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ERDőS #513 · PARTIAL

Erdős problem #513 — wave w042

Date: 2026-07-31

Claim labels

Mathematical assertions below are labelled as requested:

The main conclusion is (b+d): it uses Hadamard's three-circles theorem (equivalently He–Tang, Theorem 2.8, whose short proof is reconstructed below) and a finite Arb ball-arithmetic certificate.

Step 0: mandatory live-page check

I fetched the live page and its discussion thread through the Bright Data browser on 2026-07-31. Direct browser extraction and a full-page screenshot agreed. (d-source)

Verbatim live statement

Let \(f=\sum_{n=0}^\infty a_nz^n\) be a transcendental entire function. What is the greatest possible value of \[ > \liminf_{r\to \infty} > \frac{\max_n\lvert a_nr^n\rvert} > {\max_{\lvert z\rvert=r}\lvert f(z)\rvert}? > \]

The page calls the supremum of these values \(B\). Its displayed known results are

\[ \frac12<B,\qquad \frac47<B\leq \frac2\pi-c \]

for some absolute \(c>0\), and the more recent lower bounds

\[ B>0.5850724,\qquad B>0.5850788. \]

The page attributes these respectively to Kövári (unpublished), Clunie–Hayman, He–Tang, and a GPT-assisted computation prompted by Nat Sothanaphan. (d-source)

Gate status and every marker

The six comments were read before doing mathematics:

  1. Quanyu Tang records the Clunie–Hayman \(4/7\) lower bound, Hayman–Lingham Problem 2.14(c), and the He–Tang preprint arXiv:2602.12217, with \(B>0.58507\). (d-source)
  2. Terence Tao links an optimization-constants page. (d-source)
  3. Nat Sothanaphan links a write-up certifying the slight improvement \(B\geq0.585078819653\), rounded on the live page to \(0.5850788\), and reports that more ambitious attempts did not succeed. (d-source)
  4. Thomas Bloom notes that Clunie–Hayman claim the stronger upper bound \(2/\pi-c\) for an explicit-in-principle \(c>0\). (d-source)
  5. Kevin Barreto observes that allowing polynomials makes the limit equal to \(1\), for example with \(f(z)=z+1\), so “transcendental” is required. (a) Indeed \(\mu(r)=r\), \(M(r)=r+1\), and \(r/(r+1)\to1\).
  6. Quanyu Tang confirms that Clunie–Hayman assume \(f\) is not a polynomial. (d-source)

The page also cross-references problem #227, concerning the case where the ratio has a limit. (d-source)

There was therefore no skip condition, so I proceeded.

Literature audit

Only primary papers/preprints or official publication records were used for mathematical history.

\[ K=3.568182317714,\qquad \alpha=3.961543335688. \] I downloaded and read the note (SHA-256 de786f105f81bb2b6df99e6d846d20913316ef80ce2b156371f0610c03deb516) and independently reran its parameter pair. (d-source+d)

New explicit construction and bound

Take the exact rational numbers

\[ K=\frac{713637088446737}{200000000000000} =3.568185442233685, \qquad \alpha=\frac{990385777952557}{250000000000000} =3.961543111810228, \]

and put \(\varepsilon=e^{i\alpha}\). Define

\[ f_*(z)=\sum_{n=0}^{\infty} \frac{\varepsilon^{\,n(n-1)/2}}{K^{n(n+1)/2}}z^n. \]

Certified result.

\[ \boxed{B>0.585078819674.} \]

This improves the full-precision certified lower bound \(0.585078819653\) by

\[ 0.585078819674-0.585078819653=2.1\times10^{-11}. \]

The construction is explicit; the displayed finite decimals are exact rationals, not floating-point choices. (b+d)

This does not determine \(B\). It only raises the lower endpoint of the known interval; the live-page upper bound \(B\leq2/\pi-c\) remains untouched. (a+d-source)

From-scratch reduction

This section reconstructs the relevant He–Tang reduction so that the numerical certificate is connected to the original entire-function question.

Let

\[ T_n=\frac{n(n+1)}2,\qquad b_n=\varepsilon^{n(n-1)/2}K^{-T_n}\quad(n\in\mathbb Z), \]

and define

\[ k(z)=\sum_{n\in\mathbb Z}b_nz^n,\qquad f(z)=\sum_{n\geq0}b_nz^n. \]

1. Analyticity and scaling

For \(n\geq0\), \(|b_n|^{1/n}=K^{-(n+1)/2}\to0\), so \(f\) is entire; every \(b_n\neq0\), so it is transcendental. Both tails of \(k\) converge normally on every compact annulus, so \(k\) is holomorphic on \(\mathbb C\setminus\{0\}\). (a)

Coefficient comparison gives

\[ k(Kz)=z\,k(\varepsilon z). \]

Indeed, the coefficient of \(z^n\) on the two sides is respectively \(b_nK^n\) and \(b_{n-1}\varepsilon^{n-1}\), and these are equal by \(T_n-n=T_{n-1}\). Iteration yields

\[ k(K^mz)=K^{m(m-1)/2}\varepsilon^{m(m-1)/2}z^m k(\varepsilon^mz). \]

Therefore, with

\[ A(K,\varepsilon)=\max_{|z|=1}|k(z)|, \]

one has

\[ M(K^m,k)=K^{m(m-1)/2}A(K,\varepsilon). \]

All of these identities are exact. (a)

2. Maximum term and endpoint reduction

Since

\[ |b_n|r^n=K^{-n(n+1)/2}r^n, \qquad \frac{|b_{n+1}|r^{n+1}}{|b_n|r^n}=\frac r{K^{n+1}}, \]

the unique maximum term for \(K^m<r<K^{m+1}\) is \(n=m\); at \(r=K^m\), indices \(m-1,m\) tie. Hence

\[ \mu(K^m,f)=K^{m(m-1)/2}. \]

(a)

On each interval \(m\log K<t<(m+1)\log K\), the function \(\log\mu(e^t,f)\) is affine. Hadamard's three-circles theorem says that \(\log M(e^t,f)\) is convex. Thus

\[ t\longmapsto\log\frac{\mu(e^t,f)}{M(e^t,f)} \]

is concave on that interval, and its minimum occurs at an endpoint. Consequently the defining liminf may be taken along \(r=K^m\). (b: Hadamard three-circles)

For negative indices,

\[ k(z)-f(z)=\sum_{j\geq1}b_{-j}z^{-j}=O(|z|^{-1}) \quad(|z|\to\infty). \]

The elementary inequality

\[ |M(r,k)-M(r,f)|\leq M(r,k-f) \]

then shows that replacing \(k\) by \(f\) changes the endpoint maximum modulus by a negligible relative amount. Combining this with the preceding exact formulas gives

\[ \boxed{\displaystyle \beta(f):=\liminf_{r\to\infty}\frac{\mu(r,f)}{M(r,f)} =\frac1{A(K,\varepsilon)}.} \]

(b: Hadamard three-circles; otherwise a)

3. One-variable cosine series

Put \(z=\varepsilon e^{2i\theta}\), pair the terms of \(k\) with indices \(n\) and \(-n-1\), and multiply by \(e^{i\theta}\). Direct calculation gives

\[ e^{i\theta}k(\varepsilon e^{2i\theta}) =2\sum_{n=0}^{\infty} \varepsilon^{T_n}K^{-T_n}\cos((2n+1)\theta). \]

Since \(\theta\mapsto\varepsilon e^{2i\theta}\) parametrizes the unit circle,

\[ A(K,\varepsilon) =\max_{\theta\in\mathbb R}|P(\theta)|, \qquad P(\theta)=2\sum_{n=0}^{\infty} e^{i\alpha T_n}K^{-T_n}\cos((2n+1)\theta). \]

(a)

The frequencies are all odd, so \(P(-\theta)=P(\theta)\) and \(P(\theta+\pi)=-P(\theta)\). Therefore it is enough to maximize on \([0,\pi/2]\). (a)

Finite certificate

The standalone checker is erdos513_wavew042_reverify.py. It imports no project code and no He–Tang/Sothanaphan code. Its SHA-256 is 31b2d9840278260b1ee95e8f687f0e5a2bbea96cb94bcf324079cb57fb22b9b7.

What it certifies

Truncate \(P\) at \(n=N\):

\[ P_N(\theta)=2\sum_{n=0}^{N} e^{i\alpha T_n}K^{-T_n}\cos((2n+1)\theta). \]

Since \(T_{n+1}-T_n=n+1\), the omitted tail satisfies

\[ \sup_\theta|P(\theta)-P_N(\theta)| \leq \frac{2K^{-T_{N+1}}}{1-K^{-(N+2)}}. \]

(a)

For \(y\in[0,1]\), set \(q(y)=|P_N(\pi y/2)|^2\). If

\[ \begin{aligned} S_0&=2\sum_{n=0}^{N}K^{-T_n},\\ S_1&=\frac\pi2\,2\sum_{n=0}^{N}(2n+1)K^{-T_n},\\ S_2&=\left(\frac\pi2\right)^2 2\sum_{n=0}^{N}(2n+1)^2K^{-T_n}, \end{aligned} \]

then differentiation and the triangle inequality give

\[ |q''(y)|\leq L:=2(S_1^2+S_0S_2). \]

(a)

The program recursively subdivides \([0,1]\) into dyadic intervals. On an interval of midpoint \(m\) and radius \(r\), the Lipschitz bound

\[ q'(I)\subseteq q'(m)+[-Lr,Lr] \]

proves that the interval contains no critical point whenever the right side excludes zero. Every unpruned interval is subdivided to the requested exact dyadic depth. This is a complete critical-point cover, not a sampled-grid heuristic. (a+d)

If \(y_0\) is a critical point in a surviving interval, Taylor's theorem gives

\[ q(y_0)\leq q(m)+\frac12Lr^2. \]

The program evaluates the right side, the two endpoints, and the tail in outward-rounded Arb balls. The inputs \(K,\alpha\) are first parsed as exact FLINT rationals. Arb and python-flint provide rigorous midpoint-radius error tracking. (a+d)

Reproduction

The dependency installed and used here is python-flint==0.8.0.

/home/exedev/.venv/bin/python runs/erdos513_wavew042_reverify.py

The default run independently certifies both the new pair and the Sothanaphan baseline. Its decisive output is:

new pair:
  N / dyadic depth            = 7 / 27
  nodes / candidate intervals = 2915 / 68
  sup |q''| upper             = 180.7299558842003423426526123506259...
  max |P_N| upper             = 1.7091714250676725949970960332679...
  tail upper                  = 2.58778156653628811977894360716e-20
  A upper                     = 1.7091714250676725950229738489332...
  beta lower                  = 0.5850788196745134549506332325464...
  PASS: beta > 0.585078819674

baseline pair:
  A upper                     = 1.7091714251294884335039460052955...
  beta lower                  = 0.5850788196533528286472198685834...
  PASS: beta > 0.585078819653

These are deterministic finite ball-arithmetic results. (d)

I also reran a genuinely different certificate:

/home/exedev/.venv/bin/python runs/erdos513_wavew042_reverify.py \
  --N 8 --depth 28 --dps 100 --skip-baseline

It examined 3051 dyadic nodes, left 68 critical intervals, bounded the tail by \(2.7599265732573566\times10^{-25}\), and independently obtained

\[ A<1.709171425067672327265976434342, \qquad \frac1A>0.585078819674513546608465623886. \]

(d)

How the parameters were found

Floating-point exploration showed two active maxima: the endpoint \(\theta=0\) and an interior maximum near \(\theta=0.645541552\). Solving the equioscillation/stationarity system suggested the displayed \(K,\alpha\), after which they were rounded to exact rationals and certified from scratch. This search narrative is only (c); no part of the proof depends on the optimizer, the approximate peak location, or a claim of local/global parameter optimality.

Verified state

PARTIAL: An explicit He–Tang-family entire function and independent Arb checker rigorously improve the known lower bound to B > 0.585078819674; the exact value remains open.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger