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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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    $\begingroup$ I am curious, why do you want to calculate this by hand? $\endgroup$ – imranfat Sep 14 '13 at 3:43
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    $\begingroup$ @imranfat: Just out of curiosity! $\endgroup$ – user66733 Sep 14 '13 at 3:44
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    $\begingroup$ We have seen many biases against irrationals in numeric computations. I can calculate $\pi$ to many places easily, so if I want to calculate an expression involving $\pi$, it is as good as a rational. If we want a constant to a given accuracy, here $2^{\sqrt 2}$, any expression that gets us there is a good one. If it involves irrationals, we need a good enough approximation to keep the accuracy. $\endgroup$ – Ross Millikan Sep 14 '13 at 4:40
  • $\begingroup$ A favorite challenge of mine is that many know $\sqrt 2 \approx 1.414$. Get two more places by hand (no calculator). Even better, mentally (no paper). $\endgroup$ – Ross Millikan Sep 14 '13 at 4:48
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    $\begingroup$ But there is a much easier way. $1.414=1.4*1.01$, so $1.414^2=1.96*1.0201= 1.96+.0392+.000196= 1.999396$ so we need to increase the square by $0.000604$, so the square root by $0.00015*1.414 \approx 0.00021$, so the square root is $1.41421$. I'm not saying you should find this specific one, but working on approximation formulas is productive compared to computation. $\endgroup$ – Ross Millikan Sep 14 '13 at 5:31
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${\large\mbox{Hint}}$:

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

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    $\begingroup$ That's a nice hint. I like it. But the infinite series we'll obtain by using this approximation in the exponent will not converge exactly to $2^{\sqrt{2}}$, but to an approximation of it. $\endgroup$ – user66733 Sep 14 '13 at 4:19
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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.

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  • $\begingroup$ Well, I didn't try to define "efficiently" to avoid complications. But we don't really need to get that rigorous for this question. Do we? Do I really need to define the word "calculate"? Moreover, why do you say that we will need to involve an irrational as the result is irrational? There are many converging infinite series for $\pi$ or $e$ that have only rational terms, and both $\pi$ and $e$ are transcendental, let alone rational. $\endgroup$ – user66733 Sep 14 '13 at 3:58
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    $\begingroup$ I don't have a series at hand that will calculate $2^{\sqrt 2}$ from rationals. As the rationals are dense in the reals, we are guaranteed that there is one. I think the expression I gave is rather efficient-you can use Newton's method to double the digits of $\sqrt 2$ each time and the argument of the exponential is less than $0.06$ so the Taylor series will converge quickly. Once you use this approach, every step is rationals. $\endgroup$ – Ross Millikan Sep 14 '13 at 4:04
  • $\begingroup$ There is definitely one, even without use of analysis. Namely, the series that is obtained by writing the decimal expansion of $2^{\sqrt{2}}$, but that series doesn't really help us because it requires us to know the decimal expansion of $2^{\sqrt{2}}$ at first place. I can reword my question into something like this: Does there exist an infinite series with rational terms that converges to $2^{\sqrt{2}}$ but it is not equal to the trivial case where the series is obtained from its decimal expansion? $\endgroup$ – user66733 Sep 14 '13 at 4:09
  • $\begingroup$ Defining "calculate" is about the acceptable tools. May I use Alpha? I can get thousands of decimals easily that way. If not, some work in Excel can also get lots of places-you can write multiple precision calculations there-I have. The old time calculators had a lot of energy-William Shanks got $707$ places of $\pi, 527$ of which were correct. $\endgroup$ – Ross Millikan Sep 14 '13 at 4:10
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    $\begingroup$ A famous quote from Abraham Lincoln "Give me six hours to chop down a tree and I will spend the first four sharpening the axe." If you want to do this kind of thing, you need to work hard at finding expressions that converge quickly. Quadratic is the minimum. My exponential has a nice small argument, so will converge very quickly, but I don't have a good high accuracy $\log 2$. I would start there for improvement. $\endgroup$ – Ross Millikan Sep 14 '13 at 4:23
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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.

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  • $\begingroup$ Sorry for interrupting, but aren't you missing $\log(2)$ somewhere in the first equation? $\endgroup$ – stressed out Jul 6 '18 at 0:12
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    $\begingroup$ @stressedout - you're reading the expression as a Taylor series for $2^x$, but it's actually a Taylor series for $(1+x)^{\sqrt{2}}$, where $x=1$. $\endgroup$ – Glen O Jul 9 '18 at 7:48
  • $\begingroup$ Oh, I see. Thanks. But isn't $(1+x)^{\alpha}$ convergent only for $|x| < 1$? How do you show that it's convergent for $x=1$ too? Is it convergent at all? $\endgroup$ – stressed out Jul 9 '18 at 7:56
  • $\begingroup$ @stressedout - it's not convergent only for $|x|<1$, but it's always convergent for $|x|<1$. For $|x|=1$, convergence isn't guaranteed, but it also isn't guaranteed not to. For $(1+x)^\sqrt{2}$, it converges for $x=1$, as shown by the alternating series test (indeed, it converges for the same reason that the alternating harmonic series converges). $\endgroup$ – Glen O Jul 9 '18 at 15:53
  • $\begingroup$ Yeah. It becomes an alternating series. Cool. $\endgroup$ – stressed out Jul 9 '18 at 16:09
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$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).

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    $\begingroup$ How's using tables, calculators or slide rules "calculating by hand"? $\endgroup$ – DonAntonio Sep 14 '13 at 11:28
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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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    $\begingroup$ The problem with this approach is that powers of $\log (2)$ appear in the coefficients of the Taylor polynomials. $\endgroup$ – Pipicito Sep 14 '13 at 3:53
  • $\begingroup$ @Pipicito that can be dealt with -- see edits. $\endgroup$ – Betty Mock Sep 15 '13 at 21:10

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