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Find the Taylor polynomial of log ((1+x)/(1-x))1/2

Show that

    \[ T_{2n+1} \left( \log \sqrt{ \frac{1+x}{1-x}} \right) = \sum_{k=0}^n \frac{x^{2k+1}}{2k+1}. \]

First, we have

    \[ f(x) = \log \sqrt{ \frac{1+x}{1-x}} = \frac{1}{2} ( \log (1+x) - \log(1-x)). \]

We know from the previous exercise (Section 7.4, Exercise #6) that

    \[ T_n (\log(1+x)) = \sum_{k=1}^n \frac{(-1)^{k+1}x^k}{k}. \]

We also know (from the example on page 277) that

    \[ T_{n+1} (-\log(1-x)) = \sum_{k=1}^{n+1} \frac{x^k}{k}. \]

Therefore, using Theorem 7.2 (a), the linearity property of T_n we have

    \begin{align*}  T_{2n+1} f(x) &= \frac{1}{2} T_{2n+1} \log(1+x) + \frac{1}{2} T_{2n+1}(-\log(1-x)) \\[9pt]  &= \frac{1}{2} \left( \sum_{k=1}^{2n+1} \frac{(-1)^{k+1}x^k}{k}  + \sum_{k=1}^{2n+1} \frac{x^k}{k} \right) \\[9pt]  &= \frac{1}{2} \left( \sum_{k=1}^{2n+1} \frac{(-1)^{k+1} x^k + x^k}{k} \right) \\[9pt]  &= \frac{1}{2} \sum_{k=1}^{2n+1} \frac{x^k ((-1)^{k+1} + 1)}{k}. \end{align*}

However, (-1)^{k+1} + 1 = 0 if k is even and (-1)^{k+1} + 1 = 2 if k is odd. Therefore we can sum over just the odd values of k. Let k = 2j-1 and we have

    \begin{align*}  T_{2n+1} \sqrt{\frac{\log (1+x)}{\log(1-x)}} &= \frac{1}{2} \sum_{k=1}^{2n+1} \frac{x^k ((-1)^{k+1} + 1)}{k} \\[9pt]  &= \frac{1}{2} \sum_{j=1}^{n+1} \frac{x^{2j-1} (2)}{2j-1} \\[9pt]  &= \sum_{j=1}^{n+1} \frac{x^{2j-1}}{2j-1} \\[9pt]  &= \sum_{j=0}^n \frac{x^{2j+1}}{2j+1} \\[9pt]  &= \sum_{k=0}^n \frac{x^{2k+1}}{2k+1}. \end{align*}

Where we have renamed the index of summation in the final step so that the sum is over k as in the book.

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