the ultimate game of AI chess: Stockfish (white) vs ChatGPT (black)
ah wait this reminds me
(gian-carlo rota on alonzo church)
"Person who thinks flavour doesn't exist because they know how taste works" except it's "Person who thinks the cloud doesn't exist because they know how cloud storage works"
I am a CHAMPION! Left the watering can out front and two snails crawled right up in there. Managed to notice and coax both of them out and put them in the compost heap.
thinking about the time a prof told us that in real research mathematics it's fine to be slow, speed itself is not essential, as long as you can find it within yourself to make consistent unyielding inexorable forward progress, like the time some guy stole an M60A3 tank and terrorized a suburban neighborhood with it, said guy wasn't going that fast but plowed through cars and telephone poles and shit no problem. i'm not kidding that's what he said, that's the metaphor he used, he told us that the act of mathematics is like the 1995 san diego tank rampage
FUCKING PIECE OF SHIT HITBOXES THAT'S BULLSHIT CHESS YOU CAN'T TELL ME I GOT HIT THERE THIS LATENCY CRAP IS COMPLETELY UNFAIR
Currently TAing group theory so here's a fun little elementary problem I gave my students: Prove that any finite group with more than 2 elements has a nontrivial automorphism. The finiteness condition is not really necessary but whatever.
Feel free to reblog with your solution!
My guess is:
•If abelian and an element g has order>2 then g→g² should be fine (rather the "automorphism generated by g→g²", whatever that means).
•If abelian and all elements have order 2, let g,h distinct, then swapping g and h should work
•Otherwise, if g and h don't commute, then x→gxg^{-1} will work.
That breaks down at groups like Z/4Z, since you're sending 1 to 2, which is in turn sent to 0, so it's not an automorphism (φ(x) should have the same order as x).
But I do like the idea of separately considering Boolean and non-Boolean groups, so I came up with this
- If G contains an element g with order > 2, then φ(x) = x^{-1} is a non-trivial automorphism
- Otherwise, there are at least 2 elements g and h in G with order 2, and we can swap them like you said
This way it doesn't matter whether G is finite/abelian or not :)
Currently TAing group theory so here's a fun little elementary problem I gave my students: Prove that any finite group with more than 2 elements has a nontrivial automorphism. The finiteness condition is not really necessary but whatever.
Feel free to reblog with your solution!
My guess is:
•If abelian and an element g has order>2 then g→g² should be fine (rather the "automorphism generated by g→g²", whatever that means).
•If abelian and all elements have order 2, let g,h distinct, then swapping g and h should work
•Otherwise, if g and h don't commute, then x→gxg^{-1} will work.
Surprisingly, you need the finiteness assumption to define the swapping automorphism without choice. There are infinite Abelian groups with trivial automorphism group in some models of ZF. The finiteness means you can find a basis for it and show it's isomorphic to Z_2^n. But without a basis there isn't really any way to define a swapping automorphism.
Currently TAing group theory so here's a fun little elementary problem I gave my students: Prove that any finite group with more than 2 elements has a nontrivial automorphism. The finiteness condition is not really necessary but whatever.
Feel free to reblog with your solution!
My guess is:
•If abelian and an element g has order>2 then g→g² should be fine (rather the "automorphism generated by g→g²", whatever that means).
•If abelian and all elements have order 2, let g,h distinct, then swapping g and h should work
•Otherwise, if g and h don't commute, then x→gxg^{-1} will work.
Physicist: What do you work on?
Mathematician: Heyting algebra.
Physicist: Me too!
Blocking someone of course removes your ability to send the blog asks or messages.
But doesn't stop you from being able to gift them checkmarks.
"average academic publishes 3 papers a year" factoid actualy just statistical error. average academic publishes 0 papers per year. Papers Shelah, who lives in a cave & has published an average of 1.7 papers per month for the last 45 years, is an outlier adn should not have been counted
Can anyone come up with a function that always sends algebraic numbers to algebraic numbers, but has a local max or min at a transcendental number? I think there should be one but I can’t come up with one.
How smooth do you want this function to be?
As g-g says, big dependency on smoothness. We could say f(x)=x if x≤π and f(x)=-x if x>π but that has a pretty major discontinuity.
My instinct is that if f is differentiable and definable in a “normal way” then since it sends algebrac to algebraic, it has to only really have algebraic parameters. In that case, solving f’(x)=0 will give a roughly algebraic solution. So to find something differentiable that does this, it’ll need to be fairly pathological (and might not be constructive?).
Does that example have a local max or min? I guess if you just define it at π at all, that can be a max or a min. But yeah, I do at least want local continuity at the max or min.
And yeah, your second paragraph is sort of where I’m getting stuck. The obvious functions that send algebraics to algebraics are, like, polynomials, but then those have polynomial derivatives and so algebraic critical points. And you can expand those out to various sorts of rational functions, but those should have the same problem, I think.
And conversely I can imagine a function that like is an upside-down vee whose vertex has π as an x-coordinate, but I’m not sure if you can make both sides of that vee be algebraic and still have them meet in the middle? Like if the left half is y=x, the right half would have to be y = 2π-x or something and that’s not algebraic any more. But maybe if they have different slopes I can make it work somehow?
Couldn’t be a vee even with different slopes. You’d need y=A+Bx on the left and y=C+Dx on the right, with A+Bπ=C+Dπ, or π=(A-C)/(D-B). Since it’s algebraic to algebraic, A and C are both algebraic (x=0), A+B and C+D are algebraic (x=1), and hence B and D are algebraic too. So (A-C)/(D-B) is algebraic, if defined. So B=D, but then we’re not a vee we’re a line.
If you only want continuity at the local maximum, then it should be possible, but the examples that I’m coming up with look weird. Eg, pick a sequence of algebraic numbers (a_n) such that a_n→0 as n→∞. Then say f(π)=1, and if 1/n≤|x-π|<1/(n+1) then f(x)=(1-a_n).
I think for a counterexample to the EVT I need it to be continuous on some neighborhood of the local max, right? If it’s continuous on any neighborhood then you can stick a closed interval inside that neighborhood and your function is continuous on a closed interval but doesn’t have a local max.
I think your example is continuous at π but not in a neighborhood of π, right? But maybe if we tweak it we can get an example that’s continuous on a neighborhood.
(It’s also possible that no such thing exists, but I’m pretty sure that the EVT requires completeness, which is why I think an example should exist.)
Maybe I need to run down to the library and see if they have a copy of Counterexamples in Analysis.
Yeah, it's not continuous at π±(1/n). However, if you were to, instead of jumping up, add little slopes then I think that it works.
For each n, pick rational numbers (p_n,q_n) such that p_n<π-(1/(n-1))<q_n<π-(1/n). Then, have f(x) be as before, except in the intervals (p_n,q_n) it's linear interpolation. Since it's a straight line between two algebraic co-ordinates (p_n,a_{n-1}) to (q_n,a_n) it'll have algebraic parameters and send algebraics to algebraics. Then it's continuous everywhere and has max at π. It looks like this:
Can anyone come up with a function that always sends algebraic numbers to algebraic numbers, but has a local max or min at a transcendental number? I think there should be one but I can’t come up with one.
How smooth do you want this function to be?
As g-g says, big dependency on smoothness. We could say f(x)=x if x≤π and f(x)=-x if x>π but that has a pretty major discontinuity.
My instinct is that if f is differentiable and definable in a “normal way” then since it sends algebrac to algebraic, it has to only really have algebraic parameters. In that case, solving f’(x)=0 will give a roughly algebraic solution. So to find something differentiable that does this, it’ll need to be fairly pathological (and might not be constructive?).
Does that example have a local max or min? I guess if you just define it at π at all, that can be a max or a min. But yeah, I do at least want local continuity at the max or min.
And yeah, your second paragraph is sort of where I’m getting stuck. The obvious functions that send algebraics to algebraics are, like, polynomials, but then those have polynomial derivatives and so algebraic critical points. And you can expand those out to various sorts of rational functions, but those should have the same problem, I think.
And conversely I can imagine a function that like is an upside-down vee whose vertex has π as an x-coordinate, but I’m not sure if you can make both sides of that vee be algebraic and still have them meet in the middle? Like if the left half is y=x, the right half would have to be y = 2π-x or something and that’s not algebraic any more. But maybe if they have different slopes I can make it work somehow?
Couldn't be a vee even with different slopes. You'd need y=A+Bx on the left and y=C+Dx on the right, with A+Bπ=C+Dπ, or π=(A-C)/(D-B). Since it's algebraic to algebraic, A and C are both algebraic (x=0), A+B and C+D are algebraic (x=1), and hence B and D are algebraic too. So (A-C)/(D-B) is algebraic, if defined. So B=D, but then we're not a vee we're a line.
If you only want continuity at the local maximum, then it should be possible, but the examples that I'm coming up with look weird. Eg, pick a sequence of algebraic numbers (a_n) such that a_n→0 as n→∞. Then say f(π)=1, and if 1/n≤|x-π|<1/(n+1) then f(x)=(1-a_n).
Can anyone come up with a function that always sends algebraic numbers to algebraic numbers, but has a local max or min at a transcendental number? I think there should be one but I can’t come up with one.
How smooth do you want this function to be?
As g-g says, big dependency on smoothness. We could say f(x)=x if x≤π and f(x)=-x if x>π but that has a pretty major discontinuity.
My instinct is that if f is differentiable and definable in a "normal way" then since it sends algebrac to algebraic, it has to only really have algebraic parameters. In that case, solving f'(x)=0 will give a roughly algebraic solution. So to find something differentiable that does this, it'll need to be fairly pathological (and might not be constructive?).
I don't think this guy knows what an equation is.
I regret posting this because every time it gets a note I'm reminded of its existence.
I have one even more powerful for him:
E = mc² + AI * crypto
Imagine where we can get with Blockchain-based AIs!
BUT THEN YOU ADD KURT ANGLE TO THE MIX
