Approximation of real number to the power of irrational

Question

Say I have two numbers, $c\in\Bbb R^+\,,q\in\Bbb{R\setminus Q}$.

Is there a way to approximate the value of $c^q$?

This question bugging me because of how simple it sound in compare to how complicated it is. First I want to understand what a power of irrational number is, when talking about $\Bbb R^+$(I use this domain for sake of simplicity) we have very well defined actions for power a rational number, we can even use some theorems to get $c^{a/b}=\sqrt[b]{c^a}$, but when I am dealing with irrational number I can not do this.

I know that we have some definition for this because the solution of the equation $2^x=3$ is irrational.

So what exactly does it mean to take something to a irrational power and is there a way to approximate a numeric value for this?

Answer 1

Before I answer your question, let me ask another question: how do you compute $\pi\times e$? How do you multiply irrational numbers? If multiplication is understood as repeated addition, how can you add a thing $\pi$ times?

I submit that real exponents are no different than real multipliers.

By definition, real numbers are numbers that can be approximated by rationals. For example, $\sqrt{2}$ is the real number defined by successively better approximating rationals $1,1.4,1.41,1.414,\cdots.$ Then the multiplication $\sqrt{2}\times 2$ is defined as the real number approximated by $1\times 2,1.4\times 2,1.41\times 2,1.414\times 2,\cdots,=2,2.8,2.82,2.828,\cdots.$ Hence we conclude that $\sqrt{2}\times 2\approx 2.828,$ by doing the computation on the approximating rationals.

Similarly powers with real exponents are defined as numbers approximated by corresponding powers with rational exponents.

For example, to compute $2^{\sqrt{2}}$, we just raise $2$ to the power of each approximating rational. So our sequence approximating $2^{\sqrt{2}}$ is $2^1,2^{1.4},2^{1.41},2^{1.414},\cdots = 2,2.64,2.657,2.6647,\cdots.$ So we say that $2^{\sqrt{2}}\approx 2.6647.$

For another example, I'll use the proposed example from your question. Let $x$ be the real number represented by the sequence of rationals $1, 1.5, 1.58, 1.584, 1.5849,\dotsc$ seems to converge to a real number. No repeating decimal places, so it’s an irrational number.

Let’s exponentiate it. Then $2^x$ is the sequence $2^1, 2^{1.5}, 2^{1.58}, 2^{1.584}, 2^{1.5849},\dotsc$ and we get $2, 2.83, 2.99, 2.998, 3.00,\dotsc$

The latter sequence approaches the real number $3$, hence the former sequence $x$ is the real solution to the equation $2^x=3.$ Or in short, $x=\log_23.$

And that's what it means to exponentiate a real. Just as every real may be represented by approximating rationals, the exponential of a real is approximated by the exponentiation of its approximating rationals.

These approximating rationals can either be understood as Cauchy sequences, Dedekind cuts, or plain old strings of decimal expansions.

Answer 2

The simplest way to define what it means to take a real power of a real number is to define $$\exp(t)=\sum_{k=0}^\infty \frac{t^k}{k!},$$to define the logarithm of a positive number in any of various ways, and then define $$c^p=\exp(p\log(c))\quad(c>0, p\in\mathbb R).$$

Of course we can approximate this. One way to define $\log(c)$ is $\int_1^c\frac{dt}t$; you can use standard methods for approximating integrals. Then you can approximate $\exp(p\log(c))$ by taking finitely many terms of that power series.