slug: zhu-2020-coth-bernoulli-remainder-derivative-signs
bibkey: zhu2020completelymonotonicdegrees
doi: 10.3934/math.2020224
triage: theorem
motivation_gids:
  - D5/S3/Analytic/CothBernoulliDegree/Result.result
url: https://www.aimspress.com/article/doi/10.3934/math.2020224

Derivative signs of the power-weighted hyperbolic cotangent remainder

Problem

Zhu, Completely monotonic integer degrees for a class of special functions, Conjecture 6.4:

Let m ≥ 1, n ≥ 2, and pₙ(t) be defined as (2.2). Then ((−1)ⁿ tᵐ pₙ(t))^{(m+2n−2)} > 0 holds for all t ∈ (0, ∞) and ((−1)ⁿ tᵐ pₙ(t))^{(m+2n−1)} > 0 is not true for all t ∈ (0, ∞).

Equation (2.2) is

$$p_n(t)=\frac{\coth t}{t}-\sum _{k=0}^n\frac{2^{2k}{B}_{2k}}{(2k)!}{t}^{2k-2}.$$

Put $P(t)=(-1{)}^n{t}^m{p}_n(t)$ and $M=m+2n$. The first clause is (6.12), and the failure of universal strict positivity in the second clause is (6.13).

Formal reading

All Lean names below have prefix D5.S3.Analytic.CothBernoulliDegree.

The value assigned by real division and integer powers at zero is immaterial: PartialFraction.hasSum_iteratedDeriv_P uses neighborhood agreement at each positive argument to transfer all iterated derivatives.

Motivation

The settling declaration is D5/S3/Analytic/CothBernoulliDegree/Result.result. The source note is D5/L/Asymptotics/zhu2020completelymonotonicdegrees. The two consecutive derivative orders locate the loss of positivity for a power-weighted Bernoulli remainder throughout the full integer parameter range.

Gap

Status: Proved for all integers $m\ge 1$, $n\ge 2$ and all positive real arguments in (6.12). For (6.13), a single explicit positive argument, $t=5\pi /(m+2n)$, gives a strictly negative next derivative for each parameter pair.

Route

The cotangent partial fractions at an imaginary argument, followed by finite geometric division and the even zeta values, give

$$P(t)=2\sum _{k=1}^{\infty }\frac{t^M}{(\pi k{)}^{2n}(t^2+(\pi k{)}^2)},t>0.$$

On bounded real intervals, the quadratic reciprocal has derivative bound $|D^j(1/(t^2+a^2))|\le j!a^{-j-2}$. The product rule supplies locally uniform summable majorants for every derivative order. Hence the representation can be differentiated term by term at every positive argument.

For $a>0$, the rational kernel has certificates

$$D^N\left[\frac{t^{N+2}}{t^2+a^2}\right]=N!\left(1-\mathrm{Re}(1+it/a{)}^{-N-1}\right),$$

$$D^{N+1}\left[\frac{t^{N+2}}{t^2+a^2}\right]=\frac{(N+1)!}{a}(1+(t/a{)}^2{)}^{-(N+2)/2}\sin ((N+2)\arctan (t/a)).$$

At $N=M-2$ the first certificate is strictly positive when $t>0$, since $|1+it/a|>1$. Every weighted summand is positive, which proves (6.12).

At $t=5\pi /M$, normalize the second derivative series by $2(M-1)!/{\pi }^{2n+1}$. The first term is at most $-1/8$, because

$$(1+(5/M{)}^2{)}^{-M/2}\sin (M\arctan (5/M))\le -\frac{1}{8}(M\ge 5).$$

The remaining terms sum to at most ${\sum }_{k=2}^{\infty }{k}^{-2n-1}\le 1/16$. Multiplying back gives

$$P^{(M-1)}(5\pi /M)\le -\frac{(M-1)!}{8{\pi }^{2n+1}}<0,$$

which proves (6.13).

Falsifier

Integers $m\ge 1$, $n\ge 2$ and $t>0$ with $P^{(m+2n-2)}(t)\le 0$ would contradict the first clause. A parameter pair in this range for which the next derivative were strictly positive at every positive argument would contradict the second clause. In particular, it would contradict the explicit bound at $5\pi /(m+2n)$.

Evidence

D5/S3/Analytic/CothBernoulliDegree/Result.result proves exactly Defs.claim. Its principal dependencies are PartialFraction.P_eq_tsum, PartialFraction.hasSum_iteratedDeriv_P, PartialFraction.iteratedDeriv_P, Kernel.kernel_deriv_N, Kernel.kernel_deriv_N_pos, Kernel.kernel_deriv_succ, Kernel.phase_bound and Kernel.tail_bound. The private Result.negative_phase_sum and Result.negative_mixture combine the negative first term with the tail estimate and restore the weighted scale. The axiom closure of Result.result is {propext, Classical.choice, Quot.sound}.

The proof route was derived with the assistance of the ChatGPT Pro model. Mahmoud–Qi, Math. Inequal. Appl. 25(1) (2022), settle the related $m=1$ degree bounds for the remainder. Bouali, arXiv:2202.01801, treats $m=0$ or $n\in \{0,1\}$. No settlement of Conjecture 6.4 was found in the searched scope, which also includes Xu–Cen, Qi arXiv:2003.05300v2, Mahmoud–Qi arXiv:1912.07989v3, Bouali arXiv:2206.01527, Moustafa (2024), and Qi’s bibliography (July 2026). This bounded scope does not establish exhaustive citation-index coverage or priority.

Triage

• PROVED. Clause (6.12) holds for every integer $m\ge 1$, $n\ge 2$ and every $t>0$, by Result.result.
• PROVED in prose. Clause (6.13) has the stronger bound $P^{(M-1)}(5\pi /M)\le -(M-1)!/(8{\pi }^{2n+1})<0$, where $M=m+2n$. The private Result.negative_mixture proves this bound for the weighted derivative series; PartialFraction.iteratedDeriv_P identifies that series with the derivative of $P$, as used in Result.result.
• PROVED. The partial-fraction representation $P(t)=2{\sum }_{k\ge 1}(\pi k{)}^{-2n}{t}^M/(t^2+(\pi k{)}^2)$ for $t>0$ is PartialFraction.P_eq_tsum.
• COMPUTED. Evaluation at 120 decimal digits gives $P^{(M-1)}(5\pi /M)\approx -0.0169$ for $(m,n)=(1,2)$ and $\approx -0.949$ for $(m,n)=(3,2)$. These numerical evaluations are separate from the exact inequalities.
• OPEN. The exact completely monotonic degree of $t^m{p}_n(t)$ when it is not an integer is not determined here.

ASSUMED-UNVERIFIED

Exhaustive literature coverage, citation-index completeness and priority are unverified. The supplied bounded literature assessment has not been independently rechecked against remote sources; the formal statement and definitions are those displayed above.