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Math Help - fourier

  1. #1
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    fourier

    Hello all

    I have the function

     s(x)= sin(x)cos^2(x)

    and i need to show that it is a 2\pi-periodic trigonometric function,

    My first idea was simply something along the lines of sin(x)cos^2(x)=sin(2\pi x)cos^2(2\pi x)


    Furthermore i need to determine the fourrier coefficients for all n \in Z

    1/2pi \int_{-\pi}^{\pi} s(x)e^{-inx} dx

    Any helps or hints is very much appreciated
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  2. #2
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    Quote Originally Posted by Zaph View Post
    Hello all

    I have the function

     s(x)= sin(x)cos^2(x)

    and i need to show that it is a 2\pi-periodic trigonometric function,

    My first idea was simply something along the lines of sin(x)cos^2(x)=sin(2\pi x)cos^2(2\pi x)
    No, those are NOT the same. What you want is sin(x)cos^2(x)= sin(x+ 2\pi)cos^2(x+ 2\pi).


    Furthermore i need to determine the fourrier coefficients for all n \in Z

    1/2pi \int_{-\pi}^{\pi} s(x)e^{-inx} dx

    Any helps or hints is very much appreciated
    That will be the same as
    \frac{1}{2\pi}\int_{-\pi}^\pi s(x) cos(nx)dx= \frac{1}{2}int_{-\pi}^\pi sin(x)cos^2(x)cos(nx)dx
    and
    \frac{1}{2\pi}\int_{-\pi}^\pi s(x) sin(nx)dx= \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)sin(nx)dx

    You can use trigonometric identities for those.
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  3. #3
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    Bleh brainfart whilst tex'ing, i meant ofcourse
    sin(x)cos^2(x)= sin(2\pi+x)cos^2(2\pi+x)
    But thanx for the heads up, now i know i wasnt all that wrong

    For the other half, ill look into what you suggested, and thanx alot for the quick reply
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  4. #4
    MHF Contributor chisigma's Avatar
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    Is...

    \sin x \cos^{2} x = \sin x (1-\sin^{2} x) = \sin x - \sin^{3} x = \frac{1}{4} \sin x + \frac{1}{4} \sin 3x

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    Is...

    \sin x \cos^{2} x = \sin x (1-\sin^{2} x) = \sin x - \sin^{3} x = \frac{1}{4} \sin x + \frac{1}{4} \sin 3x

    Kind regards

    \chi \sigma
    I'm not quite following you here, care to elaborate?
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  6. #6
    MHF Contributor chisigma's Avatar
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    Is...

    \sin^{2} x + \cos^{2} x =1 \Rightarrow \cos^{2} x = 1 - \sin^{2} x (1)

    ... and...

    \sin^{3} x = \frac{3}{4} \sin x - \frac{1}{4} \sin 3x (2)

    ... so that...

    \sin x \cos^{2} x = \sin x (1-\sin^{2} x) = \sin x - \sin^{3} x =\frac{1}{4} \sin x + \frac{1}{4} \sin 3x (3)

    Kind regards

    \chi \sigma
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post


    That will be the same as
    \frac{1}{2\pi}\int_{-\pi}^\pi s(x) cos(nx)dx= \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)cos(nx)dx
    and
    \frac{1}{2\pi}\int_{-\pi}^\pi s(x) sin(nx)dx= \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)sin(nx)dx

    You can use trigonometric identities for those.
    So i tried for the first one, and i'd like someone to check if im going in the right direction.

     \frac{1}{2\pi}\int_{-\pi}^\pi s(x) cos(nx)dx=  \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)cos(nx)dx = \frac{1}{2}\int_{-\pi}^\pi e^{-ix} e^{-ix^2}e^{-inx}= \frac{1}{2}\int_{-\pi}^\pi e^{-ix} e^{-inx}= ...








    [/tex]
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    No, those are NOT the same. What you want is sin(x)cos^2(x)= sin(x+ 2\pi)cos^2(x+ 2\pi).



    That will be the same as
    \frac{1}{2\pi}\int_{-\pi}^\pi s(x) cos(nx)dx= \frac{1}{2}int_{-\pi}^\pi sin(x)cos^2(x)cos(nx)dx
    and
    \frac{1}{2\pi}\int_{-\pi}^\pi s(x) sin(nx)dx= \frac{1}{2}\int_{-\pi}^\pi sin(x)cos^2(x)sin(nx)dx

    You can use trigonometric identities for those.
    Would you care to elaborate a bit on this? Thanx in advance
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  9. #9
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    Do you need to show that the function is 2\pi-periodic and has no smaller period? Some trig functions built up out of your basic sin and cos have periods smaller than the original. |\cos(x)| is a simple one that comes to mind; the same with \sin^{2}(x).
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  10. #10
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    Quote Originally Posted by Ackbeet View Post
    Do you need to show that the function is 2\pi-periodic and has no smaller period? Some trig functions built up out of your basic sin and cos have periods smaller than the original. |\cos(x)| is a simple one that comes to mind; the same with \sin^{2}(x).
    Yes, i need to show that is is 2\pi-periodic nothing more there, and besides that i need to determine the fourier coefficients:


    \frac{1}{2}\pi \int_{-\pi}^{\pi} s(x)e^{-inx} dx

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