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Show these functions are 2 pi periodic

  1. Jun 1, 2016 #1
    • Member warned about posting without the HW template
    g(t)=½( f(t)+f(-t) ) h(t)=½( f(t)-f(-t) )
    show its 2π periodic so: g(t+2π) = ½( f(t+2π)+f(t-2π) ) why does -t become t-2π ?
    ½( f(t)+f(-t) ) = g(t)
    h(t+2π)=½( f(t+2π)-f(t-2π) )
    ½( f(t)-f(-t) ) = h(t) is this correct?
    can anybody show me some similar examples please?
    this is from a fourier series question paper.
    thanks
     
  2. jcsd
  3. Jun 1, 2016 #2

    LCKurtz

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    Of course, this problem is false unless ##f(t) = f(t+2\pi)##, which you haven't told us.

    It doesn't. If you put ##t+2\pi## in for ##t## in ##f(-t)## you get ##f(-(t+2\pi)) = f(-t -2\pi)##. Do you see why the numerator becomes ##f(t)+f(-t)##?
     
  4. Jun 3, 2016 #3

    RUber

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    An example would be the exponential definition of the cosine function.
    ##\cos(x) = \frac{1}{2} (e^{ix} + e^{-ix}) ##
    This is analogous to your problem for ##g(x)## with ##f(x) = e^{ix}##.
    ##\cos(x+2\pi) = \frac{1}{2} (e^{i(x+2\pi)} + e^{-i(x+2\pi)})\\
    = \frac{1}{2} (e^{ix}e^{i2\pi} +e^{-ix}e^{-i2\pi})\\
    = \frac{1}{2} (e^{ix} + e^{-ix})\\
    = \cos(x) ##
     
  5. Jun 3, 2016 #4

    Delta²

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    Hint: prove that if a function has period T, then ##f(x-T)=f(x)## as long as f is defined in x-T.
     
    Last edited: Jun 3, 2016
  6. Jun 4, 2016 #5

    haruspex

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    or maybe the sum of a periodic function and an odd function.
     
  7. Jun 4, 2016 #6

    LCKurtz

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    Well, I assumed, which was also unstated, that problem was to prove that if ##f## has period ##2\pi##, then when you write ##f = g + h## with ##g## even and ##h## odd, that ##g## and ##h## have period ##2\pi##. In that context I'm not sure what you are suggesting.

    And, annoyingly enough and a pet peeve of mine, the OP hasn't returned to the thread to clarify anything
     
    Last edited: Jun 5, 2016
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