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Relationships between Fourier coefficients

  1. Dec 23, 2016 #1
    1. The problem statement, all variables and given/known data

    I have ## f(t) = \sum\limits^{\infty}_0 a_{n} e^{2 \pi i n t} ## [1]
    and ## g(t) = \sum\limits^{\infty}_0 b_{n} e^{ \pi i n t} ## [2]

    I want to show that ##b_n = a _{2n} ##

    2. Relevant equations

    see above.

    3. The attempt at a solution

    So obviously you want to use the orthogonality to obtain the fourier coeffients, integrating the LHS multiplied by the negative exponential over the period. Period in [1] is ##2\pi## and the period in [2] is ##\pi ##.

    So I have:

    from [1]: ##\int^{2 \pi } _{0} f(t) e^{- 2 \pi i n t} dt= a_{n} ##

    and

    from [2]: ##\int^{\pi } _{0} g(t) e^{- \pi i n t} dt = b_{n} ##

    I'm unsure what to do next,
    Many thanks in advance.
     
    Last edited: Dec 23, 2016
  2. jcsd
  3. Dec 23, 2016 #2

    Ssnow

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    Hi, how is possible that the same function ##f## in one case has period ##2\pi## and in the other case ##\pi##?
     
  4. Dec 23, 2016 #3
    edited. ta. typo.
     
  5. Dec 23, 2016 #4

    Ray Vickson

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    If you have ##\pi##s in the exponentials you cannot have them in the integration limits. Anyway, in the integrations the exponents should go from 0 to ##2 \pi n##, so:
    $$ (1) \; a_n = c_a \int_0^1 f(t) e^{-2 \pi i n t} \, dt\\
    (2) \; b_n = c_b \int_0^2 f(t) e^{-i \pi n t} \, dt
    $$
    Here, ##c_a## and ##c_b## are some appropriate constants.

    (1) describes a function having period 1, while (2) describes a function having period 2; but of course, 2 periods of length 1 make up 1 period of length 2.
     
    Last edited: Dec 23, 2016
  6. Dec 24, 2016 #5

    Ssnow

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    Hi, as @Ray Vickson noted sometimes the definition of the fourier transform is with the factor ##2\pi## and other times is with ##\pi##, changing variable it is possible to pass from one notation to another ...

    Ssnow
     
  7. Dec 24, 2016 #6
    okay thanks, that's clarified up some basics
    so ##c_a = 1 ##, c_{b} = 1/2##
    but i sitll don't know what to do, i want to show it explicitly.
     
  8. Dec 24, 2016 #7
    e.g so I send ##n## to ##2n## in [2] and then I need to look at how my integration limits are affected by this.
    How would I go about this?
    Can I do this - does that make sense to do?
     
  9. Dec 24, 2016 #8

    Ssnow

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    You must start write what is ##b_{2n}## that is ##=c_{b}\int_{0}^{2}f(t)e^{-i\pi 2nt}dt## and by substitutions try to write the integral in the form of ##a_{n}## ...
     
  10. Dec 24, 2016 #9

    Ray Vickson

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    Note that in exponential form the Fourier series runs from ##n = -\infty## to ##n = +\infty##, not just from ##0## to ##\infty##. So anyway, we have
    $$0 = b_0 - a_0 + b_1 e^{i \pi t} + (b_2-a_1) e^{i \pi 2 t} + \cdots + \text{similar terms in } n < 0.$$
    In other words, ##\sum_n c_n e^{i \pi n t} \equiv 0##, where ##c_n = b_n## for odd ##n## and ##c_{2m} = b_{2m} - a_m## for ##n = 2m## even.
     
  11. Dec 24, 2016 #10
    So a substitution like ## t* = t/2 ## would fix the limits to ##1## and ##0## as in ##a_{n}##, however I want to keep the exponent with a ##2 \pi ## factor don't I? (rather than ##\pi## ? )
     
  12. Dec 27, 2016 #11

    SammyS

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    Are ƒ(t) and g(t) related to each other in some fashion?
     
  13. Dec 28, 2016 #12
    should the second one be ##c_b \int_0^2 g(t) e^{-i \pi n t} \, dt ## ?? where does ## f(t) ## come from here?
     
  14. Dec 28, 2016 #13
    okay thanks that's making some sense, so you have done that ##g(t)-f(t)=0##, and equated coefficients, however, I do not understand why this is ##0##...

    many thanks
     
  15. Dec 28, 2016 #14

    Ray Vickson

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    Look at my post #4, where you see your original question echoed in a panel along with my response. You will see that originally you had
    $$ (1) \; f(t) = \sum_{0}^{\infty} a_{n} e^{2 \pi i n t} \\
    (2) \; f(t) = \sum_{0}^{\infty} b_{n} e^{ \pi i n t}
    $$
    When did the second ##f(t)## become a ##g(t)##?

    It is obvious that if ##f## and ##g## are just two unrelated functions there cannot be any predictable relationship between the ##a_{2n}## and ##b_n##.
     
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