Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier series of sin(x)

  1. Sep 16, 2007 #1
    I know it's trivial....but how do you find the Fourier series of sin(x) itself? I seem to get everything going to zero...
     
  2. jcsd
  3. Sep 16, 2007 #2
    You should. The Fourier series of sin(x) is simply sin(x). This is like asking what the Taylor series of y=x^2 is.
     
  4. Sep 16, 2007 #3
    Yes, but if the Fourier series of f(x) = ao/2 + sum over n from 1 to infinity (an*cos(n*x) + bn*sin(nx)) then if ao, an, and bn go to zero, you are left with simply zero as the Fourier series. Shouldn't this method still work?
     
  5. Sep 16, 2007 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    What ever makes you think all of the terms go to zero? One term does not, the b1 term. You end up with sin(x).
     
  6. Sep 16, 2007 #5
    bn = (1/pi) int ( sin(x) * sin(n*x) dx) from -pi to pi becomes a multiple of sin (n*pi) which is always zero.
     
  7. Sep 16, 2007 #6

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

    if n = 1, then it's the integral of sin^2x, which is certainly not zero over -pi to pi.
     
  8. Sep 16, 2007 #7
    That's true. Okay, but for bn where n isn't 1, bn is zero, correct?
     
  9. Sep 17, 2007 #8

    nicksauce

    User Avatar
    Science Advisor
    Homework Helper

    Correct
     
  10. Aug 20, 2010 #9
    Sorry to gravedig, but considering this is the first result for the fourier series of sin[x]. I had to correct it.

    A fourier series is a series that has a period equal to part of a qualified f(x) from -L to L or 0 to 2l.
    Yes, the fourier series of Sin[x] with L = 2n Pi where n is an element of Z is simply Sin[x].
    But the fourier series of Sin[x] with L =/= 2n Pi is NOT Sin[x].
    Look at the graph of Sin[x] from 0 to Pi/2. If you wanted a function that repeated that period, Sin[x] would no-where near work.
    The first few terms of that fourier series would be: (4 Cos[1/2] Sin[x])/(-1 + \[Pi]^2) + (
    8 Cos[1] Sin[2 x])/(-4 + \[Pi]^2) + (
    12 Cos[3/2] Sin[3 x])/(-9 + \[Pi]^2)
    According to mathematica. (It's slightly incorrect, since the outcome of the periods are off. Working on it)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fourier series of sin(x)
  1. Derivative of x^sin(x) (Replies: 9)

  2. Integral sin(x)/x (Replies: 2)

Loading...