# Equivalence of all representations of $\exp$

I can name at least 4 different ways of representing $$\exp$$ function:

1. Taylor series: For $$x \in \mathbb{R}, \exp(x) = \sum_{k=0}^{\infty} \frac{x^k}{k!}$$.
2. Differential equation: $$f: \mathbb{R} \to \mathbb{R}$$ differentiable with $$f'(x) = f(x)$$ and $$f(0)=1$$.
3. Inverse function of $$\ln(x) = \int_1^x \frac{dt}{t}$$ for $$x>0$$.
4. Exponent: The number $$e$$ (defined e.g. as $$\sum_{k=0}^{\infty} \frac{1}{k!}$$) raised to the power $$x$$, for $$x \in \mathbb{R}$$

I managed to proved equivalence among the first $$3$$ but I am a bit puzzled by $$4.$$.

An easy way would be to look at $$a^b = \exp(b \ln(a))$$. But I am not sure that is meaningful.

Is there any other way of defining $$a^b$$ without involving $$\exp$$ that would give a more meaningful answer? Or how would you approach proving that 4. is equivalent to 1-3?

• You've already noted the problem with definition 4, which is that $a^b$ hasn't been defined for $b\notin\mathbb{Z}$. What you may want to show is that $e^k = \exp(k)$ for $k\in\mathbb{Z}$. – AlexanderJ93 Nov 16 '18 at 8:09

You could define $$\exp :\mathbb{R}\to \mathbb{R}$$ to be a continuous function that satisfies $$\exp\left(\frac{p}{q}\right) = \sqrt[q]{e^{p}}$$ for all $$p$$, $$q\in \mathbb{Z}$$, then show that such a function exists and is unique.

If you're interested, here are are two other ways of defining $$\exp$$:

1.$$\exp (x) = \lim_{n\to \infty} \left( 1 +\frac{x}{n}\right)^n$$

2.A continuous function $$\exp : \mathbb{R} \to \mathbb{R}$$ that satisfies $$\exp '(0) = 1$$ $$\exp(x+y) = \exp(x)\exp{y}$$ for all $$x,y\in \mathbb{R}$$. Of course you'll have to prove existance and uniqueness.

The only problem in defining $$a^b$$ is when $$b$$ is irrational. This can be handled as follows.

Let $$b_n$$ be a sequence of rationals tending to $$b$$ and we can define $$a^b=\lim_{n\to\infty} a^{b_n}$$. This approach is slightly difficult and presented here.

• I like that one, exactly what I was looking for. Thanks! – Othman Nejjar Nov 16 '18 at 9:42