gpt pro 5.2 attempt on Math-6044

1) FORMAL RESTATEMENT

Let f(x)=$\sum_{j=0}^d$ a_j x^j be a complex polynomial of degree d with a_0$\ne$ 0 and a_d$\ne$ 0. Let n be the number of nonzero coefficients a_j (so 2<=n<=d+1). Define

M := $\frac{\sum_{j=0}^d |a_j|}{\sqrt{|a_0 a_d|}}$.

Let z_1,...,z_d be the roots of f (counted with multiplicity), and write z_i = r_i $e^{i \theta_i}$ with $\theta_i$ $\in$ [0,2$\pi$).

Conjectured discrepancy bound: For every interval I$\subset$[0,2$\pi$],

\[ \Big| \#\{i: \theta_i \in I\} - \frac{|I|}{2\pi} d \Big| \ll \sqrt{n \log M}, \]

where the implied constant is absolute (independent of d,n,M) and |I| is the length of I.

2) QUICK LITERATURE/CONTEXT CHECK

The problem file states that Erdős--Turan proved the same inequality with n replaced by the degree d (i.e. discrepancy $\ll$ $\sqrt{d \log M}$). The question is whether one can replace d by the number n of nonzero coefficients, i.e. obtain a sparsity-sensitive bound. No further resolution is stated in the problem file.

3) ATTACK PLAN

Two complementary routes: (A) Special cases: prove the inequality for very sparse polynomials (binomials, trinomials, etc.) and see how n and M enter. (B) Disproof attempt: try to build a polynomial with small n and moderate M whose roots have angles highly concentrated in a short arc, forcing discrepancy $\gg$ d.

In this writeup I prove two elementary lemmas (one about M itself, one fully resolving the binomial case n=2), and I report small numerical sanity checks for random sparse polynomials.

4) WORK

FAST REALITY CHECK (small computations)

I sampled random sparse polynomials with a_0=a_d=1, and the remaining $n-2$ coefficients chosen as small random nonzero integers at random intermediate exponents. For each polynomial I computed its roots numerically, then computed the exact angular discrepancy max_{interval I} |#{$\theta_i\in$ I} - $|I|/(2\pi)$d|. I also computed the ratio discrepancy / sqrt(n log M).

Exact outputs for degrees 20,30 and n=3,4 (5 trials each; fixed random seed 0):

--- d=20, n=3 (a0=ad=1, other coeffs random integers) ---
trial 0: M=5.000000, disc=1.000000, disc/sqrt(n log M)=0.455
trial 1: M=4.000000, disc=0.745607, disc/sqrt(n log M)=0.366
trial 2: M=3.000000, disc=0.666667, disc/sqrt(n log M)=0.367
trial 3: M=5.000000, disc=2.000000, disc/sqrt(n log M)=0.910
trial 4: M=3.000000, disc=0.642143, disc/sqrt(n log M)=0.354

--- d=20, n=4 (a0=ad=1, other coeffs random integers) ---
trial 0: M=7.000000, disc=1.000000, disc/sqrt(n log M)=0.358
trial 1: M=6.000000, disc=1.000000, disc/sqrt(n log M)=0.374
trial 2: M=6.000000, disc=2.000000, disc/sqrt(n log M)=0.747
trial 3: M=5.000000, disc=0.887882, disc/sqrt(n log M)=0.350
trial 4: M=7.000000, disc=1.602152, disc/sqrt(n log M)=0.574

--- d=30, n=3 (a0=ad=1, other coeffs random integers) ---
trial 0: M=5.000000, disc=0.828288, disc/sqrt(n log M)=0.377
trial 1: M=5.000000, disc=0.942166, disc/sqrt(n log M)=0.429
trial 2: M=5.000000, disc=1.000000, disc/sqrt(n log M)=0.455
trial 3: M=4.000000, disc=1.000000, disc/sqrt(n log M)=0.490
trial 4: M=4.000000, disc=1.000000, disc/sqrt(n log M)=0.490

--- d=30, n=4 (a0=ad=1, other coeffs random integers) ---
trial 0: M=6.000000, disc=1.000000, disc/sqrt(n log M)=0.374
trial 1: M=8.000000, disc=1.333281, disc/sqrt(n log M)=0.462
trial 2: M=8.000000, disc=1.326786, disc/sqrt(n log M)=0.460
trial 3: M=5.000000, disc=1.260708, disc/sqrt(n log M)=0.497
trial 4: M=7.000000, disc=0.958378, disc/sqrt(n log M)=0.344

These experiments are not evidence either way (degrees are tiny and roots are numeric), but at least they do not immediately contradict the conjectured scale $\sqrt{n\log M}$.

Lemma 990.1 (basic properties of M)

(i) Scale invariance: For any nonzero complex constant c, if g(x)=c f(x) then M(g)=M(f). (ii) Lower bound: M$\ge$ 2.

Proof. (i) If g_j=c a_j then $\sum$ |g_j|=|c|$\sum$|a_j| and $\sqrt{|g_0 g_d|}$=|c|$\sqrt{|a_0 a_d|}$, so the ratio M is unchanged. (ii) Since a_0 and a_d are among the coefficients, $\sum_{j=0}^d$ |a_j| $\ge$ |a_0|+|a_d|. By AM-GM, |a_0|+|a_d| $\ge$ 2$\sqrt{|a_0 a_d|}$. Dividing by $\sqrt{|a_0 a_d|}$ gives M$\ge$ 2. $\square$

Lemma 990.2 (binomial case n=2)

Assume f(x)=a_0+a_d x^d with a_0 a_d $\ne$ 0 (so n=2). Then for every interval I$\subset$[0,2$\pi$],

\[ \Big| \#\{i: \theta_i \in I\} - \frac{|I|}{2\pi} d \Big| \le 1. \]

In particular, the conjectured inequality holds for n=2 with implied constant 1.

Proof. The roots satisfy x^d = -a_0/a_d. Write -a_0/a_d = R $e^{i\phi}$ with R>0 and $\phi\in$[0,2$\pi$). Then the d roots are z_m = R^{1/d} exp$i(\phi+2\pi m)/d$, m=0,1,...,d-1. Thus the arguments are an arithmetic progression modulo 2$\pi$ with step 2$\pi$/d; equivalently, there is exactly one root in each half-open arc of length 2$\pi$/d.

Fix an interval I of length L=|I|. The expected number under uniform distribution is d L/$2\pi$. Because the points are equally spaced with spacing 2$\pi$/d, the actual count differs from the expected count by at most 1 (a standard lattice-point-in-interval fact on the circle). Concretely: if we identify angles with residues m/d in [0,1), then counting $\theta_m$ in I is counting integers m in an interval of length d L/$2\pi$, which differs from its length by <1. Hence the discrepancy is <=1.

Now compute M. Here $\sum$ |a_j|=|a_0|+|a_d| and n=2. By Lemma 990.1(ii), M>=2, so sqrt{n log M} = sqrt{2 log M} >= sqrt{2 log 2} > 1. Therefore discrepancy <=1 <= sqrt{2 log M} = sqrt{n log M}, which is the claimed inequality with implied constant 1. $\square$

5) VERIFICATION

- Lemma 990.2 uses only the explicit root formula for a binomial; multiplicities do not occur because x^d=constant has d distinct solutions when the constant is nonzero. - The AM-GM step in Lemma 990.1(ii) is exact.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result here: - Full proof of the conjectured inequality for n=2 (binomials), with discrepancy <=1. - Structural facts: M is scale-invariant and M>=2 always. - Small numerical sanity checks for random sparse polynomials show discrepancy often of size O$1$ when coefficients are tiny, consistent with $\sqrt{n\log M}$.

(ii) First gap (crisp statement): Prove that in the Erdős--Turan method controlling angular discrepancy, the quantity d can be replaced (up to absolute constants) by the number n of nonzero coefficients. Equivalently, find a Fourier-analytic bound on the relevant trigonometric sums in terms of n and M only.

(iii) Top 3 next moves (concrete): 1. Work out the Erdős--Turan proof line-by-line and locate the precise point where the degree d appears; attempt to replace that step by a sparsity-sensitive estimate. 2. Study the trinomial family f(x)=a_0+a_m x^m + a_d x^d: attempt to prove the bound with n=3 or produce a family where angles are highly non-uniform while M stays moderate. 3. Search systematically (computationally) for sparse polynomials with small n and small M but large angular discrepancy, to guide either a proof strategy or a disproof.

(iv) Minimal counterexample structure: A family of polynomials with fixed small n (say n=3 or 4), with M growing at most moderately (e.g. log M = O(log d)), but whose root arguments concentrate in an arc of length o$1$, forcing discrepancy $\gg$ d, contradicting O$\sqrt{n\log M}$.