The numerator is

$$ D_L(z)=z\sin\frac{Lz}{2}-\frac12\cos\frac{Lz}{2}. $$

I keep looking at this little function because it is what remains after a finite arithmetic sum has been made to forget almost all of its arithmetic mess. Its zeros are real and simple. More importantly, it does not appear by hopeful analogy: Poisson summation produces it exactly once the source is flat at its support endpoint.

That is the whole generating mechanism. The rest of the construction is an increasingly fussy effort to keep the error smaller than this Dini wave farther and farther toward the center.

Why the endpoint has to be exactly flat#

Start with a real even source \(p\), supported on \([-1,1]\), and the finite arithmetic kernel

$$ K_{c,p}(x)=\frac{e^{x/2}}{\sqrt c} \sum_{1\le n

As \(x\) moves, individual summands disappear when their arguments hit \(1\). If \(p(1)\ne0\), each disappearance creates an internal value jump. Even a tiny nonzero endpoint value restores the complete jump symbol; this is not a perturbative annoyance that can be made harmless by choosing \(p(1)\) very small. The remote zeros then hear a finite Dirichlet polynomial made from all those jumps.

Set \(p(1)=0\), impose zero integral, and make the endpoint flat enough for the later integrations by parts. Poisson summation gives the exact split

$$ K_{c,p}(x) = -\frac{p(0)}{2\sqrt c}e^{x/2} +\sqrt c\,e^{-x/2} \sum_{\ell\ge1}\widehat p(\ell ce^{-x}). $$

The first term is universal: \(Ae^{x/2}\), with \(A=-p(0)/(2\sqrt c)\). Everything source-specific has been pushed into Fourier aliases. This is the lovely part. Endpoint flatness does not merely make the kernel smoother; it isolates one canonical endpoint jet.

Extend \(Ae^{x/2}\) to the left of the interval and take its centered cosine transform. One gets

$$ -\int_{-\infty}^0 Ae^{x/2} \cos\left(z(x-L/2)\right)\,dx =A\frac{D_L(z)}{z^2+1/4}. $$

So the tail model is not a bare sine wave but exactly the Dini numerator above.

Why the Dini zeros behave#

Write \(a=L/2\). The equation \(D_L(z)=0\) is the characteristic equation for

$$ -u''=z^2u,\qquad u'(0)=0,\qquad u'(a)+\frac12u(a)=0. $$

Its quadratic form is

$$ \int_0^a |u'(x)|^2\,dx+\frac12|u(a)|^2. $$

That form is positive, and the boundary conditions are separated and selfadjoint. Sturm--Liouville theory therefore gives positive simple eigenvalues. Equivalently, every zero of \(D_L\) is real and simple, with one positive zero in each expected trigonometric cell. A complicated arithmetic object has handed its remote zero geometry to an extremely well-behaved one-dimensional boundary problem. Cute.

The quantitative comparison uses

$$ B=-\partial_x^2+\frac14. $$

Apply \(B\) repeatedly and integrate by parts twice each time. After \(r\) rounds,

$$ (z^2+1/4)^rF(z) =A(z^2+1/4)^{r-1}D_L(z) +\text{endpoint defects} +\widehat{B^rK}(z). $$

The endpoint flatness kills the right-end boundary terms. The left-end defects are explicit differences between the actual jet and the universal exponential jet. On each rectangular Dini cell, lower bounds for \(D_L\) compete against the \(L^1\)-size of \(B^rK\) and those defects. Once the strict inequality holds, Rouché gives exactly one zero in the cell. Conjugation symmetry forces it onto the real axis, and uniqueness makes it simple.

There is a useful bit of bookkeeping hidden in that sentence. On a cell whose left edge is \(x\), the reference term has size essentially

$$ |A|x^{2r-1}, $$

while the transformed remainder pays \(x^{-(2r-1)}\) after division by the reference. Each lower endpoint defect pays its own visible lower power of \(x\). Nothing is bundled into a mysterious “small error”: the theorem displays the coefficient mismatches and \(\|B^rK\|_1\) separately. This is why improving the derivative estimate changes the core radius so cleanly. The Dini geometry stays fixed; only the price of manufacturing an endpoint-flat source changes.

Three core scales#

The first carrier is the Fourier-self-dual Gaussian--Hermite function

$$ h(x)=x^2(2\pi x^2-3)e^{-\pi x^2}, $$

whose Mellin transform satisfies

$$ \zeta(s)M_h(s)=\frac{\xi(s)}{2\pi}. $$

Cut it off with \((1-x^2/\lambda^2)^{2r+2}\), choose \(r\asymp\log\lambda\), and add a small Gaussian correction to restore zero integral. Polynomial--Gaussian differentiation costs roughly \((Cr)^{2r}\). The Dini reference wins once \(|\Re z|\) is a sufficiently large multiple of \(r\), giving an \(O(\log\lambda)=O(\log c)\) nonreal core.

Now conjugate the shifted Euler operator

$$ \mathcal Z=-\left(x\partial_x+\frac12\right)^2 $$

into logarithmic coordinates. It becomes \(-\partial_y^2\). Ordinary heat, \(e^{-\alpha\mathcal Z}\), inserts the zero-free Mellin multiplier \(e^{-\alpha z^2}\) and cuts the derivative growth to about \((Cr)^r\). The same comparison now closes at

$$ |\Re z|\asymp \sqrt r, $$

so the core is \(O(\sqrt{\log c})\).

Finally use superheat \(e^{-\alpha\mathcal Z^k}\) with fixed \(k\), together with a logarithmic Gevrey cutoff matched to order \(1+1/(2k)\). The transformed remainder grows only like

$$ (Cr)^{r/k+O(1)}. $$

The cutoff is performed in \(\log|x|\), not at a fixed Euclidean width. It equals one through \(|x|\le\sqrt\lambda\), dies by \(|x|=\lambda\), and vanishes to infinite order there. When Euler derivatives are distributed by Leibniz, derivatives landing on the cutoff and derivatives landing on the superheated carrier have the same leading cost, \((r/k)\log r\). I find this balancing act particularly satisfying: the cutoff is not merely smooth enough, but tuned so neither side of the product becomes the expensive side. The Fourier aliases and endpoint defects are then super-stretched small, leaving the top transformed remainder as the actual contest.

Against the \(x^{2r-1}\) Dini amplitude, this balances at

$$ |\Re z|\asymp r^{1/(2k)}, $$

giving the core \(O((\log c)^{1/(2k)})\). For any prescribed positive power of \(\log c\), one may fix \(k\) large enough to beat it.

For every fixed closed substrip \(|\Im z|\le Y<1/2\), these compact arithmetic transforms also converge uniformly across the entire real direction to

$$ \frac{\Xi(z)}{2\pi}e^{-\alpha z^{2k}}. $$

But \(k\) is fixed before \(c\to\infty\), the constants are not uniform in \(k\), and the core still diverges. The multiplier is entire and zero-free, so the limit has exactly the zero divisor of \(\Xi\); any hypothetical fixed nonreal zero eventually sits inside that expanding core. This construction gives increasingly thin real-and-simple strip tails. It does not prove RH. The gap is small in exponent and absolute in logic.

Notebook references: C-0147, C-0150, C-0153, C-0176, C-0180