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Math Help - Another Power Series

  1. #1
    Del
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    Another Power Series

    The function is represented as a power series


    Find the first few coefficients in the power series.

    I've found c0 to be 6 in this case, and the radius of convergence to be 0.1. Can someone help me find c1, c2, c3, and c4? Please help!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Del View Post
    The function is represented as a power series


    Find the first few coefficients in the power series.

    I've found c0 to be 6 in this case, and the radius of convergence to be 0.1. Can someone help me find c1, c2, c3, and c4? Please help!
    6\cdot\frac{1}{1+(10x)^2}
    and we know that \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}

    CAn you gofrom there?
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  3. #3
    Del
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    Series problems are my worst enemies. I don't know how to find coefficients.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Del View Post
    Series problems are my worst enemies. I don't know how to find coefficients.
    Put the 10x where the x^{2n} and expand
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  5. #5
    Del
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    Substitute 10x in the place of x^(2n)? Or just use x = 10x?
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Del View Post
    Substitute 10x in the place of x^(2n)? Or just use x = 10x?
    x=10x
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  7. #7
    Del
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    I still can't figure this one out. What series can I use to find the coefficients?
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  8. #8
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    Quote Originally Posted by Del View Post
    I still can't figure this one out. What series can I use to find the coefficients?
    \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}

    So,
    \frac{1}{1+(10x)^2}=\sum_{n=0}^{\infty}(-1)^{n}(10x)^{2n}

    \frac{6}{1+(10x)^2}=\sum_{n=0}^{\infty}(-1)^{n}6(10x)^{2n}

    \text{But }\sum_{n=0}^{\infty}(-1)^{n}6(10x)^{2n} = \sum_{n=0}^{n=\infty} c_n x^n


    \text{Thus }\sum_{n=0}^{\infty}6(-100)^{n}(x)^{2n} = \sum_{n=0}^{n=\infty} c_n x^n

    So clearly for an even n c_n = 6(-100)^{\frac{n}2} and for an odd n c_n = 0
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