Prove $\exp(x) = e^x$

Question

Definition $$\exp(x) := \sum^\infty_{n = 0} \frac{x ^ n}{n!}$$ and $$e := \exp(1)$$

Proposition $$\exp(x) = e^x$$ for all $x \in \mathbb{R}$

I have proved $\exp(x + y) = \exp(x)\exp(y)$ with binomial theorem and Fubini's theorem. And $\exp(0) = 1$. So the proposition holds for all natural numbers (induction).

Similarly $\exp(-x)\exp(x) = \exp(x-x) = 1$, which implies $\exp(-x) = 1 / \exp(x)$. So the proposition holds for integers as well.

How do I prove it for rational numbers? I think I have to prove that $\exp(ab) = \exp(a)^b$ but have no clue how.

Answer 1

You have already proved that $\exp(x+y) = \exp(x) \exp(y)$. We now find $\exp (\frac{1}{n})$ for all integers $n$. Note that $\exp(1) = \exp(\frac{1}{n})\exp(\frac{n-1}{n}) = \exp(\frac{1}{n})\exp(\frac{1}{n})\exp(\frac{n-2}{n}) = \ldots = \exp(\frac{1}{n})^n$, so we have $\exp(\frac{1}{n}) = e^\frac{1}{n}$. From here, you can simply induct to attain $\exp(\frac{m}{n}) = e^\frac{m}{n}$ for all integers $m, n$, which proves the rational case and allows you to finish by continuity.

Answer 2

Define $f(x)=\exp(a+b-x)\cdot\exp(x)$. Verify that $f’$ vanishes, from where we conclude that $f$ is constant, hence $f(0)=f(b)$ what was to be shown.