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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?

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  • $\begingroup$ 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 $k\in\mathbb{Z}$. $\endgroup$ – AlexanderJ93 Nov 16 '18 at 8:09
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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.

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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.

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

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