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Find the Taylor polynomial for log (1+x)

Show that

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


Since

    \[ \log (1+x) = \int \frac{1}{1+x} \, dx \]

we can apply Theorem 7.2 (c) (page 276 of Apostol) and our result from this exercise to compute,

    \begin{align*}  T_n (\log (1+x)) &= \int T_{n-1} \left( \frac{1}{1+x} \right) \, dx \\[9pt]  &= \int \sum_{k=0}^{n-1} (-1)^k x^k \, dx \\[9pt]  &= \sum_{k=0}^{n-1} (-1)^k \frac{x^{k+1}}{k+1} \\[9pt]  &= \sum_{k=1}^n \frac{(-1)^{k+1} x^k}{k}. \end{align*}

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