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I was just wondering, what does $x^\pi$ or for that matter, $x$ raised to any irrational number mean? For example, I want to represent $x^2$ then that would mean $x * x$ or if I want to do $x^\frac{2}{3}$ then that would mean, I first square it then cube root it but what does that mean for irrantional numbers such as $\pi$?


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marked as duplicate by leonbloy, Douglas S. Stones, Start wearing purple, Amzoti, Lord_Farin May 14 '13 at 14:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If $0< x$ we have: $$x^\pi=\sup\{x^r\mid r<\pi, r\text { is a rational number} \}$$

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From a very abstract viewpoint, the position $$x^\pi = e^{\pi \log x}$$ is a circular definition: after all $e$ is as much a real number as $x$. There are two ways to break this cirle:

  1. $e^{\pi \log x}$ means $\exp (\pi \log x)$, provided you have defined the exponential function in an indipendent way;
  2. we study the theory real numbers, which includes the construction of $b^x$ for $b>0$ and $x \in \mathbb{R}$. This is not trivial at all, as can be seen in W. Rudin's Principle of mathematical analysis.
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Yes. For example in may textbooks, $\log x = \int_1^x dt/t$ is a definition, then exp is the inverse function. – GEdgar May 14 '13 at 14:29
Of course we should check if the Riemann integral can be defined without any use of powers with real-valued exponents. – Siminore May 14 '13 at 14:48

It's better to think of it as rather

$$x^{\pi} = e^{\pi \log{x}}$$

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What is the rational behind that? – Jeel Shah May 14 '13 at 14:07
What is the analytics behind that? – NikolajK May 14 '13 at 14:08
Many times we wish to extend concepts that make sense only for integers. Maybe you know of $n!$ and $\Gamma(n+1)$, for example. The log is the bridge between the integers and the reals, or even complex numbers. – Ron Gordon May 14 '13 at 14:08
@NickKidman: Maybe I misunderstand your question. But, as far as I do, $x = e^{\log{x}}$. – Ron Gordon May 14 '13 at 14:09

Ultimately, it's nothing near so nice as the rational case. If we're given an irrational number $\alpha$, then $$x^\alpha:=\underset{r\to\alpha}{\lim_{r\in\Bbb Q}}x^r,$$ so long as this limit is defined (for example, it doesn't work when $x$ is negative, and it doesn't work when $x=0$ and $\alpha$ is negative).

Alternately, it is at times more beneficial to define it as $$x^\alpha:=e^{\alpha\ln x}.$$ These are equivalent definitions for $x>0$. We can even do a bit better and define $$x^\alpha:=\lim_{t\to x^+}e^{\alpha\ln t}.$$ That definition works precisely when the original definition does--namely, whenever $x>0$, and whenever $x=0$ and $\alpha>0$. Of course, both of the latter two definitions do require some independent definition of $e^w$, such as $$e^w:=\lim_{n\to\infty}\left(1+\frac wn\right)^n.$$

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Your answer is great for $x>0$. – GEdgar May 14 '13 at 14:28
Is there one that works for $x<0$? – Cameron Buie May 14 '13 at 18:36
The standard one is $x^a=\exp(a\log x)$ for nonzero complex $x$. But it is multivalued (since log is). Even when $x<0$ and $a$ are real, it is likely a non-real complex number. – GEdgar May 14 '13 at 19:00
Fair enough. I was trying to "keep it real". ;-) – Cameron Buie May 14 '13 at 20:04

Ron already gave the canonical answer: $x^\alpha = e^{\alpha \log{x}}$ by definition, and then you may refer to the very nice properties of $e^x$. If you prefer, think at two approximating series of rational numbers $\{a_n\}$ and $\{b_n\}$ converging from above and from below to $\pi$, compute $\{x^{a_n}\}$ and $\{x^{b_n}\}$ and see that they actually converge to a single value.

At least as far as I am concerned, there is not a lot of difference asking what $x^\alpha$ is and asking what an irrational number $\alpha$ is.

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