Find the Taylor polynomial for a^x

by RoRi

December 20, 2015

Show that

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

First, the $k$ th 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

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

Stumbling Robot