Prove that ax = ∑ ((log a)n / n!) xn

Assuming that $a^x$ has a power-series representation in terms of powers of $x$ , verify that it has the form

$a^x = \sum_{n=0}^{\infty} \frac{(\log a)^n}{n!} x^n, \qquad a > 0$

valid for all real $x$ .

From the definition of the exponential we have

$a^x = e^{x \log a}.$

We also know the series expansion for $e^x$ ,

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$

Therefore,

$\begin{align*} a^x &= e^{x \log a} \\ &= \sum_{n=0}^{\infty} \frac{(x \log a)^n}{n!} \\[9pt] &= \sum_{n=0}^{\infty} \frac{(\log a)^n}{n!} x^n. \end{align*}$

This converges everywhere $x \log a$ is defined (since the series expansion of $e$ converges for all real $x$ ). Hence, this is valid for all $x$ and all $a > 0$ (since that is where $\log a$ is defined).

Stumbling Robot

Prove that a^x = ∑ ((log a)ⁿ / n!) xⁿ

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