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Exercise from basic Fourier Analysis

  1. Aug 30, 2004 #1
    I really need help with this exercise (it's from a course in basic fourier analysis). It consists of two parts:

    (i) Let [Tex] s_0 = 1/2 [/Tex] and [Tex] s_n = 1/2 + \sum_{j=1}^{n}\cos(jx) [/Tex] for [Tex] n \geq 1 [/Tex]. By writing [Tex] s_n = \left(\sum_{j=-n}^{n}e^{ijx}\right)/2 [/Tex] and summing geometric series show that [Tex] (n+1)^{-1}\sum_{j=0}^{n}s_j \rightarrow 0 [/Tex] as [Tex] n \rightarrow \infty [/Tex] for all [Tex] x \neq 0~mod~2\pi [/Tex], and so

    [Tex] 0 = 1/2 + \sum_{j=1}^{\infty}\cos(jx) [/Tex] in the Cesáro sense.

    (ii) Show similarly that, if [Tex] x \neq 0~mod~2\pi [/Tex], then

    [Tex] cot(x/2) = 2\sum_{j=1}^{\infty}\sin(jx) [/Tex] in the Cesáro sense.

    In (i) I have tried to write out two geometric series and summing them, but I can't get the desired result. I have no idea on (ii).

    "in the Cesáro sense" means (i think) that the average of a given sequence [Tex] s_0,s_1,s_2,\ldots [/Tex] converges against a given limit L (the sequence itself doesn't nescessarily) - that is, the sequence [Tex] s_0, (s_0 + s_1)/2, (s_0 + s_1 + s_2)/3,\ldots \rightarrow L [/Tex].
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  3. Aug 30, 2004 #2


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    Is this how you meant it?
    (Use "tex" not "Tex")
  4. Aug 30, 2004 #3
    For part i, I would say one should first convert [tex] s_n = \left(\sum_{j=-n}^{n}e^{ijx}\right)/2 [/tex] into simpler form using the sum of geometric series formula. Then one should have no difficulty in finding [tex]\sum_{n} s_{n}[/tex]. Moreover one should note that for all integer m, [tex]|e^{imx}|=1[/tex] for all real x.
  5. Aug 31, 2004 #4
    Thanks for answering.. I have shown that [tex] (n+1)^{-1} \sum_{j=0}^{n} s_n \rightarrow 0 [/tex] as [tex] n \rightarrow \infty [/tex], but I can't see exactly how that relates to solving (i)...
  6. Aug 31, 2004 #5
    Did you mean (ii)?
  7. Aug 31, 2004 #6
    yes, i'm sorry, part (ii).. I have no idea on that one (i assume cot(x) = cos(x)/sin(x))
  8. Aug 31, 2004 #7
    hi broegger.

    I did not do part (ii), but I think all they want you to realise that [tex]sinx = \frac {1}{2i} (e^{ix}-e^{-ix}) [/tex]. Using this formula, find [tex] \sum_{j=0}^{n} sin(jx) [/tex]. Then all the steps are similar to what you did in part i), I think. Of course one should always remember [tex] cosx = \frac{1}{2} (e^{ix}+e^{-ix}) [/tex] and [tex]sinx = \frac {1}{2i} (e^{ix}-e^{-ix}) [/tex]. Using these, one may find an expression for cot(x/2) in "complex" exponential.

    I did not do it, but I think that may be the way to do it.
    Last edited: Aug 31, 2004
  9. Aug 31, 2004 #8
    Hi. Thank you very much for taking the time to help me.

    I'll try the method you advised for part (ii).. about part (i) I'm not sure my reasoning are correct; how can you apply the geometric series formula when the series in question are not infinite - ex. [tex] \sum_{j=-4}^{4}e^{ijx} = 1/2 + \sum_{j=1}^{4}\cos(jx) [/tex]. I don't know if you see what I mean (maybe I'm getting this all wrong).
  10. Aug 31, 2004 #9
    broegger, do you know that 1+r+r^2+...r^n = (1-r^(n+1))/(1-r)?
  11. Aug 31, 2004 #10
    Erm.. How can that be true?? In the limit [tex] n = \infty [/tex] the sum you are mentioning is equal to 1/(1-r).

    I'm really troubled by this :\
  12. Aug 31, 2004 #11
    Let S(r) = 1+r..+r^n, then
    r*S(r) = r+r^2+...+r^(n+1)
    (1-r)*S(r) = 1-r^(n+1)
    S(r) = (1-r^(n+1))/(1-r)

    Yes, the limit of the expression as n tends to infinity tends to 1/(1-r), *if*|r|<1.
  13. May 11, 2007 #12
    Three years old, but now I'm trying to solve the problem ((i)).

    I suppose

    [tex] 2s_n = \sum_{j=-n}^{n}e^{ijx} = \frac{1-e^{(i x)(n+1)}}{1-e^{i x}}+\frac{1-e^{(-i x)(n+1)}}{1-e^{-i x}}-1 [/tex]

    so I have to show (?) that

    [tex] \lim_{p\rightarrow\infty}\frac{1}{p+1}\sum_{n=0}^{p}s_n=\lim_{p\rightarrow\infty} \frac{1}{p+1}\sum_{n=0}^{p}\left(\frac{1-e^{(i x)(n+1)}}{1-e^{i x}}+\frac{1-e^{(-i x)(n+1)}}{1-e^{-i x}}-1\right)/2=0. [/tex]

    Apparently one should have no difficulty in showing that, but I do.
    Last edited: May 11, 2007
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