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This is motivated by an earlier question of mine, in which I realized I was never really presented a definition of $e^x$, or more generally, what it means to raise a (positive) real number to an irrational power.

I know that the definition of $a^b$ with $a \in \mathbb{R}^+, b \in \mathbb{Q}$ is pretty straightforward in terms of repeated multiplication and the property that $a^{bc}=(a^b)^c$. I also know that one can define $a^b$ where $b \in \mathbb{R} - \mathbb{Q}$ using limits. This is stated, for example, in this Math.SE question.

Other way to define exponentiation with real powers is with the function $\exp(x)$ or $e^x$, which has many equivalent definitions. For example, one may define it as $e^x = \lim\limits_{n \to \infty} (1+\frac{x}{n})^n$, or as the unique solution to $y' = y$ with $y(0)=1$. Wikipedia has a whole page stating these definitions and showing that they are equivalent to each other.

What I haven't seen is a proof that this new $e^x$ behaves just like the old way of doing exponentiation when $x \in \mathbb{Q}$. If I were to guess, I'd say it's related to Wikipedia's fifth defintion: it is the unique (with some conditions) function that satisfies $f(1) = e$ and $f(x+y)=f(x)f(y)$. However, that defintion seems to involve more advanced concepts than the other ones, concepts which I don't really understand right now.

Is there a proof of the fact that $\exp(x)$ is equivalent to the definition of exponentiation for rational powers?

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What do you find advanced about $f(1)=e$ and $f(x+y)=f(x)f(y)$? I have always looked at it in terms of being a reduction to the elementary law of exponents. $f(x+y)=e^{x+y}=e^{x}e^{y}=f(x)f(y)$ It is very possible my understanding is too elementary, however. – 000 Jun 4 '12 at 2:46
@Limitless: The advanced part is showing that such a function is unique. Wikipedia doesn't give such a proof, and makes some statements that I don't know about, such as the fact that it must be Lebesgue measurable. – Javier Jun 4 '12 at 2:49
I completely understand why you feel it's advanced now. I recall a proof demonstrating it's unique, but I don't recall the illuminating reason. – 000 Jun 4 '12 at 3:25
up vote 5 down vote accepted

Let $\exp(x)$ denote the exponential function. You can define this any way that you like, but we will assume the following facts:

Fact 1. The derivative of $\exp(x)$ is $\exp(x)$, and $\exp(0)=1$.

Fact 2. Let $f(x)$ be any differentiable function. If $f(0) = 1$ and $f'(x) = f(x)$ for all $x\in\mathbb{R}$, then $f(x) = \exp(x)$ for all $x\in\mathbb{R}$.

We will also assume the Power Rule for rational exponents. From this, we can prove the following theorem:

Theorem. Let $e = \exp(1)$. Then $e^q = \exp(q)$ for any rational number $q$.

Proof: Let $q\in\mathbb{Q}$, and let $f\colon\mathbb{R}\to\mathbb{R}$ be the function $f(x) = [\exp(x/q)]^q$. Note that $f(0) = 1^q = 1$. Furthermore, by the Power Rule and the Chain Rule, we have $$ f'(x) \;=\; q[\exp(x/q)]^{q-1} \exp(x/q)\, (1/q) \;=\; [\exp(x/q)]^q \;=\; f(x) $$ It follows that $f(x) =\exp(x)$ for all $x\in\mathbb{R}$, so $$ \exp(q) \;=\; f(q) \;=\; [\exp(q/q)]^q \;=\; e^q.\tag*{$\square$} $$

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Let's use the definition $$e^x=\lim_{n\to\infty}\left(1+{x\over n}\right)^n$$ and prove $e^{pq}=(e^p)^q$.

We have $$e^{pq}=\lim_{n\to\infty}\left(1+{pq\over n}\right)^n$$ and $$(e^p)^q=\left(\lim_{n\to\infty}\left(1+{p\over n}\right)^n\right)^q=\lim_{n\to\infty}\left(1+{p\over n}\right)^{qn}=\lim_{n\to\infty}\left(1+{pq\over qn}\right)^{qn}=\lim_{m\to\infty}\left(1+{pq\over m}\right)^m$$ which is the same thing.

But we've now proved the property that you used to define rational powers.

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Yes, we used that property to define rational powers, but we also knew how to define integer powers and roots, and we used $a^{b^c}=a^{bc}$ to define rational powers in terms of those. – Javier Jun 4 '12 at 3:11
First of all, don't write $a^{b^c}$ when what you mean is $(a^b)^c$. To show for positive integer $p$ you get $e^p$ by repeated multiplication, do $(\lim(1+(1/n))^n)^p=\lim(1+(1/n))^{np}=\lim(1+(p/np))^{np}=e^p$. – Gerry Myerson Jun 4 '12 at 6:36
Yeah, sorry about that, fixed it. I think I liked Jim Belk's answer a little more, but thanks for your help! – Javier Jun 4 '12 at 14:09

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