Timeline for Do we know how to determine the $2^{2020}$ decimal of $\sqrt{2}$?
Current License: CC BY-SA 4.0
35 events
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| May 18, 2021 at 12:05 | comment | added | Dattier | Well, I would wait 500 years for the math community to find a complex 600-page solution, like for Fermat. | |
| May 18, 2021 at 11:44 | comment | added | Yemon Choi | If you are claiming that you can answer this question, then you can either add an answer below, or write up your work as a preprint in the normal academic way, or write it in a blog post and leave a link. Saying "I can answer this, can you?" is not a constructive use of this site, nor is it collegial. | |
| May 18, 2021 at 11:44 | comment | added | Yemon Choi | I have rolled back the edits to this old question, which seem to be purely cosmetic and done for promotion. Yes the question is appropriate for MO, but that doesn't mean you need to add some kind of "RL" label | |
| May 18, 2021 at 11:44 | history | rollback | Yemon Choi |
Rollback to Revision 3
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| May 18, 2021 at 10:42 | history | edited | Dattier | CC BY-SA 4.0 |
added 53 characters in body
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| May 18, 2021 at 10:39 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
the abbreviation RL in the title is confusing and unnecessary: whether a question is research level or not is decided based on its content, not on some tag.
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| May 18, 2021 at 10:30 | history | edited | Dattier | CC BY-SA 4.0 |
edited title
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| Feb 25, 2020 at 18:03 | comment | added | John Cosgrave | When I was a student in London I used to attend some lectures by the (great and loveable) C.A. Rogers. Once - having tea and chat with him - someone asked which question he would most like to have answered. His almost immediate response was: "I'd love to know the decimal expansion of the square root of 2" | |
| Feb 24, 2020 at 16:29 | comment | added | Dattier | @KonstantinosGaitanas : why? | |
| Feb 24, 2020 at 15:06 | comment | added | Konstantinos Gaitanas | Obviously the answer is the digit $3$. | |
| Feb 24, 2020 at 14:28 | comment | added | Emil Jeřábek | @fedja No worries. You know, I first asked you about this in a comment, then later when rereading it I realized that it is indeed obviously 0 and deleted my silly comment, and then later I realized that my earlier obvious reason was off by an exponential and reposted the comment again. | |
| Feb 24, 2020 at 12:39 | comment | added | fedja | @EmilJeřábek3.0 Because my brain was totally dead in the evening :-) That happens sometimes. No, it is not. Just ignore that remark... | |
| Feb 24, 2020 at 11:35 | comment | added | Dattier | $$d=E(\dfrac{10^{2^{2020}}}{7^{800}}) \mod 10=\left( \dfrac{10^{2^{2020}}-(10^{2^{2020}}\mod 7^{800})}{7^{800}} \right) \mod 10=\dfrac{-(10^{2^{2020}}\mod 7^{800})}{7^{800}} \mod 10$$ and I find $7$. | |
| Feb 24, 2020 at 10:26 | comment | added | Emil Jeřábek | @fedja Why is the $2^{2020}$th decimal of $7^{-800}$ obviously $0$? | |
| Feb 24, 2020 at 9:03 | comment | added | joro | According to Wolfram Alpha there might be chance at base two: wolframalpha.com/input/?i=series+sqrt%28x%29+at+x%3D2 | |
| Feb 24, 2020 at 7:59 | comment | added | Emil Jeřábek | @IgorKhavkine It’s a completely different algorithm. Basically, you compute the square root by partially summing a power series, but the hard part is to compute the relevant iterated additions and multiplications without storing all the terms (recomputing some of it when needed). The basic idea of the iterated multiplication algorithm is to do it modulo many small primes and then reconstruct it from the Chinese remainder representation, but there is more to it than that. See Hesse, Allender, and Barrington doi.org/10.1016/S0022-0000(02)00025-9 . | |
| Feb 24, 2020 at 7:50 | comment | added | Igor Khavkine | @EmilJeřábek3.0, could you please elaborate? Where is the space saving for storing the digits coming from? | |
| Feb 24, 2020 at 7:20 | comment | added | Emil Jeřábek | While there is no known algorithm essentially faster than Newton iteration, one can significantly improve on the “not enough particles in the visible universe” front: the decimal expansion of $\sqrt2$ is computable in logarithmic space, which would mean here something on the order of a few KB. (Better yet, it is computable in logarithmic time on a “threshold Turing machine”, if anyone figures how to build one in the real world.) Timewise, this would be much slower than Newton iteration. | |
| Feb 24, 2020 at 6:46 | history | edited | Emil Jeřábek |
edited tags
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| Feb 24, 2020 at 6:45 | history | reopened |
Dattier Ramiro de la Vega user44143 Alexey Ustinov Emil Jeřábek |
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| Feb 24, 2020 at 0:27 | comment | added | fedja | Not voting to reopen because the question seems hopeless, but, of course, the suggestion to send it to MSE (implied by the formal closing reason) is ridiculous. | |
| Feb 24, 2020 at 0:25 | comment | added | fedja | @Dattier Yeah, but if it were slightly less than $2^{2020}$ (i.e., if the corresponding decimal digit weren't obviously $0$), it would make another interesting question... | |
| Feb 24, 2020 at 0:12 | comment | added | Dattier | It was to give the example of a rational whose period is greater than $2^{2020}$ | |
| Feb 23, 2020 at 23:27 | comment | added | Steven Landsburg | What does $7^{800}$ have to do with this? | |
| Feb 23, 2020 at 23:15 | review | Reopen votes | |||
| Feb 24, 2020 at 6:45 | |||||
| Feb 23, 2020 at 22:55 | history | closed |
Francois Ziegler user44191 Tobias Fritz Martin Brandenburg Wojowu |
Not suitable for this site | |
| Feb 23, 2020 at 22:53 | history | edited | GH from MO |
edited tags
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| Feb 23, 2020 at 22:53 | comment | added | GH from MO | This problem has nothing to do with arithmetic geometry. Arithmetic geometry is algebraic geometry over finite fields, local fields, global fields etc. | |
| Feb 23, 2020 at 22:36 | comment | added | Dattier | @Carlo : well it's the good forum, research level | |
| Feb 23, 2020 at 22:11 | comment | added | Carlo Beenakker | according to arXiv:0912.0303 no digit-extraction formula is known for $\sqrt 2$. | |
| Feb 23, 2020 at 21:58 | comment | added | fedja | @Qfwfq Newton-Rapson won't finish before the Sun explodes and won't have enough particles in the visible universe for the required memory storage, but otherwise it is fine. A legitimate answer should produce the required digit without computing all the preceding ones and that may be quite tricky since there is no obvious pattern in the digit sequence. I would say the answer is "no" but there is no proof of that either. | |
| Feb 23, 2020 at 21:32 | comment | added | Dattier | @Carlo Beenakker, I think this question requires to mobilize a strategy different from that which you propose in your link. | |
| Feb 23, 2020 at 21:25 | review | Close votes | |||
| Feb 23, 2020 at 23:00 | |||||
| Feb 23, 2020 at 21:17 | comment | added | Carlo Beenakker | the algorithm is known --- you may have to wait a long time for it to finish. | |
| Feb 23, 2020 at 20:57 | history | asked | Dattier | CC BY-SA 4.0 |