Find the Taylor series of $f(x) =2^x$
Matthew Martinez 
I'm having trouble finding the Taylor series of the function\begin{equation} f(x) = 2^x, x \in \mathbb{R} \end{equation}and showing that it converges to $f(x)$ for all $x \in \mathbb{R}$.
I've found that the derivatives of $f(x)$ are $$f'(x)=2^x\ln(2)$$ $$f''(x)=2^x\ln(2)^2$$ and $$f''(x)=2^x\ln(2)^3$$ and I obviously see the pattern with $f^{(n)}(x)=2^x\ln(2)^n$.
I thought about maybe showing it by induction but I'm trouble finding the Taylor series
$\endgroup$ 101 Answer
$\begingroup$If $f(x)=2^x$, then $f^{(n)}(x)=(\ln 2)^n\cdot 2^x$ and so $f^{(n)}(0)=(\ln2)^n$. Hence, the Taylor series of $f$ about $0$ is\begin{align} \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n&=\sum_{n=0}^{\infty}\frac{(\ln 2)^n}{n!} \\[5pt] &= 1+(\ln2)x+\frac{(\ln 2)^2}{2!}x^2+\frac{(\ln 2)^3}{3!}x^3+\ldots \end{align}But it is simpler to note that for all $t\in\mathbb{R}$,$$ e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\ldots $$Note that if $x\in\mathbb{R}$, then $x\ln2\in\mathbb{R}$, and so we can substitute $x\ln 2$ for $t$:$$ e^{x\ln2}=1+(\ln2)x+\frac{(\ln 2)^2}{2!}x^2+\frac{(\ln 2)^3}{3!}x^3+\ldots $$But since $2^x=e^{x\ln 2}$, we get that for all $x\in\mathbb{R}$,$$ 2^x=1+(\ln2)x+\frac{(\ln 2)^2}{2!}x^2+\frac{(\ln 2)^3}{3!}x^3+\ldots $$The second approach is preferable because not only does it tell us the Taylor series of $2^x$, it also tells us that $2^x$ is equal to its Taylor series for all $x\in\mathbb{R}$. Not every function is equal to its Taylor series. For example, consider the Taylor series of the following function about $0$:$$ \phi(x)=\begin{cases} e^{-1/x^2} &\text{ if $x\neq0$} \\ 0 &\text{ if $x=0$} \, . \end{cases} $$
$\endgroup$
