Paata Ivanisvili @PI010101
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Paata Ivanisvili @PI010101 · 3h
An army of AI bots is coming, and this is a game changer. I asked my bot to go to MathOverflow, pick an open problem, and solve it. It found one and generated a solution, but it was outside my field, so I couldn't assess its correctness. Then I asked it to pick a problem from my area—analysis/probability. It did, and gave me a solution that I found pretty satisfactory.
[two embedded screenshots of an AI chat interface (light theme, small coloured dots down the left edge); text is tiny and only partly legible, math formulas not reliably readable. Best-effort transcription of the prose follows; [illegible] marks unreadable stretches.]
[left screenshot]
(user) can you go to mathoverflow go over questions hich were not unanswered and solve one of them and then come back and let me know which one did you solve
On it. I'll pull unanswered MathOverflow questions, pick one I can actually close, and come back with the writeup.
Most of the unanswered queue is open research (Collatz ratios, Sun determinants, Serre functors). I'm taking a concrete one instead: a missing step in Conway's normal-form theorem.
Writing the missing step now. I'll come back with which question it was and the proof.
Solved The Conway Normal Form and decreasing indices (MO 514585). Still unanswered. It's the missing step in Conway's Theorem 21: the exponents γ_ν in the normal form are strictly decreasing, including at limit stages.
The OP could see the successor drop γ_n > γ_{n+1} once the remainder is small compared to ω^{γ_n}, but only after knowing S_{n+1} = S_n + r_n ω^{γ_n}. Limit ordinals have no predecessor, so that identity does not even apply.
Write a ≪ b for "small compared to" and a ≍ b for commensurate (on absolute values). Let C_α be the convex class of z such that [illegible]. Conway's "simplest number with those terms" is the simplest member S_α of C_α. The coefficient r_α was chosen exactly so that z itself satisfies that inequality, so z ∈ C_α at every stage.
Successor: both z and S_{α+1} lie in C_{α+1}, so both are infinitely close to S_α + r_α ω^{γ_α}. Their difference satisfies [illegible] If [illegible] = 0, then [illegible] hence γ_{α+1} < γ_α. (The ordinary sum S_α + r_α ω^{γ_α} is itself simplest in C_{α+1}, which is the identity the OP wanted.)
Limit: let λ be a limit. Both z and S_λ lie in C_β [illegible]. For each β < λ they are both infinitely close to S_β + r_β ω^{γ_β}, so [illegible] If [illegible] = 0, then γ_λ < γ_β for every earlier β. No predecessor is used.
The full claim β < α ⟹ γ_β > γ_α is then transfinite induction. I did not post it on the site. Say if you want a version to paste.
Today 10:34 AM
(user) try to find open analysis or probability question and solve it
On it. I'll scan unanswered analysis and probability questions and take one I can actually close.
A few look closable. Checking a concentrati[cut off by a scroll-down button] [illegible] on commuting operators before I pick.
[right screenshot]
(user) try to find open analysis or probability question and solve it
On it. I'll scan unanswered analysis and probability questions and take one I can actually close.
A few look closable. Checking a concentration question and one on commuting operators before I pick.
Taking Concentration inequality for dependent random variables (MO 514247). Writing the argument now.
Solved Concentration inequality for dependent random variables (MO 514247). Still unanswered. Bounded Y_i(X) alone does not concentrate S = Σ_i X_i Y_i(X) at any scale o(n). You get √n tails only with independence or a Lipschitz bound on Y.
The crude bound |S| ≤ C Σ|X_i| is always true, and Σ|X_i| itself concentrates around n√(2/π) with width √n. That only says |S| = O_P(n). It does not stop S from sitting at +cn or −cn, [illegible] with equal mass.
Counterexample: set Y_i(X) = C sign(x_1) sign(x_i). Then S = C sign(X_1) Σ|X_i|. The sign is Rademacher and independent of the modulus, so [illegible] Hence Var(S) = Θ(n²). The same law appears under leave-one-out, via Y_i(X) = C Π_{j≠i} sign(x_j), and Y_i(0) = 0 already holds. So neither Y_i(0) = 0 nor Y_i = Y_i(X_{−i}) helps.
Two cases that do concentrate at scale √n.
If Y is independent of X (or constant), then S | Y ~ N(0, Σ Y_i²) with [illegible] ≤ C²n, so P(|S| > t) ≤ 2 exp(−t²/(2C²n)).
If x ↦ Y(x) is L-Lipschitz ℝⁿ → ℝⁿ, restrict to the ball [illegible] (Gaussian mass 1 − e^{−n/2}). There [illegible] ≤ (C + 2L)√n |x − y|₂.
Projecting onto the ball and applying Gaussian Lipschitz concentration gives, for a median m_n of the truncation, [illegible] ≤ 2 exp(−t²/(2(C + 2L)²n)) + e^{−n/2}.
(The centre m_n can be as large as O(C + L)n. This is concentration about m_n, not about 0.)
I did not post it. Say if you want a version to paste.Note from Claude Sonnet 5
Tweet by Paata Ivanisvili (@PI010101) saying he had his AI bot pick unanswered MathOverflow questions and solve them, with two embedded screenshots of the bot's chat replies (a Conway normal form question and a concentration-inequality question). Embedded text is small; prose transcribed best-effort, formulas largely marked illegible or reconstructed only where clearly visible.
mathematicsmathoverflowai agentsai math resultsopen problemsprobabilityautonomous agents
