Example 4.31. Investigate the power series for the function $f(x)=a^{x}(a>0, a \neq 1)$ at the point $x_{0}=0$.
Solution. The function $f(x)=a^{x}=e^{x \ln a}$ is infinitely differentiable in the neighborhood of the point $x_{0}=0$. In this case,

$$
f^{(n)}(x)=\ln ^{n} a \cdot a^{x}, f^{(n)}(0)=\ln ^{n} a
$$

Thus,

$$
a^{x} \sim \sum_{n=0}^{\infty} \frac{\ln ^{n} a}{n!} x^{n}
$$

The remainder term of the Taylor formula, taken in the Lagrange form, is

$$
R_{n}(a)=\frac{\ln ^{n+1} a}{(n+1)!} a^{\Theta x}, 0<\Theta<1
$$

Let $n_{0}=[2|\ln a|]+1$, then for all values of $n \geqslant n_{0}$.

$$
\frac{|\ln a|}{n}<\frac{1}{2}
$$

Let $n=n_{0}+p, p \in N$ and $A=\frac{|\ln a|^{n_{0}}}{n_{0}!}$, then

$$
\left|R_{n}(x)\right|=e^{\Theta x} \cdot A \cdot \frac{|\ln a|^{p}}{\left(n_{0}+1\right) \cdot \cdots \cdot\left(n_{0}+p\right)}<e^{\Theta x} \cdot A \cdot\left(\frac{1}{2}\right)^{p}
$$

As $p \rightarrow \infty$ and also $n \rightarrow \infty$, it follows from the previous that for any $x$:

$$
\lim _{n \rightarrow \infty} R_{n}(x)=\lim _{n \rightarrow \infty}\left|R_{n}(x)\right|=0
$$

By Theorem 4.26, the formal power series converges to the function $a^{x}$. We obtain the expansion of this function in a power series at the point $x_{0}=0$.

$$
a^{x}=\sum_{n=0}^{\infty} \frac{\ln ^{n} a}{n!} x^{n}
$$

To determine the radius of convergence of the obtained power series, we apply the D'Alembert formula from Theorem 4.23:

$$
R=\lim _{n \rightarrow \infty}\left|\frac{a_{n}}{a_{n+1}}\right|=\lim _{n \rightarrow \infty} \frac{|\ln a|^{n}}{n!}: \frac{|\ln a|^{n+1}}{(n+1)!}=\infty
$$

Therefore, the series converges for all real values of $x$.