on random polynomial zeros

Overview

Edelman and Kostlan’s paper gives the clean geometric form of Kac’s formula: the expected number of real zeros is the projective length of an evaluation curve, divided by $\pi$ (Edelman & Kostlan, 1995). This explains the global fact that a degree $n$ random polynomial has only about $(2/\pi )\log n$ real zeros. The MathOverflow question about whether the largest-modulus root is real is different in character: after taking reciprocals and passing to a random power series, it becomes a local question about the first zero near the origin. The power-series reduction is explained in this MathOverflow answer.

The existing Rouché certificate proves a uniform lower bound near $8\%$ for the uniform coefficient model; see the initial certificate and the later improvement. That certificate is deliberately crude: it forces the linear term $a_{1}z$ to dominate. A more flexible mechanism is to certify a low-degree truncation, or even an argument-principle winding number, against a rigorously bounded tail. This should improve the lower-bound certificate before one tries to solve the actual limiting probability.

Edelman-Kostlan Geometry

Let

$$P_n(t)=a_0+a_{1}t+\cdots +a_n{t}^n$$

with independent standard real Gaussian coefficients, and set

$$v(t)=(1,t,\dots ,t^n{)}^T,\gamma (t)=\frac{v(t)}{\Vert v(t)\Vert }.$$

The equation $P_n(t)=0$ is the orthogonality condition $a\cdot v(t)=0$. After normalizing the coefficient vector $a/\Vert a\Vert$, Gaussian rotational invariance makes the coefficients uniformly distributed on the sphere. Thus zeros are intersections between a random point on the coefficient sphere and the family of equators $\gamma (t{)}^{\perp }$.

The Buffon-needle integral geometry in Edelman-Kostlan says that sweeping equators over a curve counts area with multiplicity. Therefore

$$\mathbb{E}{N}_{\mathbb{R}}(P_n)=\frac{1}{\pi }\mathrm{length}(\gamma ).$$

For the moment curve this gives Kac’s integral formula:

$$\mathbb{E}{N}_{\mathbb{R}}(P_n)=\frac{1}{\pi }{\int }_{-\infty }^{\infty }\sqrt{\frac{1}{(t^2-1{)}^2}-\frac{(n+1{)}^2{t}^{2n}}{(t^{2n+2}-1{)}^2}}dt.$$

In particular,

$$\mathbb{E}{N}_{\mathbb{R}}(P_n)=\frac{2}{\pi }\log n+0.6257358072\dots +\frac{2}{n\pi }+O(n^{-2}).$$

The paper’s more invariant formula is the kernel version. For a centered Gaussian random function

$$F(t)=\sum _{j=0}^n{a}_j{f}_j(t),K(x,y)=\mathbb{E}[F(x)F(y)],$$

the real-zero density is

$$\rho (t)=\frac{1}{\pi }\sqrt{{{\partial }_x{\partial }_y\log K(x,y)|}_{x=y=t}}.$$

For the Kac polynomial,

$$K_n(x,y)=1+xy+\cdots +x^n{y}^n=\frac{1-(xy{)}^{n+1}}{1-xy}.$$

For the limiting random power series with independent standard Gaussian coefficients,

$$P(z)=\sum _{k\ge 0}{a}_k{z}^k,K(x,y)=\frac{1}{1-xy},$$

and the real-zero density inside $(-1,1)$ is

$$\rho (t)=\frac{1}{\pi (1-t^2)}.$$

This density is a global statistic. It counts how many real zeros appear along the real interval. It does not directly answer which zero has the smallest or largest modulus.

Extreme Roots Become First Zeros

For i.i.d. real coefficients, the reciprocal polynomial

$$P_n^{\ast }(z)=z^n{P}_n(1/z)=a_n+a_{n-1}z+\cdots +a_0{z}^n$$

has the same distribution as $P_n$. The largest-modulus zero of $P_n$ is real exactly when the smallest-modulus zero of $P_n^{\ast }$ is real, apart from probability-zero tie events. Thus the largest-root and smallest-root probabilities agree.

As $n\to \infty$, the first few coefficients stabilize and the random polynomials converge on compact subsets of the unit disk to the random analytic function

$$P(z)=a_0+a_{1}z+a_2{z}^2+\cdots .$$

Hurwitz’s theorem then turns the finite-degree question into the first-zero question for this random power series, provided one avoids the degenerate event where the first zero is multiple or tied in modulus. This is the main conceptual move in the MathOverflow answer.

The apparent paradox now disappears. The expected number of real zeros of $P_n$ is only logarithmic, but the extreme zero is not sampled uniformly from all zeros. It is controlled by the low-order Taylor data near $0$, where the linear approximation already has a real zero. Complex zeros have many angular locations globally, but the first zero is a local competition among the first few coefficients.

The Current Rouché Certificate

Consider the uniform model $a_k\sim \mathrm{Unif}(-1,1)$ and the power series or a finite polynomial of degree at least the required tail length. The basic certificate compares $P(z)$ with $a_{1}z$ on a circle $|z|=R$:

$$P(z)=a_{1}z+\left(a_0+a_2{z}^2+a_3{z}^3+\cdots \right).$$

If

$$|a_0|+\frac{R^2}{1-R}<|a_1|R,$$

then Rouché’s theorem implies that $P$ and $a_{1}z$ have the same number of zeros in $|z|<R$, namely one. Since the coefficients are real, a single zero in the conjugation-invariant disk must be real. Optimizing over $R$ gives the sufficient condition

$$|a_1|\ge |a_0|+2\sqrt{|a_0|},$$

whose probability is

$${\int }_0^{(\sqrt{2}-1{)}^2}\left(1-x-2\sqrt{x}\right)dx=\frac{23-16\sqrt{2}}{6}\approx \mathrm{0.0621.}$$

The later improvement keeps $a_2$ random instead of burying it inside the worst-case tail. Define

$$E_R=\left\{|a_0|+|a_2|R^2+\frac{R^3}{1-R}<|a_1|R\right\}.$$

For each $R\in (0,1)$, $E_R$ again forces $a_{1}z$ to dominate the perturbation on $|z|=R$, hence certifies that the smallest-modulus zero is real. For $0<R<1/2$ the probability is explicitly

$$\mathbb{P}(E_R)=\frac{R(R^4+4R^3+4R^2-9R+3)}{6(1-R{)}^2}.$$

The displayed expression is maximized near $R=0.277$, and the rational test value gives

$$\mathbb{P}(E_{277/1000})=\frac{83544960929719}{1045458000000000}\approx \mathrm{0.0799123.}$$

What the 8% Bound Is Really Measuring

This is not evidence that the true limiting probability is near $8\%$. It is only the mass of a very special subset of the real-first-zero event: samples for which the linear monomial $a_{1}z$ is visibly dominant on a centered circle. The MO simulations and the stability experiment reported in the MathOverflow answer suggest a true uniform-model probability around $50.7\%$, so the certificate is missing most of the event.

A Quadratic-Core Theorem

The quadratic comparison can be made into a theorem with a fixed radius. The price is that the lower bound below is machine-assisted by an exact finite box count, not a closed-form integral.

Quadratic-Core Certificate

Let

$$P(z)=\sum _{k\ge 0}{a}_k{z}^k$$

where the $a_k$ are independent $\mathrm{Unif}(-1,1)$ real random variables. Then

$$\mathbb{P}\{\text{the smallest-modulus zero of }P\text{ is real}\}\ge \frac{103116}{1000000}=\mathrm{0.103116.}$$

By reciprocal invariance, the same lower bound holds for the largest-modulus zero of the finite i.i.d. uniform polynomial model for every degree $n\ge 3$.

Proof: Exact Box Certificate

Fix $R=1/3$ and write

$$Q_2(z)=a_0+a_{1}z+a_2{z}^2.$$

The remaining tail satisfies, on $|z|=R$,

$$\left|\sum _{k\ge 3}{a}_k{z}^k\right|\le \sum _{k\ge 3}{R}^k=\frac{R^3}{1-R}=\frac{1}{18}.$$

Thus it is enough to certify that $Q_2$ has exactly one zero in $|z|<1/3$ and that

$$\underset{|z|=1/3}{\min }|Q_2(z)|>\frac{1}{18}.$$

Partition $[-1,1{]}^3$ into $100^3$ boxes of side length $1/50$. The box centers are

$$\left(\frac{c_0}{100},\frac{c_1}{100},\frac{c_2}{100}\right),c_j\in \{-99,-97,\dots ,99\}.$$

Every coefficient in such a box differs from its center by at most $h=1/100$ in each coordinate. On $|z|=1/3$, the perturbation of $Q_2$ caused by moving inside the box is bounded by

$$h\left(1+R+R^2\right)=\frac{1}{100}\left(1+\frac{1}{3}+\frac{1}{9}\right)=\frac{13}{900}.$$

At a center define

$$u_{+}=9c_0+3c_1+c_2,u_{-}=9c_0-3c_1+c_2.$$

Since $Q_2(\pm 1/3)=u_{\pm }/900$, the inequalities

$$u_{+}>13,u_{-}<-13$$

or the reversed pair imply that every point in the box has $Q_2(1/3)$ and $Q_2(-1/3)$ of opposite signs. Hence $Q_2$ has exactly one real zero in $(-1/3,1/3)$. If $Q_2$ has a second zero, it is also real and is outside this interval, so $Q_2$ has exactly one zero in the disk $|z|<1/3$.

It remains to certify the boundary gap. With $x=\cos \theta$ one has the exact identity

$$810000|Q_2(e^{i\theta }/3){|}^2=A{x}^2+Bx+C,$$

where

$$A=36c_0{c}_2,B=6c_1(9c_0+c_2),C=81c_0^2+9c_1^2+c_2^2-18c_0{c}_2.$$

To survive the whole coefficient box it is enough that the center boundary minimum exceed

$$\frac{1}{18}+\frac{13}{900}=\frac{7}{100}.$$

Since $810000(7/100{)}^2=3969$, this condition is checked exactly by verifying

$$\underset{-1\le x\le 1}{\min }(A{x}^2+Bx+C)>3969.$$

The minimum of this quadratic on $[-1,1]$ is elementary: use the smaller endpoint value unless $A>0$ and $|B|\le 2A$, in which case use $C-B^2/(4A)$.

Exact integer enumeration of the $100^3$ centers gives $103116$ boxes satisfying both the robust sign condition and the robust boundary-minimum condition. Each box has probability $100^{-3}$ under the product uniform law on the first three coefficients, so the certified probability is $103116/1000000$.

On every certified box, Rouché’s theorem applies to $P=Q_2+(P-Q_2)$ on $|z|=1/3$. Thus $P$ has exactly one zero in the disk. Because $P$ has real coefficients and the disk is invariant under conjugation, that unique zero is real and is the smallest-modulus zero.

This theorem improves the fixed a_2-aware linear certificate from $0.0799123$ to $0.103116$, a gain of about $2.40$ percentage points. The improvement is conceptually the same as the numerical experiment: $a_0+a_{1}z+a_2{z}^2$ is allowed to be the local model, rather than treating the constant and quadratic terms as an adversarial perturbation of $a_{1}z$.

A stronger certificate keeps the quadratic core but does not throw away the cubic coefficient. This is the first useful instance of the arbitrary-retained-tail mechanism.

Retained-Cubic Quadratic Certificate

Let

$$P(z)=a_0+a_{1}z+a_2{z}^2+a_3{z}^3+\text{higher terms}$$

with independent $\mathrm{Unif}(-1,1)$ real coefficients. Then

$$\mathbb{P}\{\text{the smallest-modulus zero of }P\text{ is real}\}\ge \frac{62381912}{150^4}=\frac{7797739}{63281250}\approx \mathrm{0.12322353.}$$

The same bound holds for the largest-modulus zero of i.i.d. finite uniform polynomials of degree $n\ge 4$.

Proof: Exact Retained-Cubic Box Count

Fix $R=1/3$ and keep

$$Q_2(z)=a_0+a_{1}z+a_2{z}^2$$

as the core. Instead of bounding $a_3{z}^3$ by $R^3$, retain $a_3$ and use

$$|P(z)-Q_2(z)|\le |a_3|R^3+\frac{R^4}{1-R}.$$

Partition $[-1,1{]}^4$ into $150^4$ boxes with centers

$$\left(\frac{c_0}{150},\frac{c_1}{150},\frac{c_2}{150},\frac{c_3}{150}\right),c_j\in \{-149,-147,\dots ,149\}.$$

The half-width is $h=1/150$. The endpoint sign condition for $Q_2(\pm 1/3)$ is still certified by

$$u_{+}=9c_0+3c_1+c_2,u_{-}=9c_0-3c_1+c_2,$$

and requiring either $u_{+}>13,u_{-}<-13$ or $u_{+}<-13,u_{-}>13$. This gives exactly one robust real zero of $Q_2$ in $(-1/3,1/3)$: a real polynomial of degree at most two with opposite endpoint signs cannot have zero or two roots in the interval. Any second zero is real and lies outside the interval, so $Q_2$ has exactly one zero in $|z|<1/3$.

The box error from the first four coefficients on $|z|=1/3$ is

$$h\left(1+R+R^2+R^3\right)=\frac{1}{150}\cdot \frac{40}{27}.$$

Thus a center with cubic coefficient $c_3/150$ is certified if

$$\underset{|z|=1/3}{\min }|Q_2(z)|>\frac{|c_3|}{150}\cdot \frac{1}{27}+\frac{1}{54}+\frac{1}{150}\cdot \frac{40}{27}.$$

As before,

$$81\cdot 150^2|Q_2(e^{i\theta }/3){|}^2=A{x}^2+Bx+C,x=\cos \theta ,$$

with

$$A=36c_0{c}_2,B=6c_1(9c_0+c_2),C=81c_0^2+9c_1^2+c_2^2-18c_0{c}_2.$$

After clearing denominators, the boundary condition is the exact integer inequality

$$9\underset{-1\le x\le 1}{\min }(A{x}^2+Bx+C)>(|c_3|+115{)}^2.$$

As in the quadratic theorem, the minimum on $[-1,1]$ is checked from the endpoint values unless $A>0$ and $|B|\le 2A$, in which case the interior value $C-B^2/(4A)$ is used. Enumerating the $150^4$ boxes gives exactly $62381912$ certified boxes. Each has probability $150^{-4}$, and Rouché then proves the asserted lower bound.

Keeping the whole cubic as the local model gives another certified jump. The zero-count test is no longer just a sign change on the real interval; it uses an exact cubic root count in the disk. The boundary minimum is also certified without floating point minimization, by checking Bernstein coefficients of a cubic polynomial on $[-1,1]$.

Full Cubic-Core Routh-Bernstein Certificate

Let

$$P(z)=\sum _{k\ge 0}{a}_k{z}^k$$

have independent $\mathrm{Unif}(-1,1)$ real coefficients. Then

$$\mathbb{P}\{\text{the smallest-modulus zero of }P\text{ is real}\}\ge \frac{228028884}{200^4}=\frac{57007221}{400000000}\approx \mathrm{0.1425180525.}$$

The same bound holds for the largest-modulus zero of every finite i.i.d. uniform polynomial of degree $n\ge 3$.

Proof: Cubic Root Count and Bernstein Boundary Gap

Fix $R=2/5$, $M=200$, and write the center of a coefficient box as

$$q_3(z)=\frac{1}{M}\left(c_0+c_{1}z+c_2{z}^2+c_3{z}^3\right),c_j\in \{-199,-197,\dots ,199\}.$$

The box half-width is $h=1/M$. On $|z|=R$, moving within the box changes the cubic core by at most

$$h(1+R+R^2+R^3)=\frac{203}{125M},$$

while the unseen tail is bounded by

$$\sum _{k\ge 4}{R}^k=\frac{R^4}{1-R}=\frac{16}{375}.$$

Thus the center is safe if it has exactly one zero in $|z|<2/5$ and

$$\underset{|z|=2/5}{\min }|q_3(z)|>\frac{16M+609}{375M}.$$

To count the zeros of $q_3$ in the disk, use the Cayley transform

$$z=\frac{2}{5}\frac{1+s}{1-s}.$$

The disk $|z|<2/5$ corresponds to $\mathrm{Re}s<0$. After clearing denominators, set

$$H(s)=M{5}^3(1-s{)}^3{q}_3\left(\frac{2}{5}\frac{1+s}{1-s}\right)=B_0+B_{1}s+B_2{s}^2+B_3{s}^3,$$

where

$$\begin{array}{rl}{B}_0& =125c_0+50c_1+20c_2+8c_3,\\ B_1& =-375c_0-50c_1+20c_2+24c_3,\\ B_2& =375c_0-50c_1-20c_2+24c_3,\\ B_3& =-125c_0+50c_1-20c_2+8c_3.\end{array}$$

Put $G(s)=H(-s)=A_3{s}^3+A_2{s}^2+A_{1}s+A_0$. The number of zeros of $q_3$ in $|z|<2/5$ is the number of right-half-plane zeros of $G$. For a real cubic with no zero entry in the Routh first column, this number is the number of sign changes in

$$A_3,A_2,\frac{A_2{A}_1-A_3{A}_0}{A_2},A_0.$$

The enumeration keeps exactly those boxes where this sign-change count is one.

It remains to certify the boundary gap. With $x=\cos \theta$, define the integer cubic $T$ by

$$T(x)=5^6{M}^2{\left|q_3\left(\frac{2}{5}{e}^{i\theta }\right)\right|}^2.$$

Explicitly,

$$T(x)=S+D_{1}x+D_2(2x^2-1)+D_3(4x^3-3x),$$

where

$$\begin{array}{rl}S& =15625c_0^2+2500c_1^2+400c_2^2+64c_3^2,\\ D_1& =12500c_0{c}_1+2000c_1{c}_2+320c_2{c}_3,\\ D_2& =5000c_0{c}_2+800c_1{c}_3,\\ D_3& =2000c_0{c}_3.\end{array}$$

The boundary condition above is equivalent to

$$9T(x)>(16M+609{)}^2(-1\le x\le 1).$$

Let

$$F(x)=9T(x)-(16M+609{)}^2.$$

A sufficient exact positivity check is that all four Bernstein coefficients of $F$ on $[-1,1]$ are positive. For $F(x)=g_3{x}^3+g_2{x}^2+g_{1}x+g_0$, this is the integer test

$$\begin{array}{c}F(-1)>0,F(1)>0,\\ 3F(-1)+2F^{\prime }(-1)>0,3F(1)-2F^{\prime }(1)>0.\end{array}$$

Enumerating the $200^4$ boxes gives exactly $228028884$ boxes satisfying both the Routh one-zero test and the Bernstein boundary test. Each box has probability $200^{-4}$. On every certified box, Rouché’s theorem applies to the center cubic plus the box and analytic-tail perturbation, so $P$ has exactly one zero in $|z|<2/5$. By conjugation symmetry that zero is real.

Mechanisms to Improve the Certificate

The natural improvement is to stop comparing against $a_{1}z$. Let

$$Q_m(z)=a_0+a_{1}z+\cdots +a_m{z}^m.$$

The following criterion is uniform in $m$.

Uniform $m$-Core Certificate

Let

$$P(z)=\sum _{k\ge 0}{a}_k{z}^k$$

be a real-coefficient power series with $|a_k|\le 1$ for $k>m$. Fix $R\in (0,1)$ and define

$$Q_m(z)=\sum _{k=0}^m{a}_k{z}^k,{\Delta }_R(Q_m)=\underset{|z|=R}{\min }|Q_m(z)|.$$

If $Q_m$ has exactly one zero in $|z|<R$ and

$${\Delta }_R(Q_m)>\frac{R^{m+1}}{1-R},$$

then $P$ has exactly one zero in $|z|<R$. Since $P$ has real coefficients, this zero is real and is the smallest-modulus zero of $P$.

Proof

On $|z|=R$,

$$|P(z)-Q_m(z)|\le \sum _{k>m}{R}^k=\frac{R^{m+1}}{1-R}.$$

The strict boundary inequality gives $|P-Q_m|<|Q_m|$ on the circle. Rouché’s theorem gives the same number of zeros for $P$ and $Q_m$ in the disk. This number is one by assumption. A conjugation-invariant disk cannot contain exactly one nonreal zero of a real polynomial or real analytic function, so that unique zero is real.

Uniform Retained-Tail Certificate

More generally, choose integers $0\le r<m$ and write

$$Q_r(z)=\sum _{k=0}^r{a}_k{z}^k.$$

If $Q_r$ has exactly one zero in $|z|<R$ and

$$\underset{|z|=R}{\min }|Q_r(z)|>\sum _{k=r+1}^m|a_k|R^k+\frac{R^{m+1}}{1-R},$$

then the full series has exactly one zero in $|z|<R$, hence its smallest-modulus zero is real.

The proof is the same Rouché argument, with the retained coefficients $a_{r+1},\dots ,a_m$ moved to the perturbation side before the remaining geometric tail is bounded.

For rigorous probability lower bounds, use the box-stable retained-tail version. Let $B\subset [-1,1{]}^{m+1}$ be a coefficient box with center $b=(b_0,\dots ,b_m)$ and half-widths ${\epsilon }_k$. Put

$$q_r(z)=\sum _{k=0}^r{b}_k{z}^k,E_{r,m,B}(R)=\sum _{k=0}^r{\epsilon }_k{R}^k+\sum _{k=r+1}^m(|b_k|+{\epsilon }_k)R^k.$$

If $q_r$ has exactly one zero in $|z|<R$ and

$$\underset{|z|=R}{\min }|q_r(z)|>E_{r,m,B}(R)+\frac{R^{m+1}}{1-R},$$

then every series whose first $m+1$ coefficients lie in $B$ is certified. For i.i.d. $\mathrm{Unif}(-1,1)$ coefficients, any disjoint family $\mathcal{B}$ of certified boxes gives the rigorous lower bound

$$\mathbb{P}\{\text{smallest-modulus zero is real}\}\ge \sum _{B\in \mathcal{B}}\frac{\mathrm{vol}(B)}{2^{m+1}}.$$

On a uniform $M$-grid, ${\epsilon }_k=1/M$. The zero-count hypothesis for $q_r$ can be certified by the argument principle, Schur-Cohn/Jury root counting for $q_r(Rz)$, or interval root isolation. This is the arbitrary-$m$ analogue of the exact retained-cubic box count above.

Algebraic Continuation for Fixed $m$

Fix a degree $m$, a rational radius $R\in (0,1)$, and a rational coefficient box $B\subset [-1,1{]}^{m+1}$. Whether $B$ is certified by the $m$-core Rouché test can be checked using only exact rational arithmetic.

More precisely, after applying the Cayley transform

$$z=R\frac{1+s}{1-s},$$

the disk root count for the box center is a half-plane root count for a real polynomial in $s$. This count is given by the Routh-Hurwitz table, or equivalently by Schur-Cohn/Jury determinants. The boundary quantity has the form

$$|q_m(R{e}^{i\theta }){|}^2=C_0+2\sum _{\ell =1}^m{C}_{\ell }{T}_{\ell }(\cos \theta ),$$

where $T_{\ell }$ is the Chebyshev polynomial. Thus the boundary gap is positivity of one explicit degree-$m$ polynomial in $x=\cos \theta$ on $[-1,1]$, after subtracting the squared box-and-tail margin. Positivity on $[-1,1]$ can be certified by Sturm sequences, by Bernstein subdivision, or by the one-variable sum-of-squares representation for positive interval polynomials.

Therefore the proof method continues uniformly to every fixed $m$. What does not continue by hand is the measure calculation: for $m\ge 4$, an improved numerical percentage means estimating the volume of a complicated semialgebraic subset of $[-1,1{]}^{m+1}$. Without doing that volume computation, one still gets the cubic lower bound by ignoring the extra coefficients, but no sharper explicit percentage follows.

Asymptotic Completeness of the Certificate Family

Let

$$P(z)=\sum _{k\ge 0}{a}_k{z}^k$$

have independent continuous coefficients supported in $[-1,1]$. Let $C_m$ be the event that there exist integers $0\le r\le m$ and a radius $R\in (0,1)$ such that

$$Q_r(z)=\sum _{k=0}^r{a}_k{z}^k$$

has exactly one zero in $|z|<R$ and

$$\underset{|z|=R}{\min }|Q_r(z)|>\sum _{k=r+1}^m|a_k|R^k+\frac{R^{m+1}}{1-R}.$$

Then $C_m\subseteq C_{m+1}$, every $C_m$ certifies that the smallest-modulus zero of $P$ is real, and

$$\mathbb{P}(C_m)\uparrow \mathbb{P}\{\text{the smallest-modulus zero of }P\text{ is real}\},$$

up to the null event where the smallest modulus is tied or multiple.

Proof

Monotonicity follows because if the same pair $(r,R)$ certifies at level $m$, then at level $m+1$ the right-hand side changes by

$$|a_{m+1}|R^{m+1}+\frac{R^{m+2}}{1-R}-\frac{R^{m+1}}{1-R}=-\left(1-|a_{m+1}|\right)R^{m+1}\le 0.$$

Thus a valid certificate remains valid after one more coefficient is revealed.

Conversely, suppose $P$ has a unique real zero $\zeta$ of smallest modulus and no other zero with $|z|=|\zeta |$. Choose $R$ strictly between $|\zeta |$ and the second-smallest zero modulus. Then $P$ has exactly one zero in $|z|<R$ and ${\min }_{|z|=R}|P(z)|>0$. Since the partial sums $Q_m$ converge uniformly to $P$ on $|z|\le R$ and $R^{m+1}/(1-R)\to 0$, Rouché’s theorem implies that, for all sufficiently large $m$, $Q_m$ has exactly one zero in $|z|<R$ and

$$\underset{|z|=R}{\min }|Q_m(z)|>\frac{R^{m+1}}{1-R}.$$

Hence $C_m$ occurs with $r=m$. Continuous coefficient laws make multiple zeros and exact modulus ties null events, so the union of the certificate events agrees almost surely with the real-smallest-zero event.

This theorem is the clean answer to the “eventually $1/2$” question. The finite-core Rouché method is complete for this event: if the true power-series probability is $p_{\ast }>1/2$, then some finite $m$ has $\mathbb{P}(C_m)>1/2$. But proving $p_{\ast }>1/2$ is exactly the remaining probabilistic part of the MathOverflow problem; the certificate convergence does not prove that inequality by itself.

This is a strict generalization of the linear certificate. It allows the quadratic or cubic terms to help shape the first zero instead of treating them as adversarial noise. In the quadratic case the boundary minimum is still elementary. For

$$Q_2(z)=a_0+a_{1}z+a_2{z}^2,$$

one has, with $x=\cos \theta$,

$$|Q_2(R{e}^{i\theta }){|}^2=a_0^2+a_1^2{R}^2+a_2^2{R}^4+2a_{1}R(a_0+a_2{R}^2)x+2a_0{a}_2{R}^2(2x^2-1).$$

Thus ${\min }_{|z|=R}|Q_2(z)|$ is the minimum of a quadratic polynomial on $[-1,1]$. A rigorous lower bound could integrate this semialgebraic certificate, or more realistically cover coefficient space by interval boxes.

There are two further upgrades.

• Argument-principle certificates. Instead of proving dominance by a chosen $Q_m$, compute the winding number of $P(\partial D_R)$ around $0$ with interval arithmetic. Certifying winding number $1$ proves exactly one zero in the disk. If the same box also isolates a real sign-change zero, conjugation symmetry forces that zero to be the unique one.
• Adaptive finite-dimensional atlases. Sample or tile the first $m$ coefficients, choose an isolating radius between the smallest and second-smallest zero of $Q_m$, and bound the analytic tail. This is exactly the finite-box strategy suggested in the MO answer, but with the low-degree truncation and the radius chosen adaptively rather than fixed in advance.

The retained-cubic theorem gives a rigorous improvement to $12.322353\%$, and the full cubic-core Routh-Bernstein theorem raises the fixed-radius certificate to $14.25180525\%$. Less conservative adaptive-radius Monte Carlo checks go higher still, but those numbers are targets for sharper interval validation, not theorems. The important point is structural: the gap between $8\%$ and the observed $50\%+$ is not a limitation of Rouché’s theorem itself. It is a limitation of using the one-monomial comparison $a_{1}z$.

Numerical Verification

The script rouche_certificate_search.py in /codes/2026 Summer/ compares these certificates. Reproducing the full cubic count is intentionally optional because --full-cubic-m 200 enumerates $200^4$ boxes:

Certificate	Status	Typical output
Linear core with worst tail	exact formula	$\approx 0.0621$
Linear core with random $a_2$ and fixed $R=0.277$	exact formula	$\approx 0.0799123$
Quadratic core $Q_2$ with fixed $R=1/3$	exact box theorem	$0.103116$
Quadratic core $Q_2$, retaining $a_3$, fixed $R=1/3$	exact box theorem	$0.12322353$
Full cubic core $Q_3$ with fixed $R=2/5$	exact Routh-Bernstein box theorem	$0.1425180525$
Full quartic core $Q_4$ with fixed $R=2/5$	exact Routh-Bernstein box search, $M=50$	$0.1375880064$
Quadratic core $Q_2$ with adaptive radius	Monte Carlo target	about $0.14$

For a same-budget heuristic across degrees, I used $12000$ coefficient samples for each $m$, $20$ candidate radii per sample, and the Chebyshev formula

$$|Q_m(R{e}^{i\theta }){|}^2=C_0+2\sum _{\ell =1}^m{C}_{\ell }{T}_{\ell }(\cos \theta )$$

to minimize the boundary value as a one-variable polynomial on $[-1,1]$. This is not a theorem, and it is not a rigorous lower bound, because the roots and candidate radii are still selected numerically. It estimates the size of the same Rouché certificate family when the per-degree search effort is held fixed:

Full core degree $m$	Same-budget estimate	95% sampling interval
1	$0.06450$	$\pm 0.00440$
2	$0.13517$	$\pm 0.00612$
3	$0.19042$	$\pm 0.00703$
4	$0.24425$	$\pm 0.00769$
5	$0.27325$	$\pm 0.00797$
6	$0.31517$	$\pm 0.00831$
7	$0.33867$	$\pm 0.00847$
8	$0.35683$	$\pm 0.00857$
9	$0.37500$	$\pm 0.00866$
10	$0.39525$	$\pm 0.00875$

This table answers the equal-search-budget question differently from a full tensor-grid proof. If the exact box budget is fixed at the cubic theorem’s $200^4$ boxes, then the comparable grid side length in dimension $m+1$ is approximately

$$40000,1169,200,69,34,20,14,10,8,6$$

for $m=1,\dots ,10$. The box half-width therefore deteriorates quickly after $m=3$. So for $m\ge 4$, a serious proof should not use a uniform tensor grid; it should use adaptive boxes, interval root isolation, and interval or Bernstein subdivision only near the numerically discovered certificate region.

A bounded exact quartic test confirms this. At the same fixed radius $R=2/5$, using the quartic Routh table and degree-$4$ Bernstein positivity gives the following rigorous lower bounds from uniform $M^5$ grids:

Quartic grid $M$	Boxes	Exact certified lower bound
20	$20^5$	$0.1011700000$
30	$30^5$	$0.1208306173$
40	$40^5$	$0.1312355469$
50	$50^5$	$0.1375880064$

These are honest lower bounds, but the $M=50$ quartic run still falls below the cubic $M=200$ theorem. The obstruction is not the quartic core itself; the same-budget floating search suggests the quartic certificate region is larger. The obstruction is that a uniform tensor grid spends most of its budget on boxes far from the certificate boundary and loses too much margin to box half-width.

The adaptive-radius lines are deliberately labeled as targets. To turn them into theorems, replace floating-point root finding and radius search by interval polynomial root isolation and interval lower bounds for ${\min }_{|z|=R}|Q_m(z)|$ over coefficient boxes.

See Also

• on Crouzeix’s conjecture: Another setting where analytic boundary estimates certify global information about zeros or spectra.
• on degree of mapping and Lipschitz constant: The argument-principle certificate is a one-complex-variable version of degree/winding-number stability.
• on Slepian’s lemma and Gaussian comparison: Useful background for Gaussian coefficient models, where covariance kernels control zero densities.

References

🐻  Edelman, A. & Kostlan, E. 1995. How many zeros of a random polynomial are real? Bulletin of the American Mathematical Society 32(1), 1–37.