trureturingGitHub MATHEMATICAL ATLAS

RESEARCH / Focused target

Refutation of reciprocal q-gamma coefficient positivity

Literature status: not recheckedRepository record: refuted

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Problem

The question

S. I. Kalmykov and D. B. Karp, Inequalities for series in q-shifted factorials and q-gamma functions, J. Math. Anal. Appl. 460 (2018), 332–351, DOI 10.1016/j.jmaa.2017.11.041, arXiv:1702.03788v1, Conjecture 1, printed p. 12:

All conclusions of Theorem 3 are valid for all α > 0, while conditions on other parameters remain intact. In particular, µ → g(µ; x) is log-concave for each fixed 0 < x < Rg.

Theorem 3, printed p. 11:

For a real sequence {gₙ}ₙ≥₀ and fixed 0 < q < 1 define

$g(\mu;x)=\displaystyle\sum_{n=0}^{\infty}\frac{g_nx^n}{\Gamma_q(\mu+n)}.$ (16)

Suppose {gₙ}∞ₙ₌₀ is doubly positive. Then (16) has a positive radius of convergence Rg and the generalized Turánian

$\Delta_g(\alpha,\beta;x)=g(\mu+\alpha;x)g(\mu+\beta;x)-g(\mu;x)g(\mu+\alpha+\beta;x)$

is positive for 0 < x < Rg if α ∈ N, β > 0 and µ ≥ max(−β, −1). If, in addition, α ≤ β + 1, then Δg(α, β; x) has positive power series coefficients at all powers of x.

The sequence is nontrivial, nonnegative, log-concave and has no internal zeros, the definition of doubly positive on printed p. 3. Equation (6), printed p. 4, defines

$$\Gamma_q(z)=(1-q)^{1-z}\frac{(q;q)\infty}{(q^z;q)\infty}.$$

Its reciprocal is extended by zero at the poles $z\in-\mathbb N_0$. Only the coefficient-positivity conclusion is refuted here.

Formal reading

The new Lean names have prefix D5.S3.Analytic.QGammaTuran.Result.

| Source object | Lean name | Correspondence | | --- | --- | --- | | Reciprocal of (6) | rqgamma | Real infinite-product quotient, extended by zero at nonpositive integers. | | Doubly positive sequence | D5.S3.Analytic.ReciprocalGammaTuran.Defs.AdmissibleSeq | Existing nontriviality, nonnegativity, log-concavity and no-internal-zero hypotheses. | | Degree $m$ coefficient | qturanCoeff | Exact finite Cauchy sum over $0\le k\le m$. | | Coefficient clause of Conjecture 1 | claim | Every $0<q<1$, admissible sequence, permitted real shifts and natural degree. | | Refutation | result | Exactly $\neg,\texttt{claim}$. |

Motivation

Our foothold

The settling declaration is D5/S3/Analytic/QGammaTuran/Result.result. The source note is D5/L/Asymptotics/kalmykovkarp2018qgamma. The unit sequence meets the sequence hypotheses. The boundary triple $(-t,t,t)$ is permitted by the conjectured coefficient clause and exposes the same cancellation pattern as in the classical reciprocal Gamma problem.

Gap

Missing bridges

Status: Refuted for the coefficient clause. The witness is $g_n=1$, $q=(99/100)^{10}$, $\mu=-7/10$, $\alpha=\beta=7/10$, $m=2$. Its shifts satisfy $\mu=\max(-\beta,-1)$ and $\alpha\le\beta+1$. Theorem 3 itself concerns integral $\alpha$; the witness has $\alpha=7/10$, so it does not contradict that theorem. Function positivity and parameter log-concavity remain unaddressed.

Route

Proposed approach

Write $R_q(x)=1/\Gamma_q(x)$ and $[x]_q=(1-q^x)/(1-q)$. The product definition yields $R_q(x)=[x]_qR_q(x+1)$, $R_q(0)=0$ and $R_q(1)=1$. At $t=7/10$, the degree-two Cauchy sum reduces to

$$\delta_2=1-PC,\qquad P=R_q(3/10)R_q(7/10),$$

$$C=\frac{[-7/10]_q}{[7/10]_q[17/10]_q} +\frac1{[7/10]_q}+\frac1{[3/10]_q}.$$

For $r=99/100$ and $q=r^{10}$, every bracket appearing here is rational because $q^{n/10}=r^n$. The complementary exponents in $P$ give

$$P=\frac{(r^3;q)\infty(r^7;q)\infty} {(q;q)_\infty^2(1-q)}.$$

For $0\le a<1$, a partial product bounds $(a;q)_\infty$ from above. Its lower bound is the partial product multiplied by $1-aq^N/(1-q)$, using

$$\prod_{j\ge N}(1-aq^j)\ge1-\frac{aq^N}{1-q}.$$

These bounds convert $PC>1$ to a finite rational comparison and yield $\delta_2<0$.

Falsifier

What would falsify this route

One permitted sequence and parameter choice with a nonpositive coefficient contradicts the universal strict-positivity clause. The displayed unit-sequence witness has a negative second coefficient. An error in the Cauchy reduction, the product-tail bound or its rational comparison would invalidate this argument without establishing the coefficient clause.

Evidence

Evidence to collect

D5/S3/Analytic/QGammaTuran/Result.result has the type result : ¬ claim, with no additional hypotheses. The sequence hypothesis is supplied by the existing D5.S3.Analytic.ReciprocalGammaTuran.Defs.admissible_one.

The experiments are available at the immutable q-gamma experiment directory. The following readings concern the scripts at that revision.

  • COMPUTED. Command: python3 /Users/macstudio/omega-op/s3688/kk-qgamma-cert.py; exit code 0; script kk-qgamma-cert.py, SHA-256 a0b18ca90e4bfd31b2c5d3871f53212c3b53c9e66cbb3be2d04042ce85065f85. The script uses 800 product factors with outward rounding to multiples of $2^{-200}$ and the geometric tail bound. Its exact rational checks are $C>0$ and $P_{\rm lower}C>1064/1000$, hence $PC>1064/1000$. Its printed value for $1-P_{\rm lower}C$, approximately $-0.06428299060000457$, is a floating approximation; the asserted rational inequalities are exact.
  • COMPUTED. Command: python3 /Users/macstudio/omega-op/s3688/kk-qgamma-scan.py; exit code 0; script kk-qgamma-scan.py, SHA-256 e1dcb5bf944905427dbdf16ea8b07267d074c4628cd75cbe14f421619c3b15c0. This finite scan uses 30 decimal digits and truncated products, with $q\in{0.5,0.9,0.97}$, $t=k/20$ for $1\le k\le19$, $\alpha=\beta=t$, $\mu\in{-t,-t+0.05,0,0.5}$, retaining $\mu\ge\max(-t,-1)$, and degrees zero through four. It finds negative coefficients for $q=0.9$ and $q=0.97$ and none for $q=0.5$. The respective minimum values are approximately $-0.0636984$, $-0.07192$ and $0.000120102$; the negative minima occur at $(t,\mu,m)=(0.7,-0.7,2)$ and the positive minimum at $(0.05,-0.05,4)$. These are finite numerical observations.

The supplied literature scope comprises the cited-by works recorded for OpenAlex W2603745271: arXiv:1808.05105 cites the source but supplies no settlement of Conjecture 1; the 2019 journal version adds no settlement in that scope; DOI 10.1142/s1793042120400187 treats a different q-binomial question. The classical reciprocal Gamma refutation, arXiv:2607.02511 and the Zenodo preprint on failure of parameter log-concavity concern $q=1$. No settlement was found in the supplied Formal Conjectures and conjectures.io scope. This bounded search does not establish exhaustive coverage or priority.

Triage

Scope assessment

What the settlement shows

  • PROVED. The universal coefficient clause is false: D5.S3.Analytic.QGammaTuran.Result.result. The witness uses the boundary cancellation at $(-t,t,t)$, $t=7/10$, and $q=(99/100)^{10}$. Thus the boundary failure seen in the classical $q=1$ problem persists at this $q$ close to one.
  • COMPUTED. At this witness the exact rational certificate gives $PC>1064/1000$, as checked by kk-qgamma-cert.py with the command, exit code and SHA-256 in Evidence. The approximate printed bound is separate from the exact comparison.
  • COMPUTED. The finite scan in Evidence finds no negative coefficient at $q=0.5$ through degree four on its stated grid.
  • OPEN. The dependence of coefficient positivity on $q$, including any uniform interval near $q=1$ or a positive result at $q=0.5$, is not settled by this witness or finite scan.
  • OPEN. The function-positivity and parameter log-concavity clauses of Conjecture 1 are not addressed. A negative coefficient alone does not establish failure of function positivity.
  • OPEN. The remaining conjectural extensions, and any results requiring them, receive no conclusion from this coefficient counterexample. The integral-shift assertion of Theorem 3 is outside the witness's parameter choice.

ASSUMED-UNVERIFIED

Unverified assumptions

Exhaustive literature coverage, citation-index completeness and priority are unverified. The bounded literature scope is supplied bibliographic evidence; no current network search is asserted. The scan's truncated product values are numerical observations, not rigorous signs on a continuous parameter range. Its absence of negative values at $q=0.5$ does not prove coefficient positivity there.

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