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Claude Fable produced a counterexample to the Jacobian Conjecture

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Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#31
post #19

Earlier quoted context omitted.

The goal posts have moved. People generally stopped saying this stuff now. Even if you go to the ultimate anti-AI subreddit r/betteroffline, they've changed from "AI is useless" to "AI is good but the AI bubble will collapse soon" over the last 6 months.

Some quotes from a day ago, https://news.ycombinator.com/item?id=48957779 : > I hold my stance that LLMs are stochastic parrots... Making the parrots ever more complex and training > Except solving problem is probably the least (even though it's important) interesting thing in research. > Can we use AI to get a cure for cancer yet? Or is math-turbation the only thing these things are good for? > Train on enough examp…

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Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#32
The great thing about these mathematical mopping up type operations is that no person will waste their time trying to prove it to be true anymore. If anything that’s a win.

It would be great if an LLM could settle the Collatz conjecture next, god knows how many man-years have been burned on that by unsuspecting victims.

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#33
post #20
post #17

I suspect the LLM was able to synthesize a counterexample because of the availability of a lot of prior work: > The Jacobian conjecture is notorious for the large number of published and unpublished proofs that turned out to contain subtle errors. https://en.wikipedia.org/wiki/Jacobian_conjecture#cite_note-...

How does that work in this case? What do those proofs do to help find this counterexample?

In my view, it's theoretically possible for a combination of the author's iterative prompts + evaluation with Wolfram Alpha to activate the weights that encode the language that describes the constraints on these polynomials (from the faulty proofs) in such a way that the author eventually arrives at this:

> ((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#34

Earlier quoted context omitted.

https://en.wikipedia.org/wiki/Sycophancy_(artificial_intelli...

The interesting thing about using Claude Fable 5 is it's nearly as irritatingly sycophantic as past Claudes while genuinely being smarter than the previous models. So you get a kind of yo-yoing of it glazing you as a creative genius and disappointedly revealing to you that your ideas are bad and dumb.

That is easily fixable with the "Instructions for Claude" in your settings. Just put:

"be critical, but correct. I don't want affirmation, i want to actually accomplish things."

I don't get any sycophancy from it. The conversations are actually quite good, and it'll push back if it thinks I'm wrong.

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#35
post #2

Context: https://en.wikipedia.org/wiki/Jacobian_conjecture

> Jacobian conjecture [...] states that if a polynomial function from an n-dimensional space to itself has a Jacobian determinant which is a non-zero constant, then the function has a polynomial inverse. > ((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3, has jacobian determinant -2, and sends (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) to (-1/4, 0, 0…

That it has an inverse.

If f(a) = f(b) for a≠b then f can't have an inverse.

Suppose f has inverse g; then g(f(x)) must = x for all x.

But then g(f(a)) would have to equal a, and g(f(b)) would have to equal b. But they can't, because f(a)=f(b)

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#36

I asked Fable to verify and it absolutely freaked out! I have no idea what any of this stuff even means, but my AI thinks I’m a legend level mathematician!

https://en.wikipedia.org/wiki/Sycophancy_(artificial_intelli...

I mean if someone came to you with a value of n that disproved collatz, wouldn't you go crazy as well?

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#37
post #33
post #20

Earlier quoted context omitted.

How does that work in this case? What do those proofs do to help find this counterexample?

In my view, it's theoretically possible for a combination of the author's iterative prompts + evaluation with Wolfram Alpha to activate the weights that encode the language that describes the constraints on these polynomials (from the faulty proofs) in such a way that the author eventually arrives at this: > ((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3

How is that not like claiming that a "combination of iterative prompts and evaluation with Wolfram Alpha" would produce my Google Mail password? I also have to note that this response abandons the original claim, that the counterexample was found by exhaustively searching prior failed attempts.

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#38

I asked Fable to verify and it absolutely freaked out! I have no idea what any of this stuff even means, but my AI thinks I’m a legend level mathematician!

https://en.wikipedia.org/wiki/Sycophancy_(artificial_intelli...

To be fair to the sycophancy tendencies, this was an open conjecture that held for 85 years, and not for lack of trying. So, maybe a bit warranted here? :)

Re: Claude Fable produced a counterexample to the Jacobian Conjecture

#39

The great thing about these mathematical mopping up type operations is that no person will waste their time trying to prove it to be true anymore. If anything that’s a win. It would be great if an LLM could settle the Collatz conjecture next, god knows how many man-years have been burned on that by unsuspecting victims.

The reason this was "easy" is because the conjecture turned out to be false. If the collatz conjecture holds true (and most mathematicians seem to think it will), it will be much harder to prove than your average Erdos problem.
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