gjm

Mathematician. Reader. Fiftysomething. Cruciverbalist. Not very good at thinking of witty things to put in short bios.

I wasn't on Twitter before that nice Mr Musk suggested that everyone go to Mastodon, and I don't expect to be very active here, but who knows?

Public Key

npub1xeq9nj0ss36mr3ynzzym9xfcgzczcxsvju6sc4rf557q5r0utcfsydnmfq

NIP-05 Address

gjm_at_mathstodon.xyz@momostr.pink

Profile Code

nprofile1qqsrvszee8cggad3cjf3pzdjnyuypvpvrgxfwdgv235620q2ph79uycprpmhxue69uhhyetvv9ujumt0d4hhxarj9ecxjmnt389ef5

Publishing to

relay.momostr.pink

Author Public Key

npub1xeq9nj0ss36mr3ynzzym9xfcgzczcxsvju6sc4rf557q5r0utcfsydnmfq

Show more details

Last Notes

2026-08-16 22:12:23 UTC

- reply

Take a look at page 7.
The idea is to express the thing you're interested in in terms of integrals around the unit circle; the things in question are typically large near to points on the circle corresponding to angles that are simple rational multiples of pi; then you divide the circle into "major arcs" near to those points, and the "minor arcs" are everything else.
Then you want to prove something along the lines of "the integral over the major arcs is bigger than some upper bound on the size of integral over the minor arcs", and hence that the _whole_ integral is positive and hence nonzero.
(I am handwaving a lot because 1. the details are complicated and 2. I am far from expert on this and don't actually know the details. But the overall structure of this sort of "circle method" proof is pretty straightforward.)

2026-08-09 12:47:05 UTC

- reply

My immediate thought: gotta be the naan because you can frisbee it. This turns out to match the observed result but it was closer than I expected.

2026-08-04 00:39:36 UTC

- reply

A bit of pedantry: surely it's not true that the 2-adics "can go on infinitely in both directions". They can go on infinitely _to the left_ and any arbitrary finite amount to the right. Just as ordinary real numbers written in base 2 can go on infinitely to the right and any arbitrary finite amount to the left.

2026-07-25 16:55:36 UTC

- reply

I want to see the followup from Roger Penrose and Julian Jaynes about how this means the Norse gods may have been conscious even though the ancient Greek gods weren't.
(Jaynes has been dead for nearly 30 years but of course that's far from the biggest reason why that will never happen.)

2026-07-21 14:03:27 UTC

- reply

My reading of that thread is that this isn't a _nicer counterexample_ but a _nicer description of the same counterexample_. Have I misunderstood?
(It is very likely that I have. I am no algebraic geometer.)

2026-07-19 17:13:26 UTC

- reply

Israel's treatment of the West Bank is shockingly evil, but I think calling human beings "brutish animals without hearts or souls" is rather how we got into this sort of situation in the first place. I bet that's exactly how a lot of Israelis talk about the Palestinians who are fighting them.
I'm all for using very harsh language for people's _actions_ but it makes my skin crawl a little when I see such things said about _people_, even if they are terrible people doing terrible things.

2026-07-18 20:13:55 UTC

- reply

Superficial observation that might help somehow. (I suspect from Oscar's remark that he has made the same observation.)
Let A,B,C be the centres of the three half-size S. triangles making up the big S. triangle. For each 𝑛≥0, let 𝑋ₙ be the random variable (random one of A,B,C) - (random one of A,B,C), all 𝑋ₙ independent. Then the difference between any two points in the S. triangle is ∑2⁻ⁿ𝑋ₙ.
(So the expectation of the distance _squared_ is twice the expectation of |𝑋₀|² or twice (1/3)² so the RMS distance is √2/3, as per Oscar's remark.)
It feels like someone who knows more than I do about random vectors in euclidean space might find it pretty easy to use this to find the answer, or at least to find an efficient way to approximate it. Unfortunately, I know only exactly as much as I do, which isn't very much.

2026-07-18 17:42:06 UTC

- reply

And, looking more carefully, my extrapolations do seem like they're closer to one another and to the value claimed on that page than they are to 41√3/168.
(Logs of differences: last Aitken / last Wynn -16.4, last Aitken / second-last Wynn -16.3, last two Wynn -19.9; last Aitken / claim -16.4, last Wynn / claim -24.0, second-last Wynn / claim -19.9.)
So I retract what I said before: it seems _very unlikely_ that 41√3/168 is the true value, and the available evidence suggests that the programming contest problem's claimed 0.4227021810348385578571 could well be correct.

2026-07-18 17:32:11 UTC

- reply

A commenter on math.stackexchange.com at https://math.stackexchange.com/questions/4047421/what-is-the-average-distance-between-two-randomly-chosen-points-on-a-sierpi%C5%84ski links to a programming-contest site and claims that it mentions someone who's found an efficient way to compute this value. For me the link goes somewhere uninformative but archive.org has what I think the link was to. The page is in Korean so I am relying on machine translation. I think the mathematician mentioned on that page may possibly be fictional.
The page does give what it claims to be the correct answer to 22dp and it _doesn't_ match the conjecture: 0.4227021810348385578571 versus 0.4227028756566902918727.

2026-07-18 17:22:47 UTC

- reply

Some numerology:
Write 𝐷ₙ for the average distance between points in the 𝑛th-order approximation to the S. triangle, where the 1st order is just the vertices and the (𝑛+1)th order consists of three scaled copies of the 𝑛th order. (Note in particular that some points appear multiple times. I think this is OK.)
Then, aside from mere accumulated floating-point calculation errors, the first ten values of 𝐷ₙ are:
0.6666666666666667
0.4986704301902867
0.44520689777765476
0.4289567387204627
0.4243722628662094
0.42313749521839766
0.4228139755506694
0.4227306238555088
0.42270937415127596
Writing 𝑔 for the conjectured 41√3/168, the logs of the differences 𝐷ₙ−𝑔 are: -1.41, -2.58, -3.79, -5.07, -6.40, -7.74, -9.11, -10.49, -11.94 which seem to be decreasing at a pretty consistent rate.
It seems plausible that the 𝐷ₙ might converge in a clean enough way to try standard convergence-acceleration techniques. Aitken Δ² extrapolation gives things for which the logs of the differences are: -6.01, -7.08, -8.93, -10.79, -12.49, -13.65, -14.07 which are more-negative than the originals and seem to be decreasing nicely. Wynn ε extrapolation, which is a bit like an iterated version of Aitken, produces broadly similar results.
All of this seems _very consistent_ with your proposed answer being correct. None of it, of course, comes within a light-year of proving anything.

2026-04-27 22:23:18 UTC

- reply

This may well be an important and very correct and necessary book, but I really really dislike the title.
By and large, the actual nerds of Silicon Valley -- the engineers and programmers and whatnot -- aren't fascist or authoritarian at all. (Even, I _think_, the ones who work for the most blatantly evil tech companies: even a lot of employees at bloody _Palantir_ turn out to have been worried by its embracing of fascism.)
The people who are disproportionately fascist (or fascist-adjacent, or out past fascism and on the other side, or otherwise anti-democratic and generally awful) are not the nerds but the venture capitalists and big-tech-company CEOs. Some of them _started out_ as nerdy engineering types, but that's not what they are any more. (And most of them seem to have been idealistic liberals back when they were nerds. They became fascists when they _stopped_ being nerds and started being Masters of the Universe.)
It's a conspiracy of billionaires and their hangers-on, not of nerds. Nerds get enough hate and contempt without encouraging people to blur the lines between them and the bastards who own the companies some of them work for.
For the avoidance of doubt: yes, some individual nerds (e.g., Curtis Yarvin) are appalling authoritarians; yes, lots of nerds are working for companies that do evil and you can make a good case that that makes them complicit. But the people who deserve that word "Reich" attached to them are not generally the nerds.
(Declaration of interest: I am myself by any reasonable definition a nerd, I have never worked in Silicon Valley, I despise authoritarianism, and part of what bothers me is seeing People Like Me unfairly associated with fascism.)

2026-04-08 22:25:23 UTC

- reply

Au contraire, it's definitely rouge.

2026-02-10 22:38:19 UTC

- reply

Are you sure about your Juliuses and your Scaligers?
It looks to me as if:
Julius Caesar (the emperor) introduced the Julian calendar.
Julius Caesar Scaliger was an erudite chap but had nothing much to do with calendars.
Joseph Justus Scaliger, Julius Scaliger's son, introduced the "Julian period" and (kinda) the "Julian day number".
The "Julian" there is a reference to the Julian calendar, _not_ to his dad.
... But I am not an expert, I'm just reporting what I find in Wikipedia, and maybe the above is all wrong?

2026-01-20 20:40:07 UTC

- reply

I suspect DLL's comment is a deliberate reference to this, but in case (1) it isn't or (2) others didn't notice, Raymond Smullyan wrote a whole book with this premise: https://en.wikipedia.org/wiki/To_Mock_a_Mockingbird

2025-10-09 14:47:54 UTC

- reply

By "fairly small" I meant on the order of 10^-10, which is the sort of thing I would expect to be fairly easily achievable in multiple ways by things looking like linear combinations of spherical shells.

2025-10-09 12:49:30 UTC

- reply

Note that the gravitational field of _any_ spherically-symmetrical thing is the same (up to a scaling of the total mass) _everywhere_ outside the sphere containing it. So it doesn't particularly matter whether the sample points are all on a single sphere.
Of course, that's what's true for an _actually literally spherically symmetrical_ mass density, and what you've got here is just an approximation to that. I don't think I'd have predicted ahead of time that the determinant would be so very small ... though maybe what's going on isn't that there are _incredibly small eigenvectors_ but that there are _quite a lot of fairly small eigenvectors_.
Further conjecture: if you move your sample points further out, the determinant will get smaller; if you move them further in, it will get larger. (On the grounds that I expect the spherically-symmetrical approximation to get better at larger distances.)

2025-10-09 12:22:50 UTC

- reply

Have you looked to see what the almost-nullspace looks like? That is, what not-very-short vectors are mapped almost-to-zero? Or, to put it differently: your empirical finding shows that some distributions of mass have almost the same effect on your 83 red dots as others; what are they?
My best guess, which I don't trust much: your 83 red dots are all at a single distance, so any distribution of mass that looks "enough" like a spherical shell of any size smaller than that will look (up to scaling the masses) exactly like any other, and perhaps your nearly-nullspace vectors look like linear combinations of spherical shells.

2025-09-02 11:48:27 UTC

- reply

It's too bad it can't be A000001. Though it wouldn't by any means be the first known-to-be-finite OEIS sequence; for instance A100000 is another.

2025-08-22 12:48:25 UTC

- reply

Some years ago, in the UK there was an election which the Liberal Democrat candidate was expected to win, with the Conservative candidate a close second. So the Conservatives got someone to run as a "Literal Democrat", he picked up a bunch of votes from people who didn't read carefully, and the Conservatives won. (The election regulations got changed after that so we didn't get the otherwise-inevitable deluge of "Conversatives" and "Literal Democrats" fighting to see who could confuse more voters.)
Are these people hoping to attract parents who think they're getting a Montessori school? It seems kinda unlikely, so I guess they either don't know or don't care about the similarity in name. Neither is very encouraging.

2025-08-06 14:15:16 UTC

- reply

By a curious coincidence, RFK Jr. _is_ an evil troll whose actions are likely to make some children sneeze non-stop until they drop dead, or at least something in the same conceptual vicinity.

2025-08-02 18:38:45 UTC

- reply

I hope they made it not too difficult for people to leave early. I'm going to guess you weren't the only ones.

2025-08-02 12:26:50 UTC

- reply

Good luck! (I like Tavener but I don't think I would cope well with 8 hours of one Tavener piece at a stretch.)

2025-08-02 10:30:57 UTC

- reply

Pedantic note: I think you mean not "The Protecting Veil", a famous Tavener piece but only ~45 minutes long, but "The Veil of the Temple", a newer piece by the same composer that is, indeed, 8 hours long.

2025-07-22 00:04:07 UTC

- reply

This is a slightly atypical case because usually when something's named after X it's because someone else discovered the thing first and then X wrote about it later[1], whereas in this case IIUC Moessner really did get there first but didn't _prove_ it.
[1] Though my actual favourite example is Pell's equation, which so far as I can tell Pell never worked on _at all_.

2025-06-15 13:21:05 UTC

- reply

That paper also contains the line "the d-ism of Leibniz overtook the dotage of Newton" which is rather magnificent. Knuth puts it in quotes but doesn't say who it's from; it seems to be Charles Babbage, though his wording was a little different.

2024-12-02 21:18:59 UTC

- reply

Whether or not _that_ was a joke, there's something very appropriate about your photo showing Aristote's Fragme Sele.

2024-10-20 15:05:25 UTC

- reply

You don't have to expend time or money getting to and from the movie theatre. You have comfier seating. If you have friends or family, you can all watch at once for the same price. You are much less likely to catch whatever infectious diseases may be going around.
How all of that weighs against the larger screen, the maybe-better sound system, the appeal of the traditional movie-theatre atmosphere, etc., will vary from person to person. But I'm betting they wouldn't be setting their prices at 150% of cinema ticket prices if there weren't plenty of people for whom watching at home is a significantly better experience.