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Quotients on the half-gaps

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

By Exact interpolation, for every \(k\ge2.36\) there are entire functions \(H_1,H_2\) that touch \(T_k\) and \(0\) at every node. The normalized statement \((\mathrm N_k)\) follows once \(H_1\le T_k\) and \(H_2\ge0\) on \([0,\infty)\). These two inequalities are strict between the nodes and equalities of second order at the nodes, so they cannot be read off from approximate values of \(H_1,H_2\). This lesson divides out the double zeros and reduces the inequalities on the interval \([0,89.5]\) to a statement about explicit polynomials of degree \(40\).

The lesson follows [OpenAI-T, Section 5, the parts on the quotients and on the lower supports of the target, and Section 3, the part on polynomial rows].

1. Slack functions, half-gaps and quotients

Fix \(k\ge2.36\). Let \(\ell_1,\ell_2\) be the lists of Theorem 3.1 of the lesson on exact interpolation, and \(H_1,H_2\) the functions of its Corollary 3.2. The slack functions are \[ F_1=T_k-H_1,\qquad F_2=H_2 . \] With \(\sigma_1=-1\), \(\sigma_2=1\), \(\delta_1=1\), \(\delta_2=0\) they are \(F_i=\sigma_iH_i+\delta_iT_k\). They are entire, real on the real axis, and \[ F_i(m)=F_i'(m)=0\qquad(m\in\mathcal N). \] The aim is \(F_1\ge0\) and \(F_2\ge0\) on \([0,\infty)\).

The set \(\mathcal N_0\) of the nodes together with \(0\) consists of the integers \(n\ge0\) congruent to \(0,1,3,4,7,9\) modulo \(12\). The gaps between consecutive elements have the lengths \(1,2,1,3,2,3\), repeating. A half-gap is given by a centre \(m\in\mathcal N_0\) with \(m\le88\) and a neighbour \(m'\) of \(m\) in \(\mathcal N_0\): it is the interval between \(m\) and the midpoint of \(m\) and \(m'\). With \[ y=\frac{m'-m}2 \] it consists of the points \(s=m+yx\), \(0\le x\le1\). The number \(y\) is positive for the half-gap to the right of \(m\) and negative for the one to the left, and \(\tfrac12\le|y|\le\tfrac32\).

Lemma 1.1. There are \(91\) half-gaps, and their union is the interval \([0,s_0]\) with \(s_0=89.5\).

Proof. The centres are the elements of \(\mathcal N_0\) up to \(88\): six in each of the seven blocks of twelve integers from \(0\) to \(83\), and \(84,85,87,88\), together \(46\). Each has a right half-gap, and each except \(0\) has a left one; this gives \(91\). Two consecutive centres share the gap between them, which is the union of the right half-gap of the smaller and the left half-gap of the larger. The successor of \(88\) in \(\mathcal N_0\) is \(91\), so the last right half-gap is \([88,89.5]\). \(\blacksquare\)

For a function \(u\) that is twice continuously differentiable, and \(m>0\), let \[ \begin{aligned} &\mathcal D_mu(s)\\ &\quad=\frac{u(s)-u(m)-(s-m)\,u'(m)}{(s-m)^2} \end{aligned} \] for \(s\ne m\), and \(\mathcal D_mu(m)=\tfrac12u''(m)\). At the centre \(0\) nothing is divided out: \(\mathcal D_0u=u\). Let \(\nu=2\) for \(m>0\) and \(\nu=0\) for \(m=0\).

Lemma 1.2. Let \(m>0\).

  1. For all \(s\), with \(w=s-m\), \[ \begin{aligned} &\mathcal D_mu(s)\\ &\quad=\int_0^1(1-v)\,u''(m+vw)\,dv . \end{aligned} \] In particular \(|\mathcal D_mu(s)|\) is at most half the maximum of \(|u''|\) between \(m\) and \(s\).
  2. If \(u(m)=u'(m)=0\), then \(u(s)=(s-m)^2\,\mathcal D_mu(s)\).
  3. If \(u\) is entire, then for all \(x\) and \(y\) \[ \begin{aligned} &\mathcal D_mu(m+yx)\\ &\quad=\sum_{r\ge0}\frac{y^r}{(r+2)!}\,u^{(r+2)}(m)\,x^r . \end{aligned} \]

For \(m=0\) the corresponding expansion is \(u(yx)=\sum_r\tfrac{y^r}{r!}u^{(r)}(0)x^r\). So in both cases \[ \begin{aligned} &\mathcal D_mu(m+yx)\\ &\quad=\sum_{r\ge0}\frac{y^r}{(r+\nu)!}\,u^{(r+\nu)}(m)\,x^r . \end{aligned} \]

Proof. (1) is the formula of Taylor with the remainder as an integral, and \(\int_0^1(1-v)\,dv=\tfrac12\). (2) is the definition. (3) follows from the Taylor series of \(u\) at \(m\), in which the terms of degree \(0\) and \(1\) are removed and the others divided by \((yx)^2\). \(\blacksquare\)

By (2), \(F_i(s)=(s-m)^2\,\mathcal D_mF_i(s)\) for every centre \(m>0\). So \(F_i\ge0\) on a half-gap as soon as \(\mathcal D_mF_i\ge0\) there. The quotient is formed for each function separately, \[ \mathcal D_mF_i=\sigma_i\,\mathcal D_mH_i+\delta_i\,\mathcal D_mT_k , \] and each term is then approximated. In this way no approximation error is ever divided by \((s-m)^2\).

2. Three comparisons

Let \(\ell_i^{\mathrm{ref}}=X_iq(k)\) be the reference lists, \(p_i^{\mathrm{ref}}\) their cardinal functions, and \[ \begin{gathered} H_1^{\mathrm{ref}}=p_1^{\mathrm{ref}}+K_{p_2^{\mathrm{ref}}},\\ H_2^{\mathrm{ref}}=p_2^{\mathrm{ref}}+K_{p_1^{\mathrm{ref}}}, \end{gathered} \] formed with the exact spectral integrals. These functions need not have any zeros.

Proposition 2.1. On every half-gap, with centre \(m\), \[ |\mathcal D_m(H_i-H_i^{\mathrm{ref}})(s)|<0.00011 . \]

Proof. Let \(\Delta_i=\ell_i-\ell_i^{\mathrm{ref}}\). Its extra entries are \(0\), and \(\|\Delta_1\|_1+\|\Delta_2\|_1<2\cdot10^{-8}\) by Theorem 3.1 of the lesson on exact interpolation. By linearity \(H_i-H_i^{\mathrm{ref}}=p_{\Delta_i}+K_{p_{\Delta_j}}\), with \(j\) the other index. For a list with \(C=0\) and \(W=\|\Delta\|_1\), Corollary 2.3 of the lesson on cardinal functions and Theorem 3.2 of the lesson on Gaussian damping give, for real \(s\ge0\) and \(a\ge0\), \[ \begin{gathered} |p_\Delta^{(a)}(s)|\le36\,\pi^a\,\|\Delta\|_1,\\ |K_{p_\Delta}^{(a)}(s)|\le36\cdot3\,(3\pi)^a\,\|\Delta\|_1 . \end{gathered} \] So for \(s\ge0\) \[ \begin{aligned} &|(H_i-H_i^{\mathrm{ref}})^{(a)}(s)|\\ &\quad<108\,(3\pi)^a\cdot2\cdot10^{-8}. \end{aligned} \] For \(m=0\) the claim is the case \(a=0\): the bound is \(2.16\cdot10^{-6}\). For \(m>0\), Lemma 1.2(1) with \(a=2\) gives the bound \(\tfrac12\cdot108\,(3\pi)^2\cdot2\cdot10^{-8}<9.6\cdot10^{-5}\), because \((3\pi)^2<88.83\). \(\blacksquare\)

Next the quotient of \(H_i^{\mathrm{ref}}\) is replaced by a polynomial. Fix a half-gap, with \(m\), \(y\) and \(\nu\). By Lemma 1.2 and the remark after it, \[ \begin{gathered} \sigma_i\,\mathcal D_mH_i^{\mathrm{ref}}(m+yx)=\sum_{r\ge0}\beta_{i,r}\,x^r,\\ \beta_{i,r}=\sigma_i\,y^{-\nu}\,\gamma_{i,r},\\ \gamma_{i,r}=\frac{y^{r+\nu}}{(r+\nu)!}\,(H_i^{\mathrm{ref}})^{(r+\nu)}(m). \end{gathered} \] The scaled derivative \(\gamma_{i,r}\) is the sum of the scaled derivatives of order \(a=r+\nu\) of \(p_i^{\mathrm{ref}}\) and of \(K_{p_j^{\mathrm{ref}}}\) at \(s=m\). Let \(\widetilde\beta_{i,r}\) be the number obtained when both are replaced by their quadrature forms (A positive quadrature rule, Section 3). The quadrature forms are linear in the lists, and the lists are \(X_1q(k)\) and \(X_2q(k)\). So there is a row \(l_{i,r}\) of ten real numbers, not depending on \(k\), with \[ \widetilde\beta_{i,r}=l_{i,r}\,q(k). \] The row \(l_{i,r}\) is obtained by applying the same operations to the ten columns of \(X_i\) and of \(X_j\) in place of the lists. It depends on the half-gap, on \(i\) and on \(r\).

Proposition 2.2. On every half-gap, for \(i=1,2\) and \(0\le x\le1\), \[ \begin{gathered} \Bigl|\sigma_i\,\mathcal D_mH_i^{\mathrm{ref}}(m+yx)-\sum_{r=0}^{40}l_{i,r}\,q(k)\,x^r\Bigr|\\ <10^{-6}. \end{gathered} \]

Proof. The coefficients with \(r\le40\). By Theorem 3.3 of the lesson on the quadrature rule, which applies because \(0\le m\le88\) and \(|y|\le\tfrac32\), and by linearity, the scaled derivatives of \(p_i^{\mathrm{ref}}\) and \(K_{p_j^{\mathrm{ref}}}\) differ from their quadrature forms by less than \(10^{-31}\) times \(\|\ell_i^{\mathrm{ref}}\|_1\) and \(\|\ell_j^{\mathrm{ref}}\|_1\). The sum of these norms is less than \(4.8\), and \(|y|^{-\nu}\le4\). So \(|\beta_{i,r}-\widetilde\beta_{i,r}|<20\cdot10^{-31}\), and the \(41\) coefficients together contribute less than \(10^{-28}\).

The coefficients with \(r\ge41\). Let \(Y=|y|\) and \(l=41+\nu\). Then \[ \begin{aligned} &\sum_{r\ge41}|\beta_{i,r}|\\ &\quad\le Y^{-\nu}\sum_{a\ge l}\frac{Y^a}{a!}\,\bigl|(H_i^{\mathrm{ref}})^{(a)}(m)\bigr| . \end{aligned} \] The function \(H_i^{\mathrm{ref}}\) is a superposition of exponentials \(e^{i\pi ts}\), from \(p_i^{\mathrm{ref}}\), and \(\lambda(t)e^{i\pi z(t)s}\), from \(K_{p_j^{\mathrm{ref}}}\), with frequencies \(|t|\le1\) and with total weight at most \(36\,(\|\ell_1^{\mathrm{ref}}\|_1+\|\ell_2^{\mathrm{ref}}\|_1)<180\); a derivative of order \(a\) multiplies them by \((i\pi t)^a\) and \((i\pi z(t))^a\). For one exponential of the first kind, \[ \begin{gathered} \sum_{a\ge l}\frac{Y^a}{a!}\,(\pi|t|)^a\le\frac{(\pi Y)^l}{l!\,(1-\rho)},\\ \rho=\frac{3\pi Y}{l+1}, \end{gathered} \] because the ratio of consecutive terms is at most \(\pi Y/(l+1)\le\rho\); here \(\rho<0.34\). For one of the second kind let \(t_{j-1}\le|t|\le t_j\) with \(1\le j\le6\) and \(t_j=j/6\). Since \(|z(t)|\) and \(\operatorname{Im}z(t)\) decrease in \(|t|\), and \(|\lambda|<3\), \(|z|<3\), the sum is at most \[ 3\,e^{-\pi m\operatorname{Im}z(t_j)}\,\frac{(\pi|z(t_{j-1})|\,Y)^l}{l!\,(1-\rho)} . \] Hence \(\sum_{r\ge41}|\beta_{i,r}|\le E(m,Y,\nu)\), where \[ \begin{gathered} E(m,Y,\nu)=\frac{180\,Y^{-\nu}}{l!\,(1-\rho)}\,\bigl((\pi Y)^l+3\,\varepsilon\bigr),\\ \varepsilon=\sum_{j=1}^6e^{-\pi m\operatorname{Im}z(t_j)}\bigl(\pi|z(t_{j-1})|\,Y\bigr)^l . \end{gathered} \] For fixed \(\nu\) this expression decreases when \(m\) increases, and increases with \(Y\), because \(Y^{-\nu}Y^l=Y^{41}\) and \(\rho\) increases with \(Y\). For \(m=0\) we have \(Y=\tfrac12\); for \(m=1\) and \(m=3\) the half-gaps have \(Y\le1\); for \(m\ge4\), \(Y\le\tfrac32\). So \(E\) is at most the largest of the four numbers \(E(0,\tfrac12,0)\), \(E(1,1,2)\), \(E(3,1,2)\), \(E(4,\tfrac32,2)\). The numbers \(|z(t_j)|\) for \(j=0,\dots,5\) are at most \(2.934\), \(2.713\), \(2.268\), \(1.859\), \(1.548\), \(1.320\), and the numbers \(\pi\operatorname{Im}z(t_j)\) for \(j=1,\dots,6\) are at least \(7.666\), \(4.923\), \(2.829\), \(1.515\), \(0.704\), \(0.187\). With these values the four numbers are less than \(3.1\cdot10^{-20}\), \(2.5\cdot10^{-12}\), \(5.7\cdot10^{-17}\) and \(2\cdot10^{-11}\). \(\blacksquare\)

The third comparison concerns the target. The parameter range is divided by the endpoints \[ 2.36,\quad2.65,\quad3.2,\quad4,\quad6,\quad\infty . \] For two consecutive endpoints \(K<J\) and a half-gap, define numbers \(a_0^*,\dots,a_{40}^*\) as follows:

Proposition 2.3. Let \(K<J\) be consecutive endpoints and \(K\le k\le J\), or \(k\ge K\) if \(J=\infty\). On every half-gap, for \(0\le x\le1\), \[ \mathcal D_mT_k(m+yx)\ge\sum_{r=0}^{40}a_r^*\,x^r-10^{-6}. \]

Proof. The centre \(0\). Here \(0\le s\le\tfrac12\). Since \(T_k(s)=k^{-1}e^{-k(s-1)}\), \[ \frac{\partial T_k(s)}{\partial k}=T_k(s)\Bigl(1-s-\frac1k\Bigr)>0 \] for \(k>2\). So \(T_k(s)\ge T_K(s)=\tfrac{e^K}Ke^{-Ks}\), and with \(s=yx\) the Taylor coefficients of the right side are the numbers \(a_r^*\). The terms of degree above \(40\) are bounded in absolute value by \[ \begin{aligned} \frac{e^K}K\sum_{r\ge41}\frac{(K|y|)^r}{r!}&\le\frac{e^6}6\cdot\frac{3^{41}}{41!\,(1-\frac3{42})}\\ &<8\cdot10^{-29}, \end{aligned} \] because \(K\le6\) for every pair of consecutive endpoints, \(K|y|\le3\), and \(e^K/K\) increases for \(K>1\).

The centre \(1\). With \(u=s-1\) we have \(T_k(1)=1/k\), \(T_k'(1)=-1\), and \[ \mathcal D_1T_k(1+u)=\frac{e^{-ku}-1+ku}{k\,u^2}. \] Its derivative with respect to \(k\) is \[ \frac{1-(1+ku)\,e^{-ku}}{k^2u^2}\ge0, \] by \(e^{ku}\ge1+ku\); this holds for positive and negative \(u\), and the value \(k/2\) at \(u=0\) increases with \(k\) as well. So the quotient is at least its value for \(k=K\), whose expansion in powers of \(u=yx\) has the coefficients \(K(-Ky)^r/(r+2)!\). Here \(|y|\le1\), and the terms of degree above \(40\) are bounded by \[ \begin{aligned} \sum_{r\ge41}\frac{K\,(K|y|)^r}{(r+2)!}&\le\frac{6\cdot6^{41}}{43!\,(1-\frac6{44})}\\ &<10^{-20}. \end{aligned} \]

All centres \(m>0\). By Lemma 1.2(1) and \(T_k''(s)=k\,e^{-k(s-1)}\), \[ \begin{aligned} &\mathcal D_mT_k(m+u)\\ &\quad=k\int_0^1(1-v)\,e^{-k(m-1+vu)}\,dv\ge0 . \end{aligned} \] This is the claim in all cases in which the numbers \(a_r^*\) are \(0\).

The centre \(3\) with \(J<\infty\). Here \(-1\le u\le\tfrac12\), so the exponent is \(-ka\) with \(a=2+vu\ge1\). The function \(k\mapsto k\,e^{-ak}\) decreases for \(k\ge1/a\), in particular for \(k\ge2.36\). So the quotient is at least \[ J\int_0^1(1-v)\,e^{-J(2+vu)}\,dv . \] The function \(2(1-v)\) is a probability density on \([0,1]\) with mean \(\tfrac13\), and the exponential function is convex. By the inequality of Jensen the integral is at least \(\tfrac12e^{-J(2+u/3)}\), and \(2+u/3\le\tfrac{13}6\). \(\blacksquare\)

3. The reduction

For a half-gap, \(i=1,2\), and \(k\) in the parameter interval from \(K\) to \(J\), let \[ \Phi_i(x;k)=\sum_{r=0}^{40}\bigl(l_{i,r}\,q(k)+\delta_i\,a_r^*\bigr)\,x^r . \] This is a polynomial of degree at most \(40\) in \(x\). Its coefficients are given by explicit finite formulas: the rows \(l_{i,r}\) through the matrices \(X_1,X_2\) and the quadrature rule, the numbers \(a_r^*\) through \(K\) and \(J\), and \(q(k)\) through \(k\).

Theorem 3.1. Let \(k\ge2.36\), and let \(K<J\) be consecutive endpoints with \(K\le k\le J\), or \(k\ge K\) if \(J=\infty\). On every half-gap, for \(i=1,2\) and \(0\le x\le1\), \[ \mathcal D_mF_i(m+yx)>\Phi_i(x;k)-0.000112 . \] Consequently, if \(\Phi_i(x;k)\ge0.001\) for all \(0\le x\le1\), on all half-gaps and for \(i=1,2\), then \(F_1\ge0\) and \(F_2\ge0\) on \([0,89.5]\).

Proof. By the formula for \(\mathcal D_mF_i\) at the end of Section 1, \[ \begin{aligned} \mathcal D_mF_i={}&\sigma_i\,\mathcal D_m(H_i-H_i^{\mathrm{ref}})\\ &+\sigma_i\,\mathcal D_mH_i^{\mathrm{ref}}+\delta_i\,\mathcal D_mT_k . \end{aligned} \] By Proposition 2.1 the first term is larger than \(-0.00011\). By Proposition 2.2 the second is larger than \(\sum_{r\le40}l_{i,r}q(k)x^r-10^{-6}\). By Proposition 2.3 the third is at least \(\delta_i(\sum_{r\le40}a_r^*x^r-10^{-6})\). Adding gives the inequality. If \(\Phi_i\ge0.001\), then \(\mathcal D_mF_i>0\) on the half-gap. For \(m=0\) this is \(F_i>0\); for \(m>0\) it gives \(F_i(s)=(s-m)^2\mathcal D_mF_i(s)\ge0\). The half-gaps cover \([0,89.5]\) by Lemma 1.1. \(\blacksquare\)

So the two sign conditions on \([0,89.5]\), for all \(k\ge2.36\), are reduced to lower bounds for \(91\cdot2\) families of polynomials of degree \(40\), each family depending on the parameter \(k\) through the ten numbers \(q(k)\).

4. Exercises

4.1. List the half-gaps with the centres \(0,1,3,4\), with their numbers \(y\).

4.2. Let \(u(s)=(s-m)^2g(s)\) with a twice continuously differentiable function \(g\). Show \(\mathcal D_mu=g\).

4.3. Verify the formula \[ \mathcal D_1T_k(1+u)=\frac{e^{-ku}-1+ku}{k\,u^2} \] and its expansion in powers of \(u\).

4.4. Why is the quotient \(\mathcal D_m\) applied to \(H_i-H_i^{\mathrm{ref}}\), which need not vanish at \(m\), and not the division of \(H_i-H_i^{\mathrm{ref}}\) by \((s-m)^2\)?

5. Solutions

4.1. Centre \(0\): the half-gap \([0,\tfrac12]\), \(y=\tfrac12\). Centre \(1\): \([\tfrac12,1]\) with \(y=-\tfrac12\) and \([1,2]\) with \(y=1\). Centre \(3\): \([2,3]\) with \(y=-1\) and \([3,3.5]\) with \(y=\tfrac12\). Centre \(4\): \([3.5,4]\) with \(y=-\tfrac12\) and \([4,5.5]\) with \(y=\tfrac32\).

4.2. \(u(m)=0\) and \(u'(m)=0\), so \(\mathcal D_mu(s)=u(s)/(s-m)^2=g(s)\) for \(s\ne m\). At \(s=m\), \(u''(m)=2g(m)\), so \(\mathcal D_mu(m)=g(m)\).

4.3. \(T_k(1+u)=e^{-ku}/k\), \(T_k(1)=1/k\) and \(T_k'(1)=-1\), so the numerator of the quotient is \((e^{-ku}-1+ku)/k\). Since \(e^{-ku}-1+ku=\sum_{j\ge2}(-ku)^j/j!\), division by \(ku^2\) gives \(\sum_{r\ge0}K_r\,u^r\) with \(K_r=k\,(-k)^r/(r+2)!\).

4.4. The difference \(H_i-H_i^{\mathrm{ref}}\) is small, of the order \(10^{-7}\), but it does not vanish at \(m\); divided by \((s-m)^2\) it would be unbounded near \(m\). The quotient \(\mathcal D_m\) first removes the value and the slope at \(m\), and is bounded by half the second derivative. The exact slack \(F_i\) has zero value and slope at \(m\), so for it the quotient is the true division, and \(\mathcal D_m\) is linear.

References