How do you actually calculate numbers like $2^{\pi}$ or $\sqrt[\leftroot{-2}\uproot{2}160]{2}$

Question

My question is very simple, but I started to wonder how does one calculate numbers like $2^{\pi}$ or $\sqrt[\leftroot{-2}\uproot{2}160]{2}$?

For example I know that:

$$2^{3/2} = \sqrt{2^3}=\sqrt{8}\approx2.83,$$

which is easy to calculate. But what about the cases I gave as an example? How would one go about and calculate those numbers using nothing else but a pencil and paper without calculator allowed?

Would we use some kind of series here to approximate these numbers?

It is also a bit unclear to me what $2^{\pi}$ means. For example to me, $2^3$ means in words: "Multiply number $2$ three times by itself", so multiplying $2$ by $\pi$ times by itself feels a bit weird when you're used to having integers in the exponent.

Thank you for any help and clarifications :)

Answer 1

Let $a>0$.

1) One starts defining $a^n$ for $n$ integer: $$ a^0:=1,\quad a^{n+1}:=a\cdot a^n. $$ 2) Then one defines $y=a^{1/n}$ as the unique positive solution of $y^n=a$.

3) Next $$ a^{m/n}=\left(a^{1/n}\right)^m, $$ obtaining a general definition of power with rational exponent.

4) Finally, for any $x\in \mathbb R$, one can find $q_i=m_i/n_i\in \mathbb Q$ such that $q_i\to x$. Therefore one defines $$ a^x=\lim_ia^{q_i}. $$ This is well defined because one can prove that the result is independent of the approximating sequence (continuity) i.e. $$ \lim_iq_i=\lim_iq'_i\Rightarrow \lim_ia^{q_i}=\lim_ia^{q'_i} $$ (This proof is not difficult, being the exponential monotone).

Answer 2

High school example. The binomial theorem.

$$(1+x)^n=1+nx + \frac{n(n-1)x^2}{2!}+\frac{n(n-1)x^3}{3!}+\ldots$$ here, $x=1$ and $n=\pi$ thus \begin{align} (1+1)^\pi &= 1 + \pi.1 + \frac{\pi(\pi-1)}{2}.1^2+\ldots \\ &= 1+ \pi + \frac{\pi^2-\pi}{2}+\ldots \\ &= 1+\frac{\pi}{2}+\frac{\pi^2}{2}\\ &\approx 1 + \frac{3}{2}+\frac{9}{2}\\ &= 7 \end{align} admittedly a very very crude approximation.

Answer 3

Consider $$2^\pi=2^3\times 2^{\pi-3}=8\times 2^{\pi-3}=8e^{(\pi-3)\log(2)}$$ Now, let us use the truncated series $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)$$ and use $(\pi-3)\log(2)\approx 0.0981445\approx \frac 1{10}$. So $$e^{(\pi-3)\log(2)}\approx 1+\frac 1{10}+\frac 1{200}+\frac 1{6000}=\frac{6631}{6000}$$ So, $$2^\pi\approx \frac{8 \times 6631}{6000}=\frac{6631}{750}\approx 8.84133$$ while the "exact" value should be $8.82498$.

For sure, this implies that you know the value of $\pi$ and the value of $\log(2)$.

For the second one $$\sqrt[\leftroot{-2}\uproot{2}160]{2}=e^{\frac{\log(2)}{160}}\approx e^{\frac{43}{1000}}$$ and use again the development of $e^x$ for two terms (I suppose).

Answer 4

How I would think about $2^{\pi}$ is the following:

Consider an infinite sequence $(a_{0},a_{1},a_{2}\ldots)$ such that: $$\sum_{n=0}^{\infty}a_{n}=\pi$$ Then $$2^{\pi}=2^{\sum_{n=0}^{\infty}a_{n}}=2^{a_0+a_1+a_2+\cdots}=2^{a_{0}}2^{a_{1}}2^{a_{2}}\cdots$$ Now $2^{\pi}$ can be represented as a infinite product: $$2^{\pi}=\prod_{n=0}^{\infty}2^{a_{n}}$$ There are infinite number of equations for $a_{n}$ but here's a link for a few, if you want to know them. http://mathworld.wolfram.com/PiFormulas.html