How to show that the infinite sum of $\frac{x^n}{n!}$ is $e^x$ without knowledge of the derivative of $e^x$

Question

I am aware that this is the Taylor expansion of $e^x$ but how would people have known that the taylor expansion whose derivative is itself is the exponential function $e^x$.

Answer 1

Hope this will help a lot. Here are answers different from using Taylor expansion.

Why does the sum of the reciprocals of factorials converge to $e$?

Answer 2

HINT: let $f(x):=\sum_{k=0}^\infty\frac{x^k}{k!}$, then start showing, using binomial expansion and absolute convergence, that $f(x+y)=f(x)\cdot f(y)$ for all $x,y\in\Bbb R$.

Then show that $f(1)=e$, hence $f(n)=e^n$ for all $n\in\Bbb N$. Do the same for the cases $f(1/n)$ noticing that $f(n)\cdot f(1/n)=f(1)$ and that $f(-n)=\frac1{f(n)}$.

Then you knows now that $f(q)=e^q$ for all $q\in\Bbb Q$. Now use the locally uniform convergence of the series to show that $f$ is continuous, hence $$f(x)=\lim_{n\to\infty}f(q_n)=\lim_{n\to\infty}e^{q_n}$$ for any sequence of rationals $(q_n)$ that converge to $x$. By last you will need that to show that

$$\lim_{n\to\infty}e^{q_n}=e^x$$