Power Series Representation

Sep 2012
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Los Angeles, California
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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MHF Helper
Aug 2008
12,883
4,999
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

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

Also notice that

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


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

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

and so

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

Therefore \(\displaystyle \displaystyle \begin{align*} \ln{ \left( \sqrt{ 4 - x^2 } \right) } = -\sum_{ k = 0}^{\infty} \frac{ x^{2k+2} }{ \left( 2k + 2 \right) 2^{2k + 2} } \end{align*}\). It is obvious that this is convergent where \(\displaystyle \displaystyle \begin{align*} |x| < 2 \end{align*}\) and divergent where \(\displaystyle \displaystyle \begin{align*} |x| > 2 \end{align*}\). You will need to use some other tests to determine the convergence of this series where \(\displaystyle \displaystyle \begin{align*} |x| = 2 \end{align*}\).
 
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MHF Helper
Aug 2008
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Also, you should know that \(\displaystyle \displaystyle \begin{align*} \sqrt{ 4 - x^2} \end{align*}\) is NOT \(\displaystyle \displaystyle \begin{align*} \sqrt{4} - \sqrt{x^2} \end{align*}\), and \(\displaystyle \displaystyle \begin{align*} \log{ (a + b)} \end{align*}\) is NOT \(\displaystyle \displaystyle \begin{align*} \log{(a)} + \log{(b)} \end{align*}\), so you can NOT "pull out" \(\displaystyle \displaystyle \begin{align*} \ln{(2)} \end{align*}\).