Generalized Binomial Biroot Conjecture

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Let x>0x>0, let nn be a positive integer, and let c>0c>0. Define

βmn(xn,c)=∑k=0⌈m/n⌉xkcm−nk(mnk)∑k=0⌈m/n⌉−1xkcm−nk−1(mnk+1).\beta_m^n(x^n,c)=\frac{\displaystyle\sum_{k=0}^{\lceil m/n\rceil}x^k c^{m-nk}\binom{m}{nk}}{\displaystyle\sum_{k=0}^{\lceil m/n\rceil-1}x^k c^{m-nk-1}\binom{m}{nk+1}}.

Generalized Binomial Biroot Conjecture. The approximations converge according to

lim⁡m→∞βmn(x,c)=xn.\lim_{m\to\infty}\beta_m^n(x,c)=\sqrt[n]{x}.

This is the paper’s principal generalization from square roots to arbitrary positive integer roots. The conclusion is supported by computational experimentation, while the paper identifies proving it as an open question.

References

Primary source

Isaac Wolford, “Combinatorial and Gaussian Foundations of Rational Nth Root Approximations: Theorems and Conjectures”, arXiv:2508.14095 (2025).

Progress summary

Refreshed
Claimed solved

The original paper leaves the general root problem open, but an unverified posted proof now claims to settle it for every positive integer root index.

Isaac Wolford’s 2025 paper conjectures that these binomial ratios approach the positive nnth root of xx for all positive parameters. It proves the square-root case and presents computational evidence for higher roots, while identifying the general claim as open.

Known results

  • Square-root case, with optimal parameter conditions, proved by Isaac Wolford (2025); higher-root convergence supported computationally.

Posted attempt

A reader-written argument claims a complete proof for every n≥1n\ge1: a roots-of-unity filter isolates the dominant term (c+x1/n)m(c+x^{1/n})^m, yielding exponential convergence, with a separate treatment of n=1n=1. The argument has not been independently verified.

Current status (as of August 2026): The square-root case is proved, while a complete general proof has been claimed but remains unverified; absent confirmation, the higher-root cases are not settled.

Sources

Solutions 1

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Proof for every positive integer root index, with an explicit exponential rate. Use the definition in the original source:

βm(n)(x,c)=∑k=0⌈m/n⌉(mnk)xkcm−nk∑k=0⌈m/n⌉−1(mnk+1)xkcm−nk−1,x,c>0,m≥1.\beta_m^{(n)}(x,c) = \frac{\displaystyle \sum_{k=0}^{\lceil m/n\rceil} \binom m{nk}x^k c^{m-nk}} {\displaystyle \sum_{k=0}^{\lceil m/n\rceil-1} \binom m{nk+1}x^k c^{m-nk-1}}, \qquad x,c>0,\quad m\ge1.

The denominator is positive because its initial summand is mcm−1mc^{m-1}. Terms with binomial index exceeding mm are zero.

First assume n≥2n\ge2, set t=x1/n>0t=x^{1/n}>0, and let ζ=e2πi/n\zeta=e^{2\pi i/n}. The numerator and denominator are, respectively,

Nm=∑0≤j≤m\j≡0(modn)(mj)tjcm−j,N_m= \sum_{\substack{0\le j\le m\j\equiv0\;(\mathrm{mod}\;n)}} \binom mj t^jc^{m-j},

and

Dm=1t∑0≤j≤m\j≡1(modn)(mj)tjcm−j.D_m= \frac1t \sum_{\substack{0\le j\le m\j\equiv1\;(\mathrm{mod}\;n)}} \binom mj t^jc^{m-j}.

Indeed,

⌈mn⌉−1=⌊m−1n⌋,\left\lceil\frac mn\right\rceil-1 = \left\lfloor\frac{m-1}{n}\right\rfloor,

so the denominator includes every and only nonzero residue-11 term. Applying the roots-of-unity filter gives the exact formulas

Nm=1n∑s=0n−1(c+tζs)m,N_m=\frac1n\sum_{s=0}^{n-1}(c+t\zeta^s)^m, Dm=1nt∑s=0n−1ζ−s(c+tζs)m.D_m=\frac1{nt} \sum_{s=0}^{n-1}\zeta^{-s}(c+t\zeta^s)^m.

The s=0s=0 summand has strictly larger modulus than all the others, because

∣c+tζs∣2=c2+t2+2ctcos⁡(2πs/n)<(c+t)2(1≤s<n).|c+t\zeta^s|^2 = c^2+t^2+2ct\cos(2\pi s/n) < (c+t)^2 \qquad(1\le s<n).

Define

ρ=c2+t2+2ctcos⁡(2π/n)c+t<1.\rho= \frac{\sqrt{c^2+t^2+2ct\cos(2\pi/n)}}{c+t} <1.

Then

Nm=(c+t)mn(1+Um),Dm=(c+t)mnt(1+Vm),N_m=\frac{(c+t)^m}{n}(1+U_m), \qquad D_m=\frac{(c+t)^m}{nt}(1+V_m),

where

∣Um∣, ∣Vm∣≤(n−1)ρm.|U_m|,\ |V_m| \le(n-1)\rho^m.

It follows immediately that

βm(n)(x,c)=t1+Um1+Vm⟶t=x1/n.\beta_m^{(n)}(x,c) = t\frac{1+U_m}{1+V_m} \longrightarrow t=x^{1/n}.

More precisely, whenever (n−1)ρm<1(n-1)\rho^m<1,

∣βm(n)(x,c)−x1/n∣≤2x1/n(n−1)ρm1−(n−1)ρm.\boxed{\displaystyle \left|\beta_m^{(n)}(x,c)-x^{1/n}\right| \le \frac{2x^{1/n}(n-1)\rho^m} {1-(n-1)\rho^m}. }

For fixed n,tn,t, the optimal spectral rate occurs precisely at c=tc=t: indeed

ρ2=1−2ct(1−cos⁡(2π/n))(c+t)2,\rho^2 = 1-\frac{2ct(1-\cos(2\pi/n))}{(c+t)^2},

and ct/(c+t)2≤1/4ct/(c+t)^2\le1/4, with equality exactly when c=tc=t. The minimum contraction factor is cos⁡(π/n)\cos(\pi/n).

The index n=1n=1 must be handled separately because residues 00 and 11 then coincide, whereas the denominator omits its constant term. Directly,

Nm=(c+x)m,Dm=(c+x)m−cmx,N_m=(c+x)^m, \qquad D_m=\frac{(c+x)^m-c^m}{x},

so

βm(1)(x,c)=x1−(c/(c+x))m⟶x.\beta_m^{(1)}(x,c) = \frac{x}{1-(c/(c+x))^m} \longrightarrow x.

Thus the conjecture holds for its entire stated range n≥1n\ge1, x>0x>0, c>0c>0.

Source: Isaac Wolford, Combinatorial and Gaussian Foundations of Rational Nth Root Approximations, §3.1.3, Conjecture 3.4, https://arxiv.org/abs/2508.14095. The defining expression in that source is βm(n)(x,c)\beta_m^{(n)}(x,c); an occurrence of βm(n)(xn,c)\beta_m^{(n)}(x^n,c) in the problem display is a transcription mismatch.