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.
People have definitely not stopped saying this stuff.
Claude Fable produced a counterexample to the Jacobian Conjecture
41–50 of 562 posts
Re: Claude Fable produced a counterexample to the Jacobian Conjecture
#42Re: Claude Fable produced a counterexample to the Jacobian Conjecture
#43Maybe not? https://en.wikipedia.org/w/index.php?title=Jacobian_conjectu...
UPD: The edit got reverted and there's this on the talk page now: https://en.wikipedia.org/wiki/Talk:Jacobian_conjecture#c-DaR...
UPD2: There are edit wars happening now: https://en.wikipedia.org/w/index.php?title=Jacobian_conjectu... https://en.wikipedia.org/wiki/Talk:Jacobian_conjecture#c-Sea...
Re: Claude Fable produced a counterexample to the Jacobian Conjecture
#44It's reasoning from a flawed premise that math universally requires intelligence and creativity. It does not. Anyone that's proved things via "diagram chasing" can affirm that. The conclusion you should draw is that math (at least the kind they excel at) isn't actually a creative endeavor.
Re: Claude Fable produced a counterexample to the Jacobian Conjecture
#45The 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.
Re: Claude Fable produced a counterexample to the Jacobian Conjecture
#46Re: Claude Fable produced a counterexample to the Jacobian Conjecture
#47Earlier quoted context omitted.
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.
I'm yet to see anyone cite the prior work. If it's true I think some credit is due for the authors Fable/Sol are branching from.
Re: Claude Fable produced a counterexample to the Jacobian Conjecture
#48Maybe not? https://en.wikipedia.org/w/index.php?title=Jacobian_conjectu...
Let this be a lesson to those who fell for such AI psychosis and to not believe everything you see on the internet as real.
Re: Claude Fable produced a counterexample to the Jacobian Conjecture
#49Anybody can ELI5?
If so, you've probably heard of the determinant. It's a certain way of "summarizing" a matrix with one value.
The determinant in this case is of the Jacobian, which is a matrix you can construct from a multi-variable function. Each term is the partial derivative with respect to each variable (x, y, z, etc.), with one line per output variable (vector element).
The Jacobian of a polynomial function is, in general, going to be a matrix where every term is some polynomial expression. And the determinant of that will also be a complicated expression. But in some cases all the variable terms cancel out and you're left with a single constant (0 or some other value).
The conjecture says that if the Jacobian determinant is constant (i.e., all the terms cancel out), then there must be a polynomial inverse. And the key condition for an inverse is that there must not be two input points that evaluate to the same output. It's just like y=x^2. It's not invertible, because both +2 and -2 square to +4.
So if you can find a function where the Jacobian determinant is constant and also find two or more points that evaluate to the same output, then you've found a counterexample to the conjecture. And that's what's been done. And remarkably, the counterexample is pretty simple. It would be tedious but a bright high school student could verify it.
Re: Claude Fable produced a counterexample to the Jacobian Conjecture
#50Maybe not? https://en.wikipedia.org/w/index.php?title=Jacobian_conjectu...
I'd rather wait for independent seasoned mathematicians to verify such claims first before someone at said AI lab posting a claim about solving a proof online. Let this be a lesson to those who fell for such AI psychosis and to not believe everything you see on the internet as real.
The author is a Princeton math doctorate.