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Math Help - determine interval of convergence

  1. #1
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    determine interval of convergence

    I'm supposed to use the power series:
    \frac{1}{1+x} = \sum^ \infty_ {n=0} (-1)^nx^n to determine a power series, centered at 0, for the function. Indentify the interval of convergence.

    Did I do this right?

    f(x) = ln(1-x^2) = \int \frac{1}{1+x}dx - \int \frac{1}{1-x}dx

    ln(1-x^2) = \int (\sum^ \infty_ {n=0}(-1)^n(x^2)^n)dx

    => C + \sum^ \infty_ {n=0} \frac{(-1)^n(x^{2n+2})}{n+1}
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by saiyanmx89 View Post
    I'm supposed to use the power series:
    \frac{1}{1+x} = \sum^ \infty_ {n=0} (-1)^nx^n to determine a power series, centered at 0, for the function. Indentify the interval of convergence.

    Did I do this right?

    f(x) = ln(1-x^2) = \int \frac{1}{1+x}dx - \int \frac{1}{1-x}dx

    ln(1-x^2) = \int (\sum^ \infty_ {n=0}(-1)^n(x^2)^n)dx

    => C + \sum^ \infty_ {n=0} \frac{(-1)^n(x^{2n+2})}{n+1}
    You have missed out the step:

    f(x) = \ln(1-x^2) = \int_0^x \frac{-2\zeta}{1-\zeta^2}d\zeta

    and you don't explain how the rest follows from the first line, there should be no arbitary constant in the answer, the final answer is neater if you change the variable of summation from n to m=n+1, check the (-1)^n term, also you have not answered the part about the interval of convergence.

    CB
    Last edited by CaptainBlack; April 12th 2009 at 11:50 PM.
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