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The Fourier Series of Sin^5(x)

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data

    So I have to find the fourier series for [tex]sin^{5}(x)[/tex].

    2. Relevant equations

    I know the [tex]a_{n}[/tex] in:
    [tex]\frac{a_{0}}{2} + \sum^{\infty}_{n=1}a_{n}cos_{n}x + \sum^{\infty}_{n=1}b_{n}sin_{n}x[/tex]
    goes to zero, which leaves me with taking the [tex]b_{n}[/tex].

    3. The attempt at a solution

    So what I got so far is trying to integrate to find [tex]b_{n}[/tex].

    [tex]b_{n} = \frac{1}{\pi} \int^{\pi}_{-\pi} sin^{5}(x)sin(nx)[/tex]

    But I am not sure how to proceed from here, do I make use of [tex]sin^{2}(x)=1/2(1-cos2x)[/tex] and [tex]cos^{2}(x)=1/2(1+sin2x)[/tex]?

    Am I even going in the right direction?


    I just plugged it into Maple and got:

    [tex]\frac{1}{\pi}\left( \frac{sin((n-5)x}{32(n-5)} - \frac{sin((n+5)x}{32(n+5)} - \frac{5sin((n-3)x}{32(n-3)} + \frac{5sin((n-1)x}{16(n-1)} - \frac{5sin((n+1)x}{16(n+1)} \right)^{\pi}_{-\pi}[/tex]

    is this the direction I need to go in?
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 11, 2008 #2
    Euler's Formula

    I did not attempt this, but have you tried using Euler's formula to help simplify before plugging in the series?
  4. Mar 11, 2008 #3


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    Staff Emeritus
    Science Advisor

    Have you tried using trig identies to write sin5 in terms of cos(nx) and sin(nx)?

    For example, Since cos(2x)= cos2(x)- sin2[/sup= (1- sin(x))- sin2(x)= 1- 2sin2(x), sin2(x)= (1/2)(1- cos(2x)), sin5(x)= (sin2(x))(sin2(x))(sin(x))= (1/2)(1- cos(2x))(1/2)(1- cos(2x))(sin(x))= (1/4)sin(x)- (1/2)cos(2x)sin(x)+ (1/4)cos2(x).
    Replacing sin2(x) in the first identity with 1- cos2(x) will give you cos2(x)= (1/2)(1+ cos(2x)) and you can use sin(a+ b)= cos(a)sin(b)+ sin(a)cos(b), sin(a-b)= -cos(a)sin(b)+ sin(a)cos(b) so that, adding the two equations. 2cos(a)sin(b)= sin(a+b)+ sin(a-b) with a= 2x, b= x, to resolve that cos(2x)sin(x). When you have reduced all products to sin(nx) and cos(nx), you have your Fourier series.
  5. Mar 11, 2008 #4

    I ended up going with
    \frac{1}{\pi}\left( \frac{sin((n-5)x}{32(n-5)} - \frac{sin((n+5)x}{32(n+5)} - \frac{5sin((n-3)x}{32(n-3)} + \frac{5sin((n-1)x}{16(n-1)} - \frac{5sin((n+1)x}{16(n+1)} \right)^{\pi}_{-\pi}

    and I just plugged in [itex]\pi[/itex] and [itex]-\pi[/itex]. I talked to one of the "smart" kids in class about it and I was on the right track, which was quite a relief.

    Thanks for your help!
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