Addendum: A separation in modulus property of the zeros of a partial theta function

TL;DR

This paper investigates the zeros of the partial theta function, establishing their precise locations and uniqueness within specific annuli and disks for certain ranges of the parameter q.

Contribution

It provides new results on the distribution and uniqueness of zeros of the partial theta function for complex parameters q within specified domains.

Findings

Exactly one simple zero in each specified annulus for q in D(0.55)

Unique zero in the punctured disk for k=1 when q in D(0.55)

Results extend to q in D(0.6) for certain values of k

Abstract

We consider the partial theta function $\theta (q,z):=\sum _{j=0}^{\infty}q^{j(j+1)/2}z^j$, where $z\in \mathbb{C}$ is a variable and $q\in \mathbb{C}$, $0<|q|<1$, is a parameter. Set $D(a):=\{ q\in \mathbb{C}$, $0<|q|\leq a$, $\arg (q)\in [\pi /2,3\pi /2]\}$. We show that for $k\in \mathbb{N}$ and $q\in D(0.55)$, there exists exactly one zero of $\theta (q,.)$ (which is a simple one) in the open annulus $|q|^{-k+1/2}<z<|q|^{-k-1/2}$ (if $k\geq 2$) or in the punctured disk $0<z<|q|^{-3/2}$ (if $k=1$). For $k=1$, $4$, $5$, $6$, $\ldots$, this holds true for $q\in D(0.6)$ as well.

Equations

$$|{\xi }_k|\in U_{k-1/2,k+1/2},\mathrm{if}k\ge 2,\mathrm{and}{\xi }_k\in ({\mathbb{D}}_{|q{|}^{-3/2}}\setminus 0),\mathrm{if}k=1,$$

$$(1-1/({\alpha }_{0}k){)}^{-k+1/2}=:m_k\le |{\xi }_k|\le M_k:=(1-1/({\alpha }_{0}k){)}^{-k-1/2}.$$

$$\begin{array}[]{ccrrrrrrrrrr}n&&5&6&7&8&9&10&15&20&25&30\\ \\ \tau_{n}&&0.27&0.39&0.48&0.54&0.59&0.63&0.75&0.81&0.85&0.87\\ \\ m_{n}&&336.2&164.5&115.2&92.8&80.2&72.2&55.3&49.5&46.5&44.7\\ \\ M_{n}&&1225.1&416.1&239.1&169.8&134.4&113.4&73.0&60.5&54.4&50.9\\ \\ \end{array}$$

$$Q:=\prod _{j=1}^{\infty }(1-q^j),U:=\prod _{j=1}^{\infty }(1+z{q}^j)\mathrm{and}R:=\prod _{j=1}^{\infty }(1+q^{j-1}/z).$$

$$|Q_1|\ge \prod _{j=12}^{\infty }(1-|q{|}^j)\ge \prod _{j=12}^{\infty }(1-0.6^j)=\gamma :=0.9945691384\dots .$$

$$|Q_0|\ge 1.2/\gamma =1.206552620\dots$$

$$\begin{array}[]{llllcll}|U|&\geq&\prod_{j=1}^{\infty}||zq^{j}|-1|&=&\prod_{j=1}^{\infty}||q|^{-k+j+1/2}-1|&\geq&\prod_{j=1}^{\infty}|0.6^{-k+j+1/2}-1|\\ \\ &\geq&\prod_{j=1}^{\infty}|0.6^{-9/2+j}-1|&=&\eta&:=&0.2411047426\ldots\end{array}$$

$$\begin{array}[]{llllcll}|R|&\geq&\prod_{j=0}^{\infty}|1-|q^{j}/z||&=&\prod_{j=0}^{\infty}|1-|q|^{k+j-1/2}|&\geq&\prod_{j=0}^{\infty}|1-|q|^{9/2+j}|\\ \\ &\geq&\prod_{j=0}^{\infty}|1-0.6^{9/2+j}|&=&\xi&:=&0.7715882456\ldots~{}.\end{array}$$

$$|QRU|\ge 1.2\cdot \eta \cdot \xi =0.2232403024\dots .$$

$$\begin{array}[]{lllllll}|G|&\leq&\sum_{j=1}^{\infty}|q|^{j(j-1)/2}/|z|^{j}&=&\sum_{j=1}^{\infty}|q|^{j(j-1)/2+(k-1/2)j}&&\\ \\ &\leq&\sum_{j=1}^{\infty}|q|^{j(j-1)/2+9j/2}&\leq&\sum_{j=1}^{\infty}0.6^{j(j-1)/2+9j/2}&=&0.1066576686\ldots~{}.\end{array}$$

$$|G|\le \sum _{j=1}^{\infty }|q{|}^{j(j-1)/2+7j/2}\le \sum _{j=1}^{\infty }0.6^{j(j-1)/2+7j/2}=0.1851580824\dots .$$

$$S_j:=\{\arg (z)\in [(j-1)\pi /4,j\pi /4)\},j=1,2,3,S_4:=\{\arg (z)\in [3\pi /4,\pi ]\},$$

$$|1-\zeta {|}^2=1+|\zeta {|}^2-2|\zeta |\cos (\psi ).$$

$$|1-\zeta |\ge {\mu }_j(|\zeta |):=(1+|\zeta {|}^2-2|\zeta |\cos ((j-1)\pi /4){)}^{1/2}.$$

$$1+|q{|}^{-7/2+k}{\omega }_k,|{\omega }_k|=1.$$

$${\mu }_3(0.6^{-5/2})\cdot {\mu }_2(0.6^{-3/2})\cdot {\mu }_1(0.6^{-1/2})=1.742379963\dots =:{\chi }_0.$$

$$\begin{array}[]{lll}\tilde{u}_{j}:=|u_{3j+1}|\cdot|u_{3j+2}|\cdot|u_{3j+3}|\geq\min(A_{j},B_{j})&,&{\rm where}\\ \\ A_{j}:=\mu_{3}(|q|^{-5/2+3j})\cdot\mu_{2}(|q|^{-3/2+3j})\cdot\mu_{1}(|q|^{-1/2+3j})&&{\rm and}\\ \\ B_{j}:=\mu_{1}(|q|^{-5/2+3j})\cdot\mu_{4}(|q|^{-3/2+3j})\cdot\mu_{1}(|q|^{-1/2+3j})&.&\end{array}$$

$$0.1749135662\dots ,0.7772399345\dots ,0.9492771959\dots ,0.9889171980\dots$$

$$1.2\cdot {\chi }_0\cdot {\chi }_1\cdot {\chi }_{\ast }^2=0.1930636291\dots >0.1851580824\dots \ge |G|$$

$$C:=1+{\rho }^{6j+6},E:=1+{\rho }^{6j+4}-\sqrt{2}{\rho }^{3j+2},F:=1+{\rho }^{6j+2}-2{\rho }^{3j+1}.$$

$$\begin{array}[]{ccl}(CF)^{\prime}&=&\rho^{3j}((6j+6)\rho^{3j+5}+(6j+2)\rho^{3j+1}+(12j+8)\rho^{9j+7}\\ \\ &&-(6j+2)-(18j+14)\rho^{6j+6})~{}.\end{array}$$

$$M:=(1-{\rho }^{3j+1}{)}^2,N:=1+{\rho }^{6j+4}+\sqrt{2}{\rho }^{3j+2},W:=(1-{\rho }^{3j+3}{)}^2.$$

$$\begin{array}[]{lccclc}K_{1}:=(6j+2)\rho^{6j+1}&,&K_{2}:=\sqrt{2}(3j+2)\rho^{3j+1}&,&L_{1}:=(6j+2)\rho^{3j}&,\\ \\ K_{3}:=(6j+4)\rho^{6j+3}&,&K_{4}:=\sqrt{2}(9j+4)\rho^{9j+3}&,&L_{2}:=2\sqrt{2}(6j+3)\rho^{6j+2}&,\\ \\ K_{5}:=(12j+6)\rho^{12j+5}&&{\rm and}&&L_{3}:=(18j+10)\rho^{9j+4}&.\end{array}$$

$$b_0:=1,b_1:=z,b_2:=q{z}^2\mathrm{and}{r}_k:=\sum _{j=k}^{\infty }{q}^{j(j-1)/2}{z}^j.$$

$$|r_2|\le \sum _{j=2}^{\infty }|q{|}^{j(j-1)/2-j/2}=:{\Phi }_{♭}(|q|),|r_3|\le \sum _{j=3}^{\infty }|q{|}^{j(j-1)/2-j/2}={\Phi }_{♭}(|q|)-1$$

$$|1+z|\ge (1+|z{|}^2{)}^{1/2}=(1+|q{|}^{-1}{)}^{1/2}=:{\Phi }_{\ast }(|q|)>{\Phi }_{♭}(|q|).$$

$${\Phi }_{\ast }(0.6)=1.632993162\dots >1.618354488\dots ={\Phi }_{♭}(0.6).$$

$$|z+q{z}^2|=|z|\cdot |1+qz|\ge |q{|}^{-1/2}\cdot (1+|q|{)}^{1/2}={\Phi }_{\ast }(|q|).$$

$$\begin{array}[]{lllllllll}|1+z+qz^{2}|&\geq&{\rm Im}\,(1+z+qz^{2})&\geq&{\rm Im}\,(z)&\geq&\sin(3\pi/4)\cdot|q|^{-1/2}&\geq&\\ \\ (2\cdot 0.6)^{-1/2}&=&0.9128709292\ldots&>&0.618354488\ldots&=&\Phi_{\flat}(0.6)-1&\geq&|r_{3}|~{},\end{array}$$

$$\mathrm{Im}(1+z+q{z}^2)\ge \tau (|q|)\mathrm{and}\mathrm{Re}(1+z+q{z}^2)\ge 1-\tau (|q|)$$

Full text

Keywords: partial theta function; separation in modulus

AMS classification: 26A06

1 The new results

In this addendum to [3] we consider the partial theta function $\theta(q,z):=\sum_{j=0}^{\infty}q^{j(j+1)/2}z^{j}$. The series converges for $q\in\mathbb{D}_{1}$, $z\in\mathbb{C}$, where $\mathbb{D}_{a}$ denotes the open disk centered at [math] and of radius $a$. We regard $q$ as a parameter and $z$ as a variable.

Definition 1**.**

For $0<a<b$ and for $q\in\mathbb{D}_{1}\setminus 0$ fixed, denote by $U_{a,b}\subset\mathbb{C}$ the open annulus $\{|q|^{-a}<|z|<|q|^{-b}\}$. For $q$ fixed, we say that strong separation (in modulus) of the zeros of $\theta$ takes place for $k\geq k_{0}$, if for any $k\geq k_{0}$, there exists a unique zero $\xi_{k}$ of $\theta(q,.)$ which is a simple one and**

[TABLE]

while the remaining $k_{0}-1$ zeros (counted with multiplicity) are in $\mathbb{D}_{|q|^{-k_{0}+1/2}}\setminus 0$. If $k_{0}=1$, then we say that the zeros of $\theta$ are strongly separated. (One can notice that the cases $k_{0}=1$ and $k_{0}=2$ are identical.)**

It is shown in [3] that the zeros of $\theta$ are strongly separated for $0<|q|\leq c_{0}:=0.2078750206\ldots$, see Lemma 1 and its proof therein. Set $D(a):=\{q\in\mathbb{C}$, $0<|q|\leq a$, $\arg(q)\in[\pi/2,3\pi/2]\}$.

Theorem 2**.**

(1) For $q\in D(0.55)$, the zeros of $\theta$ are strongly separated.

(2) For $q\in D(0.6)$, conditions (1) hold true for $k=1$, $4$, $5$, $6$, $\ldots$, and there are exactly two zeros of $\theta$ (counted with multiplicity) in $U_{3/2,7/2}$.

Remarks 3**.**

(1) The domain $D(0.55)\cup(\mathbb{D}_{c_{0}}\setminus 0)$ is much larger than the domain $\mathbb{D}_{c_{0}}\setminus 0$ in which strong separation of the zeros of $\theta$ is guaranteed by Lemma 1 in [3]. It is impossible to extend Theorem 2 to the whole of $\overline{\mathbb{D}_{0.55}\setminus 0}$, because for certain values of $q$ in the right half of $\mathbb{D}_{0.55}\setminus 0$, the function $\theta(q,.)$ has double zeros, see [4] and [3].

(2) Numerical computations suggest that in part (1) of Theorem 2 the number $0.55$ can be replaced by $0.6$, see Remark 14.**

2 Comments

Theorem 2 can be compared with the results of [3]. Set $\alpha_{0}:=\sqrt{3}/2\pi=0.2756644477\ldots$. Parts (1) and (4) of Theorem 5 in [3] read:

Theorem 4**.**

(1) For $n\geq 5$ and for $|q|\leq 1-1/(\alpha_{0}n)$, strong separation of the zeros $\xi_{k}$ of $\theta$ occurs for $k\geq n$.

(2) For $0<|q|\leq 1/2$, strong separation of the zeros of $\theta$ occurs for $k\geq 4$.

Part (1) of the theorem implies that for $k\geq n\geq 5$, one has

[TABLE]

The following table gives an idea how the numbers $m_{n}$ and $M_{n}$ decrease as $n$ increases (both of them tend to $e^{1/\alpha_{0}}=37.62236657\ldots$ as $n\rightarrow\infty$). We list the truncations up to the second decimal for the numbers $\tau_{n}:=1-1/(\alpha_{0}n)$ and up to the first decimal for $m_{n}$ and $M_{n}$. The numbers $m_{n}$ and $M_{n}$ are the internal and external radius of the open annulus to which the zero $\xi_{n}$ belongs for $|q|=\tau_{n}$.

[TABLE]

3 Proof of Theorem 2

We use the same method of proof as the one of the proof of Theorem 5 in [3], but with more accurate estimations. We remind that (see [3]) $\theta=\Theta^{*}-G$, where $\Theta^{*}(q,z):=\sum_{j=-\infty}^{\infty}q^{j(j+1)/2}z^{j}$ and $G(q,z):=\sum_{-\infty}^{-1}q^{j(j+1)/2}z^{j}=\sum_{j=1}^{\infty}q^{j(j-1)/2}z^{-j}$. Set

[TABLE]

We recall that by formula (4) in [3] (resulting from the Jacobi triple product), one has $\Theta^{*}=QUR$.

Lemma 5**.**

For $q\in D(0.6)$, one has $|Q|\geq 1.2$.

Proof.

We set $Q:=Q_{0}Q_{1}$, where $Q_{0}:=\prod_{j=1}^{11}(1-q^{j})$ and $Q_{1}:=\prod_{j=12}^{\infty}(1-q^{j})$. For the quantity $|Q_{1}|$ one obtains the minoration

[TABLE]

Hence to prove the lemma it suffices to show that for $q\in D(0.6)$, the inequality

[TABLE]

holds true. This can be proved using the maximum principle. The polynomial $Q_{0}$ has no zeros in $D(0.6)$ hence $\min_{\overline{D(0.6)}}|Q_{0}|=1/\max_{\overline{D(0.6)}}|1/Q_{0}|$ is attained on the border $\partial D(0.6)$ of the domain $D(0.6)$. The restrictions of $|Q_{0}|^{2}$ to the segment (resp. to the arc) of $\partial D(0.6)$ are a polynomial and a trigonometric polynomial respectively and the latter displayed inequality is readily checked numerically. The check can be limited to the upper half-plane, because all coefficients of $Q$ are real hence $Q(\bar{q})=\overline{Q(q)}$. ∎

Lemma 6**.**

For $q\in D(0.6)$, $k\in\mathbb{N}$, $k\geq 5$ and $|z|=|q|^{-k+1/2}$, one has $\theta(q,z)\neq 0$.

Proof.

One can minorize $|U|$ and $|R|$ as follows:

[TABLE]

and

[TABLE]

Thus using Lemma 5 one obtains the inequality

[TABLE]

On the other hand the quantity $|G|$ can be majorized as follows:

[TABLE]

The lemma follows from inequalities (2) and (3) – for $|z|=|q|^{-k+1/2}$, one has at the same time $\theta=\Theta^{*}-G$, $|\Theta^{*}|>0.22$ and $|G|<0.11$, so $\theta=0$ is impossible. ∎

Lemma 7**.**

For $q\in D(0.6)$ and $|z|=|q|^{-7/2}$, one has $\theta(q,z)\neq 0$.

Lemma 8**.**

For $q\in D(0.6)$ and $|z|=|q|^{-3/2}$, one has $\theta(q,z)\neq 0$.

Lemma 9**.**

For $q\in D(0.55)$ and $|z|=|q|^{-5/2}$, one has $\theta(q,z)\neq 0$.

The last three lemmas are proved in Sections 4, 5 and 6 respectively. We explain how Theorem 2 results from them and from Lemma 6. We recall that for $|q|\leq 0.108$, all zeros of $\theta(q,.)$ are simple (see [2]). For any $k\in\mathbb{N}$ fixed and for $|q|$ sufficiently small, there exists a single zero of $\theta(q,.)$ which is $\sim-q^{-k}$ as $q\rightarrow 0$, see Proposition 10 in [1]. Hence for $|q|$ sufficiently small, this zero satisfies the conditions (1). These conditions hold true as $q$ varies along any segment $S$ belonging to a half-line passing through the origin and such that $S\subset(D(0.55)\cup\mathbb{D}_{c_{0}})$. Hence these conditions hold true for $q\in D(0.55)$. In the same way one sees that, with the possible exception of $\xi_{2}$ and $\xi_{3}$, the zeros of $\theta$ satisfy conditions (1) for $q\in D(0.6)$.

4 Proof of Lemma 7

For $|G|$ we obtain the majoration

[TABLE]

We minorize the quantity $|Q|$ using Lemma 5. There remains to minorize $|U|$ and $|R|$.

We observe that $\theta(\bar{q},\bar{z})=\overline{\theta(q,z)}$, therefore when $|\theta|$ is majorized, we can assume that $\arg(q)\in[0,\pi]$. Suppose that $q\in D(0.6)$, $\arg(q)\in[3\pi/4,\pi]$. We define the sectors $S_{j}$ in $\mathbb{C}$ by the fomula

[TABLE]

and $S_{-j}$ as the symmetric of $S_{j}$ w.r.t. the real axis. Suppose that $\zeta\in S_{\pm j}$. Set $\psi:=\arg(\zeta)$. By the cosine theorem

[TABLE]

The function $\cos(t)$ being even and decreasing on $[0,\pi]$ one obtains the minoration

[TABLE]

Remarks 10**.**

(1) For $\zeta\neq 0$, one has $\mu_{4}(|\zeta|)>\mu_{3}(|\zeta|)>\mu_{2}(|\zeta|)>\mu_{1}(|\zeta|)$.

(2) For $j$ fixed and for $|\zeta|\geq 1$, the right-hand side of (4) is an increasing function in $|\zeta|$.

(3) One can prove by straightforward computation that for $1\geq|\zeta_{1}|>|\zeta_{2}|$ and for $3\geq\ell>m\geq 1$, one has $\mu_{\ell}(|\zeta_{1}|)\cdot\mu_{m}(|\zeta_{2}|)>\mu_{\ell}(|\zeta_{2}|)\cdot\mu_{m}(|\zeta_{1}|)$.* *

Consider for $|z|=|q|^{-7/2}$ (i.e. for $z=|q|^{-7/2}\omega$, $|\omega|=1$), the moduli of three consecutive factors $u_{k}:=1+zq^{k}$ of $U$, for $k=k^{*}$, $k^{*}+1$ and $k^{*}+2$. (The role of $\zeta$ will be played by the numbers $-zq^{k}$.) Notice that the numbers $u_{k}$ are of the form

[TABLE]

Remarks 11**.**

(1) At least one of the three numbers $-zq^{k}$ belongs to the left half-plane, because $\arg(q)\in[\pi/2,\pi]$. Hence to the corresponding modulus $|u_{k}|$ minoration $\mu_{3}$ is applicable. If at most one of the other two numbers $-zq^{k}$ belongs to $S_{1}\cup S_{-1}$, then to the corresponding modulus $|u_{k}|$ minoration $\mu_{1}$, and to the third modulus minoration $\mu_{2}$ are applicable respectively.

(2) Suppose that at least two of the three numbers $-zq^{k}$ belong to $S_{1}\cup S_{-1}$. Then these correspond to $k=k^{*}$ and $k=k^{*}+2$, because $\arg(q)\in[\pi/2,\pi]$. Moreover, $-zq^{k^{*}+1}\in S_{4}\cup S_{-4}$, so minoration $\mu_{4}$ is applicable to $|u_{k^{*}+1}|$ and minoration $\mu_{1}$ to $|u_{k^{*}}|$ and $|u_{k^{*}+2}|$.**

Consider the three numbers $u_{1}$, $u_{2}$, $u_{3}$. When represented in the form (5), their exponents $-7/2+k$ are negative, so $|-zq^{k}|>1$. For fixed $\omega_{k}$, the modulus $|-zq^{k}|$ decreases as $|q|$ increases. Hence one can apply part (2) of Remarks 10 and minorize $|u_{k}|$ by its value for $|q|=0.6$. Making use of Remarks 11 one finds that the product $|u_{1}|\cdot|u_{2}|\cdot|u_{3}|$ is minorized by the least of the $7$ numbers $\mu_{1}(0.6^{-5/2})\cdot\mu_{4}(0.6^{-3/2})\cdot\mu_{1}(0.6^{-1/2})$ and $\mu_{i_{1}}(0.6^{-5/2})\cdot\mu_{i_{2}}(0.6^{-3/2})\cdot\mu_{i_{3}}(0.6^{-1/2})$, where $(i_{1},i_{2},i_{3})$ is a permutation of $(1,2,3)$. This is the number

[TABLE]

Now consider for $j=1$, $2$, $\ldots$, a triple $|u_{3j+1}|$, $|u_{3j+2}|$, $|u_{3j+3}|$, $j\in\mathbb{N}$. Hence $|-zq^{k}|<1$ and by Remarks 10 one has

[TABLE]

Set $\rho:=|q|$. We prove Lemma 7 with the help of the following result (the proof is given at the end of this section):

Lemma 12**.**

The quantities $A_{j}(\rho)$ and $B_{j}(\rho)$ are decreasing in $\rho$ for $\rho\in(0,0.6]$.

This means that one can minorize the product $\tilde{u}_{j}$ by $\chi_{j}:=\min(A_{j}(0.6),B_{j}(0.6))$. For $j=1$, $2$, $3$ and $4$, the values of $\chi_{j}$ are respectively

[TABLE]

(in all cases they equal $A_{j}(0.6)$). For $k\geq 16$, the factors $|u_{k}|$ can be minorized by $|1-|q|^{-7/2+k}|$, and then by $|1-0.6^{-7/2+k}|$. We set $\chi_{5}:=\prod_{k=16}^{\infty}|1-0.6^{-7/2+k}|=0.9957913379\ldots$. Thus we minorize $|U|$ by $\chi_{0}\cdot\chi_{1}\cdot\chi_{2}\cdot\chi_{3}\cdot\chi_{4}\cdot\chi_{5}$.

To minorize the product $R$ one can observe that for $|z|=|q|^{-7/2}$, each number $1+q^{j-1}/z$ is of the form (5) with $k\geq 7$. Therefore when minorizing $|R|$ one can use the same reasoning as for $|U|$ and obtain a minoration by $\chi_{2}\cdot\chi_{3}\cdot\chi_{4}\cdot\chi_{5}$ (the first value of the index $k$ being $7$, not $1$, one has to skip the analogs of the minorations of $|u_{k}|$ for $k=1$, $\ldots$, $6$, i.e. to skip $\chi_{0}$ and $\chi_{1}$). Set $\chi_{*}:=\chi_{2}\cdot\chi_{3}\cdot\chi_{4}\cdot\chi_{5}$. Thus one can minorize the product $|Q|\cdot|U|\cdot|R|$ by

[TABLE]

from which the lemma follows.

Proof of Lemma 12:.

One has $A_{j}^{2}=CEF$, where

[TABLE]

For $\rho\in(0,0.6]$, each factor $C$, $E$ and $F$ is positive-valued. Clearly $E^{\prime}=(3j+2)\rho^{3j+1}(2\rho^{3j+2}-\sqrt{2})<0$, because $2\rho\leq 1.2<\sqrt{2}$. We show that $(CF)^{\prime}<0$ from which and from $A_{j}>0$ follows that $A_{j}^{\prime}<0$. A direct computation shows that

[TABLE]

For $j\geq 1$ and $\rho\in(0,0.6]$, the first three summands inside the brackets are majorized respectively by $(6j+6)\cdot 0.6^{8}$, $(6j+2)\cdot 0.6^{4}$ and $(12j+8)\cdot 0.6^{16}$. Their sum is less than $6j+2$, so $(CF)^{\prime}<0$.

We set $B_{j}^{2}=MNW$, where

[TABLE]

Obviously $W^{\prime}<0$. We show that $(MN)^{\prime}<0$ which together with $M>0$, $N>0$, $W>0$ and $B_{j}>0$ proves that $B_{j}^{\prime}<0$. One has $(MN)^{\prime}=(K_{1}+K_{2}-L_{1})+(K_{3}+K_{4}-L_{2})+(K_{5}-L_{3})$, where

[TABLE]

The inequality $K_{5}<L_{3}$ is evident. One has $K_{4}/L_{2}\leq(3/4)\rho^{3j+1}\leq(3/4)\cdot 0.6^{4}=0.0972$ and $K_{3}/L_{2}\leq(10/9)\cdot(0.6/2\sqrt{2})=0.2357022603\ldots$, so $K_{3}+K_{4}<L_{2}$. Finally, $K_{1}/L_{1}\leq 0.6^{3j+1}\leq 0.6^{4}=0.1296$ and $K_{2}/L_{1}\leq\sqrt{2}\cdot 0.6=0.8485281372\ldots$ hence $K_{1}+K_{2}<L_{1}$ and $(MN)^{\prime}<0$. ∎

5 Proof of Lemma 8

We set $\theta^{\dagger}(q,z):=\theta(q,z/q)=\sum_{j=0}^{\infty}q^{j(j-1)/2}z^{j}=1+z+qz^{2}+q^{3}z^{3}+\cdots$. We show that $\theta^{\dagger}(q,z)\neq 0$ for $|z|=|q|^{-1/2}$ from which the lemma follows. As $\theta^{\dagger}(\bar{q},\bar{z})=\overline{\theta^{\dagger}(q,z)}$, we consider only the case $\arg(q)\in[\pi/2,\pi]$. We assume that $|q|\in[c_{0},0.6]$, because for $|q|\leq c_{0}$, the zeros of $\theta$ are strongly separated in modulus, see [3]. Set

[TABLE]

To show that $\theta^{\dagger}\neq 0$ we prove that the modulus of the sum of some or all of the terms $b_{0}$, $b_{1}$ and $b_{2}$ (we denote the set of the chosen terms by $S$) is larger than the modulus of the sum of the remaining terms of the series of $\theta^{\dagger}$. We distinguish the following cases according to the intervals to which $\arg(z)$ and $\arg(q)$ belong:

Case 1) Re $z\geq 0$, i.e. $\arg(z)\in[-\pi/2,\pi/2]$, and $\arg(q)\in[\pi/2,\pi]$. We set $S:=\{b_{0},b_{1}\}$. One has

[TABLE]

and

[TABLE]

The last inequality follows from $\Phi_{*}(|q|)$ and $\Phi_{\flat}(|q|)$ being respectively decreasing and increasing on $[c_{0},0.6]$ and

[TABLE]

Case 2) $\arg(z)\in[\pi,3\pi/2]$ and $\arg(q)\in[\pi/2,\pi]$. Then $\arg(qz)\in[3\pi/2,5\pi/2]$, i.e. Re $(qz)\geq 0$. We set $S:=\{b_{1},b_{2}\}$. Hence

[TABLE]

The other terms of $\theta^{\dagger}$ are $1$, $q^{3}z^{3}$, $q^{6}z^{4}$, $\ldots$. For $|z|=|q|^{-1/2}$, their moduli equal respectively $1$, $|q|^{3/2}$, $|q|^{4}$, $\ldots$, which are precisely the terms of the series $\Phi_{\flat}$, so as in Case 1) one concludes that $|z+qz^{2}|\geq\Phi_{*}(|q|)>\Phi_{\flat}(|q|)\geq 1+|r_{3}|$.

Case 3) $\arg(z)\in[\pi/2,3\pi/4]$ and $\arg(q)\in[\pi/2,\pi]$. The case is subdivided into five subcases in all of which we set $S:=\{b_{0},b_{1},b_{2}\}$.

Case 3A) $\arg(z)\in[\pi/2,3\pi/4]$ and $\arg(q)\in[3\pi/4,\pi]$. Then $\arg(qz^{2})\in[7\pi/4,5\pi/2]$. If $\arg(qz^{2})\in[2\pi,5\pi/2]$, then

[TABLE]

see (6). Set $\tau(|q|):=(2\cdot|q|)^{-1/2}-2^{-1/2}=\sin(3\pi/4)\cdot|q|^{-1/2}+\sin(7\pi/4)=-\cos(3\pi/4)\cdot|q|^{-1/2}-\cos(7\pi/4)$. If $\arg(qz^{2})\in[7\pi/4,2\pi]$, then

[TABLE]

(one can observe that $1-\tau(|q|)>0$ for $q\geq c_{0}$). Thus

[TABLE]

The latter quantity is minimal for $\tau=1/2$, i.e. for $|q|=0.3431457506\ldots$, when it equals $1/\sqrt{2}=0.7071067814\ldots>\Phi_{\flat}(0.6)-1\geq|r_{3}|$.

Case 3B) $\arg(z)\in[5\pi/8,3\pi/4]\subset[\pi/2,3\pi/4]$ and $\arg(q)\in[\pi/2,3\pi/4]$. Then $\arg(qz^{2})\in[7\pi/4,9\pi/4]\subset[7\pi/4,5\pi/2]$ and one proves as in Case 3A) that $|1+z+qz^{2}|>|r_{3}|$.

Case 3C) $\arg(z)\in[\pi/2,5\pi/8]$ and $\arg(q)\in[5\pi/8,3\pi/4]$. Then $\arg(qz^{2})\in[13\pi/8,2\pi]$. If $|q|\in[0.3,0.6]$, then

[TABLE]

If $|q|\in[c_{0},0.3]$, then

[TABLE]

Case 3D) $\arg(z)\in[9\pi/16,5\pi/8]$ and $\arg(q)\in[\pi/2,5\pi/8]$. Then $\arg(qz^{2})\in[13\pi/8,15\pi/8]$. Just as in Case 3C) one obtains $|1+z+qz^{2}|>|r_{3}|$.

Case 3E) $\arg(z)\in[\pi/2,9\pi/16]$ and $\arg(q)\in[\pi/2,5\pi/8]$. Then $\arg(qz^{2})\in[3\pi/2,7\pi/4]$ and

[TABLE]

so $|1+z+qz^{2}|>(0.5712^{2}+0.2661^{2})^{1/2}>0.63>\Phi_{\flat}(0.6)-1\geq|r_{3}|$.

Case 4) $\arg(z)\in[3\pi/4,\pi]$ and $\arg(q)\in[\pi/2,\pi]$. Then $\arg(qz^{2})\in[2\pi,3\pi]$, i.e. Im $(qz^{2})\geq 0$. We consider four subcases in all of which we set $S:=\{b_{0},b_{1},b_{2}\}$:

Case 4A) $\arg(z)\in[3\pi/4,5\pi/6]$ and $\arg(q)\in[\pi/2,\pi]$. In this case Im  $(z)\geq 0.6^{-1/2}\cdot\sin(5\pi/6)$ and

[TABLE]

Case 4B) $\arg(z)\in[5\pi/6,11\pi/12]$ and $\arg(q)\in[\pi/2,\pi]$. Hence $\arg(qz^{2})\in[13\pi/6,17\pi/6]$ (with $\sin(13\pi/6)=\sin(17\pi/6)=1/2$) and

[TABLE]

Case 4C) $\arg(z)\in[11\pi/12,\pi]$ and $\arg(q)\in[\pi/2,5\pi/6]$ hence $\arg(qz^{2})\in[7\pi/3,17\pi/6]$. If $\arg(qz^{2})\in[7\pi/3,5\pi/2]$, then

[TABLE]

If $\phi:=\arg(qz^{2})\in[5\pi/2,17\pi/6]$, then

[TABLE]

The function $T(\phi)$ takes only values larger than $1$ (hence larger than $|r_{3}|$) for $\phi\in[5\pi/2,17\pi/6]$.

Case 4D) $\arg(z)\in[11\pi/12,\pi]$ and $\arg(q)\in[5\pi/6,\pi]$, so $\arg(qz^{2})\in[8\pi/3,3\pi]$ and

[TABLE]

Hence $|1+z+qz^{2}|\geq|{\rm Re}\,(1+z+qz^{2})|>\Phi_{\flat}(0.6)-1\geq|r_{3}|$.

6 Proof of Lemma 9

We set $z:=q\xi$ hence $|\xi|=|q|^{-3/2}$. Thus $\theta=\sum_{j=0}^{\infty}q^{j(j-1)/2}\xi^{j}$. Next we set $A:=1+\sum_{j=4}^{\infty}q^{j(j-1)/2}\xi^{j}$ and $B:=1+q\xi+q^{3}\xi^{2}$, so $\theta=A+\xi B$. Finally we set $\xi:=q\zeta$, thus $B=1+\zeta+q\zeta^{2}$ and $|\zeta|=|q|^{-1/2}$.

Lemma 13**.**

(1) For $|q|\leq 0.6$ and $|\xi|\leq|q|^{-3/2}$, one has $|A|\leq a_{0}:=1+\sum_{j=4}^{\infty}0.6^{j(j-1)/2-3j/2}=2.330487021\ldots$.

(2) For $|q|\leq 0.6$, $\arg(q)\in[2\pi/3,\pi]$ and $|\xi|\leq|q|^{-3/2}$, one has $|\xi B|>a_{0}\geq|A|$ hence $\theta\neq 0$.

(3) For $|q|\leq 0.55$ and $|\xi|\leq|q|^{-3/2}$, one has $|\xi B|>a_{0}\geq|A|$ hence $\theta\neq 0$.

Remark 14**.**

For $\arg(q)\in[\pi/2,2\pi/3]$, it is possible to show numerically that $|\xi B|>a_{0}\geq|A|$ also for $|q|\in[0.55,0.6]$. To this end one can compute the quantities $|\xi B|$ and $|A|$ with sufficiently small steps in $\arg(q)$ and $|q|$. To obtain a majoration of $|A|$ one can set $A:=A^{*}+A^{**}$ with $A^{*}:=1+\sum_{j=4}^{7}q^{j(j-1)/2}\xi^{j}$ and $A^{**}:=\sum_{j=8}^{\infty}q^{j(j-1)/2}\xi^{j}$. For $|\xi|=|q|^{-3/2}$ and $|q|\leq 0.6$, one has $|A^{**}|\leq\sum_{j=8}^{\infty}0.6^{j(j-1)/2-3j/2}=0.0002925303367\ldots$. Thus there remains to estimate $|\xi B|$ and $|A^{*}|$ which contain finitely-many terms.**

Proof of Lemma 13.

Part (1) follows from $|A|\leq 1+\sum_{j=4}^{\infty}|q|^{j(j-1)/2-3j/2}\leq a_{0}$. To prove parts (2) and (3) we set $q:=\rho e^{i\omega}$, $\rho\geq 0$, $\omega\in[\pi/2,\pi]$, and $\zeta:=\rho^{-1/2}e^{i\psi}$, $\psi\in[0,2\pi]$. Observe that $|q\zeta^{2}|=1$. Thus

[TABLE]

Set $a:=\cos(\psi+\omega/2)\in[-1,1]$ and $b:=\cos(\omega/2)\in[0,\sqrt{2}/2]$ (because $\omega/2\in[\pi/4,\pi/2]$). The polynomial $4a^{2}+4\rho^{-1/2}ab+\rho^{-1}$ (considered as a polynomial in $a$ with $b$ as a parameter) takes its minimal value for $a=-\rho^{-1/2}b/2$. This value is $(1-b^{2})\rho^{-1}$. Therefore for $b\leq 1/2$ (hence for $\omega\in[2\pi/3,\pi]$), one has $|B|^{2}\geq 3\cdot\rho^{-1}/4\geq 3\cdot 0.6^{-1}/4$. So for $|\xi|=|q|^{-3/2}\geq 0.6^{-3/2}$, one has

[TABLE]

For $\rho=|q|\leq 0.55$, one has $|B|^{2}\geq\rho^{-1}/2$ and $|\xi B|\geq 0.55^{-2}/\sqrt{2}=2.337543079\ldots>a_{0}\geq|A|$. ∎