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Math Help - Fourier Series

  1. #1
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    Exclamation Fourier Series

    Hi, i really need help with this question

    find the fourier series of the function f(t)= 3sin(t) + sin(4t).

    I know that the function is odd (so an=0), and has period 2pi, but when i try finding bn, integrating between -pi and pi, i keep just getting 0. This cant be right, where am I going wrong?

    Please help!
    Katy
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by harkapobi View Post
    Hi, i really need help with this question

    find the fourier series of the function f(t)= 3sin(t) + sin(4t).

    I know that the function is odd (so an=0), and has period 2pi, but when i try finding bn, integrating between -pi and pi, i keep just getting 0. This cant be right, where am I going wrong?

    Please help!
    Katy
    It is its own fourier series on the interval (-\pi,\pi)(It is already a triginometric polynomial)

    \sin(x) is orthogonal to every \cos(nx)

    and the only non zero coeffients of the sine sine series will be when n=1 or 4.

    This is exactly like expanding the Taylor series of a polynomial centered at 0. You will get the same thing back.
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    It is its own fourier series on the interval (-\pi,\pi)(It is already a triginometric polynomial)

    \sin(x) is orthogonal to every \cos(nx)

    and the only non zero coeffients of the sine sine series will be when n=1 or 4.

    This is exactly like expanding the Taylor series of a polynomial centered at 0. You will get the same thing back.
    I'm sorry, i'm probably being really dumb but i dont understand. So does this mean i should integrate over a different interval? Its just i need this fourier series to work out the particular integral of a second order differential equation.
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  4. #4
    Behold, the power of SARDINES!
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    You are trying to find a fourier series on the interval (-\pi,\pi).

    f(t)=3\sin(t)+\sin(4t)

    To calculate the a_n we use the formula

    a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)dt

    pluggint in f(t) we get

    a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\cos(nt)dt

    This will equal ZERO for all n since sine and cosine are orthongonal

    Now we do the same thing for the sine terms

    b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt)dt

    b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(nt)dt

    Since sine is an orthogonal basis the only non zero terms will be when n=1 or when n=4.


    b_1=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(t)dt=3

    b_4=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(4t)dt=1
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    You are trying to find a fourier series on the interval (-\pi,\pi).



    f(t)=3\sin(t)+\sin(4t)



    To calculate the a_n we use the formula



    a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)dt



    pluggint in f(t) we get



    a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\cos(nt)dt



    This will equal ZERO for all n since sine and cosine are orthongonal



    Now we do the same thing for the sine terms



    b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt)dt



    b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(nt)dt



    Since sine is an orthogonal basis the only non zero terms will be when n=1 or when n=4.





    b_1=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(t)dt=3



    b_4=\frac{1}{\pi}\int_{-\pi}^{\pi}[3\sin(t)+\sin(4t)]\sin(4t)dt=1

    Ahh right, i understand, because the integral of sin(nt)sin(mt) over -pi to pi is pi if m doesn't equal n. is that right?

    So now if I have the differential equation y'' + 5y = 3sin(t) + sint(4t)
    and i have
    b_4= 1 and b_1 = 3, using the trial particular integral:

    y= c_0/2 + \Sigma( c_ncos(nt)+ b_nsin(nt))

    I get that d_1 = 3/4 and d_4= -1/9 so is the particular integral just:

    y = 3/4sin(t) when n=1
    ... -1/9sin(4t) when n=4 ? is that the answer?
    Last edited by harkapobi; March 15th 2009 at 11:12 AM.
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