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Find the Taylor polynomial for ax

Show that

    \[ T_n (a^x) = \sum_{k=0}^n \frac{(\log a)^k}{k!} x^k. \]

First, the kth derivative of a^x is

    \[ f^{(k)} (x) = a^x (\log a)^k. \]

This follows since

    \[ f(x) = a^x = e^{x \log a} \quad \implies \quad f'(x) = a^x (\log a). \]

Then if f^{(n)} (x) = a^x (\log a)^n we have

    \[ f^{(n+1)}(x) = \left( f^{(n)}(x) \right)' = \left( a^x (\log a)^n \right)' = a^x (\log a)^{n+1}. \]

Therefore, by induction, the formula holds for all positive integers k. Next,

    \[ f^{(k)} (0) = a^0 (\log a)^k = (\log a)^k. \]

So, the Taylor polynomial is then

    \[ T_n (a^x) = \sum_{k=0}^n \frac{f^{(k)} (0)}{k!} x^k = \sum_{k=0}^n \frac{(\log a)^k}{k!} x^k.\]

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