Timeline for Proof that $\pi$ has infinitely many $1$s
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| May 2, 2022 at 8:22 | comment | added | Kurisuto Asutora | Compare a related problem for $\sqrt{2}$, which is a bit more accessible: arxiv.org/abs/1711.01722. | |
| Apr 30, 2022 at 17:27 | comment | added | Dave L Renfro | @Gerry Myerson: Yes. I wondered whether I should be writing something like this after 4 hours of sleep, but I did it anyway. What's small (measure zero & meager, even $\sigma$-porous) is the set of non-disjunctive numbers. There is a dichotomy of Lebesgue measure and Baire category for normal numbers: Most numbers ARE normal (Lebesgue measure) and most numbers are NOT normal (Baire category; indeed, "NOT" in a very strong way -- see Olsen and Stylianou). However, most (both ways) numbers are disjunctive. | |
| Apr 30, 2022 at 12:46 | comment | added | Gerry Myerson | @Dave, if a number is normal, then every finite digit sequence appears, so it's disjunctive. The normals are a subset of the disjunctives, no? | |
| Apr 30, 2022 at 6:28 | comment | added | Dave L Renfro | @Gerry Myerson: Means every finite digit sequence appears (equivalently, appears infinitely often) -- see here. Related is "lexicon" -- see The typical number is a lexicon by Calude/Zamfirescu (1998). While normal numbers are full measure (large in one way) and meager (small in another way), the disjunctive numbers are measure zero and meager and smaller still -- see this 19 February 2003 sci.math post. | |
| Apr 30, 2022 at 4:57 | comment | added | Gerry Myerson | What does "disjunctive" mean in this context? | |
| Apr 29, 2022 at 20:26 | comment | added | Alessandro Della Corte | @mathworker21 Can you please add a reference? | |
| Apr 29, 2022 at 19:10 | comment | added | Z. M | @SamHopkins In base 2, every real number has a binary expansion which contains infinitely many 1's, since $\sum_{k=1}^\infty2^{-k}=1$. | |
| Apr 29, 2022 at 19:01 | comment | added | Dave L Renfro | We do know that there exists a $10^{100}$-length sequence of digits that appears infinitely often in the decimal expansion of $\pi$ (and $10^{100}$ can be replaced by any other positive integer), but this is true for any infinite sequence of decimal digits . . . | |
| Apr 29, 2022 at 17:52 | comment | added | mathworker21 | It's not known whether the decimal digits of $\pi$ are eventually all $0$s and $2$s. | |
| Apr 29, 2022 at 16:43 | review | Close votes | |||
| May 3, 2022 at 14:56 | |||||
| Apr 29, 2022 at 16:28 | comment | added | Timothy Chow | Does this answer your question? What is the state of our ignorance about the normality of pi? or Does pi contain 1000 consecutive zeroes (in base 10)? | |
| Apr 29, 2022 at 16:01 | history | edited | kodlu | CC BY-SA 4.0 |
added 25 characters in body
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| Apr 29, 2022 at 14:12 | comment | added | Wojowu | No, no such results have been proven. | |
| Apr 29, 2022 at 14:06 | comment | added | Sam Hopkins | Well, I can think of a proof in base 2.... | |
| Apr 29, 2022 at 14:04 | history | asked | skytect | CC BY-SA 4.0 |