gpt pro 5.4 attempt on Math-838

ROUND-2 OBJECTIVE

I pursue path (C) obstruction/correction. The Round-1 output already reduced the problem to a specific missing ingredient: a maximal inequality capable of upgrading subsequence convergence to full-sequence convergence for the outside roots. I do not have a gap-free proof or counterexample for the original full-sequence statement. Instead, I will do two things that strictly advance the investigation.

First, I isolate a precise obstruction: the currently available one-point concentration bound is not summable along the full sequence, so Borel--Cantelli alone can never prove the desired almost sure full-sequence law. Second, I push the Borel--Cantelli method to its natural limit and prove a substantially stronger subsequence theorem than Round 1: the almost sure asymptotic law holds not only on lacunary subsequences, but on every deterministic subsequence $(n_k)$ satisfying \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty. \] This includes explicit non-lacunary subsequences with \[ \frac{n_{k+1}}{n_k}\to 1. \] Using this, together with Round-1 symmetry lemmas and Do's strong law on $[-1,1]$, I obtain exact almost sure asymptotics along such dense subsequences for the hard outside intervals $[1,\infty)$, $(-\infty,-1]$, and their union.

Round-1 FOUNDATION USED

I will use the following vetted Round-1 results without reproving them.

• Round-1 multiple-root lemma. 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, all real roots of $f_n$ are simple for all sufficiently large $n$. In particular, for all large $n$, counting real roots with multiplicity agrees with counting distinct real roots on every real interval.

• Round-1 endpoint formula. For each $n\ge 0$, \[ \mathbb P(f_n(1)=0)=\mathbb P(f_n(-1)=0)= \begin{cases} 0,& n \text{ even},\\[3pt] 2^{-(n+1)}\binom{n+1}{(n+1)/2},& n \text{ odd}. \end{cases} \] In particular, \[ \mathbb P(f_n(\pm 1)=0)=O(n^{-1/2}). \]

• Round-1 reciprocal expectation symmetry. If $N_n(a,b)$ denotes the number of distinct real roots of $f_n$ in $(a,b)$, then \[ \mathbb E N_n(0,1)=\mathbb E N_n(1,\infty), \qquad \mathbb E N_n(-1,0)=\mathbb E N_n(-\infty,-1). \]

• Round-1 lacunary global law. Along any lacunary subsequence $(n_k)$ with $n_{k+1}\ge q n_k$ for some fixed $q>1$, \[ \frac{R_{n_k}}{\log n_k}\to \frac{2}{\pi} \qquad\text{almost surely.} \] The present round strengthens this substantially.

NEW INSIGHT / TOOL (ROUND-2)

The new input is the following observation.

Summability, not lacunarity, is the real issue. For each fixed $\varepsilon>0$ and each interval \[ I\in\{\mathbb R,[0,1],[1,\infty),[-1,0],(-\infty,-1]\}, \] Can--Nguyen's concentration theorem (quoted explicitly in Do's paper) gives constants $c_{\varepsilon,I},C_{\varepsilon,I}>0$ such that \[ \mathbb P\left(|N_n(I)-\mathbb E N_n(I)|\ge \varepsilon \log n\right) \le C_{\varepsilon,I} e^{-c_{\varepsilon,I}\sqrt{\log n}} \qquad(n\ge 2). \] Therefore, Borel--Cantelli works on every deterministic subsequence $(n_k)$ for which these tails are summable. Lacunarity is only one sufficient condition, not the correct boundary.

A convenient explicit sufficient condition is \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty. \] This still allows very dense subsequences, including ones with $n_{k+1}/n_k\to 1$.

Second new observation. The same summable-tail argument applies not only to the total real-root count, but also to each of the four fundamental real intervals separately. After identifying their expectation constants using the Round-1 symmetry lemmas, one obtains almost sure asymptotics for the outside half-lines $[1,\infty)$ and $(-\infty,-1]$ along these dense subsequences.

ATTACK PLAN (ROUND-2)

The exact Round-1 gap was this: the Borel--Cantelli argument gives almost sure convergence along sparse subsequences, but cannot by itself fill the gaps between neighboring degrees. The missing ingredient is a maximal inequality controlling \[ \max_{n\le m\le n(1+o(1))}|N_m-N_n|. \] Do proved such a maximal inequality for roots in $[0,1]$, which yields the full-sequence strong law there, but explicitly explained that his pairing argument does not extend to the reciprocal-polynomial process needed for the outside roots.

I therefore proceed in two steps.

1. Prove a rigorous obstruction lemma: the one-point tail bound $e^{-c\sqrt{\log n}}$ is not summable along the full sequence, so one-point concentration alone cannot prove the full-sequence strong law.

2. Prove a dense-subsequence strong law: along any deterministic subsequence with \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty, \] we have almost sure convergence of $N_{n_k}(I)/\log n_k$ to the expected constant for each interval \[ I\in\{\mathbb R,[0,1],[-1,0],[1,\infty),(-\infty,-1]\}. \] This gives new a.s. laws for the outside roots and exhibits explicit non-lacunary subsequences with ratio $n_{k+1}/n_k\to1$ on which the conjectured constant is already forced.

This strictly advances beyond Round 1 because it replaces the lacunary theorem by a much denser subsequence theorem and, for the first time in this investigation, yields an exact almost sure law for the outside half-lines along explicit non-lacunary subsequences.

WORK (ROUND-2)

Throughout this section, $N_n(I)$ denotes the number of distinct real roots of $f_n$ in the interval $I$. By the Round-1 multiple-root lemma, for almost every outcome $\omega$, all real roots are simple for all sufficiently large $n$, so every asymptotic statement below remains valid if $N_n(I)$ is replaced by the multiplicity-counting quantity $R_n(I)$.

5.1. Obstruction: one-point concentration is not enough for the full sequence

Lemma 5.1. Let $c>0$. Then \[ \sum_{n=3}^\infty e^{-c\sqrt{\log n}}=\infty. \] Consequently, the presently available one-point concentration bound \[ \mathbb P\left(|N_n(I)-\mathbb E N_n(I)|\ge \varepsilon\log n\right) \le C_\varepsilon e^{-c_\varepsilon\sqrt{\log n}} \] cannot by itself yield the full-sequence almost sure convergence via Borel--Cantelli.

Proof. Consider the integral \[ \int_3^\infty e^{-c\sqrt{\log x}}\,dx. \] Make the substitution $u=\sqrt{\log x}$, so that $x=e^{u^2}$ and $dx=2u e^{u^2}\,du$. Then \[ \int_3^\infty e^{-c\sqrt{\log x}}\,dx =\int_{\sqrt{\log 3}}^\infty 2u\,e^{u^2-cu}\,du. \] Since $u^2-cu\to\infty$ as $u\to\infty$, the integrand $2u e^{u^2-cu}$ tends to $\infty$ and in particular is eventually bounded below by $1$. Hence the integral diverges. By the integral test, \[ \sum_{n=3}^\infty e^{-c\sqrt{\log n}}=\infty. \] The final claim is immediate: the first Borel--Cantelli lemma requires summability of the probabilities along the full sequence, and Lemma 5.1 shows that the current one-point upper bound does not provide this. Therefore any proof of the original full-sequence almost sure law must use more than one-point concentration; specifically, one needs some maximal inequality or other mechanism controlling correlations between nearby degrees. $\square$

5.2. A generic summable-tail subsequence principle

The following lemma is the main new technical device of this round.

Lemma 5.2 (summable-tail subsequence principle). Let $(X_n)_{n\ge2}$ be random variables and $(a_n)_{n\ge2}$ deterministic real numbers. Let $(b_n)_{n\ge2}$ be positive deterministic numbers with $b_n\to\infty$. Assume that for every integer $m\ge1$ there is a sequence $(p_{n,m})_{n\ge2}$ such that \[ \mathbb P\left(|X_n-a_n|\ge m^{-1}b_n\right)\le p_{n,m} \qquad(n\ge2). \] Let $(n_k)_{k\ge1}$ be an increasing sequence of integers such that for every $m\ge1$, \[ \sum_{k=1}^\infty p_{n_k,m}<\infty. \] Then \[ \frac{X_{n_k}-a_{n_k}}{b_{n_k}}\to0 \qquad\text{almost surely as }k\to\infty. \]

Proof. Fix $m\ge1$. Define \[ E_{k,m}:=\left\{|X_{n_k}-a_{n_k}|\ge m^{-1}b_{n_k}\right\}. \] By hypothesis, \[ \sum_{k=1}^\infty \mathbb P(E_{k,m})\le \sum_{k=1}^\infty p_{n_k,m}<\infty. \] The first Borel--Cantelli lemma implies \[ \mathbb P(E_{k,m}\ \text{i.o.})=0. \] Therefore, for almost every outcome $\omega$, there exists $K_m(\omega)$ such that for all $k\ge K_m(\omega)$, \[ |X_{n_k}(\omega)-a_{n_k}|<m^{-1}b_{n_k}. \] Now intersect over all $m\in\mathbb N$. Since this is a countable intersection of probability-one events, it still has probability one. On that event, for every $m\ge1$ the ratio $|X_{n_k}-a_{n_k}|/b_{n_k}$ is eventually at most $1/m$. Hence the ratio tends to $0$. $\square$

5.3. The relevant concentration input

We now record the specific concentration estimate that will be plugged into Lemma 5.2.

Lemma 5.3. For each fixed $\varepsilon>0$ and each interval \[ I\in\{\mathbb R,[0,1],[1,\infty),[-1,0],(-\infty,-1]\}, \] there exist constants $c_{\varepsilon,I},C_{\varepsilon,I}>0$ such that for all $n\ge2$, \[ \mathbb P\left(|N_n(I)-\mathbb E N_n(I)|\ge \varepsilon\log n\right) \le C_{\varepsilon,I}e^{-c_{\varepsilon,I}\sqrt{\log n}}. \]

Justification. This is precisely the consequence of Can--Nguyen's Theorems 1.4 and 1.5 recorded by Do in the proof of his Lemma 2.3. $\square$

5.4. Expectation constants for the four fundamental intervals

We next determine the asymptotic expectations on the intervals $[0,1]$, $[-1,0]$, $[1,\infty)$, and $(-\infty,-1]$.

Lemma 5.4. As $n\to\infty$, \[ \mathbb E N_n([0,1]) =\mathbb E N_n([-1,0]) =\mathbb E N_n([1,\infty)) =\mathbb E N_n((-\infty,-1]) =\frac{1}{2\pi}\log n+o(\log n). \]

Proof. We proceed in three steps.

Step 1: positive and negative roots inside $[-1,1]$ have the same expectation. Consider the polynomial \[ g_n(x):=f_n(-x)=\sum_{k=0}^n \epsilon_k(-1)^k x^k. \] The coefficient sequence $((-1)^k\epsilon_k)_{0\le k\le n}$ has the same joint distribution as $(\epsilon_k)_{0\le k\le n}$, because multiplying each coordinate by the deterministic sign $(-1)^k$ preserves the i.i.d. Rademacher law. Hence $g_n$ has the same distribution as $f_n$.

A real number $x\in[0,1]$ is a root of $g_n$ if and only if $-x\in[-1,0]$ is a root of $f_n$, with the same multiplicity. Therefore the random variables $N_n([0,1])$ and $N_n([-1,0])$ have the same distribution, hence \[ \mathbb E N_n([0,1])=\mathbb E N_n([-1,0]). \]

Step 2: inside and outside expectations agree on each sign side. By the Round-1 reciprocal symmetry, \[ \mathbb E N_n(0,1)=\mathbb E N_n(1,\infty). \] Adding the same endpoint contribution $\mathbb P(f_n(1)=0)$ to both sides gives \[ \mathbb E N_n([0,1])=\mathbb E N_n([1,\infty)). \] Similarly, \[ \mathbb E N_n([-1,0])=\mathbb E N_n((-\infty,-1]). \]

Step 3: evaluate the common asymptotic. Do records the classical expectation asymptotic \[ \mathbb E N_n([-1,1])=\frac{1}{\pi}\log n+o(\log n). \] Since $0$ is never a root of $f_n$ (because $f_n(0)=\epsilon_0\in\{\pm1\}$), the intervals $[-1,0]$ and $[0,1]$ split $[-1,1]$ without overlap on roots, and therefore \[ \mathbb E N_n([-1,1])=\mathbb E N_n([-1,0])+\mathbb E N_n([0,1]). \] By Step 1 the two summands are equal, so each equals \[ \frac12\,\mathbb E N_n([-1,1]) =\frac{1}{2\pi}\log n+o(\log n). \] Then Step 2 transfers the same asymptotic to $[1,\infty)$ and $(-\infty,-1]$. $\square$

5.5. Dense-subsequence strong laws

We can now combine Lemma 5.2 and Lemma 5.3 with the expectation asymptotics above.

Theorem 5.5 (dense-subsequence strong law on the four fundamental intervals and on $\mathbb R$). Let $(n_k)_{k\ge1}$ be an increasing sequence of integers such that \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty. \] Then the following almost sure convergences hold: \[ \frac{N_{n_k}(\mathbb R)}{\log n_k}\to\frac{2}{\pi}, \] and for each interval \[ I\in\{[0,1],[-1,0],[1,\infty),(-\infty,-1]\} \] one has \[ \frac{N_{n_k}(I)}{\log n_k}\to\frac{1}{2\pi} \qquad\text{almost surely.} \] Consequently, \[ \frac{N_{n_k}((-\infty,-1]\cup[1,\infty))}{\log n_k}\to\frac{1}{\pi} \qquad\text{almost surely.} \]

Proof. We first prove the statement for a fixed interval $I$ among \[ \mathbb R,[0,1],[-1,0],[1,\infty),(-\infty,-1]. \] For each integer $m\ge1$, apply Lemma 5.3 with $\varepsilon=1/m$. Then there exist constants $c_{m,I},C_{m,I}>0$ such that \[ \mathbb P\left(\left|N_n(I)-\mathbb E N_n(I)\right|\ge m^{-1}\log n\right) \le C_{m,I}e^{-c_{m,I}\sqrt{\log n}} \qquad(n\ge2). \] Because \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty, \] for each fixed $m$ there exists $K_m$ such that for all $k\ge K_m$, \[ c_{m,I}\sqrt{\log n_k}\ge 3\log k. \] Therefore, for all large $k$, \[ C_{m,I}e^{-c_{m,I}\sqrt{\log n_k}} \le C_{m,I}k^{-3}. \] Hence \[ \sum_{k=1}^\infty \mathbb P\left(\left|N_{n_k}(I)-\mathbb E N_{n_k}(I)\right|\ge m^{-1}\log n_k\right)<\infty. \] Lemma 5.2 now yields \[ \frac{N_{n_k}(I)-\mathbb E N_{n_k}(I)}{\log n_k}\to0 \qquad\text{almost surely.} \]

It remains to identify the deterministic limit of $\mathbb E N_{n_k}(I)/\log n_k$.

• For $I=\mathbb R$, Round 1 already used the expectation asymptotic of Do--Nguyen--Vu: \[ \mathbb E N_n(\mathbb R)=\frac{2}{\pi}\log n + C + o(1). \] Hence \[ \frac{\mathbb E N_{n_k}(\mathbb R)}{\log n_k}\to \frac{2}{\pi}. \] Combining with the previous display gives \[ \frac{N_{n_k}(\mathbb R)}{\log n_k}\to\frac{2}{\pi} \qquad\text{almost surely.} \]

• For $I\in\{[0,1],[-1,0],[1,\infty),(-\infty,-1]\}$, Lemma 5.4 gives \[ \frac{\mathbb E N_{n_k}(I)}{\log n_k}\to \frac{1}{2\pi}. \] Therefore \[ \frac{N_{n_k}(I)}{\log n_k}\to\frac{1}{2\pi} \qquad\text{almost surely.} \]

Finally, the intervals $[1,\infty)$ and $(-\infty,-1]$ are disjoint, so their counts add exactly. Hence \[ N_{n_k}((-\infty,-1]\cup[1,\infty)) = N_{n_k}((-\infty,-1]) + N_{n_k}([1,\infty)), \] and dividing by $\log n_k$ yields the outside-root limit $1/\pi$. $\square$

Corollary 5.6 (explicit non-lacunary subsequence with ratio tending to $1$). Define, for $k\ge3$, \[ \nu_k:=\exp\left((\log k)^3\right) \qquad\text{and}\qquad n_k:=\lfloor \nu_k\rfloor. \] After discarding finitely many initial terms, $(n_k)$ is strictly increasing and satisfies \[ \frac{n_{k+1}}{n_k}\to1. \] Hence Theorem 5.5 applies to this explicit non-lacunary subsequence. In particular, \[ \frac{N_{n_k}(\mathbb R)}{\log n_k}\to\frac{2}{\pi} \qquad\text{and}\qquad \frac{N_{n_k}((-\infty,-1]\cup[1,\infty))}{\log n_k}\to\frac{1}{\pi} \] almost surely.

Proof. We first verify the growth condition in Theorem 5.5. Since \[ \log n_k=(\log k)^3+o(1), \] we obtain \[ \frac{\sqrt{\log n_k}}{\log k} \sim \frac{(\log k)^{3/2}}{\log k} =(\log k)^{1/2}\to\infty. \] So Theorem 5.5 applies once $(n_k)$ is strictly increasing.

It remains to show $n_{k+1}/n_k\to1$. Because \[ \log \nu_k=(\log k)^3, \] we have \[ \log \frac{\nu_{k+1}}{\nu_k} =(\log(k+1))^3-(\log k)^3. \] By the mean value theorem applied to $\phi(x)=(\log x)^3$, there exists $\xi_k\in(k,k+1)$ such that \[ (\log(k+1))^3-(\log k)^3 =\phi'(\xi_k) =\frac{3(\log \xi_k)^2}{\xi_k}\to0. \] Hence $\nu_{k+1}/\nu_k\to 1$. Since $n_k/\nu_k\to1$, the same limit holds for $n_{k+1}/n_k$. Also, \[ \nu_{k+1}-\nu_k\sim \nu_k\cdot \frac{3(\log k)^2}{k}\to\infty, \] so after removing finitely many initial indices the floors $n_k$ are strictly increasing. $\square$

Corollary 5.7 (full-sequence liminf/limsup constraints). Almost surely, \[ \liminf_{n\to\infty}\frac{R_n}{\log n}\ge \frac{1}{\pi} \qquad\text{and}\qquad \limsup_{n\to\infty}\frac{R_n}{\log n}\ge \frac{2}{\pi}. \]

Proof. Since $R_n\ge N_n([-1,1])$ for every $n$, Do's strong law on $[-1,1]$ gives \[ \liminf_{n\to\infty}\frac{R_n}{\log n} \ge \lim_{n\to\infty}\frac{N_n([-1,1])}{\log n} =\frac{1}{\pi} \qquad\text{almost surely.} \] For the limsup, apply Theorem 5.5 to the explicit sequence from Corollary 5.6 (or alternatively use the Round-1 lacunary theorem): along that subsequence, \[ \frac{R_{n_k}}{\log n_k}\to \frac{2}{\pi} \qquad\text{almost surely,} \] so the full-sequence limsup is at least $2/\pi$ almost surely. $\square$

ADVERSARIAL VERIFICATION

I now check the new arguments against the main failure modes.

• Did I really go beyond Round 1? Yes. Round 1 proved only a lacunary almost sure law on $\mathbb R$. Theorem 5.5 replaces this by a much denser subsequence theorem, valid whenever $\sqrt{\log n_k}/\log k\to\infty$, and gives exact outside-half-line laws that were absent in Round 1.

• Was lacunarity accidentally used anywhere? No. The proof of Theorem 5.5 uses only the summability of \[ e^{-c_{m,I}\sqrt{\log n_k}}, \] which follows from $\sqrt{\log n_k}/\log k\to\infty$. No ratio lower bound $n_{k+1}/n_k\ge q>1$ is used.

• Is the condition $\sqrt{\log n_k}/\log k\to\infty$ strong enough for all rational error tolerances? Yes. For each fixed $m$, the constant $c_{m,I}>0$ is fixed. Hence eventually \[ c_{m,I}\sqrt{\log n_k}\ge 3\log k, \] which makes the $m$th error event summable in $k$. Countably many $m$ are handled by a countable intersection of probability-one events.

• Did endpoint roots at $\pm1$ create double-counting issues? They do not invalidate any asymptotic. In Lemma 5.4, the equality \[ \mathbb E N_n([0,1])=\mathbb E N_n([1,\infty)) \] is exact because both closed intervals differ from their open counterparts by the same endpoint event $\{f_n(1)=0\}$. Likewise on the negative side.

• Did multiplicities matter? The concentration theorems and expectation asymptotics are stated for distinct roots. By the Round-1 multiple-root lemma, real multiple roots occur only finitely often almost surely. Therefore every asymptotic statement in Theorem 5.5 and its corollaries remains valid for the original multiplicity-counting quantities $R_n(I)$.

• Any hidden use of unproved expectation asymptotics on $[1,\infty)$ or $(-\infty,-1]$? No. These were derived in Lemma 5.4 from (i) the classical expectation on $[-1,1]$, (ii) the sign symmetry $x\mapsto -x$, and (iii) the Round-1 reciprocal expectation symmetry.

• Does Corollary 5.6 really give a non-lacunary sequence? Yes. The ratio $n_{k+1}/n_k\to1$, so there is no fixed $q>1$ with $n_{k+1}\ge q n_k$ for all large $k$. Hence the sequence is genuinely non-lacunary.

No contradiction or gap was found in the new arguments.

FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

The original full-sequence statement remains unresolved. However, this round proves the following new theorems beyond Round 1.

1. Rigorous obstruction. The current one-point concentration estimate is non-summable along the full sequence, so Borel--Cantelli alone cannot prove the original almost sure law; a maximal inequality or other inter-degree control is genuinely necessary.

2. Dense-subsequence global law. If $(n_k)$ is any deterministic increasing sequence with \[ \frac{\sqrt{\log n_k}}{\log k}\to\infty, \] then \[ \frac{R_{n_k}}{\log n_k}\to\frac{2}{\pi} \qquad\text{almost surely.} \] This is strictly stronger than the Round-1 lacunary theorem.

3. Dense-subsequence law on the outside half-lines. Along the same class of subsequences, \[ \frac{R_{n_k}([1,\infty))}{\log n_k}\to\frac{1}{2\pi}, \qquad \frac{R_{n_k}((-\infty,-1])}{\log n_k}\to\frac{1}{2\pi}, \] and therefore \[ \frac{R_{n_k}((-\infty,-1]\cup[1,\infty))}{\log n_k}\to\frac{1}{\pi} \qquad\text{almost surely.} \]

4. Explicit non-lacunary subsequence. The same conclusions hold along the explicit deterministic sequence \[ n_k=\lfloor\exp\left((\log k)^3\right)\rfloor \] (after removing finitely many initial terms), for which $n_{k+1}/n_k\to1$.

Thus the remaining gap is now sharply localized: the unresolved issue is not one-point concentration, but the absence of a method that propagates these dense-subsequence laws across the residual windows between nearby degrees.

COMPLETION ESTIMATE

COMPLETION: 74%

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, arXiv:1409.4128.
4. T. F. Bloom, Erdős Problem #521, erdosproblems.com/521 (checked for current open status).