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Fourier series complex numbers

  1. Jun 17, 2012 #1
    1. The problem statement, all variables and given/known data


    3. The attempt at a solution

    I don't understand this equation. 2pi, 4pi, 6pi only = 0 when there is a sine function before it, so I don't see how the evens = 0. I don't see why the e vanishes. I also can't get the i's to vanish since one of them is in exponential form and the other is not. i * 2^i does not get rid of the i in my book. really lost here.
  2. jcsd
  3. Jun 17, 2012 #2


    Staff: Mentor

    No. None of the numbers 2[itex]\pi[/itex], 4[itex]\pi[/itex], 6[itex]\pi[/itex], etc. is equal to zero. It's true that sin([itex]2\pi[/itex]) = 0, but the argument is not equal to zero.

    ##e^{-i\pi n}## happens to be equal to 1 for any even integer n. This is because of Euler's formula, eix = cos(x) + isin(x). Replace x by any even multiple of ##\pi## to see this.
  4. Jun 17, 2012 #3
    I'm watching this lecture,

    so if I still don't understand after watching it, I'll get back to you.
    Last edited by a moderator: Sep 25, 2014
  5. Jun 17, 2012 #4
    I can't figure out how to combine cos(x) + isin(x) to get a real number. For instance 2 + i is still imaginary.
  6. Jun 17, 2012 #5
    Think what happens when sin(x) is zero....(equivalent to x=n∏)

    In simpler terms, what happens to (a+ib) if b is zero :wink:
  7. Jun 17, 2012 #6
    if b is 0 in a + ib then we just get a line along the x axis, y = 0. I don't see that helps me with my problem.
  8. Jun 17, 2012 #7
    No, you don't. You get 'a', a real number, which is constant. Not a line.

    By Euler's formula, [itex]e^{i\theta} = cos\theta + i\cdot sin\theta[/itex] which is the same as [itex]z = a+ib[/itex]

    See now?
  9. Jun 17, 2012 #8
    I don't see how euler's formula gets the equation I listed in the OP
  10. Jun 17, 2012 #9
    Well....the equation you have listed in the OP contains [tex]e^{-in\pi}[/tex] and Euler's formula gives an alternate expression for [tex]e^{i\theta}[/tex] Is there any resemblance? :smile:
  11. Jun 17, 2012 #10
    no resemblance. even if there was it wouldn't tell me why 1/πin, n odd, nor would it tell me why (e^-inπ -1)/-2πin results in 1/πin, n odd or 0 if n is even
  12. Jun 17, 2012 #11

    But but...its all laid out to resemble :wink:

    Another hint : put θ = -n∏, in Euler's equation....

    Also, try to forget about the (1/-2∏in) term for the time being.
  13. Jun 17, 2012 #12
    never mind, i'm tired of your hints. i give up.
  14. Jun 17, 2012 #13
    Good idea. You should take a break and try again when you are fresh, sometime. :tongue2:

    Its very, very simple :smile:
  15. Jun 17, 2012 #14


    Staff: Mentor

    No, 2 + i is complex, which means it has a real part (2) and an imaginary part (1, the coefficient of i). 2 + 0i is purely real. 2i = 0 + 2i is purely imaginary, and its imaginary part is 2.
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