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Tough problem

  1. Feb 9, 2006 #1
    The problem states:

    For -1<p<1 prove that,

    a)[tex] \sum p^n \cos nx = (1 - p\cos(x))/(1-2p\cos(x) + p^2)[/tex]

    b)[tex] \sum p^n \sin nx = (1 - p\sin(x))/(1-2p\cos(x) + p^2)[/tex]


    What I have been doing is searching for some type of expanding series for P^n\sin nx, however, I am pretty stumped
     
  2. jcsd
  3. Feb 9, 2006 #2
    if there is anything lacking in my question lemme know.

    I thought 1 out of 20 views might have been able to help! :smile:
     
  4. Feb 9, 2006 #3

    quasar987

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    This is a suggestion; I haven't pushed my thoughts farther than this:

    Can you show that the right hand side has the left hand side for a fourier series? Then convergence for p in (-1,1) is assured by Weierstrass M test. Remains to show that it converges towards the right hand side.
     
    Last edited: Feb 9, 2006
  5. Feb 9, 2006 #4
    hmmm I could try, but we haven't covered forier series in class, I'd be suprised if I had to use one.
     
  6. Feb 9, 2006 #5

    quasar987

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    k forget it then. mmmhh....
     
  7. Feb 9, 2006 #6
    sorry:frown:
     
  8. Feb 10, 2006 #7

    matt grime

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    De Moivre's theorem springs to mind. What is the real part of (pe^{ix})^n? And the imaginary part.
     
  9. Feb 10, 2006 #8

    quasar987

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    With this new look on the problem, I can invoque a theorem that assures the convergence of the fourier serie towards the function that spawned it. So would only remain to show that the RHS effectively has the LHS has a Fourier serie. But the resulting integral would probably be a pain, and fourier theory is probably not necessary anyway.
     
  10. Feb 10, 2006 #9

    matt grime

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    why would you invoke fourier series? fourier series have nothing to do with this question at all. it is just a question about geometric progressions. anyone who knows what a complex number and a geometric progression is can do it.
     
  11. Feb 10, 2006 #10

    quasar987

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    Why do you say it has nothing to do with fourier series? Is it not true that the LHS is the fourier serie of the RHS?
     
  12. Feb 10, 2006 #11

    matt grime

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    Just because one can give an interpretation of it as a question in fourier series does not make it a question on fourier series. as the OP doesn't know about them and does not need to know about them let's forget them. The question was not asked in the context of fourier series, and the question is therefore not about fourier series (hence my opinion that it has nothing to do with fourier series, certainly as far as the OP is concerned) any more than the question 'show x^2=1 has two roots in R*' is about group theory just because R* happens to be a group.

    One should also withhold from equating functions with their fourier series unnecessarily.
     
    Last edited: Feb 10, 2006
  13. Feb 10, 2006 #12

    quasar987

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    One will take note of that... :uhh:
     
  14. Feb 10, 2006 #13

    krab

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    b) looks not even true, since the LHS is odd, but the RHS is neither even nor odd. Or, more specifically, the LHS is zero for x=0, but the RHS is not.

    Ah, found it. It's

    [tex] \sum p^n \sin nx = p\sin(x)/(1-2p\cos(x) + p^2)[/tex]

    This can be found in:

    G. Chrystal, "Textbook of Algebra - Part 2" p. 273.
     
    Last edited: Feb 10, 2006
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