How to calculate $e^\pi$ by hand?

Question

There are some ways to find the values of the form "rational power of rational number" what about nonrecurring numbers? Is there any way to calculate that?

Answer 1

There are many ways you could compute such a number. One way could be to take a close rational approximation: $\pi=\frac{a}{b}$. Your goal would then be to find the number $x$, such that $x=e^{a/b}$. This equation could be rearranged to $x^b-e^a=0$, which you could tackle with root-finding algorithms and an appropriate approximation for $e$.

For example, you could use $\pi\approx\frac{22}{7}$, to get $x^{7}-e^{22}=0$, and then use a sequence such as $x_{n+1}=x_n-\frac{x_n^7-e^{22}}{7x_n^{6}}$. With a perfect approximation for $e^{22}$, this sequence gives you an answer with just $0.13\%$ relative error. Using more precise rational approximations would improve your result but would be harder to compute.

Answer 2

Just to make user190080's point explicit, if $k_0:=\frac{1}{\sqrt{2}},\,k_{n+1}:=\frac{1-\sqrt{1-k_n^2}}{1+\sqrt{1-k_n^2}}$ then $(4/k_{n+1})^{2^{-n}}\to e^\pi$ as $n\to\infty$. The square-rooting can be done with a famous method that's a special case of Newton-Raphson; if $y\approx\sqrt{z}$, a better approximation of $\sqrt{z}$ is $(y+z/y)/2$.

But what is you wanted something other than literally the example of $e^\pi$? You can use the Taylor series of $\exp x$, ensuring the argument is small enough for rapid convergence by using $\exp x = (\exp x/2)^2$ repeatedly.

What if you don't want $\exp x$ but $a^x=\exp (x\ln a)$? You can compute logarithms in a similar way; $\ln (1+x)$ has a Taylor series for $|x|<1$, so use $\ln x=2\ln\sqrt{x}$ if need be. (I discussed square-rooting above.)