gjm

gjm, 3 days ago

@cstross Turing's paper does set up the "imitation game" in a way that suggests a machine specifically trying to imitate a woman. But it's pretty clear that the sex difference isn't what's on Turing's mind -- the rest of the paper is full of "it will be assumed that the best strategy is to try to provide answers that would naturally be given by a man" and "Is it true that [...] C can be made to play satisfactorily the part of A in the imitation game, the part of B being taken by a man?"

Which is all pretty sexist, of course, but a different kind of sexist. I think the business about men and women in the setup is there only for the sake of vividity and Turing isn't really supposing that the machine is specifically trying to pretend to be a man as opposed to a woman, or a woman as opposed to a man.

(The intentions of today's AI companies are a different matter.)

gjm, 15 days ago

@gregeganSF @johncarlosbaez So, he's done best possible packings and worst-possible packings. Next up: most mediocre possible packings!

gjm, 1 month ago

@christianp @ColinTheMathmo Eh, I thought they were kinda fun. In every case it was an exercise in "can you spot the trick?" and enough of the tricks were at least semi-interesting to hold my interest.

(Except maybe for 16, which feels like it should be straightforward but somehow isn't working out for me so far. I suspect I've done something stupid or missed something easy, and it does fall under the rubric of "spot the trick that makes it simple" even though I either haven't spotted it or have screwed it up.)

I'm old and slow (at least compared with when I was younger and quicker) so I certainly didn't do them all in 20 minutes.

The "finals" problems at https://math.mit.edu/~yyao1/pdf/2024_finals.pdf are ... somewhat tougher.

gjm, 1 month ago

@ColinTheMathmo What I have on my shelf is Hartley & Hawkes, "Rings, Modules and Linear Algebra". Fairly short. Looks pretty clear from a quick glance. Seems to be out of print but available second-hand at reasonable prices.

Part I (~90pp): rings and modules: definitions and examples of rings; subrings, homomorphisms, ideals; direct sums, polynomial rings, etc.; integral domains and factorization; modules (inc. homomorphisms, quotients, direct sums); a few special classes of modules.

Part II (~45pp): decomposition theorems for f.g. modules over a p.i.d.

Part III (~50pp): applications to groups and matrices: f.g. abelian groups as ℤ-modules, linear transformations, matrices, canonical forms; computation of canonical forms.

I somehow don't have a copy of Herstein to compare with.

gjm, 1 month ago

@christianp House-buying is just such a miserable business in every way.

First there's all the searching. Everything is too expensive. Nothing is quite what you need. And the process of house-searching corrupts your soul, so that for the next six months every time you go into someone else's house part of your brain will be thinking "hmm, it's probably worth this much, and we'd need to put some more bookshelves there", etc.

If you're buying with a mortgage, which almost everyone is, then you have to arrange one, and they're designed to be confusing and make it hard to know just what you will actually end up paying, and the mortgage providers will want to dig into your finances to verify that you're rich enough to be fit to do business with, and it's tedious and demeaning.

Then you have to actually buy the damn thing, which involves endless lawyering and the constant feeling that if someone makes a trivial slip then it could have life-ruining consequences. And it takes sooo long, and an annoying amount of getting multiple sets of ducks all in a row.

Then you move all your stuff, and possibly have to buy a load of new stuff, and you just know that some of those boxes are going to take a year before you get around to unpacking them.

And at the end of the process, yes, you have a hopefully-nice new house. But you also have much less money, or a mountain of debt, or both.

(But at least you aren't just pouring rent money into an everlasting black hole. However miserable house-buying is, house-renting is even worse.)

gjm, 2 months ago

@johncarlosbaez Knapp's book on elliptic curves kinda-sorta has your definition (but not as a definition, and not stated very explicitly) in section X.2. The other elliptic curve books I have (of which there aren't many, and they're all fairly low-level) don't seem to talk about the relationship between L-functions and zeta functions.

gjm, 3 months ago

@christianp That's surprising to me. I'm pretty sure the great majority of images I see here have alt text. (I just did a quick scroll down my feed and it's definitely doing better than 47%, and most of the alt-less images I see are actually from other servers.)

gjm, 3 months ago

@christianp I'm pretty sure the denominator has to be images rather than posts or users.

gjm, 3 months ago

@christianp Ha! There's a third possibility (which maybe you're deliberately ignoring for comic effect): if 3yo is asking that question because earlier you said something like "we can do X later", then they may be asking "is it time to do X yet?", and that answer is neither trivially yes nor trivially no.

gjm, 3 months ago

@christianp Hooray! I am successfully able to think at the level of a 3-year-old. :-)

gjm, 3 months ago

@siderea I'd expect one to begin "gr[a]ec-" and the OED contains no such word with that meaning. (There's also "Hellenist[ic]" but that refers to a particular type of Greek and also doesn't appear to have the right shade of meaning.)

So it looks like the answer might be no.

(For the specific phrase there, "classical" might kinda do the job of "latinate and [that-word-that-maybe-doesn't-exist]", though I don't think it exactly fits.)

gjm, 3 months ago

@fanf You ask whether there's a formula; I think there is.

Let's put the corner at (0,0) and consider the "horizontal" ellipse, which has to pass horizontally through, say, (-a,-b). (I'm choosing signs to try to make parameters usually-positive.)

An "axis-aligned" ellipse is 𝑝(𝑋−𝑢)²+𝑞(𝑌−𝑣)²=1 where (𝑢,𝑣) is the centre. In this case we know 𝑢=−𝑎. So this is 𝑝(𝑋+𝑎)²+𝑞(𝑌−𝑣)²=1.

Differentiating, we get 𝑝(𝑋+𝑎)𝑑𝑋+𝑞(𝑌−𝑣)𝑑𝑌=0, so the requirement 𝑑𝑋=𝑑𝑌 at (0,0) says 𝑝𝑎=𝑞𝑣.

Passing through (0,0) says 𝑝𝑎²+𝑞𝑣²=1. Passing through (−𝑎,−𝑏) says 𝑞(𝑏+𝑣)²=1.

The first two of these equations say 𝑝=1/𝑎(𝑎+𝑣) and 𝑞=1/𝑣(𝑎+𝑣). Then the third says (𝑏+𝑣)²=𝑣(𝑎+𝑣) and hence 𝑣=𝑏²/(𝑎−2𝑏).

Sanity-check: your diagram here looks like it has roughly (a,b)=(4,1). This gives v=1/2, p=1/18, q=4/9 meaning centre at (-4,1/2), semimajor axis 3 sqrt(2), semiminor axis 3/2, all of which looks plausible.

gjm, 3 months ago

@fanf It looks like you have a very different equation for q from mine; you've got let q = b+v; where I'd have let q = 1/(v*(a+v));.

And then I think you're treating p,q as semiaxes, but they aren't that, they're 1 / semiaxis^2. So I think you want to use 1/sqrt(p) and 1/sqrt(q) as your semiaxes.

Try this:

 function gjm(a,b) {  
 let v = b*b / (a - 2*b);  
 let q = 1 / (v*v + a*v);  
 let p = 1 / (a*a + a*v);  
 spot(0, Y - v, 10);  
 return [p,q];  
 }  
 (function() {  
 style(1, "#800", "#8008");  
 let [a,b] = gjm(X, B - Y);  
 a = 1/Math.sqrt(a); b = 1/Math.sqrt(b);  
 console.log(A,B)  
 console.log(X,Y)  
 console.log(a,b)  
 console.log(B-b)  
 ellipse(0,B-b,a,b);  
 })()  

gjm, 3 months ago

@fanf (ah, triple-backquote formatting doesn't do what I hoped it might, at least not in the web UI, but no matter)

gjm, 3 months ago

@christianp No recollection from when I was at school, but my brain-autocomplete strongly prefers "lady" to "man".

gjm, 4 months ago

@johncarlosbaez @mattmcirvin Becoming useless?

gjm, 4 months ago

@ColinTheMathmo Also, and much less interestingly, you're my uncle. (My PhD supervisor was one of Bela's students.)

gjm, 4 months ago

@ColinTheMathmo Yup. Always went by Keith rather than Thomas, so far as I know. Very nice chap.

gjm, 4 months ago

@cstross It would probably be possible (which is not to say it would be a good idea) to set up a mechanism where a reader can commit that if all the books in a promised series get published then they will immediately buy all of them, in a way that lets the publisher (and author) see that the demand is there even though the earlier books aren't being bought yet.

I can't see it actually working very well: it would be difficult, or at least more trouble than it'd be worth, to force readers to pay up, and publishers and authors still wouldn't be getting the money until the whole series was finished.

You could layer on further complexity to deal with the second of those problems: no doubt there would be financial institutions willing to buy "$X+$Y+$Z once all three books are done" and sell "something less than $X when the first book is done, something less than $Y when the second book is done, and something less than $Z when the third is done". But I expect everyone involved would hate it.

gjm, 4 months ago

@rakyat When Uber came along, it didn't involve any technological advances to speak of. Today's AI systems do. The relevant difference is that technological progress sometimes continues.

So the pro-Uber narrative was something like "we're doing a new thing, and it may turn out to be a huge win". The pro-AI narrative is more like "this stuff is improving fast; 5 years ago it was hopeless, now it's arguably comparable with humans for a lot of things we wouldn't have thought machines could do; in 5 years it might be much better than humans at all those things."

It might turn out that no such progress materializes. It might turn out that it does and it's terrible (we all lose our jobs, or lots of wealth gets transferred to people who are already fabulously wealthy at everyone else's expense, or the robots kill all the humans, or whatever). But it's a different situation from Uber -- there's actual progress to extrapolate from which might or might not pan out.

gjm, 5 months ago

@ColinTheMathmo This has been discussed a bit on a mailing list I frequent, where (1) someone has observed that it looks like each way of choosing which variable is which sort of number leads (using the usual parameterization of Pythagorean triples) to an algebraic curve of genus 3, and there's a famous theorem saying that such a curve has only finitely many rational solutions, so we'd expect there only to be a finite number of solutions, and (2) someone else has done the number-crunching for solutions with largest number up to about 10^14, and found no new solutions.

So my guess is that there aren't any more.

(I think proving that one's found all the possible solutions to a thing like this is generally difficult, but I'm not an expert.)

gjm, 6 months ago

@ColinTheMathmo yeah, seems OK to me.

gjm, 3 months ago

@gregeganSF @johncarlosbaez Does the paper actually explain anywhere how that figure was drawn?

It doesn't look like it to me, but I haven't read the paper carefully.

It's a plot of roots of minimal polynomials of "expansion constants" (= exp(entropy)) associated with certain dynamical systems on the unit interval.

There's an explicitly stated theorem of the form "x is an expansion constant for such a system iff it's a positive real algebraic integer whose absolute value is bigger than that of any of its Galois conjugates".

Except that that's for a more general class of dynamical systems, in which the function being iterated can wlog be taken to be a polynomial of arbitrary degree, and the "entropy bagel" picture (which Thurston doesn't call by that name) shows all the conjugates of expansion constants derived from quadratic polynomials.

The paper talks quite a bit about what's going on when the polynomials have bounded degree, but if there's an explicit criterion or algorithm given then I've failed to see it.

(Most likely it's there, or at least implicitly so, and I have failed to see it.)

gjm, 3 months ago

@gregeganSF @johncarlosbaez Well, the paper does say how to find all the expansion constants for postcritically finite maps. But what it doesn't (at least on my cursory reading) clarify is how you find specifically the expansion constants for postcritically finite quadratic maps.

(The entropy, hence the expansion constant, is invariant under a fairly large class of transformations of the maps, which allegedly means it's enough to consider only polynomials, and "only quadratic maps" here is apparently equivalent to "only maps with a single turning point".)