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The definition of the exponential with integer exponents is straightforward to define: $x^n=\underbrace{x\cdot\ldots\cdot x}_{n-\text{times}}$.

These days I've been thinking about the formal definition of the exponential with real exponents.

My question:

What is the definition of $a^{b}$, where $a,b\in \mathbb R$.


marked as duplicate by Travis, skyking, Lee Mosher, Benjamin Dickman, tilper Apr 25 '17 at 12:58

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  • $\begingroup$ I've edited your title and a bit of your question to reflect what I think you're really asking. $\Bbb R$ is real numbers. If you really wanted to ask about rational exponents, that would be $a, b \in \Bbb Q$. But I assumed you were asking about real exponents because that's more interesting. Short answer is they're defined in terms of limits, if I recall correctly. But I'll let someone else provide a more detailed answer.. $\endgroup$ – tilper Apr 25 '17 at 11:37
  • $\begingroup$ Dan's answer to the duplicate question should have what you need. It's not the accepted answer but it does have more rigor. $\endgroup$ – tilper Apr 25 '17 at 12:58
  • $\begingroup$ @tilper so following his definition $4^{\frac{1}{2}}=2$, instead of $\pm2$, is that right? $\endgroup$ – user42912 Apr 25 '17 at 13:26
  • $\begingroup$ I think technically $-2$ will also satisfy the description he laid out, since the 2nd power of $-2$ is $(-2)^2 = 4$. But it is a convention that for $b > 0$ we take $b^{1/n}$ to always be positive and $-b^{1/n}$ to always be negative. He may have been sweeping that under the rug in order to avoid getting too sidetracked. Also, in general there are $n$ numbers whose $n$th power is $b$. But sometimes (most of the time?) not all of these numbers are rational or even real. For example, there are three distinct numbers $x$ such that $x^3 = 1$. One of them is $1$, and two of them are imaginary. $\endgroup$ – tilper Apr 25 '17 at 13:33

If both $a,b\in \mathbb R$ then we have to define $a^b$ by means of complex functions:

\begin{align*} a^b = e^{bLn(a) } = e^{b\ln(|a|)+i b\arg(a)+2\pi pb i} , \end{align*} $p \in \mathbb Z$. Now the exponential function is defined by the series: \begin{align*} e^x = \sum_{n=0}^\infty \frac{x^n}{n!}. \end{align*}

Noting that $\frac{d}{dx}e^x = e^x >0$ (not as easy as it sounds) we know that $f(x) = e^x$ is monotonically increasing on $\mathbb R$. Thus it has an inverse which is called $\ln$. Now this defines $a^b$. Calculating an approximate value of say $(-2)^\pi$ from this is hard (or at least boring).

  • $\begingroup$ You only have to introduce complex numbers if $a$ is negative, otherwise it's perfectly feasible to define it without them (normally the real non-negativ solution is taken then). Furthermore your definition lacks the definition of $\ln$ so it just pushes the problem onto another problem... $\endgroup$ – skyking Apr 25 '17 at 12:30
  • $\begingroup$ I'm aware of your point. But $a \in \mathbb R$. Furthermore I did define $\ln$ as the inverse of $\exp$ on the reals which suffice. $\endgroup$ – Martin Apr 25 '17 at 12:33

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