Timeline for Two (probably) equal real numbers which are not proved to be equal?
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50 events
| when toggle format | what | by | license | comment | |
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| Oct 22 at 16:42 | comment | added | Jade Vanadium | The equality "$x=y$" between computable numbers is computably reducible to the halting problem and vice versa. Select arbitrary computable $x$, arbitrary Turing machine $M$, define $b_t\in\{0,1\}$ where $b_t=1$ iff $M$ halts at time $t$, define $z=\sum_t b_t2^{-n}$, and let $y=x+z$. We observe $x=y$ iff $M$ doesn't halt, so we get nice examples when $M$ is conjectured (but not proven) to never halt. Every nontrivial instance of your problem is obtained in exactly this way, by letting $M$ be the machine which halts iff there exists a proof of $x=y$ over Peano Arithmetic. | |
| May 28 at 3:40 | history | edited | Gerry Myerson |
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| May 30, 2020 at 18:01 | answer | added | Jesús Guillera | timeline score: 4 | |
| May 27, 2019 at 5:21 | comment | added | Nicola Gigante | I see, thanks for the clarification | |
| May 26, 2019 at 22:21 | comment | added | Robert Furber | @gigabytes When people say "equality is undecidable for the reals" they are usually referring to computable real numbers. There are several equivalent formulations of them, but one is real numbers that have a (Turing) computable sequence of rational approximations. Perhaps you can see how if we could decide equality for these things we could solve the halting problem. | |
| May 26, 2019 at 7:40 | comment | added | Nicola Gigante | I might be mistaken (I’m from CS, not a mathematician), but it seems to me that this has not much to do with the reals themselves, but rather to the language you use to express your identities. Your examples use transcendental functions and integration, but if you restrict the syntax to just polynomials, then equality is decidable on the reals (Tarski theorem) but not on integers (Hilbert’s tenth problem?). So are we talking about a specific set of operations? Is there an undecidability result regarding polynomials augmented with transcendental functions and/or integration? | |
| May 25, 2019 at 22:24 | answer | added | Kevin Buzzard | timeline score: 14 | |
| May 12, 2019 at 20:10 | vote | accept | Kevin Buzzard | ||
| May 12, 2019 at 18:50 | answer | added | Timothy Chow | timeline score: 25 | |
| May 11, 2019 at 15:34 | vote | accept | Kevin Buzzard | ||
| May 12, 2019 at 20:10 | |||||
| May 10, 2019 at 5:28 | comment | added | none | How about the de Bruijn-Newman constant $\Lambda$? It's long been known that RH is true iff $\Lambda\le 0$. We now know that $0\le\Lambda\le 0.22$ which means RH is equivalent to $\Lambda=0$. So according to some, $\Lambda$ is probably 0, but it's obviously hard to prove. I wouldn't quite say this is silly in Kevin's sense. | |
| May 9, 2019 at 13:58 | comment | added | Kevin Buzzard | I don't know a rigorous definition of "silly", but on the other hand I am 100% convinced that this example falls under the "let's encode a well-known conjecture into a possibly slightly artificial real number" umbrella :D | |
| May 9, 2019 at 10:04 | comment | added | Don Hatch | I'm not sure you've defined "silly" in a useful way. For example, here's a case where we know how to compute the two numbers to gazillions of decimal places and yet it's still pretty silly: A = π, B = π with the n'th bit in its binary expansion flipped if Goldbach's conjecture fails at n. | |
| May 9, 2019 at 9:06 | comment | added | none | @KevinBuzzard per Fedor Petrov and EGME, it looks like that formula has been proven, so it's not such a good answer after all. Oh well. It is sure cool! | |
| May 9, 2019 at 6:32 | comment | added | Asaf Karagila♦ | Take $r$ to be a real number that is probably, but not provably, $0$. Then $\pi$ and $\pi+r$ fit your bill. | |
| May 9, 2019 at 1:14 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| May 8, 2019 at 13:12 | answer | added | Yoav Kallus | timeline score: 17 | |
| May 8, 2019 at 11:28 | comment | added | EGME | @none : Mathematica (12) calculates none’s example exactly, so we can considered that proved! | |
| May 8, 2019 at 10:30 | comment | added | Fedor Petrov | @none I believe this identity with $\pi^4$ is actually proven, Borwein in the article (see the reference in Wikipedia) says that "most but not all" identities of this type are proved, and this specific identity looks the simplest among them. | |
| May 8, 2019 at 9:50 | answer | added | François Brunault | timeline score: 16 | |
| May 8, 2019 at 9:26 | comment | added | François Brunault | The Clausen function $\mathrm{Cl}_2(\theta)$ is just the Bloch-Wigner dilogarithm $D(e^{i\theta})$, which can be computed very efficiently e.g. in Pari/GP with polylog(2,exp(I*theta),2) | |
| May 8, 2019 at 8:46 | comment | added | Kevin Buzzard | @Zidane I guess it's hard (but not impossible) to make the ell curve example "explicit" -- the L-function doesn't converge for s=1 (even if written as the L-function of a grossencharacter -- one would have to turn the sum into an explicit integral which converges there, which should be possible by the techniques in Tate's thesis I guess, although "explicit" is now becoming less and less like a sensible adjective to use here...) | |
| May 8, 2019 at 8:42 | comment | added | Yaakov Baruch | @none: I think your example is both beautiful and closest to the spirit of the question. it would make a very nice answer. | |
| May 8, 2019 at 8:41 | comment | added | Kevin Buzzard | @none that is the best answer I have yet seen! Please post as an answer! | |
| May 8, 2019 at 8:38 | comment | added | Kevin Buzzard | @Zidane what Matt F. said. I believe that Stein did some explicit calculations with the curve of algebraic rank 4 and smallest conductor (he also calculated the analytic value of Sha to hundreds of decimal places IIRC, although that might have been for the rank 2 curve of smallest conductor). | |
| May 8, 2019 at 6:47 | comment | added | none | The article en.wikipedia.org/wiki/Experimental_mathematics has some nice true, false, and unsolved examples. $ \sum_{k=1}^\infty \frac{1}{k^2}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k}\right)^2 = \frac{17\pi^4}{360}$ was verified to 100+ decimal digits (see reference linked from the wiki page) but might still be unproven. | |
| May 8, 2019 at 6:40 | answer | added | none | timeline score: 3 | |
| May 8, 2019 at 1:46 | comment | added | user44143 | @EGME, indeed, I was just providing the computation (up to a sign error) in a form that anyone can play with freely. | |
| May 8, 2019 at 1:36 | comment | added | user44143 | @Zidane, will you write that up as an answer? A version that also explains it without the words “rank”, “twist” and “L-function” would be especially welcome, and perphaps if you write the bare-bones answer, others can fill in the exposition. | |
| May 7, 2019 at 23:47 | answer | added | Jason Rute | timeline score: 50 | |
| May 7, 2019 at 21:00 | comment | added | Zidane | You can take an elliptic curve over ${\mathbb Q}$ with rank 4. Then the second derivative of its L-function at s=1 is computable (if you start from a twist of $y^2=x^3-x$, you can even write an explicit series), should be 0, but no one knows how to prove it. | |
| May 7, 2019 at 19:40 | comment | added | Sam Hopkins | It is now known that the connective constant (en.wikipedia.org/wiki/Connective_constant) of the honeycomb lattice is $\sqrt{2+\sqrt{2}}$, but if that were not rigorously known, would be the kind of thing you're looking for? | |
| May 7, 2019 at 19:29 | history | became hot network question | |||
| May 7, 2019 at 19:08 | answer | added | Robert Israel | timeline score: 10 | |
| May 7, 2019 at 18:48 | comment | added | Andreas Blass | @KevinBuzzard Probably the best way is to just require that we know (and can prove) what algorithms compute the two numbers. Essentially equivalent: ask people to give you an example of two algorithms such that both are known to compute real numbers and those two real numbers are believed but not known to be equal. | |
| May 7, 2019 at 18:33 | comment | added | Kevin Buzzard | PS @Wojowu I am completely happy to move the question to Math.SE (but don't know how to); I no longer have a good feeling as to what belongs where and am happy to trust yours. | |
| May 7, 2019 at 18:32 | comment | added | Kevin Buzzard | @AndreasBlass can you tell me how to rephrase the question to make it conform better to the meaning that I want to give it? | |
| May 7, 2019 at 18:30 | comment | added | Kevin Buzzard | @Wojowu inspired by your links one could even suggest $\Sigma_z|z|^{-100}$ where the sum is over all non-trivial zeros of $\zeta(s)$ which are not on the critical line :-/ Whilst I admit that this might well satisfy the requirements, it's somehow not what I was after :-/ | |
| May 7, 2019 at 17:17 | comment | added | Andreas Blass | As one of the pedants mentioned in the last paragraph of the question, I must point out that "0 if the Birch and Swinnerton-Dyer conjecture is true and 1 if not" is a computable real number. Admittedly, we don't know which of two obvious algorithms computes it, but we (using classical logic) do know that there exists an algorithm computing it, and that's what "computable" means. | |
| May 7, 2019 at 17:15 | review | Close votes | |||
| May 8, 2019 at 9:31 | |||||
| May 7, 2019 at 16:57 | comment | added | EGME | The LHS around -11 seems consistent with what I found ... | |
| May 7, 2019 at 16:47 | comment | added | user44143 | rextester.com/KMHKJ81925 will execute the following R code, which gets LHS ~ 11 and RHS ~ 0: f <- function(s){log(abs(2*sin(s)))}; Cl2 <- function(t){stats::integrate(f,0,t,abs.tol=1e-6)$value}; t2 <- 2*atan(sqrt(2)); lhs <- 27*Cl2(t2) - 9*Cl2(2*t2) + Cl2(3*t2); rhs <- 8*Cl2(pi/4) + 8*Cl2(3*pi/4); c(lhs,rhs) | |
| May 7, 2019 at 16:43 | comment | added | EGME | I stand corrected. The numerical integration produces, for the LHS, something not close to 0. I read my result incorrectly (need reading glasses) ... | |
| May 7, 2019 at 16:16 | comment | added | EGME | Hi Kevin, Mathematica gives a negligibly large number, practically 0, on both sides with numerical integrate. It has trouble with $3\pi/4$ in the fifth term so I made that 2.9999 ... It computes the RHS exactly as 0, and it produces a complex number with exact integration for the left hand side which is nowhere near 0. I will try to see if I can get something better for the LHS, I can use M pretty well. Pity about the discouraging comment ... I haven’t tried the latest version, will also do that ... | |
| May 7, 2019 at 15:50 | comment | added | Aeryk | I know of rational quantities related to algebraic curves that were originally recognized by their decimal expansions through analytic methods and where later computed/proven by explicitly constructing the prime factorizations. Cf: arxiv.org/abs/0711.4316 | |
| May 7, 2019 at 15:20 | comment | added | Wojowu | As much as I like the question, I feel like it would fit better on Math.SE. | |
| May 7, 2019 at 15:16 | comment | added | Wojowu | Example 1, example 2, example 3. All identities involve computable numbers and are equivalent to RH. | |
| May 7, 2019 at 15:15 | history | edited | user44143 |
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| May 7, 2019 at 15:02 | comment | added | Kevin Buzzard | I tried computing both sides using pari-gp but I've never used it for numerical integration before, and pari suggested that the right hand side was zero and the left hand side was about -10 :-( | |
| May 7, 2019 at 15:01 | history | asked | Kevin Buzzard | CC BY-SA 4.0 |