How do I calculate $2^\pi$?

Question

We are taught that $$2^3 = 8$$ Because $$ 2\times2\times2 = 8$$ However, how would we calculate something like: $$2^\pi = \space?$$

Answer 1

Hint: use that $$2^\pi=e^{\pi\ln(2)}$$

Answer 2

There is a sequence of rational numbers $a_1,a_2,...$ that converge to $\pi$. Then $$ \displaystyle 2^{\pi} = 2^{\lim_{n\to \infty} a_n}$$

and you know how to calculate $2^{\rm rational\;number}$

Answer 3

Irrational exponents can be expressed as finding increasingly accurate rational approximations of the exponent. For our purposes, we shall use estimates derived from the continued fraction of $\pi$: $$3, \frac{22}{7}, \frac{333}{106}, \frac{355}{113}, \frac{103993}{33102}, \frac{104348}{33215}, \dots$$ This gives the estimates $$2^3 = 8$$ $$2^{22/7} = \sqrt[7]{2^{22}} = 8.832716109\dots$$ $$2^{333/106} = \sqrt[106]{2^{333}} = 8.824468786\dots$$ $$2^{355/113} = \sqrt[113]{2^{355}} = 8.824979458\dots$$ $$2^{103993/33102} = \sqrt[33102]{2^{103993}} = 8.824977823\dots$$ $$2^{104348/33215} = \sqrt[33215]{2^{104348}} = 8.824977829\dots$$ $$\vdots$$ $$2^\pi = 8.824977827\dots$$ Alternatively, in the case that we do not have a continued fraction, we could instead just use an increasing number of digits of $\pi$ to approximate this: $$3, \frac{31}{10}, \frac{314}{100}, \frac{3141}{1000}, \frac{31415}{10000}, \frac{314159}{100000}, \dots$$ This does converges at a slower rate, but still gives the same value. $$2^3 = 8$$ $$2^{31/10} = \sqrt[10]{2^{31}} = 8.574187700\dots$$ $$2^{314/100} = \sqrt[100]{2^{314}} = 8.815240927\dots$$ $$2^{3141/1000} = \sqrt[1000]{2^{3141}} = 8.821353304\dots$$ $$2^{31415/10000} = \sqrt[10000]{2^{31415}} = 8.8244110824\dots$$ $$2^{314159/100000} = \sqrt[100000]{2^{314159}} = 8.824961595\dots$$ $$\vdots$$ $$2^\pi = 8.824977827\dots$$

Answer 4

Knowing the value of $\pi$, then $$2^\pi=2^3 \times 2^{\pi-3}=8\times 2^{\pi-3}=8\times e^{(\pi-3)\log(2)}$$ Now, using $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$ with $x=(\pi-3)\log(2)$ would converge quite fast. Adding terms up to $n=p$,we should get $$\left( \begin{array}{cc} p & \text{value} \\ 0 & 8.00000000000000 \\ 1 & 8.78515638899013 \\ 2 & 8.82368579868838 \\ 3 & 8.82494628252957 \\ 4 & 8.82497720993398 \\ 5 & 8.82497781700521 \\ 6 & 8.82497782693533 \\ 7 & 8.82497782707456 \\ 8 & 8.82497782707627 \\ 9 & 8.82497782707629 \end{array} \right)$$ which is the correct value for fifteen significant figures.