Growth and the geometry of the large zeros

Proof of Proposition 11.2

The coefficient of \(w^n\) in \(\Delta_N\) has modulus \(1/(Nn)!\). Stirling's formula and Standard fact 11.1 give order \(1/N\). Since \(M_{f(\,\cdot\,^N)}(r)=M_f(r^N)\) for the maximum modulus, composition with \(z^N\) multiplies the order by \(N\) and gives order one for \({\mathsf D}_N\).

The finite exponential representation gives

so the type is at most one. Choose a root \(\eta_0\) and put \(z=t\overline{\eta_0}\) with \(t>0\). The term \(e^{\eta_0z}=e^t\) has strictly larger real exponent than every other term. Hence

The type is therefore one.

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Proof of Theorem 11.4

For \(z=re^{i\theta}\), the real part of the exponent \(\eta_kz\) is

The triangle inequality gives the upper indicator bound \(h_N(\theta)\).

Away from a tie direction, the maximizing vertex is unique. It is constant on each connected component of the complement of the tie directions. A compact set avoiding all ties meets only finitely many such components, and compactness gives one uniform positive lower bound for every relevant projection gap. Division by the maximizing exponential gives the stated uniform asymptotic and the zero-free conclusion.

At a tie direction exactly two adjacent vertices attain the maximum. After rotation it is enough to take \(\theta=0\). With \(c_N=\cos(\pi/N)\) and \(s_N=\sin(\pi/N)\), the two dominant terms give

and every other term is exponentially smaller. Along a sequence on which \(|\cos(s_Nr)|\geq1/2\), the lower indicator bound is \(c_N=h_N(0)\). Rotational symmetry treats every tie direction.

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Proof of Lemma 11.5

A direct projection estimate gives the stronger bound

for every zero in the sector, without a large-radius assumption. Write \(z=re^{i\theta}\) with \(0\leq\theta\leq a_N\) and \(r\geq0\). For \(0\leq\ell\leq N-2\), the angle \(\theta+(2\ell+1)a_N\) lies between \(a_N+\theta\) and \(2\pi-a_N-\theta\). Monotonicity of cosine on \([0,\pi]\), followed by reflection about \(\pi\), therefore gives

The remaining vertex \(e^{-ia_N}\) has projection \(r\cos(a_N-\theta)\). Their difference is at least \(2s_Nr\sin\theta=2s_Ny\). If \({\mathsf D}_N(z)=0\), divide its finite exponential sum by the term with exponent \(e^{-ia_N}z\) and apply the triangle inequality:

Thus \(y\leq Y\). Conjugation gives the same bound for \(-y\) on the lower half sector. Since \(N\geq3\) and \(s_N>0\), this \(Y\) is positive.

On the resulting strip, the roots \(e^{ia_N}\) and \(e^{-ia_N}\) have real part \(c_N\), while each remaining root \(\eta=u+iv\) satisfies \(u\leq\cos(3a_N)=c_N-\gamma_N\). Hence

Division by \(e^{e^{-ia_N}z}\) gives the displayed normalization. Each differentiation multiplies a remainder term by \(\eta-e^{-ia_N}\), whose modulus is at most two. Thus, for \(q=0,1,2\),

One may take \(R=1\) and \(C=4(N-2)e^{2Y}\).

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Proof of Lemma 11.6

In fact, the constants \(\varepsilon_M=1/2\) and \(C_M=3/2\) work for every complex \(w\), without the strip restriction. Put \(u=-(1+e^w)\) and \(L=\log(1+u)=\log(-e^w)\), using the principal logarithm. The elementary logarithm estimate on \(|u|\leq1/2\) gives \(|L|\leq(3/2)|u|\). Since \(e^{L+\pi i}=e^w\), there is an integer \(m\) such that

This gives the required distance bound. Distinct centers are separated by at least \(2\pi\), whereas two centers satisfying the bound would be separated by at most \(3|1+e^w|\leq3/2\). Thus the integer is unique. The formal proof uses the convergent logarithm estimate and the exact periodicity theorem for the complex exponential.

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Proof of Theorem 11.7

Work first in the sector \(|\arg z|\leq a_N\). By Lemma 11.5, every sufficiently large zero lies in a fixed strip about the positive real axis. Put

Then

with an exponentially small remainder and derivative.

If \(G(z)=0\), then

The real part of \(2is_Nz\) is bounded in the strip. Apply Lemma 11.6 with \(w=2is_Nz\). There is a unique integer \(m\) such that

where

In particular, \(|\Re z-t_m|=o(1)\). For all sufficiently large zeros we have \(|\Re z-t_m|\leq1\), and hence

with a constant independent of \(m\). Because \(\Re z\to+\infty\) in this argument, the corresponding integers \(m\) are eventually nonnegative and tend to infinity.

Fix a large \(m\), put \(a=2is_N\), and consider the closed disk \(B_m=\{z:|z-t_m|\leq 1/(2s_N)\}\). The remainder bounds hold uniformly on these fixed-radius disks. For sufficiently large \(m\), \(|R_N|\leq 1/2\) and \(|R_N'|\leq s_N/2\) there. Define

using the principal logarithm. The estimate \(|\log(1+u)|\leq(3/2)|u|\) shows that \(\Phi_m\) maps \(B_m\) into itself. The derivative bound \(|\Phi_m'|\leq1/2\) makes it a contraction. Banach's fixed-point theorem gives a unique fixed point. On \(B_m\), \(|a(z-t_m)|\leq1<\pi\), so the principal logarithm inverts the exponential and the fixed-point equation is equivalent to \(G(z)=0\). Moreover,

At this zero,

so the zero is simple. The disks are mutually disjoint because their diameter \(1/s_N\) is smaller than the center spacing \(\pi/s_N\). Enlarging the strip if necessary, these disks lie inside the fundamental sector and the fixed strip for large \(m\).

The localization estimate puts every sufficiently large zero in the fundamental sector into one of these disks. The zero sets of \(G\) and \({\mathsf D}_N\) coincide because their prefactor is nonzero. Each disk is invariant under complex conjugation, and \({\mathsf D}_N\) has real coefficients. The conjugate of its unique zero is therefore the same zero. It is real, positive for large \(m\), and simple by the derivative bound above. Denote it by \(\tau_m\).

The parts of the fundamental sector away from the positive tie ray, including its boundary rays, were already excluded uniformly. Rotational invariance now gives every remaining large zero as \(\omega^k\tau_m\). Distinct rays and the disjoint disks give uniqueness of \(k\) and \(m\). Taking \(\gamma=\gamma_N\) proves the zero location in the statement.

Differentiate the normalized expression. At \(z=\tau_m\),

Since \(G(\tau_m)=0\),

Taking absolute values gives the derivative asymptotic. Differentiating \({\mathsf D}_N(\omega z)={\mathsf D}_N(z)\) gives the rotation formula.

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Proof of Theorem 11.8

Put \(\delta_m=\tau_m-t_m\). The large zero theorem gives \(\delta_m=O(e^{-\gamma_Nt_m})\). Since \(e^{2is_Nt_m}=-1\),

The remainder estimate through the first derivative gives

Substitution in the zero equation gives the displacement formula.

At the real point \(t_m\), the normalized value

is purely imaginary. The two dominant terms cancel there, so \(R_N(t_m)\) is purely imaginary and the displayed correction is real.

For the derivative, use the formula in the proof of Theorem 11.7 and

The stated real sign follows.

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Proof of Corollary 11.9

The only nondominant exponent is \(-1\). Hence

At \(t=t_m\), its phase is \(i(-1)^m\). Insert this value into Theorem 11.8.

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Proof of Corollary 11.10

This is the standard zero set of the cosine function.

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Proof of Corollary 11.11

Use Theorem 11.7 for \(N\geq3\) and Corollary 11.10 for \(N=2\).

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Proof of Corollary 11.13

The identity \({\mathsf D}_N(z)=\Delta_N(z^N)\) identifies each orbit of \(N\) nonzero zeros of \({\mathsf D}_N\) with one zero of \(\Delta_N\). For \(N\geq3\), the large orbit representatives are the positive zeros \(\tau_m\) from Theorem 11.7; for \(N=2\), use Corollary 11.10. Simplicity is preserved because

and \(z\ne0\). The asymptotic follows by expanding \(\tau_m^N\) about \(t_m\).

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Proof of Theorem 11.14

Each of the \(N\) rays contains one large zero near

Since \(\tau_m-t_m=o(1)\) and the model spacing is the positive constant \(A_N\), replacing \(\tau_m\leq R\) by \(t_m\leq R\) changes the count by at most a bounded amount. The number of such indices with \(t_m\leq R\) is

Multiplication by \(N\) and addition of the finite exceptional set give the first formula. The second follows from

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Proof of Corollary 11.15

The asymptotic counting formulas in Theorem 11.14 give the two exponents by partial summation.

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