Show that $f(x)=e^x$

Question

In this case $f(x)=1+x+x^2/2!+x^3/3!+x^4/4! + ... = \sum_{n=0}^\infty \frac{x^n}{n!}$.

I understand it conceptually in terms of the Taylor series, but I have no idea how to prove it rigorously.

Answer 1

You have: $$f(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$ and $f(0)=1$. Also, $$f'(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots =f(x)$$ This is a simple differential equation with the solution: $$f(x)=Ae^x$$ From the condition $f(0)=1$, you get $A=1$ i.e $f(x)=e^x$.

$\blacksquare$

Answer 2

By the binomial theorem, $$\begin{align}(1+\frac xp)^p&=\sum_{n=0}^p\frac{p!}{(p-n)!\ n!}\frac{x^n}{p^n}\\&=\sum_{n=0}^p\frac{p!}{(p-n)!\ p^n}\frac{x^n}{n!}\\&=\sum_{n=0}^p\frac pp\frac{p-1}p\frac{p-2}p...\frac{p-n+1}p\frac{x^n}{n!}.\end{align}$$ In the limit when $p\rightarrow\infty$, every $n$-th term tends to $\frac{x^n}{n!}$ so that $$\lim_{p\rightarrow\infty}{(1+\frac xp)}^p=\sum_{n=0}^{\infty}\frac{x^n}{p^n}$$

As $\frac{t_n}{t_{n-1}}=\frac{\frac{x^n}{n!}}{\frac{x^{n-1}}{(n-1)!}}=\frac xn$, convergence is guaranteed everywhere.

On another hand, we have for $x\neq 0$ $$\lim_{p\rightarrow\infty}{(1+\frac xp)}^p=\lim_{p\rightarrow\infty}{(1+\frac xp)}^{\frac pxx}=[\lim_{\frac px\rightarrow\infty}{(1+\frac xp)}^{\frac px}]^x=e^x.$$

Also observe that $$\frac d{dx}\lim_{p\rightarrow\infty}{(1+\frac xp)}^p=\lim_{p\rightarrow\infty}{\frac d{dx}(1+\frac xp)}^p=\lim_{p\rightarrow\infty}{\frac pp(1+\frac xp)}^{p-1}=\lim_{p\rightarrow\infty}{(1+\frac xp)}^p.$$