider the series
    ##\sum_1^\infty \frac{\cos n \theta} {\sqrt {n+1}}##

    The same argument you gave in the first post using Dirichlet's test shows that this series is pointwise convergent (except in 0). But Parseval's theorem implies that there is no ## L^2(-\pi,\pi)## norm convergence.
     
  6. Oct 23, 2015 #5
    Right that's a good point. I know that for a sequence of of functions pointwise convergence doesn't imply convergence in norm nor the other way around. I just never seen the concept used for series before.
    Parsevel's equation was a good idea. Perhaps I could use the same for the functions I have
    ##\frac{1}{\pi}\int_{-\pi}^\pi |f(\theta)|^2d\theta = \sum_1^\infty \frac{1}{(n+1)^2}## which is convergent so it's also ##L^2## convergent. and identical for the alternating series. Is this what you meant?

    Any ideas how I could go about the absolute convergence of the series? An informal argument by me was that since it's not alternating we probably end up with an average of ##\sum \frac{c}{n+1}## ,where ##c## is some positive constant between ##0## and ##1##, which is divergent but that doesn't really prove it.
     
  7. Oct 25, 2015 #6

    Samy_A

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    Yes, that is what I meant.
    I also think the series doesn't converge absolutely.

    However, my proof is elementary but rather tedious. I suspect that there must be a more elegant way to prove it, so maybe someone else could help with that.

    Essentially, for the series to converge absolutely, ##\cos n \theta## must get close to 0 for large ##n##. That means that for large ##n##, ##n \theta## must get close to the values where cosine is 0, but there is no particular reason for that to happen.
    More precisely, you can easily prove absolute divergence when ##\theta## is a rational multiple of ##\pi##. The general case is a little more tedious, but you can prove that if ##|\cos n \theta|## is "very small", then ##|\cos (n+1) \theta|## will be larger than some constant (depending on ##\theta## but not on ##n##), proving the series doesn't converge absolutely.

    But again, this may not be the best way to prove this, so let's see if someone else has a more elegant solution.
     
    Last edited: Oct 25, 2015
  8. Oct 25, 2015 #7
    That's a neat way of doing it. Similar to the my "intuition" that I didn't know how to expand upon.

    Another way of doing it may (or may not) be too use the integral test
    ##\sum_1^\infty \frac{|\cos nx|}{n}## converges if and only if ##I = \int_1^\infty \frac{|\cos nx|}{n}## converges. Suppose ##\theta > 0## which isn't any limitation since ##\cos## is even. Making a change of variables ##t = n\theta## we have
    ##I = \int_\theta^\infty \frac{|\cos t|}{t}dt = \sum_1^\infty \int_{2\pi(n-1)+\theta}^{2\pi n+\theta} \frac{|\cos| t}{t}dt \ge \sum_1^\infty \frac{1}{2\pi (n-1) +\theta)} \int_{2\pi(n-1)+\theta}^{2\pi n+\theta} |\cos t| dt = \sum_1^\infty \frac{2}{\pi (n-1) + \theta} \to \infty ##

    I did this rather quick so there's probably some error in here that makes it all wrong. The approach isn't entirely mine either, I found the approach for the integral in thread over at math.stackexchange http://math.stackexchange.com/quest...converges-absolutely-conditionaly-or-diverges
     
  9. Oct 25, 2015 #8

    Samy_A

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    I'm not sure: doesn't the integral test have a condition that the function must be monotone decreasing?
     
  10. Oct 25, 2015 #9
    Sadly it seems it does require that. I guess I'm back at the argument where I need to remove all those (infinitely many) points that's in the way of it being monotone decreasing.
     
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