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Math Help - Power series question

  1. #1
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    Power series question

    We just started power series in class and I'm somewhat confused about them. I'm stuck on this particular problem. It says to make a power series for f(x) = x/(2x^2+1). What I did was to base it on geometric series so I convered the equation to look like 1/(1+x) so now it looks like x/[1-(-2x^2)]. What I got for the power series was x multiplied by the sum of (-2x^2)^n from n=0 to infinity. Is this right?
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  2. #2
    Super Member Aryth's Avatar
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    You're close, except you forgot to add a coefficient, rational functions have this relation:

    \frac{x}{2x^2 + 1} = \sum_{n=0}^{\infty} a_nx^n

    It's that simple, now you have to find the coefficients:

    Multiply the denominator through:

    x = (2x^2 + 1)\sum_{n=0}^{\infty} a_nx^n

    Now we multiply the sum through:

    x = 2\sum_{n=0}^{\infty}a_nx^{n+2} + \sum_{n=0}^{\infty}a_nx^n

    Adjust the indices:

    x = 2\sum_{n=2}^{\infty}a_{n-2}x^n + \sum_{n=0}^{\infty}a_nx^n

    Now we combine like terms:

    x = a_0 + a_1x + \sum_{n=2}^{\infty}(2a_{n-2} + a_n)x^n

    Since there are no coefficients on the left side, then:

    a_0 = 0

    Since the coefficient of the x term on the left side is 1, then the coefficient of the x term on the right side is also 1:

    a_1 = 1

    This gives you:

    x = x + \sum_{n=2}^{\infty}(2a_{n-2} + a_n)x^n

    Which gives you an interesting result:

    a_n = -2a_{n-2}

    Now you have your power series:

    f(x) = \sum_{n=2}^{\infty}-2a_{n-2}x^n

    This gives you:

    f(x) = \sum_{n=0}^{\infty}-2a_nx^{n+2}
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