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Math Help - Power Series Representation

  1. #1
    Newbie lessthanthree's Avatar
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    Power Series Representation

    I need help finding the power series representation of ln(root(4-x2)).

    I have gathered that I pull out the root 4, so I'll have ln(2)+something, but I'm not sure where to go from there.

    I know that ln(1-x) is -xk/k, but I don't know how to manipulate it as far as the root part.
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  2. #2
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    Re: Power Series Representation

    Quote Originally Posted by lessthanthree View Post
    I need help finding the power series representation of ln(root(4-x2)).

    I have gathered that I pull out the root 4, so I'll have ln(2)+something, but I'm not sure where to go from there.

    I know that ln(1-x) is -xk/k, but I don't know how to manipulate it as far as the root part.
    Notice that

    \displaystyle \begin{align*} f(x) &= \ln{ \left( \sqrt{ 4 - x^2 } \right) }  \\ &= \ln{ \left[ \left( 4 - x^2 \right) ^{ \frac{1}{2} } \right] } \\ &= \frac{1}{2} \ln{ \left( 4 - x^2 \right) }  \end{align*}.

    Also notice that

    \displaystyle \begin{align*} f'(x) &= -\frac{x}{4 - x^2} \\ &= -x \left( \frac{1}{4 - x^2} \right) \\ &= -\frac{x}{4} \left( \frac{1}{1 - \frac{x^2}{4}} \right) \\ &= -\frac{x}{4} \left[ \frac{1}{1 - \left( \frac{ x}{2} \right) ^2 } \right] \end{align*}


    Now remember that in a geometric series, \displaystyle \begin{align*} \sum_{k = 0}^{\infty} r^k = \frac{1}{1 - r} \end{align*} as long as \displaystyle \begin{align*} |r| < 1 \end{align*}. Do you see that \displaystyle \begin{align*} f'(x) \end{align*} has a geometric series?

    \displaystyle \begin{align*} f'(x) &= -\frac{x}{4} \left[ \frac{1}{1 - \left( \frac{x}{2} \right) ^2 } \right] \\ &= -\frac{x}{4} \sum_{ k = 0}^{\infty} \left[ \left( \frac{x}{2} \right) ^2 \right] ^k \textrm{ with } \left| \left( \frac{x}{2} \right) ^2 \right| < 1 \\ &= -\frac{x}{4} \sum_{k =0}^{\infty} \frac{x^{2k}}{2^{2k}} \textrm{ with } |x| < 2 \\ &= -\sum_{k = 0}^{\infty} \frac{ x^{2k+1} }{ 2^{2k + 2} } \end{align*}

    and so

    \displaystyle \begin{align*} f(x) &= \int{ -\sum_{k = 0}^{\infty} \frac{x^{2k + 1}}{2^{2k+2}} \, dx} \\ &= -\sum_{k=0}^{\infty} \int{ \frac{x^{2k+1}}{2^{2k+2}} \, dx } \\ &= -\sum_{k = 0}^{\infty} \frac{x^{2k + 2}}{\left( 2k + 2 \right) 2^{2k + 2}} \end{align*}

    Therefore \displaystyle \begin{align*} \ln{ \left( \sqrt{ 4 - x^2 } \right) } = -\sum_{ k = 0}^{\infty} \frac{ x^{2k+2} }{ \left( 2k + 2 \right) 2^{2k + 2} } \end{align*}. It is obvious that this is convergent where \displaystyle \begin{align*} |x| < 2 \end{align*} and divergent where \displaystyle \begin{align*} |x| > 2 \end{align*}. You will need to use some other tests to determine the convergence of this series where \displaystyle \begin{align*} |x| = 2 \end{align*}.
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    Re: Power Series Representation

    Also, you should know that \displaystyle \begin{align*} \sqrt{ 4 - x^2} \end{align*} is NOT \displaystyle \begin{align*} \sqrt{4} - \sqrt{x^2} \end{align*}, and \displaystyle \begin{align*} \log{ (a + b)}  \end{align*} is NOT \displaystyle \begin{align*} \log{(a)} + \log{(b)} \end{align*}, so you can NOT "pull out" \displaystyle \begin{align*} \ln{(2)} \end{align*}.
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