#Toeplitz determinants and minors
Toeplitz minors connect Schur functions with polynomial recurrences and real-rootedness. We first describe the determinant identities and limiting-zero criteria, then give finite-size proofs for two five-diagonal band shapes.
For a Laurent polynomial \(b(z)=\sum_{k=-r}^s b_kz^k,\) where \(r,s\geq1\) and \(b_{-r}b_s\ne0,\) the Toeplitz sections are \[T_n(b)=(b_{i-j})_{i,j=1}^n,\qquad P_n(t)=\det(tI-T_n(b)),\qquad P_0=1.\] The Laurent polynomial \(b\) is the symbol. A Toeplitz minor is the determinant of a square submatrix obtained by selecting equally many rows and columns.
#Schur functions as minors
Let \(h_m=\completeH_m(\xvec)\) be the complete homogeneous symmetric function, with \(h_0=1\) and \(h_m=0\) for \(m\lt{}0.\) For increasing nonnegative row indices \(i_1\lt{}\dotsb\lt{}i_k\) and column indices \(j_1\lt{}\dotsb\lt{}j_k\) with \(i_\ell\geq j_\ell,\) put \[\lambda_p=i_{k+1-p}+p-k,\qquad \mu_p=j_{k+1-p}+p-k\qquad(1\leq p\leq k).\] These are partitions with \(\mu\subseteq\lambda,\) and the skew Jacobi–Trudi identity gives \[\det(h_{i_p-j_q})_{p,q=1}^k=\schurS_{\lambda/\mu}.\] Indeed, reversing both index orders gives the matrix \((h_{\lambda_p-\mu_q-p+q})\) and leaves its determinant unchanged. If some \(i_\ell\lt{}j_\ell,\) the lower-triangular zero pattern makes the minor zero. The dual identity gives analogous formulas with elementary symmetric functions.
Nonnegative specializations therefore give totally non-negative Toeplitz matrices. For the general connection with coefficient sequences and real zeros, see Pólya frequency sequences; for banded minors, determinant formulas, and recurrences, see [Ale12]. The rectangular Schur formula below makes the recurrence explicit for a balanced pentadiagonal matrix.
#Beraha–Kahane–Weiss and limiting zeros
The characteristic roots of a linear recurrence also control where zeros accumulate as \(n\to\infty.\) This is the content of the Beraha–Kahane–Weiss theorem [BKW75], also used in the transfer-matrix setting. We now let \(x\) be complex and work on a domain where the characteristic roots and amplitudes are single-valued analytic functions.
Theorem (Beraha–Kahane–Weiss, analytic form).
Let \(d\geq2\) and let \(a_i,\lambda_i\) be analytic on a connected open set \(\Omega\subseteq\setC,\) with none identically zero. Suppose \(\lambda_i\not\equiv\omega\lambda_j\) for \(i\ne j\) and every constant \(|\omega|=1.\) Set \[P_n(x)=\sum_{i=1}^d a_i(x)\lambda_i(x)^n.\] Call \(i\) dominant at \(x\) if \(|\lambda_i(x)|=\max_j|\lambda_j(x)|.\) A point \(x\in\Omega\) is a limit of zeros \(x_k\) of \(P_{n_k}\) with \(n_k\to\infty\) if and only if either
exactly one index \(i\) is dominant and \(a_i(x)=0\); or
at least two indices are dominant.
Every neighborhood of such a point contains a zero of \(P_n\) for all sufficiently large \(n.\)
In [Sok04, Thm. 1.5], the condition on constant unit-modulus ratios is imposed on branches that are dominant on a set with nonempty interior.
Example
For \(P_n(x)=1+x^n+(x+1)^n,\) all amplitudes equal \(1,\) so there are no isolated limits arising from amplitude zeros. The limiting set is \[\begin{aligned} &\{x:|x|=1,\ |x+1|\leq1\}\\ {}\cup{}&\{x:|x+1|=1,\ |x|\leq1\}\\ {}\cup{}&\{x:|x|=|x+1|\geq1\}. \end{aligned}\] The first two sets are circular arcs. The third consists of the two rays \[\operatorname{Re}x=-\tfrac12,\qquad |\operatorname{Im}x|\geq\tfrac{\sqrt3}{2}.\] All three branches meet at \((-1\pm i\sqrt3)/2.\) These equations follow by comparing the moduli of \(1,x,x+1.\) In particular, the nonreal limiting points force nonreal zeros for all sufficiently large \(n.\)
The derivative-recurrence criterion uses the values and logarithmic derivatives of the characteristic roots; BKW uses their relative moduli. Both depend on the amplitudes, which are determined by the initial polynomials.
For repeated poles the amplitudes \(a_i(n,x)\) depend polynomially on \(n,\) so the displayed theorem does not apply directly. A common scalar factor in \(n\) can be removed for large \(n\); general polynomial-in-\(n\) amplitudes require an extension.
#Real spectra of banded Toeplitz matrices
The limiting spectrum is \[\Lambda(b)=\{t\in\setC:\lim_{n\to\infty} \operatorname{dist}(t,\operatorname{spec}T_n(b))=0\}.\] Order the roots of \(z^r(b(z)-t)\) by modulus, with multiplicity. The Schmidt–Spitzer theorem [SS60] gives \[\Lambda(b)=\{t:|z_r(t)|=|z_{r+1}(t)|\}.\] See also the recurrence and BKW interpretation.
For real coefficients, B. Shapiro and F. Štampach [SS19, SS23] considered three conditions: all finite spectra are real; \(\Lambda(b)\subseteq\setR\); and \(b^{-1}(\setR)\) contains a Jordan curve surrounding \(0.\) The first implies the second, and the second implies the third. The claimed converse involving the Jordan curve was withdrawn in the correction. F. Rohner and C. Thalhammer [RT26, Thm. 1.3] disprove even the implication from the Jordan-curve condition to a real limiting spectrum, using the real symbol \[b(z)=\left(z^2+97z+103-\frac{10201}{z}\right)^2.\] Their example does not settle whether a real limiting spectrum forces every finite spectrum to be real. The following proposition settles this implication for balanced pentadiagonal symbols with positive product of the outer coefficients. The argument was communicated by P. Alexandersson in 2026.
#Balanced pentadiagonal symbols
Suppose \(b_{-2},\dotsc,b_2\) are real and \(b_{-2}b_2\gt{}0.\) Replacing \(b(z)\) by \((b(\rho z)-b_0)/c,\) where \[\begin{aligned} \rho&=(b_{-2}/b_2)^{1/4}\gt{}0,\\ c&=\operatorname{sgn}(b_2)\sqrt{b_{-2}b_2}, \end{aligned}\] preserves reality of the finite and limiting spectra: the substitution is a diagonal similarity, followed by a real shift and nonzero real scaling. We may therefore use the normalized symbol \[b(z)=z^2+Az+Bz^{-1}+z^{-2},\qquad A,B\in\setR.\]
Proposition
For this symbol the following conditions are equivalent:
\(P_n\) is real-rooted for every \(n\geq1\);
\(P_n\) is real-rooted for every sufficiently large \(n\);
\(\Lambda(b)\subseteq\setR\);
\(A=B,\) or \((AB-16)^3\geq27(A^2-B^2)^2.\)
Proof
The first two implications follow from the definition of \(\Lambda(b).\) If \(A=B,\) every section is real symmetric. Assume henceforth that \(A\ne B.\) We first show that a nonreal critical point \(w\) gives a nonreal point \(b(w)\in\Lambda(b).\) Put \[q_t(z)=z^4+Az^3-tz^2+Bz+1,\qquad g(z)=z^3b'(z)=2z^4+Az^3-Bz-2.\] If \(t=b(w)\) were real, the double root \(w\) of \(q_t\) would imply \(q_t(z)=(z-w)^2(z-\bar w)^2.\) Its constant coefficient gives \(|w|=1,\) and comparison of its cubic and linear coefficients gives \(A=B.\) Thus \(t\) is nonreal.
At this critical value, \[q_t(z)=(z-w)^2\bigl(z^2+(A+2w)z+w^{-2}\bigr).\] The remaining roots divided by \(w\) are the roots \(y_1,y_2\) of \(y^2+py+d,\) where \(p=2+A/w\) and \(d=w^{-4}.\) Write \(R=|w|^2\) and \(x=\operatorname{Re}w.\) Taking real and imaginary parts of \(g(w)=0,\) for \(x\ne0,\) gives \[A=\frac{R-R^{-1}}x-4x,\qquad B=-\frac{R^2-1}x-\frac{4x}R.\] Also \(p-\bar p d=(B-A/R)/w^3.\) If \(R\ne1,\) these identities imply \[\frac{|p-\bar p d|}{|1-|d|^2|}=\frac{\sqrt R}{|x|}\gt{}1.\] Now use the elementary identity \[(1-|d|^2)^2-|p-\bar p d|^2 =(1-|y_1|^2)(1-|y_2|^2)|1-y_1\bar y_2|^2.\] Its left side is negative, so exactly one \(y_j\) lies inside the unit circle and the other lies outside. The case \(R=1,\) \(x\ne0\) would give \(A=B.\) If \(x=0,\) criticality gives \(w=\pm i,\) \(B=-A,\) and the same left side is \(-4A^2\lt{}0.\) Thus the double root occupies the middle two modulus ranks: \[|z_1|\lt{}|z_2|=|z_3|\lt{}|z_4|.\] The Schmidt–Spitzer theorem gives \(b(w)\in\Lambda(b).\) Consequently, a real limiting spectrum forces all four roots of \(g\) to be real.
Conversely, suppose all four critical points are real. Their product is \(-1,\) so three have one sign and one has the other, counting multiplicity. Replacing \(z\) by \(-z\) is an alternating diagonal similarity; we may assume three critical points are negative. On the negative axis, \(b(z)\) tends to \(+\infty\) at both ends. If these critical points are distinct, the two local minima and intervening maximum give a level \(t_0\) with four negative solutions of \(b(z)=t_0.\) A double negative critical point \(w\) instead gives a triple root of \(q_{b(w)},\) whose remaining root is \(w^{-3}\lt{}0\); a triple critical point gives a quadruple root. In all cases, \[b(z)-t_0=z^{-2}\prod_{j=1}^4(z+\alpha_j),\qquad \alpha_j\gt{}0.\] Each lower bidiagonal matrix \(T_{n+2}(z+\alpha_j)\) is totally non-negative. The submatrix of their product in rows \(3,\dotsc,n+2\) and columns \(1,\dotsc,n\) is \(T_n(b)-t_0I,\) hence is totally non-negative. By the dual Jacobi–Trudi identity, \[\det(T_n(b)-t_0I)=\schurS_{(n,n)}(\alpha_1,\dotsc,\alpha_4)\gt{}0.\] Its first upper and lower diagonals are positive. The oscillatory-matrix criterion therefore gives positive simple eigenvalues, as in the valid total-nonnegativity argument of [SS23]. Adding back \(t_0I\) proves reality of every finite spectrum.
Finally, \[\operatorname{Disc}(g)=4\bigl((AB-16)^3-27(A^2-B^2)^2\bigr).\] Since \(g(0)=-2\) and \(g\) is positive at both real infinities, it always has a positive and a negative real root. For nonzero discriminant, positive discriminant is therefore equivalent to four real roots. At zero discriminant, a double real root and a nonreal conjugate pair, or two conjugate double roots, would have positive product, contrary to \(-1.\) This proves the parameter criterion, including its boundary.
#Toeplitz determinants and reciprocal roots
For the balanced pentadiagonal symbol \[b(z)=z^2+Az+Bz^{-1}+z^{-2},\qquad A,B\in\setR,\] put \(P_n(t)=\det(tI-T_n(b))\) and \(P_0=1.\) Let \(z_1,\dotsc,z_4\) be the roots of \(q_t(z)=z^4+Az^3-tz^2+Bz+1.\) The dual Jacobi–Trudi identity gives \(P_n(t)=(-1)^n\schurS_{(n,n)}(z_1,\dotsc,z_4).\) The Toeplitz determinant formula and Schur recurrence in [Ale12] therefore give the six characteristic roots \[\lambda_{ij}=-z_i z_j,\qquad 1\leq i\lt{}j\leq4.\] For distinct \(z_i,\) the corresponding expansion is \[P_n(t)=\sum_{1\leq i\lt{}j\leq4} \frac{(z_i z_j)^2(-z_i z_j)^n} {\prod_{k\notin\{i,j\}}(z_i-z_k)(z_j-z_k)}.\] Thus the amplitudes are nonzero away from collisions of the symbol roots. The characteristic polynomial is \[\begin{aligned} D(u)&=\prod_{i\lt{}j}(u+z_i z_j)\\ &=u^6-tu^5+(AB-1)u^4+(2t+A^2+B^2)u^3\\ &\qquad +(AB-1)u^2-tu+1. \end{aligned}\] Equivalently, summing the expansion gives \[G(u)\coloneqq\sum_{n\geq0}P_n(t)u^n=\frac{1-u^2}{D(u)}.\] These polynomial identities extend to repeated symbol roots by continuity. In particular, \(D\) specifies an order-six recurrence for \(n\geq6,\) with initial polynomials determined by this generating function; special parameters can reduce the minimal order.
Since \(z_1z_2z_3z_4=1,\) complementary pairs of rates multiply to \(1.\) Writing \(v=u+u^{-1}\) reduces \(D(u)=0\) to \[(v^2-4)(v-t)+ABv+A^2+B^2=0.\] Moreover, the rational generating function satisfies \(G(1/u)=-u^4G(u).\) Extending the order-six recurrence backwards gives \(P_{-1}=P_{-2}=P_{-3}=0\) and \(P_{-4-n}=P_n\) for \(n\geq0.\) Indeed, the expansion of \(G\) at infinity is \(-\sum_{n\lt{}0}P_nu^n=-u^{-4}\sum_{n\geq0}P_nu^{-n}.\)
Order the symbol roots by modulus. The two largest pair-product moduli are \(|z_3z_4|\) and \(|z_2z_4|,\) so a tie for dominance occurs precisely when \(|z_2|=|z_3|.\) On analytic domains satisfying the BKW nondegeneracy hypotheses, the expansion above explains this equal-modulus locus as a limiting zero set. The Schmidt–Spitzer theorem supplies the global statement, including collisions and exceptional parameter choices. Reciprocal pairing gives the same locus for the backward recurrence; it does not by itself prove reality of finite spectra. For positive outer product, the critical-point and total-nonnegativity proof does prove that a real limiting spectrum forces every \(P_n\) to be real-rooted.
#Five-term Hessenberg recurrences
Consider \[P_n(t)=tP_{n-1}(t)+\alpha P_{n-2}(t)+\beta P_{n-3}(t)+\gamma P_{n-4}(t), \qquad P_0=1,\quad P_{-1}=P_{-2}=P_{-3}=0,\] where \(\alpha,\beta,\gamma\in\setR\) and \(\gamma\ne0.\) These are characteristic polynomials of the Toeplitz sections with symbol \[b(z)=-z^{-1}+\alpha z-\beta z^2+\gamma z^3,\qquad G(z)=z^2b'(z)=3\gamma z^4-2\beta z^3+\alpha z^2+1.\] The matrix is lower Hessenberg: entries above the first superdiagonal vanish. It has one upper and three lower diagonals, unlike the balanced pentadiagonal case.
Theorem (Sufficiency in [Ndi22, Thm. 5]).
Every \(P_n\) is real-rooted if either of the following holds:
\(G\) has four real roots, counted with multiplicity;
\(G\) has two distinct real roots of opposite signs and a nonreal conjugate pair, and \[24\alpha\gamma-5\beta^2\geq0,\qquad 64\gamma^3+(4\alpha\gamma-\beta^2)^2\geq0.\]
The published proof invokes the Jordan-curve implication that fails in general. We give direct proofs of these sufficiency statements below. The coefficient \(24\alpha\gamma-5\beta^2\) is the published condition; the earlier preprint has a different inequality.
#Four real critical points
We first prove case (a). Replacing \(z\) by \(-z\) gives the equivalent symbol \[f(z)=z^{-1}+az+dz^2-\gamma z^3,\qquad a=-\alpha,\quad d=-\beta.\]
Proof (The case \(\gamma\lt{}0\)).
Put \(c=-\gamma\gt{}0.\) The critical quartic \(G(-z)\) has root product \(-1/(3c),\) so three critical points have one sign and one has the other. Replacing \(f(z)\) by \(-f(-z),\) if needed, preserves reality of spectra and makes three critical points negative. On the negative axis \(f\) tends to \(-\infty\) at both ends. If its three critical points there are distinct, two local maxima and the intervening minimum give a level \(t_0\) with four negative solutions. At a double critical point \(w\lt{}0,\) the kernel has a triple root \(w\); its remaining root is \(1/(cw^3)\lt{}0.\) A triple critical point gives a quadruple root. Thus in every case \[f(z)-t_0=cz^{-1}\prod_{j=1}^4(z+\delta_j),\qquad \delta_j\gt{}0.\] The bidiagonal-product argument from the balanced case, now selecting rows \(2,\dotsc,n+1\) and columns \(1,\dotsc,n,\) makes \(T_n(f)-t_0I\) totally non-negative. Its determinant is \(c^n\schurS_{(n,n,n)}(\delta_1,\dotsc,\delta_4)\gt{}0,\) and its first upper and lower diagonals are positive. The oscillatory-matrix criterion gives real simple eigenvalues.
Proof (The case \(\gamma\gt{}0\)).
Put \(g=\gamma\gt{}0,\) so that \[f(z)=z^{-1}+az+dz^2-gz^3.\] Then \[\sum_{n\geq0}P_n(t)u^n =\frac1{1-tu+au^2+du^3-gu^4}.\] The polynomial \(-z^2f'(z)=1-az^2-2dz^3+3gz^4=G(-z)\) has four nonzero real roots. Their product is positive and the sum of their reciprocals is zero, so two are positive and two negative. Also \(a\gt{}0\): the second elementary symmetric function of the reciprocal roots equals \(-\frac12\) times the sum of their squares, and equals \(-a.\)
Choose the smallest positive root \(R\) of \[H_*(r)=1-ar^2+2|d|r^3+3gr^4.\] Such a root exists because \(H_*\) is one of \(G(r),G(-r).\) Since \(H_*(0)\gt{}0,\) its derivative at its first root is nonpositive. Using \(H_*(R)=0\) in \(RH_*'(R)\leq0\) gives \[|d|R^3+3gR^4\leq1. \tag{1}\]
For \(-1\lt{}x\lt{}1,\) set \[H_x(r)=1-ar^2-2dxr^3+g(4x^2-1)r^4.\] We have \(H_x(R)\lt{}0,\) since \[H_x(R)-H_*(R)=-2(dx+|d|)R^3+4g(x^2-1)R^4\lt{}0.\] Moreover, on \(0\lt{}r\leq R,\) \[\left(\frac{H_x(r)}{r^2}\right)'' =\frac6{r^4}+2g(4x^2-1)\gt{}0\] by (1), which implies \(R^4\leq1/(3g).\) Since \(H_x(r)/r^2\) tends to positive infinity at zero and is negative at \(R,\) strict convexity gives exactly one zero \(r(x)\) in \((0,R),\) with negative derivative there. This zero varies continuously with \(x.\)
For \(0\lt{}\theta\lt{}\pi,\) take \(x=\cos\theta,\) \(r=r(x),\) and \(t(\theta)=f(re^{i\theta}).\) The equation \(H_x(r)=0\) says exactly that \(t(\theta)\) is real. The two other roots of \(1-tz+az^2+dz^3-gz^4\) are the roots of \[L(z)=z^2+(2rx-d/g)z-\frac1{gr^2}.\] They have opposite signs. In fact both lie outside \([-r,r],\) since \[gr^2 L(\pm r) =gr^4(1\pm2x)\mp dr^3-1 \leq3gr^4+|d|r^3-1\lt{}0.\] Write them as \(s\gt{}r\) and \(\ell\lt{}-r,\) and put \(\eta=r/s\in(0,1),\) \(\zeta=r/\ell\in(-1,0).\)
The map \(\theta\mapsto t(\theta)\) is continuous and injective: the quartic at a fixed real \(t\) has just one nonreal conjugate pair, which uniquely determines \(\theta\in(0,\pi).\) Thus it is strictly monotone.
It remains to count finite-size zeros. Define \[U_j(x)=\frac{\sin((j+1)\theta)}{\sin\theta},\qquad S_n(v)=\sum_{j=0}^n v^jU_{n-j}(x).\] The generating function factorization gives \[r^nP_n(t(\theta)) =\frac{\eta S_n(\eta)-\zeta S_n(\zeta)}{\eta-\zeta}.\] A finite geometric-sum identity gives \[S_n(v)=\frac{U_n(x)-vU_{n+1}(x)+v^{n+2}}{1-2vx+v^2}.\] At \(\theta_k=k\pi/(n+2),\) for \(k=1,\dotsc,n+1,\) \(U_{n+1}=0\) and \(U_n=(-1)^{k+1}.\) Since \(|v|\lt{}1,\) both \(S_n(\eta)\) and \(S_n(\zeta)\) have this sign. The weights \(\eta\) and \(-\zeta\) are positive, so \[\operatorname{sgn}P_n(t(\theta_k))=(-1)^{k+1}.\] Strict monotonicity provides \(n\) disjoint real intervals with sign changes. Since \(P_n\) is monic of degree \(n,\) all its roots are real and simple. The argument includes repeated roots of \(G\); only the strict interior inequalities \(|x|\lt{}1\) and \(r\lt{}R\) were used.
#A quartic-kernel sufficient condition
For case (b), we prove a sufficient condition without a critical-root hypothesis. The two conjugate root pairs have different moduli, and a residue estimate gives alternating signs at suitable real values of \(t.\)
Proposition
Suppose \(a,d\in\setR\) and \(c\gt{}0,\) and define the monic polynomials \(Q_n\) by \[\sum_{n\geq0}Q_n(t)u^n=\frac1{1-tu+au^2+du^3+cu^4}.\] If \(a\geq2\sqrt c+d^2/(4c),\) then every \(Q_n\) is real-rooted. If the inequality is strict, all its roots are simple.
Proof
Replacing \(u\) by \(c^{-1/4}u\) reduces to \(c=1,\) with parameters \(a/\sqrt c,d/c^{3/4}\) and variable \(t/c^{1/4}.\) We first assume \(a\gt{}2+d^2/4.\) Put \[f(z)=z^{-1}+az+dz^2+z^3.\] For \(-1\leq x\leq1,\) consider \[H_x(r)=1-ar^2-2dxr^3+(1-4x^2)r^4.\] We have \[\begin{aligned} H_x(1)&=2-a+d^2/4-(2x+d/2)^2\lt{}0,\\ (H_x(r)/r^2)''&=6/r^4+2(1-4x^2)\gt{}0\quad(0\lt{}r\lt{}1). \end{aligned}\] Since \(H_x(r)/r^2\) tends to positive infinity at zero, strict convexity gives a unique zero \(r(x)\in(0,1),\) with negative derivative there. It depends continuously on \(x,\) including the endpoints.
Take \(x=\cos\theta,\) \(0\lt{}\theta\lt{}\pi,\) and \(r=r(x).\) Then \(t(\theta)=f(re^{i\theta})\) is real, since \(\operatorname{Im}f(re^{i\theta})=-H_x(r)\sin\theta/r.\) The kernel \(1-tz+az^2+dz^3+z^4\) has roots \(re^{\pm i\theta}.\) Write \(\eta,\zeta\) for \(r\) divided by its other two roots, and set \(\rho=r^2.\) Comparing the constant and cubic coefficients gives \[\eta\zeta=\rho^2,\qquad \eta+\zeta=-dr^3-2\rho^2x.\] The identity \(H_x(r)=0\) and \(a\gt{}2+d^2/4\) imply \[(1-\rho)^2\gt{} \left(\frac{dr^3}{2\rho}+2\rho x\right)^2.\] Define \(y=-dr^3/(2\rho)-\rho x.\) Thus \[|y-\rho x|\lt{}1-\rho,\qquad |y|\lt{}1.\] Consequently \(\eta,\zeta=\rho e^{\pm i\phi},\) where \(\phi=\arccos y\in(0,\pi).\) Moreover, \[|y|\leq\rho|x|+1-\rho \quad\Longrightarrow\quad \sin\phi\geq\rho\sin\theta.\] Indeed, subtracting \(\rho^2(1-x^2)\) from \(1-(\rho|x|+1-\rho)^2\) gives \(2\rho(1-\rho)(1-|x|)\geq0.\)
The two smaller kernel roots have modulus \(r\lt{}1,\) and the other two have modulus \(r/\rho=1/r\gt{}1.\) Thus \(t(\theta)\) is injective: its value determines its unique smaller conjugate pair and hence \(\theta.\) It is therefore strictly monotone.
The coefficient of \(v^n\) in \[\frac1{(1-e^{i\theta}v)(1-e^{-i\theta}v) (1-\rho e^{i\phi}v)(1-\rho e^{-i\phi}v)}\] is \(r^nQ_n(t(\theta)).\) For four distinct numbers \(\lambda_j,\) this coefficient equals \(\sum_j\lambda_j^{n+3}/\prod_{k\ne j}(\lambda_j-\lambda_k).\) All four numbers here are distinct. Put \[\begin{aligned} A&=(e^{i\theta}-\rho e^{i\phi})(e^{i\theta}-\rho e^{-i\phi}),\\ \psi&=\arg\bigl((1+\rho^2)x-2\rho y +i(1-\rho^2)\sin\theta\bigr). \end{aligned}\] The argument lies in \((0,\pi),\) and \(A=e^{i\theta}|A|e^{i\psi}.\) Combining the residues of the first conjugate pair gives \[\begin{aligned} r^nQ_n(t(\theta)) &=\frac{\sin((n+2)\theta-\psi)}{\sin\theta\,|A|}+E_n(\theta),\\ |E_n(\theta)|&\leq\frac{\rho^{n+2}}{\sin\phi\,|A|}\\ &\leq\frac{\rho^{n+1}}{\sin\theta\,|A|}. \end{aligned}\] For the error bound, the absolute product of the two cross-pair differences at \(\rho e^{i\phi}\) equals \(|A|\); its within-pair difference has absolute value \(2\rho\sin\phi.\) The last inequality uses \(\sin\phi\geq\rho\sin\theta.\)
The function \(\psi\) extends continuously to \([0,\pi],\) with \(\psi(0)=0\) and \(\psi(\pi)=\pi.\) At the first endpoint its real part is \(1+\rho^2-2\rho y\gt{}0,\) and at the second it is \(-1-\rho^2-2\rho y\lt{}0.\) Hence the continuous phase \((n+2)\theta-\psi\) runs from \(0\) to \((n+1)\pi.\) We can select successively ordered angles where it equals \((k+1/2)\pi,\) for \(k=0,\dotsc,n.\) At these angles the residue estimate, with \(\rho\lt{}1,\) shows that \(Q_n(t(\theta))\) has sign \((-1)^k.\) Strict monotonicity of \(t(\theta)\) now gives \(n\) distinct real roots. Finally, replacing \(a\) by \(a+\varepsilon\) and taking \(\varepsilon\downarrow0\) proves the weak-inequality case, since the fixed-degree monic real-rooted polynomials form a closed set.
#The published two-real-critical-point region
In case (b) of [Ndi22, Thm. 5], \(G\) has exactly two distinct real roots of opposite signs and one nonreal conjugate pair. Its root product is negative, so \(\gamma\lt{}0.\) Set \[c=-\gamma\gt{}0,\qquad a=-\alpha,\qquad d=-\beta.\] The published inequalities are \[24ac\geq5d^2,\qquad (4ac-d^2)^2\geq64c^3.\] After the preceding normalization, these read \(24a\geq5d^2\) and \(|4a-d^2|\geq8.\) We show that the negative branch \(4a-d^2\leq-8\) is impossible.
Put \(v=d^2.\) On that branch, \[4a+8\leq v\leq24a/5,\qquad a\geq10.\] The critical quartic is \(1-az^2-2dz^3-3z^4,\) up to a real nonzero scaling of its argument. Its discriminant is \(16D(a,v),\) where \[D(a,v)=-27v^2+(a^3+108a)v-3a^4-72a^2-432.\] For fixed \(a,\) this is concave in \(v,\) and its endpoint values are \[\begin{aligned} D(a,4a+8)&=(a-10)(a+6)^3,\\ D(a,24a/5)&=\frac9{25}(a^2-100)(5a^2+12). \end{aligned}\] Both are nonnegative for \(a\geq10.\) This contradicts the negative discriminant of a real quartic with two simple real roots and a nonreal conjugate pair. Therefore \(4a-d^2\geq8,\) which is precisely \(a\geq2+d^2/4.\) The proposition applies.
In the original parameters, the sufficient condition used is \[-\alpha\geq2\sqrt{-\gamma}+\frac{\beta^2}{-4\gamma}.\] This proves case (b) of the sufficiency theorem.
Bibliography
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