Number raised to power of irrational number

Question

What is the consequence of raising a number to the power of irrational number?

Ex: $2^\pi , 5^\sqrt2$

1. Does this mathematically makes sense? (Are there any problems in physics world where we encounter such a calculation?)
2. How does one calculate or maybe estimate its value ? (I want to know if there is an infinite summation formula, instead of simply rounding $\pi$ to 3.14)

Answer 1

Formally, we have $a^b = e^{b \ln(a)}$ and

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$

$$\ln x = \int_1^x \frac{dt}{t}$$

And for integer $n$, we define $x^n$ as

$$\prod^n_{i=1} x$$

This is needed because we don't want to define the powers in $e^x$ circulary.

Also note that since we use $\ln a$ in this definition, we must have $a>0$.

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You can also just approximate the exponent with a rational number.

A good approximation for $\pi$ is $\frac{355}{113}$.

Answer 2

Observe that $$ a^r=e^{r\ln a} $$ and use the Taylor series for $e^x$.

Answer 3

Let $$x=a.a_1a_2a_3... and$$ $$y=b.b_1b_2d_3...>0$$. You can approximate x^y as a limit of the sequence $$a^b,a.a_1^{(b.b_1)},a,a_1a_2^{(b.b_1b_2)},....$$ As for physics application suppose you are dealing with an experiment which is governed by the differential equation $$yy'-(y')^2=(y^2)/x$$ One of the solutions is $$x^x$$ and if $$x=2^.5$$ then you get to evaluate an expression of the type you have mentioned.

Answer 4

If the number is positive, then raising it to the power of an irrational number is well defined. This is because an irrational number can be defined as a converging sequence of rational numbers (like 3, 3.1, 3.14, 3.141, 3.1415 etc), and as these approach the irrational number then the power of these rational numbers also converges to a fixed value, which is the power of the irrational number.

If the number is negative, then the power is not defined. This is because whilst the irrational number can be defined a converging sequence of rational numbers, the power of these numbers does not converge.