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Thread: Fourier cosine series

  1. #1
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    Fourier cosine series

    Can someone help me with the following: (I'm getting lost with the integrals and I can't really type them here)

    f(x) = x(a-x), 0<= x <= a Find this in a Fourier cosine series.

    I have the answers, I'm just having trouble computing it.

    Thanks so much in advance.
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  2. #2
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    Quote Originally Posted by BenWong View Post
    Can someone help me with the following: (I'm getting lost with the integrals and I can't really type them here)

    f(x) = x(a-x), 0<= x <= a Find this in a Fourier cosine series.

    I have the answers, I'm just having trouble computing it.

    Thanks so much in advance.
    You need to use these formulas:

    $\displaystyle f\left( x \right) = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos \left( {\frac{{\pi nx}}{l}} \right)} {\text{ }}\left( {{\text{0}} \leqslant x \leqslant l} \right)$,

    where $\displaystyle {a_0} = \frac{1}{l}\int\limits_0^l {f\left( x \right)} dx$ and $\displaystyle {a_n} = \frac{2}{l}\int\limits_0^l {f\left( x \right)\cos \left( {\frac{{\pi nx}}{l}} \right)} dx{\text{ }}\left( {n \geqslant 1} \right)$.

    In your case (problem) $\displaystyle l = a$.

    What you can not?
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  3. #3
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    Yeah, how do you integrate

    x(a-x)cos(pi*n*x/a) dx?
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  4. #4
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    Quote Originally Posted by DeMath View Post
    $\displaystyle {a_0} = \frac{1}{l}\int\limits_0^l {f\left( x \right)} dx$
    Surely it's:

    $\displaystyle {a_0} = \frac{2}{l}\int\limits_0^l {f\left( x \right)} dx$?
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by skamoni View Post
    Surely it's:

    $\displaystyle {a_0} = \frac{2}{l}\int\limits_0^l {f\left( x \right)} dx$?
    Look at it again carefully

    $\displaystyle f\left( x \right) = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos \left( {\frac{{\pi nx}}{l}} \right)} {\text{ }}\left( {{\text{0}} \leqslant x \leqslant l} \right)$

    $\displaystyle {\text{where }}{a_0} = \frac{1}{l}\int\limits_0^l {f\left( x \right)} dx{\text{ and }}{a_n} = \frac{2}{l}\int\limits_0^l {f\left( x \right)\cos \left( {\frac{{\pi nx}}{l}} \right)} dx{\text{ }}\left( {n \geqslant 1} \right)
    $

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  6. #6
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    Sorry. Should have read that more carefully.

    I would suggest Integrating by parts, i'm guessing you know how to if you're studying Fourier series.
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  7. #7
    Senior Member DeMath's Avatar
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    Quote Originally Posted by BenWong View Post
    Can someone help me with the following: (I'm getting lost with the integrals and I can't really type them here)

    f(x) = x(a-x), 0<= x <= a Find this in a Fourier cosine series.

    I have the answers, I'm just having trouble computing it.

    Thanks so much in advance.
    So, Find a Fourier cosine series for $\displaystyle f\left( x \right) = x\left( {a - x} \right)$, $\displaystyle 0 \leqslant x \leqslant a$.

    $\displaystyle f\left( x \right) = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos \left( {\frac{{\pi nx}}{l}} \right)} {\text{ }}\left( {{\text{0}} \leqslant x \leqslant l} \right)$,

    where $\displaystyle {a_0} = \frac{1}{l}\int\limits_0^l {f\left( x \right)} dx{\text{ and }}{a_n} = \frac{2}{l}\int\limits_0^l {f\left( x \right)\cos \left( {\frac{{\pi nx}}{l}} \right)} dx{\text{ }}\left( {n \geqslant 1} \right).$

    In your case (problem) $\displaystyle l = a$.

    $\displaystyle {a_0} = \frac{1}{a}\int\limits_0^a {x\left( {a - x} \right)} dx = \frac{1}{a}\left. {\left( {\frac{{a{x^2}}}{2} - \frac{{{x^3}}}{3}} \right)} \right|_0^a = \frac{{{a^2}}}{2} - \frac{{{a^2}}}{3} = \frac{{{a^2}}}{6}.$

    $\displaystyle {a_n} = \frac{2}{a}\int\limits_0^a {x\left( {a - x} \right)\cos \left( {\frac{{\pi nx}}{a}} \right)} dx =$

    $\displaystyle = 2\int\limits_0^a x \cos \left( {\frac{{\pi nx}}{a}} \right)dx - \frac{2}{a}\int\limits_0^a {{x^2}} \cos \left( {\frac{{\pi nx}}{a}} \right)dx = {I_1} - {I_2}.$

    $\displaystyle {I_1} = 2\int\limits_0^a x \cos \left( {\frac{{\pi nx}}{a}} \right)dx = \frac{{2a}}{{\pi n}}\int\limits_0^a x d\left( {\sin \left( {\frac{{\pi nx}}{a}} \right)} \right) =$

    $\displaystyle = \left. {\frac{{2ax}}{{\pi n}}\sin \left( {\frac{{\pi nx}}{a}} \right)} \right|_0^a - \frac{{2a}}{{\pi n}}\int\limits_0^a {\sin \left( {\frac{{\pi nx}}{a}} \right)dx} =$

    $\displaystyle = \left. {\frac{{2ax}}{{\pi n}}\sin \left( {\frac{{\pi nx}}{a}} \right)} \right|_0^a - \frac{{2{a^2}}}{{{\pi ^2}{n^2}}}\int\limits_0^a {\sin \left( {\frac{{\pi nx}}{a}} \right)d} \left( {\frac{{\pi nx}}{a}} \right) =$

    $\displaystyle = \left. {\left[ {\frac{{2ax}}{{\pi n}}\sin \left( {\frac{{\pi nx}}{a}} \right) + \frac{{2{a^2}}}{{{\pi ^2}{n^2}}}\cos \left( {\frac{{\pi nx}}{a}} \right)} \right]} \right|_0^a =$

    $\displaystyle = \frac{{2{a^2}}}{{\pi n}}\sin \left( {\pi n} \right) + \frac{{2{a^2}}}{{{\pi ^2}{n^2}}}\cos \left( {\pi n} \right) - \frac{{2{a^2}}}{{{\pi ^2}{n^2}}} =$

    $\displaystyle = \frac{{2{a^2}}}{{{\pi ^2}{n^2}}}{\left( { - 1} \right)^n} - \frac{{2{a^2}}}{{{\pi ^2}{n^2}}} = \frac{{2{a^2}}}{{{\pi ^2}{n^2}}}\left[ {{{\left( { - 1} \right)}^n} - 1} \right].$

    $\displaystyle {I_2} = \frac{2}{a}\int\limits_0^a {{x^2}} \cos \left( {\frac{{\pi nx}}{a}} \right)dx = \frac{2}{{\pi n}}\int\limits_0^a {{x^2}d\left( {\sin \left( {\frac{{\pi nx}}{a}} \right)} \right)} =$

    $\displaystyle = \left. {\frac{{2{x^2}}}{{\pi n}}\sin \left( {\frac{{\pi nx}}{a}} \right)} \right|_0^a - \frac{4}{{\pi n}}\int\limits_0^a {x\sin \left( {\frac{{\pi nx}}{a}} \right)dx} =$

    $\displaystyle = \left. {\frac{{2{x^2}}}{{\pi n}}\sin \left( {\frac{{\pi nx}}{a}} \right)} \right|_0^a + \frac{{4a}}{{{\pi ^2}{n^2}}}\int\limits_0^a {xd\left( {\cos \left( {\frac{{\pi nx}}{a}} \right)} \right)} =$

    $\displaystyle = \left. {\left[ {\frac{{2{x^2}}}{{\pi n}}\sin \left( {\frac{{\pi nx}}{a}} \right) + \frac{{4ax}}{{{\pi ^2}{n^2}}}\cos \left( {\frac{{\pi nx}}{a}} \right)} \right]} \right|_0^a - \frac{{4a}}{{{\pi ^2}{n^2}}}\int\limits_0^a {\cos \left( {\frac{{\pi nx}}{a}} \right)} dx =$

    $\displaystyle = \left. {\left[ {\frac{{2{x^2}}}{{\pi n}}\sin \left( {\frac{{\pi nx}}{a}} \right) + \frac{{4ax}}{{{\pi ^2}{n^2}}}\cos \left( {\frac{{\pi nx}}{a}} \right)} \right]} \right|_0^a - \frac{{4{a^2}}}{{{\pi ^3}{n^3}}}\int\limits_0^a {\cos \left( {\frac{{\pi nx}}{a}} \right)} d\left( {\frac{{\pi nx}}
    {a}} \right) =$

    $\displaystyle = \left. {\left[ {\frac{{2{x^2}}}{{\pi n}}\sin \left( {\frac{{\pi nx}}
    {a}} \right) + \frac{{4ax}}{{{\pi ^2}{n^2}}}\cos \left( {\frac{{\pi nx}}
    {a}} \right) - \frac{{4{a^2}}}{{{\pi ^3}{n^3}}}\sin \left( {\frac{{\pi nx}}
    {a}} \right)} \right]} \right|_0^a =$

    $\displaystyle = \frac{{2{a^2}}}{{\pi n}}\sin \left( {\pi n} \right) + \frac{{4{a^2}}}{{{\pi ^2}{n^2}}}\cos \left( {\pi n} \right) - \frac{{4{a^2}}}{{{\pi ^3}{n^3}}}\sin \left( {\pi n} \right) = \frac{{4{a^2}}}{{{\pi ^2}{n^2}}}{\left( { - 1} \right)^n}.$

    So, we have

    $\displaystyle {a_n} = \frac{{2{a^2}}}{{{\pi ^2}{n^2}}}\left[ {{{\left( { - 1} \right)}^n} - 1} \right] - \frac{{4{a^2}}}{{{\pi ^2}{n^2}}}{\left( { - 1} \right)^n} = \frac{{2{a^2}}}{{{\pi ^2}{n^2}}}\left[ {{{\left( { - 1} \right)}^{n + 1}} - 1} \right].$

    $\displaystyle {a_1} = 0,{\text{ }}{a_2} = - \frac{{4{a^2}}}{{4{\pi ^2}}},{\text{ }}{a_3} = 0,{\text{ }}{a_4} = - \frac{{4{a^2}}}{{16{\pi ^2}}},{\text{ }} \ldots ,{\text{ }}{a_{2k - 1}} = 0,{\text{ }}{a_{2k}} = - \frac{{4{a^2}}}{{{{\left( {2k} \right)}^2}{\pi ^2}}}{\text{.}}$

    Finally, we have

    $\displaystyle f\left( x \right) = \frac{{{a^2}}}{6} - \frac{{4{a^2}}}{{{\pi ^2}}}\sum\limits_{k = 1}^\infty {\frac{1}{{{{\left( {2k} \right)}^2}}}\cos \left( {\frac{{2\pi kx}}{a}} \right)} {\text{ }}\left( {0 \leqslant x \leqslant a} \right).$
    Last edited by DeMath; Jan 18th 2009 at 11:26 AM.
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  8. #8
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    Quote Originally Posted by DeMath View Post
    You need to use these formulas:

    $\displaystyle f\left( x \right) = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos \left( {\frac{{\pi nx}}{l}} \right)} {\text{ }}\left( {{\text{0}} \leqslant x \leqslant l} \right)$,

    where $\displaystyle {a_0} = \frac{1}{l}\int\limits_0^l {f\left( x \right)} dx$ and $\displaystyle {a_n} = \frac{2}{l}\int\limits_0^l {f\left( x \right)\cos \left( {\frac{{\pi nx}}{l}} \right)} dx{\text{ }}\left( {n \geqslant 1} \right)$.

    In your case (problem) $\displaystyle l = a$.

    What you can not?
    If you define your Fourier cosine sereis as

    $\displaystyle f\left( x \right) = \frac{a_0}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos \left( {\frac{{\pi nx}}{l}} \right)} {\text{ }}\left( {{\text{0}} \leqslant x \leqslant l} \right)$,

    then $\displaystyle {a_n} = \frac{2}{l}\int\limits_0^l {f\left( x \right)\cos \left( {\frac{{\pi nx}}{l}} \right)} dx{\text{ }} n = 0, 1, 2, \cdots$.

    Just one integral formula to remember.
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  9. #9
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    Quote Originally Posted by danny arrigo View Post
    If you define your Fourier cosine sereis as

    $\displaystyle f\left( x \right) = \frac{a_0}{2} + \sum\limits_{n = 1}^\infty {{a_n}\cos \left( {\frac{{\pi nx}}{l}} \right)} {\text{ }}\left( {{\text{0}} \leqslant x \leqslant l} \right)$,

    then $\displaystyle {a_n} = \frac{2}{l}\int\limits_0^l {f\left( x \right)\cos \left( {\frac{{\pi nx}}{l}} \right)} dx{\text{ }} n = 0, 1, 2, \cdots$.

    Just one integral formula to remember.
    I understood you.
    This is the same, just in a more general way.
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