I've been trying to calculate $2^{\sqrt{2}}$ by hand efficiently, but whatever I've tried to do so far fails at some point because I need to use many decimals of $\sqrt{2}$ or $\log(2)$ to get a roughly good approximation.
Is it even possible to do so without facing irrational expressions like $\sqrt{2}$ or $\log(2)$ in our calculations?
EDIT
It seems like no one is paying attention to the requirements in my question at all : ( You are not allowed to use use $\log(2)$ or $\sqrt{2}$ in your answers. Use of continued fractions is allowed. Let me phrase my question in this way: Find an infinite series $\displaystyle \sum_{n=0}^{\infty}a_n$ such that $a_n \in \mathbb{Q}$. There exists at least one such series, namely, the series that is obtained by writing the decimal expansion of $2^{\sqrt{2}}$, but that series is good for nothing because if we already knew the decimal expansion of $2^{\sqrt{2}}$ then we didn't need to be after approximating $2^{\sqrt{2}}$ by using infinite series.
Look at the following series:
$\displaystyle e = \sum_{n=0}^{\infty}\frac{1}{n!} = 2 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24}+\frac{1}{120}+\frac{1}{720} + \cdots$
$\displaystyle \pi = \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \frac{4}{9} - \frac{4}{11} + \cdots$
$\displaystyle \pi = 3 + \frac{4}{2\times 3 \times 4} - \frac{4}{ 4 \times 5 \times 6} + \frac{4}{6 \times 7 \times 8} - \frac{4}{8 \times 9 \times 10} + \cdots $
Both $e$ and $\pi$ are irrational transcendental numbers. But we have found non-trivial infinite series with rational terms for them. Can someone possibly find a similar series for $2^{\sqrt{2}}$? This is something I proposed as a challenge to myself and I failed, now I wonder if someone on here could tackle it.
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$\begingroup$${\large\mbox{Hint}}$:
$$ \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} $$
$\endgroup$ 1 $\begingroup$In this question "calculate" and "efficiently" are not well defined. You will need to involve an irrational as the result will be irrational. How many places are you looking for? What resources are acceptable? Alpha will give you $\sqrt 2\approx 1.414213562373095048801688724209698078569671875376948073176679...$and $\log 2\approx 0.693147180559945309417232121458176568075500134360255254120680...$ and more places if you want. For your problem it gives $2^{\sqrt 2} \approx 2.665144142690225188650297249873139848274211313714659492835979...$ If you want truly pencil and paper, I would use $2^{\sqrt 2}=(2\sqrt 2)2^{1.5-\sqrt 2}=(2\sqrt 2)\exp((1.5-\sqrt 2)\log 2)$, evaluate the $\sqrt 2$ by the old-time procedure (see Digit-by-digit calculation here)and the exponential by the Taylor series as the argument is small.
$\endgroup$ 7 $\begingroup$Notice that $$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)!!} $$ (where $n!!$ is the double factorial), and change the order of summation for the product of terms. The specific choice of sum will influence the speed of convergence. Taken directly, the sum is alternating, so convergence acceleration should be particularly useful.
$\endgroup$ 5 $\begingroup$$2^{\sqrt{2}} =e^{(\ln 2)\sqrt{2}} =e^{e^{(\ln\ln 2)+(\ln 2)/2}} $.
Get a table of $\ln(x)$ and $e^x$.
Look up $a = \ln(2)$.
Look up $b = \ln(a)$.
Set $c = b+a/2$.
Set $d = e^c$.
Set $h = e^d$.
If this isn't good enough, I can loan you one of my slide rules (circular or straight).
$\endgroup$ 1 $\begingroup$What you can do is expand the function $2^x$ into its Taylor's series around x = 0 and use the first several terms of the series. Now you are evaluating a polynomial, which is much easier.
The series does not converge uniformly, so the farther x is from 0 the more terms you are going to need. But $ \sqrt 2$ is not all that far out. You can get estimates of the remainder (lots of textbooks discuss these remainders) and see how many terms you need to get the convergence you want.
Noted as per comment below that the coefficients of this series involve powers of log2. However, log2 is the sum of the alternating harmonic series, so some number of terms of that can be used to approximate it.
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