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Math Help - maclaurin series

  1. #1
    Member tinspire's Avatar
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    maclaurin series

    find the maclaurin series for f(x)=ln(1-x) and its radius of convergence. how do i do this? is it similar to that of ln(1+x)?
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  2. #2
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    Re: maclaurin series

    Taking the derivative, you find f'(x) = -\dfrac{1}{1-x} = -\sum_{n\ge 0}x^n. Integrating both sides, we get f(x) = -\sum_{n\ge 0}\dfrac{x^{n+1}}{n+1}+C. Since f(0) = 0 = C, this becomes \ln(1-x) = -\sum_{n\ge 1}\dfrac{x^n}{n} (after you reindex to start at one instead of zero).
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    Re: maclaurin series

    Let $g(x)=\ln(1+x)$

    $f(x)=g(-x)$

    We know (or can easily find out) that

    $g(x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}\dfrac{x^k}{k}$

    $f(x)=g(-x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}\dfrac{(-x)^k}{k}$

    $f(x)=\displaystyle{\sum_{k=1}^\infty}(-1)^{k-1}(-1)^k\dfrac{x^k}{k}$

    $f(x)=\displaystyle{\sum_{k=1}^\infty}-\dfrac{x^k}{k}$
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  4. #4
    Member tinspire's Avatar
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    Re: maclaurin series

    thank you! and how do i find R?
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    Re: maclaurin series

    Quote Originally Posted by tinspire View Post
    thank you! and how do i find R?
    The radius of convergence is the same as that of ln(1+x) (or if you use the method I gave you, it is the radius of convergence of the geometric sum, which is 1).
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