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Someone on Art of Problem Solving claims to know how to calculate the $2^{2020}$th decimal place of $\sqrt{2},$ and will tell us if everyone gives up. Brute force will not work, nor will a BBP style formula (for the reason that one does not exist, and the ones known so far are for base $2^k$ expansions; $10$ is not a power of $2$).

Is there a feasible solution, or am I being trolled? Trying to search various queries relating to my question online results in nothing except spigot algorithms which spit out the digits one by one from the start. As you may know, they will not work. I am operating under the assumption that if a clever solution exists, it is not a new discovery, for such a simple question has surely been considered before.

Update: The problem poster has promised to post the solution in 2021 if no one finds it before then. This is not the first time they've done this, but every time they've done this in previous years, they have delivered on the promises.

2nd Update: 2 weeks left and I've forgotten to show the source. Soon we'll see if we have been trolled or not.

Display name
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    My suspicion is you are likely being trolled. Such an algorithm will likely be publish-worthy. – Aryabhata Jan 24 '20 at 03:03
  • you only need to know $2^{n+1}$ digits of an eighth root to find the $2^{n+3}$ th digit of the sqrt. –  Jan 24 '20 at 03:07
  • @RoddyMacPhee Good idea, but how do you find the $8$th root? – Display name Jan 24 '20 at 03:07
  • I'll admit not easily but you could use a similar trick to what I started with that would allow you to calculate more digits of it. which means higher precision in your square root answer after cubing. –  Jan 24 '20 at 03:12
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    @RoddyMacPhee By "similar trick to what I started with", do you mean taking the square root of $2 \cdot 10^{2^{2021}}$? That's unfeasible. – Display name Jan 24 '20 at 03:18
  • no but you can apply the same trick to a smaller region and work up to it. newtons method is quadratic in nature for approximations. https://math.stackexchange.com/questions/3116326/is-there-a-faster-method-to-calculate-1-x-x-an-integer-than-this for example is close to this. –  Jan 24 '20 at 03:20
  • @RoddyMacPhee What smaller region? I'm not sure what you mean. – Display name Jan 24 '20 at 03:21
  • know the first 16 you can predict the 32nd. etc. for 8th roots to sqrt you can use 16 to predict 48 of the sqrt or more. –  Jan 24 '20 at 03:22
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    @RoddyMacPhee But then you need to know the first $2^{2019}$ digits, which is still unfeasible. How would you even store them? – Display name Jan 24 '20 at 03:23
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    The iteration $x_0=1$, $x_{n+1}=\dfrac12(x_n+2/x_n)$ does (roughly) double the number of correct decimals in each iteration, but this won't let you calculate an individual decimal far out. And as pointed out by many, calculating all of them up to that point is a non-starter. – Jyrki Lahtonen Jan 24 '20 at 13:08
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    We have

    $$ \sqrt2=\frac12\left(3-\sum_{n=0}^\infty\prod_{k=1}^n\frac1{a_k}\right) $$

    with $a_1=6$ and $a_{n+1}=a_n^2-2$. The number of digits in the denominator grows exponentially, so you only need to include about $2000$ terms to get the $2^{2020}$-th digit right. That leaves the problem of how to extract the relevant digits of each term. I don’t know a feasible way to do that. The decimal periods seem to be relatively short (e.g. the third term, $\frac1{235416}$, has decimal period $576$), but I doubt they’re short enough.

     – joriki Jan 24 '20 at 17:39
  • (For more on this series, see Wikipedia and OEIS sequences A082405, A001109 and A228931.) – joriki Jan 24 '20 at 17:40
  • @Mourad: The terms in that series only decrease as $10^{-8k}$; that's not nearly enough to make it feasible to obtain the $2^{2020}$-th digit – you'd have to sum something like $2^{2000}$ terms. I don't see from what you conclude that the series I suggested can only provide $1$ or $2$ decimal places. – joriki Jan 25 '20 at 05:22
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    @Aryabhata Perhaps this person is the victim of a fallacy. – Peter Jan 25 '20 at 08:36
  • @joriki Good luck to calculate the $2000$ th term of this sequence. – Peter Jan 25 '20 at 08:37
  • @Displayname The formula you mentioned ($\pi$ in base $2$) would also be useless to calculate the $2^{2020}$-th digit. This formula does not allow to calculate then $n$-th bit of $\pi$ "immediately", it only can do it without calculating all the previous digits. – Peter Jan 25 '20 at 08:46
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    I wonder if one can say anything about the number whose $n$-th digit is the $2^n$-th digit of $\sqrt{2}$. Probably not. – Julian Rosen Jan 26 '20 at 00:21
  • @Displayname "they have delivered on the promises" ? Maybe, but this time it seems to be a hoax, but the reactions to this question shows that such hoaxes are successful. Seriousness does not matter, the main point is to mention something "interesting". Religion in mathematics ! – Peter Jan 27 '20 at 08:45
  • If there is a solution given in 2021, kindly copy (or link to) it here. – GEdgar Jan 28 '20 at 19:17
  • @Peter. Look up BBP style formulas which do allow extracting a digit without computing earlier digits. The famous case is $\pi$ in base 16 (hence base 2 too). – Aryabhata Jan 28 '20 at 22:46
  • @Displayname, If we take log, we get $0.5 log 2=log a-log(10^n) $, where a is the number its (n+1)th is what you want. $a=10^n\times10^{0.5 log 2}$. $10^{0.5log 2}=1.1414213562 $ which is the value of $\sqrt 2$. So this is like asking to find the nth decimal digit of $10^{0.5 log 2}$ which seems to be simpler to find. – sirous Feb 17 '20 at 14:36
  • To motivate people that this isn't a hoax it would be good to include a link to the AOPS source and perhaps some examples of "promises they've delivered on". – Thomas Apr 19 '20 at 15:03
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    Why don't we see the statistical distribution of the digits of $√2$. That may give some insight. – Awe Kumar Jha May 20 '20 at 08:55
  • Things I found: a) If the denominator $m$ has at most a few million digits, it is possible to calculate the $2^{2020}$-th digit of $1/m$ using for example pari/gp. This is done by repeatedly squaring $10$ modulo $10m$... In this way, one can treat the first 20 terms of joriki's series. b) There are rational approximations of $\sqrt2$ only containing powers of 2 in the denominator (see here) Problem: What to do if the denominator cannot be represented in the computer? – Helmut May 24 '20 at 16:31
  • I notice the original author of the question has posted a similar puzzle here without answering the first one. This makes me suspect they don't have an answer. – Oscar Cunningham Jan 25 '21 at 08:03
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    @OscarCunningham The thread was locked after the author refused to share their work. I'm disappointed by the outcome, but in hindsight, something was likely to go wrong with such a difficult and seemingly impossible problem. – Display name Jun 12 '21 at 00:31

1 Answers1

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Required digits equals to $d_n=\bar d(2^n),\;$ where $$\bar d(k)=\left\lfloor 10^{k}\sqrt 2\right\rfloor\hspace{-10pt}\mod 10,\tag1$$ $$\{d_n\}=\{{1, 2, 6, 0, 9, 9, 5, 3, 6, 5, 9, 4, 7, 4, 0, 9, 4, 9, 0, 6, \dots}\}$$

Alternative way is to use the expression $$d_n=\left\lfloor 10^{2^n}a(n)\right\rfloor\hspace{-10pt}\mod 10,\quad n=1,2\dots\tag2$$ where $a(n)$ can be defined via recurrence relation $$a(0)=\dfrac32,\quad a(n+1)=\frac12a(n)+a(n)^{-1},\tag3$$ or from the closed form $$a(n)=\sqrt2\coth\left(2^{n}\ln\left(3+2\sqrt2\right)\right),\quad n=1,2,\dots\tag4$$ Formulas $(2),(3)$ allow to operate with the rational numbers. This provides exact result.

EDIT of 30.07.23

At the same time, is possible improved algorithm in the form of $$\begin{align} &d_n=b_n\hspace{-8pt} \mod 10,\quad n=0, 1, 2\dots,\\[4pt] &b_0=14,\quad b_{n+1} = \dfrac12 10^{2^n}\, b_n+\genfrac\lfloor\rfloor{}{}{1000^{2^n}}{b_n}, \end{align}\tag5$$ which stores the intermediate results in the integer format.

Gary
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Yuri Negometyanov
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