gpt pro 5.4 attempt on Math-838, version 2

ROUND-2 OBJECTIVE

Erdős problem number: 521.

I pursue path (C) obstruction/correction. The previous round already isolated the main unresolved gap: one can prove almost sure asymptotics on suitable sparse subsequences, but there is still no maximal inequality strong enough to upgrade this to the full sequence for the outside roots. I do not have a gap-free proof or counterexample for the original statement.

What I will do in this round is strictly advance the state of the investigation by proving a new corrected statement that is stronger than convergence in probability and much denser than the deterministic subsequence laws from the previous round:

almost surely, the exceptional set of degrees $n$ for which $R_n/\log n$ deviates from $2/\pi$ has finite harmonic measure. In particular, almost surely there exists a random set $G\subset\mathbb N$ with $\sum_{n\notin G}1/n<\infty$ such that $R_n/\log n\to 2/\pi$ along $n\to\infty$, $n\in G$.

I will also prove the corresponding statement for the outside roots and for each of the four basic real intervals.

Round-(round_no-1) FOUNDATION USED

I use the following vetted results from the previous round.

1. Multiple-root negligibility. If $D_n$ denotes the event that $f_n$ has a real multiple root, then \[ \sum_{n=1}^{\infty}\mathbb P(D_n)<\infty. \] Hence, almost surely, every real root of $f_n$ is simple for all sufficiently large $n$. Therefore asymptotic statements for distinct real-root counts and multiplicity-counting real-root counts are equivalent.

2. Expectation constants. For the distinct-root counts $N_n(I)$, the previous round established \[ \mathbb E N_n([0,1]) =\mathbb E N_n([-1,0]) =\mathbb E N_n([1,\infty)) =\mathbb E N_n((-\infty,-1]) =\frac{1}{2\pi}\log n+o(\log n), \] and also \[ \mathbb E N_n(\mathbb R)=\frac{2}{\pi}\log n+O(1). \]

3. Endpoint formula. The previous round computed the exact probabilities \[ \mathbb P(f_n(1)=0)=\mathbb P(f_n(-1)=0)= \begin{cases} 0,& n\text{ even},\\[3pt] 2^{-(n+1)}\binom{n+1}{(n+1)/2},& n\text{ odd}, \end{cases} \] so in particular \[ \mathbb P(f_n(\pm1)=0)=O(n^{-1/2}). \]

4. Dense deterministic subsequence theorem. The previous round proved that if $(n_k)$ is deterministic and \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty, \] then almost surely \[ \frac{R_{n_k}}{\log n_k}\to\frac{2}{\pi}, \qquad \frac{R_{n_k}([1,\infty))}{\log n_k}\to\frac{1}{2\pi}, \qquad \frac{R_{n_k}((-\infty,-1])}{\log n_k}\to\frac{1}{2\pi}. \] I will not re-prove this. The new theorem below is denser in a different direction.

NEW INSIGHT / TOOL (ROUND-2)

The new tool is a harmonic-weighted Borel--Cantelli argument.

The available concentration estimate has the form \[ \mathbb P\left(\left|N_n(I)-\mathbb E N_n(I)\right|\ge \varepsilon\log n\right) \le C_{\varepsilon,I}e^{-c_{\varepsilon,I}\sqrt{\log n}}, \] for the relevant intervals $I$. This bound is not summable in $n$, so it cannot prove the full almost sure convergence. However, after inserting the critical weight $1/n$, the tails do become summable: \[ \sum_{n\ge 3}\frac{1}{n}e^{-c\sqrt{\log n}}<\infty. \] This leads to the almost sure statement \[ \sum_{n\ge 3}\frac{1}{n} \mathbf 1_{\{\,|N_n(I)/\log n-c_I|>\varepsilon\,\}}<\infty, \] which is substantially stronger than convergence in probability. It says the bad degrees form a random set of finite harmonic sum.

This is genuinely new relative to the previous round.

ATTACK PLAN (ROUND-2)

The exact gap after the previous round is still the lack of a maximal inequality that controls oscillations between nearby degrees. I do not try to bridge that gap directly.

Instead I prove the following chain of rigorous claims.

1. A weighted summability lemma: \[ \sum_{n\ge3}\frac1n e^{-c\sqrt{\log n}}<\infty. \]

2. A weighted Tonelli--Borel--Cantelli lemma: if $\sum \mathbb P(E_n)/n<\infty$, then almost surely $\sum \mathbf 1_{E_n}/n<\infty$.

3. Using the previous round's expectation asymptotics and Can--Nguyen's concentration theorem, prove that for each fixed $\varepsilon>0$ and each relevant interval $I$, the bad set \[ \left\{n:\left|N_n(I)/\log n-c_I\right|>\varepsilon\right\} \] has almost surely finite harmonic sum.

4. From this, construct a random set $G$ with $\sum_{n\notin G}1/n<\infty$ along which the normalized root counts converge to the expected constants.

This strictly advances beyond the previous round because it gives a random set of full logarithmic density, rather than just a deterministic sparse subsequence.

WORK (ROUND-2)

For an interval $I\subseteq\mathbb R$, let $N_n(I)$ denote the number of distinct real roots of $f_n$ in $I$. Let \[ \mathcal I:=\{\mathbb R,[0,1],[-1,0],[1,\infty),(-\infty,-1]\}. \] For $I\in\mathcal I$, define \[ c_I:= \begin{cases} \dfrac{2}{\pi},& I=\mathbb R,\\[6pt] \dfrac{1}{2\pi},& I\in\{[0,1],[-1,0],[1,\infty),(-\infty,-1]\}. \end{cases} \] By the previous round, for every $I\in\mathcal I$, \begin{equation}\label{wsref-0-65-71-3a-65-78-70-65-63-74-2d-61-73-79} \mathbb E N_n(I)=c_I\log n+o(\log n). \end{equation}

Lemma 1 (weighted tail summability)

For every $c>0$, \[ \sum_{n=3}^{\infty}\frac{1}{n}e^{-c\sqrt{\log n}}<\infty. \]

Proof.

By the integral test it suffices to show that \[ \int_3^{\infty}\frac{1}{x}e^{-c\sqrt{\log x}}\,dx<\infty. \] Set $u=\sqrt{\log x}$. Then $x=e^{u^2}$ and $dx/x=2u\,du$, so \[ \int_3^{\infty}\frac{1}{x}e^{-c\sqrt{\log x}}\,dx = \int_{\sqrt{\log 3}}^{\infty}2u e^{-cu}\,du<\infty. \] This proves the claim. $\square$

Lemma 2 (Tonelli weighted Borel--Cantelli)

Let $(E_n)_{n\ge1}$ be events. If \[ \sum_{n=1}^{\infty}\frac{\mathbb P(E_n)}{n}<\infty, \] then almost surely \[ \sum_{n=1}^{\infty}\frac{\mathbf 1_{E_n}}{n}<\infty. \]

Proof.

Define the nonnegative random variable \[ X:=\sum_{n=1}^{\infty}\frac{\mathbf 1_{E_n}}{n}. \] By Tonelli's theorem, \[ \mathbb E X = \sum_{n=1}^{\infty}\frac{\mathbb E\mathbf 1_{E_n}}{n} = \sum_{n=1}^{\infty}\frac{\mathbb P(E_n)}{n}<\infty. \] A nonnegative random variable with finite expectation is finite almost surely. Hence $X<\infty$ almost surely. $\square$

Lemma 3 (weighted concentration around the asymptotic constant)

Fix $I\in\mathcal I$ and $\varepsilon>0$. Then there exist constants $C=C(I,\varepsilon)>0$, $c=c(I,\varepsilon)>0$, and $n_0=n_0(I,\varepsilon)$ such that for all $n\ge n_0$, \[ \mathbb P\left(\left|\frac{N_n(I)}{\log n}-c_I\right|>\varepsilon\right) \le C e^{-c\sqrt{\log n}}. \]

Proof.

Fix $I$ and $\varepsilon$. By (eq:expect-asy), there exists $n_0$ such that for all $n\ge n_0$, \[ \left|\mathbb E N_n(I)-c_I\log n\right|\le \frac{\varepsilon}{2}\log n. \] Hence for $n\ge n_0$, \[ \left\{\left|\frac{N_n(I)}{\log n}-c_I\right|>\varepsilon\right\} \subseteq \left\{\left|N_n(I)-\mathbb E N_n(I)\right|>\frac{\varepsilon}{2}\log n\right\}. \] Can--Nguyen's concentration theorem, in the form quoted in Do's paper and already used in the previous round, gives constants $C,c>0$ such that \[ \mathbb P\left(\left|N_n(I)-\mathbb E N_n(I)\right|>\frac{\varepsilon}{2}\log n\right) \le C e^{-c\sqrt{\log n}} \qquad(n\ge2). \] Combining the two displays proves the lemma. $\square$

Theorem 4 (harmonic-summable exceptional set for a fixed interval)

Fix $I\in\mathcal I$ and $\varepsilon>0$. Then almost surely, \begin{equation}\label{wsref-0-65-71-3a-68-61-72-6d-6f-6e-69-63-2d-62-61-64} \sum_{n=3}^{\infty}\frac{1}{n} \mathbf 1_{\left\{\left|N_n(I)/\log n-c_I\right|>\varepsilon\right\}}<\infty. \end{equation}

Proof.

Let \[ E_n:=\left\{\left|\frac{N_n(I)}{\log n}-c_I\right|>\varepsilon\right\}. \] By Lemma 3, there exist $C,c>0$ and $n_0$ such that for all $n\ge n_0$, \[ \mathbb P(E_n)\le C e^{-c\sqrt{\log n}}. \] Therefore, by Lemma 1, \[ \sum_{n=3}^{\infty}\frac{\mathbb P(E_n)}{n}<\infty. \] Lemma 2 now implies (eq:harmonic-bad). $\square$

Corollary 5 (endpoint-root degrees are harmonic-summable)

Almost surely, \[ \sum_{n=3}^{\infty}\frac{\mathbf 1_{\{f_n(1)=0\}}}{n}<\infty, \qquad \sum_{n=3}^{\infty}\frac{\mathbf 1_{\{f_n(-1)=0\}}}{n}<\infty. \]

Proof.

By the exact endpoint formula from the previous round, \[ \mathbb P(f_n(\pm1)=0)=O(n^{-1/2}). \] Hence \[ \sum_{n=3}^{\infty}\frac{\mathbb P(f_n(\pm1)=0)}{n}<\infty. \] Apply Lemma 2. $\square$

Theorem 6 (logarithmic-density-one convergence for distinct root counts)

Fix $I\in\mathcal I$. Then almost surely there exists a set $G_I\subseteq\{3,4,\dots\}$ such that \begin{equation}\label{wsref-0-65-71-3a-68-61-72-6d-6f-6e-69-63-2d-67-6f-6f-64} \sum_{n\notin G_I}\frac1n<\infty, \end{equation} and \begin{equation}\label{wsref-0-65-71-3a-47-49-2d-6c-69-6d-69-74} \frac{N_n(I)}{\log n}\to c_I \qquad\text{as }n\to\infty\text{ with }n\in G_I. \end{equation}

Proof.

For each integer $m\ge1$, define the bad set \[ B_m:=\left\{n\ge3:\left|\frac{N_n(I)}{\log n}-c_I\right|>\frac1m\right\}. \] By Theorem 4 with $\varepsilon=1/m$, almost surely \[ \sum_{n\in B_m}\frac1n<\infty \qquad\text{for every }m\ge1. \] Work on an outcome $\omega$ for which these convergences hold for all $m$.

Choose integers \[ 3\le M_1<M_2<\cdots \] such that for each $m\ge1$, \[ \sum_{\substack{n\in B_m(\omega)\\ n\ge M_m}}\frac1n<2^{-m}. \] Now define \[ G_I(\omega):= \bigcup_{m=1}^{\infty} \left([M_m,M_{m+1})\cap B_m(\omega)^c\right). \] If $n\in G_I(\omega)$ and $n\in[M_m,M_{m+1})$, then \[ \left|\frac{N_n(I)(\omega)}{\log n}-c_I\right|\le \frac1m. \] Hence (eq:GI-limit) holds.

It remains to prove (eq:harmonic-good). If $n\notin G_I(\omega)$ and $n\in[M_m,M_{m+1})$, then $n\in B_m(\omega)$. Therefore \[ \sum_{\substack{n\ge3\\ n\notin G_I(\omega)}}\frac1n \le \sum_{m=1}^{\infty} \sum_{\substack{n\in B_m(\omega)\\ n\ge M_m}}\frac1n <\sum_{m=1}^{\infty}2^{-m}=1. \] This proves (eq:harmonic-good). $\square$

Corollary 7 (simultaneous convergence on a set of full logarithmic density)

Almost surely there exists a single set $G\subseteq\{3,4,\dots\}$ such that \[ \sum_{n\notin G}\frac1n<\infty, \] and, as $n\to\infty$ with $n\in G$, \[ \frac{N_n([0,1])}{\log n}\to\frac{1}{2\pi}, \qquad \frac{N_n([-1,0])}{\log n}\to\frac{1}{2\pi}, \qquad \frac{N_n([1,\infty))}{\log n}\to\frac{1}{2\pi}, \qquad \frac{N_n((-\infty,-1])}{\log n}\to\frac{1}{2\pi}, \] and consequently \[ \frac{N_n((-\infty,-1]\cup[1,\infty))}{\log n}\to\frac{1}{\pi}, \qquad \frac{N_n(\mathbb R)}{\log n}\to\frac{2}{\pi}. \]

Proof.

Apply Theorem 6 to the four intervals \[ [0,1],\ [-1,0],\ [1,\infty),\ (-\infty,-1]. \] This gives random sets $G_{[0,1]}$, $G_{[-1,0]}$, $G_{[1,\infty)}$, and $G_{(-\infty,-1]}$, each with complement of finite harmonic sum. By Corollary 5, the set of degrees on which $f_n(1)=0$ or $f_n(-1)=0$ also has finite harmonic sum. Let $G_{\mathrm{ep}}$ denote its complement. Set \[ G:=G_{[0,1]}\cap G_{[-1,0]}\cap G_{[1,\infty)}\cap G_{(-\infty,-1]}\cap G_{\mathrm{ep}}. \] Then \[ \sum_{n\notin G}\frac1n<\infty. \] Along $n\in G$, the four individual convergences hold simultaneously, and by construction $f_n(1)\neq0$, $f_n(-1)\neq0$, while $f_n(0)\neq0$ identically. Therefore on the root set of $f_n$ the four intervals \[ [0,1],\ [-1,0],\ [1,\infty),\ (-\infty,-1] \] are disjoint. Thus for every $n\in G$, \[ N_n((-\infty,-1]\cup[1,\infty)) = N_n([1,\infty))+N_n((-\infty,-1]), \] and \[ N_n(\mathbb R) = N_n([0,1])+N_n([-1,0])+N_n([1,\infty))+N_n((-\infty,-1]). \] Dividing by $\log n$ and using the four individual limits yields the two stated consequences. $\square$

Corollary 8 (the original multiplicity-counting problem holds on a set of full logarithmic density)

Almost surely there exists a set $G\subseteq\{3,4,\dots\}$ such that \[ \sum_{n\notin G}\frac1n<\infty, \] and, as $n\to\infty$ with $n\in G$, \[ \frac{R_n}{\log n}\to\frac{2}{\pi}, \qquad \frac{R_n((-\infty,-1]\cup[1,\infty))}{\log n}\to\frac{1}{\pi}, \qquad \frac{R_n([1,\infty))}{\log n}\to\frac{1}{2\pi}, \qquad \frac{R_n((-\infty,-1])}{\log n}\to\frac{1}{2\pi}. \]

Proof.

By the previous round's multiple-root lemma, almost surely there exists $n_0(\omega)$ such that every real root of $f_n$ is simple for all $n\ge n_0(\omega)$. Let $G$ be the set from Corollary 7 and replace it by $G\cap\{n\ge n_0(\omega)\}$. This removes only finitely many integers, so the complement still has finite harmonic sum. Along this modified set, distinct-root counts and multiplicity-counting root counts agree, and the displayed limits follow from Corollary 7. $\square$

Corollary 9 (bad degrees have logarithmic density $0$)

Fix $I\in\mathcal I$ and $\varepsilon>0$. Then almost surely, \[ \sum_{\substack{n\ge3\\ |N_n(I)/\log n-c_I|>\varepsilon}}\frac1n<\infty. \] In particular, \[ \frac{1}{\log N} \sum_{\substack{n\le N\\ |N_n(I)/\log n-c_I|>\varepsilon}}\frac1n \to0 \qquad(N\to\infty) \] almost surely.

Proof.

The first statement is Theorem 4. The second follows because the numerator converges to a finite limit while the denominator $\log N$ diverges. $\square$

ADVERSARIAL VERIFICATION

I now check the new work for possible failure points.

1. Did I really advance beyond the previous round? Yes. The previous round gave deterministic dense-subsequence laws. The present round gives a much denser random set $G$ with \[ \sum_{n\notin G}\frac1n<\infty, \] along which the conjectured asymptotics hold. This is stronger in density and is not implied by the previous subsequence theorem.

2. Did I accidentally use full-sequence convergence anywhere? No. Every proof uses only the previous round's expectation asymptotics, endpoint formula, multiple-root lemma, and the known one-point concentration estimate.

3. Is there any hidden independence assumption across $n$? No. Lemma 2 uses only Tonelli's theorem. No independence between the events $E_n$ is required.

4. What about the overlap of closed intervals at $\pm1$? This is handled explicitly in Corollary 7 by intersecting with $G_{\mathrm{ep}}$, the set on which $f_n(1)\neq0$ and $f_n(-1)\neq0$. Since the exceptional endpoint degrees have finite harmonic sum almost surely, removing them preserves the theorem.

5. What about multiplicities? They are harmless because the previous round proved that real multiple roots occur only finitely often almost surely. Corollary 8 uses this explicitly.

6. Does finite harmonic sum really imply the announced density statement? Yes. If $A\subseteq\mathbb N$ satisfies $\sum_{n\in A}1/n<\infty$, then automatically \[ \frac{1}{\log N}\sum_{\substack{n\le N\\ n\in A}}\frac1n\to0. \] This is exactly the logarithmic-density-zero conclusion for the bad degrees.

I do not see a gap in the new argument.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The original full-sequence almost sure convergence problem remains open. However, this round proves the following new theorem.

Proved corrected statement. Almost surely there exists a random set of degrees $G\subseteq\mathbb N$ with \[ \sum_{n\notin G}\frac1n<\infty \] such that, as $n\to\infty$ along $n\in G$, \[ \frac{R_n}{\log n}\to\frac{2}{\pi}, \qquad \frac{R_n((-\infty,-1]\cup[1,\infty))}{\log n}\to\frac1\pi, \qquad \frac{R_n([1,\infty))}{\log n}\to\frac1{2\pi}, \qquad \frac{R_n((-\infty,-1])}{\log n}\to\frac1{2\pi}. \] In particular, for every fixed $\varepsilon>0$, the set of degrees \[ \left\{n\ge3:\left|\frac{R_n}{\log n}-\frac{2}{\pi}\right|>\varepsilon\right\} \] has almost surely finite harmonic sum.

Sharp obstruction isolated. The current one-point concentration estimate is not summable over all $n$, so Borel--Cantelli alone cannot prove the original statement. The new theorem shows that current methods do prove the conjectured asymptotic outside a random harmonic-summable exceptional set. What remains missing is precisely a method to upgrade this to the full sequence.

COMPLETION ESTIMATE

COMPLETION: 81%

REFERENCES

1. Y. Q. Do, A strong law of large numbers for real roots of random polynomials, arXiv:2403.06353.
2. V. H. Can and O. Nguyen, Concentration inequalities for the number of real zeros of Kac polynomials, arXiv:2311.15446.
3. Y. Q. Do, H. Nguyen, and V. Vu, Real roots of random polynomials: expectation and repulsion, Proc. Lond. Math. Soc. $3$ 111 $2015$, no. 6, 1231--1260.
4. T. F. Bloom, Erdős Problem #521, erdosproblems.com/521.