How to calculate $2^{\sqrt{2}}$ by hand efficiently?

Solution 1:

${\large\mbox{Hint}}$:

$$ \sqrt{\vphantom{\large A}2\,} = \sqrt{98 \over 49} = {1 \over 7}\,\sqrt{\vphantom{\large A}100 - 2} = {10 \over 7}\,\sqrt{\vphantom{\large A}1 - {1 \over 50}} \approx {10 \over 7}\,\left(1 - {1 \over 2}\,{1 \over 50}\right) = {10 \over 7} - {1 \over 70} $$

Solution 2:

In this question "calculate" and "efficiently" are not well defined. You will need to involve an irrational as the result will be irrational. How many places are you looking for? What resources are acceptable? Alpha will give you $\sqrt 2\approx 1.414213562373095048801688724209698078569671875376948073176679...$and $\log 2\approx 0.693147180559945309417232121458176568075500134360255254120680...$ and more places if you want. For your problem it gives $2^{\sqrt 2} \approx 2.665144142690225188650297249873139848274211313714659492835979...$ If you want truly pencil and paper, I would use $2^{\sqrt 2}=(2\sqrt 2)2^{1.5-\sqrt 2}=(2\sqrt 2)\exp((1.5-\sqrt 2)\log 2)$, evaluate the $\sqrt 2$ by the old-time procedure (see Digit-by-digit calculation here)and the exponential by the Taylor series as the argument is small.

Solution 3:

Notice that $$2^\sqrt2=\sum_{n=0}^\infty \frac{a_n}{n!}$$ where $$a_n = \prod_{k=0}^{n-1} (\sqrt2-k)$$

We can express $a_n$ in the form $b_n+c_n\sqrt2$, through the iterative scheme,

\begin{align} b_{n+1} &= 2c_n-nb_n&\qquad c_{n+1} &= b_n-nc_n\\ b_0&=1&\qquad c_0&=0 \end{align}

So we have $$ 2^\sqrt2 = \sum_{n=0}^\infty \frac{b_n}{n!}+\sqrt2\sum_{n=0}^\infty \frac{c_n}{n!} $$ Note that $b_n$ and $c_n$ each grow faster than $n!$; the sums must be evaluated together. From here, you can replace $\sqrt2$ with your choice of infinite sum that evaluates to $\sqrt2$, such as

$$ \sqrt{2} = \sum_{m=0}^\infty (-1)^{m+1}\frac{(2m-3)!!}{(2m)!!} $$ (where $n!!$ is the double factorial), and change the order of summation for the product of terms. The specific choice of sum will influence the speed of convergence. Taken directly, the sum is alternating, so convergence acceleration should be particularly useful.