# Odd unimodal sequences

Kathrin Bringmann  University of Cologne, Department of Mathematics and
Computer Science, Weyertal 86-90, 50931 Cologne, Germany<EMAIL_ADDRESS>koeln.de and  Jeremy Lovejoy  CNRS, Université Paris Cité, Bâtiment Sophie
Germain, Case courier 7014, 8 Place Aurélie Nemours, 75205 Paris Cedex 13,
FRANCE<EMAIL_ADDRESS>
###### Abstract.

In this paper we study odd unimodal and odd strongly unimodal sequences. We
use $q$-series methods to find several fundamental generating functions.
Employing the Euler–Maclaurin summation formula we obtain the asymptotic main
term for both types of sequences. We also find families of congruences modulo
$4$ for the number of odd strongly unimodal sequences.

###### Key words and phrases:

asymptotics, Euler–Maclaurin summation formula, mock theta functions, modular
forms, partitions, Tauberian theorems, unimodal sequences

###### 2020 Mathematics Subject Classification:

11F03, 11F37, 11P82, 11P83, 33D15

##  1\. Introduction and statement of results

A sequence is unimodal if it is weakly increasing up to a point and then
weakly decreasing thereafter. Let $u(n)$ denote the number of unimodal
sequences of natural numbers having the form

$a_{1}\leq\cdots\leq a_{r}\leq\overline{c}\geq b_{1}\geq\cdots\geq b_{s},$
(1.1)

with

$n=c+\sum_{j=1}^{r}a_{j}+\sum_{j=1}^{s}b_{j}.$

The distinguished point $c$ is called the peak of the sequence and the sum of
the entries $n$ is called the weight. For example, we have $u(4)=12$, the
twelve unimodal sequences of weight $4$ being

$\begin{gathered}\left(\overline{4}\right),\left(1,\overline{3}\right),\left(\overline{3},1\right),\left(2,\overline{2}\right),\left(1,1,\overline{2}\right),\left(1,\overline{2},1\right),\left(\overline{2},2\right),\left(\overline{2},1,1\right),\\\
\left(1,1,1,\overline{1}\right),\left(1,1,\overline{1},1\right),\left(1,\overline{1},1,1\right),\left(\overline{1},1,1,1\right).\end{gathered}$
(1.2)

A unimodal sequence is strongly unimodal if the inequalities in (1.1) are
strict. Let $u^{*}(n)$ denote the number of strongly unimodal sequences of
natural numbers with weight $n$. For example, we have $u^{*}(4)=4$, the four
strongly unimdodal sequences of weight $4$ being111For strongly unimodal
sequences we drop the overline notation since the peak cannot repeat.

$(4),(1,3),(3,1),(1,2,1).$ (1.3)

Unimodal sequences and strongly unimodal sequences have been the subject of a
considerable amount of research, especially over the last two decades. The
generating functions for these sequences are related to number-theoretic
objects like mock theta functions, false theta functions, and quantum modular
forms, whose theories can then be applied to deduce many interesting results
(see Subsections 14.4, 15.7, and 21.4 of [11]; for some more recent work, see
[13, 16]).

In this paper we initiate the study of odd unimodal sequences and odd strongly
unimodal sequences, wherein all numbers must be odd.222These were briefly
treated in [4], where they were called convex and strictly convex compositions
with odd parts. Let ${\rm ou}(n)$ and ${\rm ou}^{*}(n)$ denote the number of
odd unimodal and odd strongly unimodal sequences of weight $n$. Continuing the
example in (1.2) and (1.3), we have ${\rm ou}(4)=6$ and ${\rm ou}^{*}(4)=2$.

We begin by computing the generating functions upon which all of our work is
based. We use two-variable generating functions, where the second variable
tracks the rank of the unimodal sequence. In the notation of (1.1), the rank
is defined to be $r-s$, or the number of parts of the sequence to the left of
the peak minus the number to the right. From the perspective of modular forms,
this second variable is the “Jacobi variable”. We use the standard
$q$-hypergeometric notation,

$(a;q)_{n}:=\prod_{j=0}^{n-1}\left(1-aq^{j}\right),\qquad(a_{1},\dots,a_{\ell};q)_{n}:=(a_{1};q)_{n}\cdots(a_{\ell};q)_{n}$
(1.4)

valid for $\ell\in\mathbb{N}$, $a,a_{1},\dots,a_{\ell}\in\mathbb{C}$ and
$n\in\mathbb{N}_{0}\cup\\{\infty\\}$. Let ${\rm ou}(m,n)$ denote the number of
odd unimodal sequences of weight $n$ with rank $m$.

######  Theorem 1.1.

We have

$\displaystyle\sum_{\begin{subarray}{c}n\geq 0\\\
m\in\mathbb{Z}\end{subarray}}\textnormal{ou}(m,n)\zeta^{m}q^{n}=\sum_{n\geq
0}\frac{q^{2n+1}}{\left(\zeta q,\zeta^{-1}q;q^{2}\right)_{n+1}}$ (1.5)
$\displaystyle\hskip 28.16838pt=\sum_{n\geq
0}(-1)^{n+1}\zeta^{3n+1}q^{3n^{2}+2n}\left(1+\zeta
q^{2n+1}\right)+\frac{1}{\left(\zeta
q,\zeta^{-1}q;q^{2}\right)_{\infty}}\sum_{n\geq
0}(-1)^{n}\zeta^{2n+1}q^{n^{2}+n}$ (1.6) $\displaystyle\hskip
28.16838pt=\frac{q}{\left(q^{2};q^{2}\right)_{\infty}}\left(\sum_{n,r\geq
0}-\sum_{n,r<0}\right)\frac{(-1)^{n+r}q^{n^{2}+3n+4rn+r^{2}+3r}}{1-\zeta
q^{2r+1}}.$ (1.7)

Let $\mathrm{ou}^{*}(m,n)$ denote the number of odd strongly unimodal
sequences of weight $n$ with rank $m$.

######  Theorem 1.2.

We have, with
$Q(r,s):=\frac{r^{2}}{4}+\frac{7rs}{2}+\frac{s^{2}}{4}+\frac{3r}{2}+\frac{5s}{2}+1$,

$\displaystyle\sum_{\begin{subarray}{c}n\geq 0\\\
m\in\mathbb{Z}\end{subarray}}\mathrm{ou}^{*}(m,n)\zeta^{m}q^{n}$
$\displaystyle=\sum_{n\geq 0}\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{n}q^{2n+1}$ (1.8)
$\displaystyle=-\frac{1}{\left(q^{2};q^{2}\right)_{\infty}}\sum_{n\in\mathbb{Z}}\frac{(-1)^{n}q^{3n^{2}+3n+1}}{1+\zeta
q^{2n+1}}+\frac{1}{\left(q^{2};q^{2}\right)_{\infty}}\sum_{n\in\mathbb{Z}}\frac{\zeta^{-n}q^{n^{2}+2n+1}}{1+\zeta
q^{2n+1}}$ (1.9)
$\displaystyle=\frac{q}{\left(q^{2};q^{2}\right)_{\infty}}\left(\sum_{n,r\geq
0}-\sum_{n,r<0}\right)(-1)^{n}\zeta^{r}q^{3n^{2}+3n+4nr+r^{2}+2r}$ (1.10)
$\displaystyle=\frac{\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{\infty}}{\left(q^{2};q^{2}\right)_{\infty}^{2}}\left(\sum_{\begin{subarray}{c}r,s\geq
0\\\ r\equiv s\
(\mathrm{mod}\,2)\end{subarray}}-\sum_{\begin{subarray}{c}r,s<0\\\ r\equiv s\
(\mathrm{mod}\,2)\end{subarray}}\right)\frac{(-1)^{\frac{r-s}{2}}q^{Q(r,s)}}{1+\zeta
q^{r+s+1}}.$ (1.11)

######  Remark.

The analogue of Theorem 1.1 for ordinary unimodal sequences is [22,
Proposition 2.1]. For strongly unimodal sequences, the corresponding
generating functions are dispersed in the literature. The analogues of (1.8)
and (1.9) are [11, equation (14.18)] and [12, Lemma 3.1], the analogues of
(1.10) and (1.11) are Theorems 4.1 and 1.3 of [19].

Our first use of these generating functions is to find asymptotic estimates
for ${\rm ou}(n)$ and ${\rm ou}^{*}(n)$.

######  Theorem 1.3.

We have, as $n\to\infty$,

${\rm
ou}(n)\sim\frac{e^{\pi\sqrt{\frac{2n}{3}}}}{2^{\frac{13}{4}}3^{\frac{1}{4}}n^{\frac{3}{4}}}.$
(1.12)

######  Theorem 1.4.

We have, as $n\to\infty$,

${\rm
ou}^{*}(n)\sim\frac{e^{\pi\sqrt{\frac{n}{3}}}}{2^{\frac{5}{2}}3^{\frac{1}{4}}n^{\frac{3}{4}}}.$
(1.13)

######  Remark.

By [7, 27], the analogue of (1.12) for the number of ordinary unimodal
sequences $u(n)$ is

$u(n)\sim\frac{e^{\pi\sqrt{\frac{4n}{3}}}}{2^{3}3^{\frac{3}{4}}n^{\frac{5}{4}}}.$

The analogue of (1.13) for the number of strongly unimodal sequences
$u^{*}(n)$ is

$u^{*}(n)\sim\frac{e^{\pi\sqrt{\frac{2n}{3}}}}{2^{\frac{13}{4}}3^{\frac{1}{4}}n^{\frac{3}{4}}},$
(1.14)

due to [26]. Note that the asymptotics in (1.12) and (1.14) agree.

As a second result, we prove congruences modulo $4$ for the number of odd
strongly unimodal sequences of weight $n$. Here we are motivated by a
corresponding result for strongly unimodal sequences, which says that if
$\ell\equiv 7,11,13,17\ (\mathrm{mod}\,24)$ is prime and
$(\frac{j}{\ell})=-1$, then

$u^{*}\left(\ell^{2}n+\ell j-\left(\tfrac{\ell^{2}-1}{24}\right)\right)\equiv
0\ (\mathrm{mod}\,4).$

This was conjectured by Bryson, Ono, Pitman, and Rhoades [14] and proved by
Chen and Garvan [16]. Our analogue for $\mathrm{ou}^{*}(n)$ is as follows; see
Theorem 6.2 for a more general theorem.

######  Theorem 1.5.

Let $\ell\geq 5$ be prime. If $\ell\equiv 7,13\ (\mathrm{mod}\,24)$ and
$(\frac{3j}{\ell})=-1$ or if $\ell\not\equiv 7,13\ (\mathrm{mod}\,24)$ and
$\ell\nmid j$, then we have:

  1. (1)

If $j$ is odd, then

$\mathrm{ou}^{*}\left(4\ell^{2}n+2j\ell+\left(\tfrac{8\ell^{2}+1}{3}\right)\right)\equiv
0\ (\mathrm{mod}\,4).$

  2. (2)

If $j$ is even, then

$\mathrm{ou}^{*}\left(4\ell^{2}n+2j\ell+\left(\tfrac{2\ell^{2}+1}{3}\right)\right)\equiv
0\ (\mathrm{mod}\,4).$

The paper is organized as follows: In Section 2, we gather some necessary
background on asymptotic methods and indefinite theta functions. In Section 3
we prove Theorems 1.1 and 1.2. Sections 4 and 5 contain proofs of Theorems 1.3
and 1.4. In Section 6, we show Theorem 6.2, which contains the congruences in
Theorem 1.5 as a special case. We close in Section 7 with some open problems.

## Acknowledgements

The authors thank Caner Nazaroglu for help with numerical calculations. The
first author has received funding from the European Research Council (ERC)
under the European Union’s Horizon 2020 research and innovation programme
(grant agreement No. 101001179).

##  2\. Preliminaries

###  2.1. A Tauberian Theorem

Recall the following333The second condition is often dropped in (2.1) which
makes the proposition unfortunately incorrect (see [10]). special case
$\alpha=0$ of Theorem 1.1 of [10], which follows from Ingham’s Theorem [20].

######  Proposition 2.1.

Suppose that $B(q)=\sum_{n\geq 0}b(n)q^{n}$ is a power series with non-
negative real coefficients and radius of convergence at least one and that the
$b(n)$ are weakly increasing. Assume that $\lambda$, $\beta$,
$\gamma\in\mathbb{R}$ with $\gamma>0$ exist such that

$B\left(e^{-t}\right)\sim\lambda t^{\beta}e^{\frac{\gamma}{t}}\quad\text{as
}t\to 0^{+},\qquad
B\left(e^{-z}\right)\ll|z|^{\beta}e^{\frac{\gamma}{|z|}}\quad\text{as }z\to
0,$ (2.1)

with $z=x+iy$ ($x,y\in\mathbb{R},x>0$) in each region of the form
$|y|\leq\Delta x$ for $\Delta>0$. Then

$b(n)\sim\frac{\lambda\gamma^{\frac{\beta}{2}+\frac{1}{4}}}{2\sqrt{\pi}n^{\frac{\beta}{2}+\frac{3}{4}}}e^{2\sqrt{\gamma
n}}\qquad\text{as }n\to\infty.$

###  2.2. The Euler–Maclaurin summation formula

For simplicity we only state the versions of the Euler–Maclaurin summation
formula that we use in this paper; see [10] for a more general version for all
dimensions. Let $D_{\theta}:=\\{re^{i\alpha}:r\geq 0\text{ and
}|\alpha|\leq\theta\\}$. A multivariable function $f$ in $\ell$ variables is
of sufficient decay in $D$ if there exist
$\varepsilon_{1},\dots,\varepsilon_{\ell}>0$ such that (we write vectors in
bold letters)
$f(\bm{x})\ll(x_{1}+1)^{-1-\varepsilon_{1}}\cdots(x_{\ell}+1)^{-1-\varepsilon_{\ell}}$
uniformly as $|x_{1}|+\ldots+|x_{\ell}|\to\infty$ in $D$. We first require a
one-dimensional version of the Euler–Maclaurin summation formula (see [28]).

######  Proposition 2.2.

Suppose that $0\leq\theta<\frac{\pi}{2}$. Let $f:\mathbb{C}\to\mathbb{C}$ be
holomorphic in a domain containing $D_{\theta}$, so that in particular $f$ is
holomorphic at the origin, and assume that $f$ and all of its derivatives are
of sufficient decay. Then for $a\in\mathbb{R}$ and $N\in\mathbb{N}_{0}$, we
have, uniformly as $z\to 0$ in $D_{\theta}$,

$\sum_{m\geq
0}f((m+a)z)=\frac{1}{z}\int_{0}^{\infty}f(w)dw-\sum_{n=0}^{N-1}\frac{B_{n+1}(a)f^{(n)}(0)}{(n+1)!}z^{n}+O_{N}\left(z^{N}\right),$

where $B_{n}(x)$ denotes the $n$-th Bernoulli polynomial.

We also require the two-dimensional case of the Euler–Maclaurin summation
formula (see [10]).

######  Proposition 2.3.

Suppose that $0\leq\theta_{j}<\frac{\pi}{2}$ for $1\leq j\leq 2$, and that
$f:\mathbb{C}^{2}\to\mathbb{C}$ is holomorphic in a domain containing
$D_{\bm{\theta}}:=D_{\theta_{1}}\times D_{\theta_{2}}$. If $f$ and all of its
derivatives are of sufficient decay in $D_{\bm{\theta}}$, then for
$\bm{a}\in\mathbb{R}^{2}$ and $N\in\mathbb{N}_{0}$ we have uniformly, as $w\to
0$ in $D_{\bm{\theta}}$,

$\displaystyle\sum_{\bm{m}\in\mathbb{N}_{0}^{2}}f((\bm{m}+\bm{a})z)$
$\displaystyle=\frac{1}{z^{2}}\int_{0}^{\infty}\int_{0}^{\infty}f(\bm{w})dw_{1}dw_{2}-\frac{1}{z}\sum_{n_{1}=0}^{N-1}\frac{B_{n_{1}+1}(a_{1})}{(n_{1}+1)!}z^{n_{1}}\int_{0}^{\infty}f^{(n_{1},0)}(0,w_{2})dw_{2}$
$\displaystyle\qquad-\frac{1}{z}\sum_{n_{2}=0}^{N-1}\frac{B_{n_{2}+1}(a_{2})}{(n_{2}+1)!}z^{n_{2}}\int_{0}^{\infty}f^{(0,n_{2})}(w_{1},0)dw_{1}$
$\displaystyle\qquad+\sum_{n_{1}+n_{2}<N}\frac{B_{n_{1}+1}(a_{1})B_{n_{2}+1}(a_{2})f^{(n_{1},n_{2})}(\bm{0})}{(n_{1}+1)!(n_{2}+1)!}z^{n_{1}+n_{2}}+O_{N}\left(z^{N}\right).$

###  2.3. Indefinite theta functions

In this subsection, we recall results from Zwegers’ thesis [29]. Fix a
quadratic form $Q$ of signature $(n,1)$ with associated matrix $A$, so that
$Q(\bm{x})=\frac{1}{2}\bm{x}^{T}A\bm{x}$. Let
$B(\bm{x},\bm{y}):=Q(\bm{x}+\bm{y})-Q(\bm{x})-Q(\bm{y})$ denote the
corresponding bilinear form. The set of vectors $\bm{c}\in\mathbb{R}^{\ell}$
with $Q(\bm{c})<0$ splits into two connected components. Two vectors
$\bm{c_{1}}$ and $\bm{c_{2}}$ lie in the same component if and only if
$B(\bm{c_{1}},\bm{c_{2}})<0$. We fix one of the components and denote it by
$C_{Q}$. Picking any vector $\bm{c_{0}}\in C_{Q}$, we have

$C_{Q}=\left\\{\bm{c}\in\mathbb{R}^{\ell}:Q(\bm{c})<0,\
B(\bm{c},\bm{c_{0}})<0\right\\}.$

The cusps are elements from

$S_{Q}:=\left\\{\bm{c}\in\mathbb{Z}^{\ell}:\gcd(c_{1},c_{2},\dots,c_{\ell})=1,\
Q(\bm{c})=0,\ B(\bm{c},\bm{c_{0}})<0\right\\}.$

Let $\overline{C}_{Q}:=C_{Q}\cup S_{Q}$ and define for,
$\bm{c}\in\overline{C}_{Q}$

$R(\bm{c}):=\begin{cases}\mathbb{R}^{\ell}&\text{if }\bm{c}\in C_{Q},\\\
\left\\{\bm{a}\in\mathbb{R}^{\ell}:B(\bm{c},\bm{a})\notin\mathbb{Z}\right\\}&\text{if
}\bm{c}\in S_{Q}.\end{cases}$

Let $\bm{c_{1}},\bm{c_{2}}\in\overline{C}_{Q}$. We define the theta function
with characteristic $\bm{a}\in R(\bm{c_{1}})\cap R(\bm{c_{2}})$ and
$\bm{b}\in\mathbb{R}^{\ell}$ by

$\vartheta_{\bm{a},\bm{b}}(\tau):=\sum_{\bm{n}\in\mathbb{Z}^{\ell}+\bm{a}}\varrho(\bm{n};\tau)e^{2\pi
iB(\bm{b},\bm{n})}q^{Q(\bm{n})},$

where

$\varrho(\bm{n};\tau)=\varrho_{Q}^{\bm{c_{1}},\bm{c_{2}}}(\bm{n};\tau):=\varrho^{\bm{c_{1}}}(\bm{n};\tau)-\varrho^{\bm{c_{2}}}(\bm{n};\tau)$

with ($\tau=u+iv$)

$\varrho^{\bm{c}}(\bm{n};\tau):=\begin{cases}E\left(\frac{B(\bm{c},\bm{n})\sqrt{v}}{\sqrt{-Q(\bm{c})}}\right)&\text{if
}\bm{c}\in C_{Q},\\\ \operatorname{sgn}(B(\bm{c},\bm{n}))&\text{if }\bm{c}\in
S_{Q}.\end{cases}$

Here the odd function $E$ is defined as

$E(w):=2\int_{0}^{w}e^{-\pi t^{2}}dt$

with the usual convention that $\operatorname{sgn}(w):=\frac{w}{|w|}$ for
$w\in\mathbb{R}\setminus\\{0\\}$ and $\operatorname{sgn}(0):=0$. Note that

$E(x)=\operatorname{sgn}(x)\left(1-\beta\left(x^{2}\right)\right),\text{ where
}\beta(x):=\int_{x}^{\infty}w^{-\frac{1}{2}}e^{-\pi w}dw.$ (2.2)

This in particular yields that $E(x)\sim\operatorname{sgn}(x)$ as
$|x|\to\infty$.

The theta function satisfies the following transformation law.

######  Theorem 2.4.

If $\bm{a},\bm{b}\in R(\bm{c_{1}})\cap R(\bm{c_{2}})$, then

$\vartheta_{\bm{a},\bm{b}}\left(-\frac{1}{\tau}\right)=\frac{1}{\sqrt{-\det(A)}}(-i\tau)^{\frac{\ell}{2}}e^{2\pi
iB(\bm{a},\bm{b})}\sum_{\bm{\ell}\in
A^{-1}\mathbb{Z}^{\ell}/\mathbb{Z}^{\ell}}\vartheta_{\bm{b}+\bm{\ell},-\bm{a}}(\tau).$

##  3\. Generating functions and the proofs of Theorems 1.1 and 1.2

In this section we establish in Theorems 1.1 and 1.2. We begin with Theorem
1.1.

###### Proof of Theorem 1.1.

Equation (1.5) is a straightforward consequence of the fact that $(\zeta
q;q^{2})_{n+1}^{-1}$ is the generating function for partitions into odd parts
of size at most $2n+1$, with the exponent of $\zeta$ counting the number of
parts. Namely, in the notation of (1.1), the term $q^{2n+1}$ generates the
peak $\overline{c}$, the term $(\zeta q;q^{2})_{n+1}^{-1}$ generates the odd
parts $(a_{1},\dots,a_{r})$ to the left of the peak, and the term
$(\zeta^{-1}q;q^{2})_{n+1}^{-1}$ generates the odd parts $(b_{1},\dots,b_{s})$
to the right of the peak. The exponent of $\zeta$ is $r-s$. Equation (1.6) is
a result in Ramanujan’s lost notebook [6, Entry 6.3.4].

Equation (1.7) requires more work. Fot its proof, we require so-called Bailey
pairs (for background see [25]). A pair of sequences
$(\alpha_{n},\beta_{n})_{n\geq 0}$ is called a Bailey pair relative to $(a,q)$
if

$\beta_{n}=\sum_{k=0}^{n}\frac{\alpha_{k}}{(q;q)_{n-k}(aq;q)_{n+k}}.$

If $(\alpha_{n},\beta_{n})$ is a Bailey pair relative to $(a,q)$, then by [24,
equation (1.5)]

$\sum_{n\geq 0}q^{n}\beta_{n}=\frac{1}{(aq,q;q)_{\infty}}\sum_{n,r\geq
0}(-a)^{n}q^{\frac{n(n+1)}{2}+(2n+1)r}\alpha_{r}.$ (3.1)

The following sequences form a Bailey pair relative to $(q^{2},q^{2})$ [23,
pp. 727–728]:

$\alpha_{n}=\frac{(-1)^{n}q^{n^{2}+n}\left(1-q^{4n+2}\right)}{\left(1-q^{2}\right)\left(1-\zeta
q^{2n+1}\right)\left(1-\zeta^{-1}q^{2n+1}\right)},\quad\beta_{n}=\frac{1}{\left(\zeta
q,\zeta^{-1}q;q^{2}\right)_{n+1}}.$ (3.2)

Inserting (3.2) into (3.1) and using the fact that

$\frac{1-q^{4r+2}}{\left(1-\zeta
q^{2r+1}\right)\left(1-\zeta^{-1}q^{2r+1}\right)}=\frac{1}{1-\zeta
q^{2r+1}}+\frac{\zeta^{-1}q^{2r+1}}{1-\zeta^{-1}q^{2r+1}},$ (3.3)

we compute

$\displaystyle\sum_{n\geq 0}\frac{q^{2n+1}}{\left(\zeta
q,\zeta^{-1}q;q^{2}\right)_{n+1}}=\frac{q}{\left(q^{2};q^{2}\right)_{\infty}^{2}}\sum_{n,r\geq
0}\frac{(-1)^{n+r}q^{n^{2}+3n+4nr+r^{2}+3r}\left(1-q^{4r+2}\right)}{\left(1-\zeta
q^{2r+1}\right)\left(1-\zeta^{-1}q^{2r+1}\right)}$ $\displaystyle\hskip
28.45274pt=\frac{q}{\left(q^{2};q^{2}\right)_{\infty}^{2}}\left(\sum_{n,r\geq
0}\frac{(-1)^{n+r}q^{n^{2}+3n+4nr+r^{2}+3r}}{1-\zeta
q^{2r+1}}+\zeta^{-1}\sum_{n,r\geq
0}\frac{(-1)^{n+r}q^{n^{2}+3n+4nr+r^{2}+5r+1}}{1-\zeta^{-1}q^{2r+1}}\right)$
$\displaystyle\hskip
28.45274pt=\frac{q}{\left(q^{2};q^{2}\right)_{\infty}^{2}}\left(\sum_{n,r\geq
0}-\sum_{n,r<0}\right)\frac{(-1)^{n+r}q^{n^{2}+3n+4rn+r^{2}+3r}}{1-\zeta
q^{2r+1}}.$

In the last step we let $(n,r)\mapsto(-n-1,-r-1)$ and simplify. This completes
the proof. ∎

We now turn to the proof of Theorem 1.2.

###### Proof of Theorem 1.2.

We use the fact that $(-\zeta q;q^{2})_{n}$ is the generating function for
partitions into distinct odd parts of size at most $2n-1$, with the exponent
of $\zeta$ counting the number of parts.

For (1.9) we require two identities,

$\displaystyle\sum_{n\geq 0}\frac{q^{2n^{2}+2n+1}}{\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{n+1}}$
$\displaystyle=\frac{1}{\left(q^{2};q^{2}\right)_{\infty}}\sum_{n\in\mathbb{Z}}\frac{(-1)^{n}q^{3n^{2}+3n+1}}{1+\zeta
q^{2n+1}},$ (3.4)
$\displaystyle\sum_{n\in\mathbb{Z}}\frac{(a,b;q)_{n}w^{n}}{(c,d;q)_{n}}$
$\displaystyle=\frac{\left(aw,\frac{d}{a},\frac{c}{b},\frac{dq}{abw};q\right)_{\infty}}{\left(w,d,\frac{q}{b},\frac{cd}{abw};q\right)_{\infty}}\sum_{n\in\mathbb{Z}}\frac{\left(a,\frac{abw}{d};q\right)_{n}\left(\frac{d}{a}\right)^{n}}{(aw,c;q)_{n}}.$
(3.5)

Equation 3.4 may be found in [1, p. 397], while equation (3.5) is a bilateral
transformation of Bailey [18, example 5.20 (i)]. The notation in (1.4) is
extended to all integers via

$(a;q)_{n}=\frac{(a;q)_{\infty}}{(aq^{n};q)_{\infty}}.$

Note that we have the identity [18, (I.2)]

$(a;q)_{-n}=\frac{(-1)^{n}q^{\frac{n(n+1)}{2}}}{a^{n}\left(\frac{q}{a};q\right)_{n}}.$
(3.6)

We begin by letting $(a,b,w,q)=(-\zeta q,-\zeta^{-1}q,q^{2},q^{2})$ in (3.5)
and then letting $c,d\to 0$. Simplifying and exchanging left- and right-hand
sides gives

$\displaystyle\frac{1}{\left(q^{2};q^{2}\right)_{\infty}}\sum_{n\in\mathbb{Z}}\frac{\zeta^{-n}q^{n^{2}+2n+1}}{1+\zeta
q^{2n+1}}=\sum_{n\in\mathbb{Z}}\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{n}q^{2n+1}$ $\displaystyle\hskip
8.5359pt=\sum_{n\geq 0}\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{n}q^{2n+1}+\sum_{n\leq-1}\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{n}q^{2n+1}$ $\displaystyle\hskip
8.5359pt=\sum_{\begin{subarray}{c}n\geq 0\\\
m\in\mathbb{Z}\end{subarray}}\mathrm{ou}^{*}(m,n)\zeta^{m}q^{n}+\sum_{n\geq
1}\left(-\zeta q,-\zeta^{-1}q;q^{2}\right)_{-n}q^{-2n+1}$ $\displaystyle\hskip
8.5359pt=\sum_{\begin{subarray}{c}n\geq 0\\\
m\in\mathbb{Z}\end{subarray}}\mathrm{ou}^{*}(m,n)\zeta^{m}q^{n}+\sum_{n\geq
1}\frac{q^{2n^{2}-2n+1}}{\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{n}}=\sum_{\begin{subarray}{c}n\geq 0\\\
m\in\mathbb{Z}\end{subarray}}\mathrm{ou}^{*}(m,n)\zeta^{m}q^{n}+\sum_{n\geq
0}\frac{q^{2n^{2}+2n+1}}{\left(-\zeta q,-\zeta^{-1}q;q^{2}\right)_{n+1}}$
$\displaystyle\hskip 8.5359pt=\sum_{\begin{subarray}{c}n\geq 0\\\
m\in\mathbb{Z}\end{subarray}}\mathrm{ou}^{*}(m,n)\zeta^{m}q^{n}+\frac{1}{\left(q^{2};q^{2}\right)_{\infty}}\sum_{n\in\mathbb{Z}}\frac{(-1)^{n}q^{3n^{2}+3n+1}}{1+\zeta
q^{2n+1}}.$

Here the final equality uses (3.4) and the antepenultimate equality uses
(3.6). Comparing the extremes in this string of equations gives (1.9).

We now turn to (1.10). For this require the fact that if
$(\alpha_{n},\beta_{n})$ is a Bailey pair relative to $(a,q)$, then we use
[24, Corollary 1.3]

$\sum_{n\geq 0}(aq;q)_{2n}q^{n}\beta_{n}=\frac{1}{(q;q)_{\infty}}\sum_{n,r\geq
0}(-a)^{n}q^{\frac{3n(n+1)}{2}+(2n+1)r}\alpha_{r},$ (3.7)

along with the following Bailey pair relative to $(1,q)$ [3, Lemma 3]:

$\alpha_{n}=\begin{cases}(-1)^{n}\left(w^{n}q^{\frac{n(n-1)}{2}}+w^{-n}q^{\frac{n(n+1)}{2}}\right)&\text{if
}n\geq 1,\\\ 1&\text{if
}n=0,\end{cases}\qquad\beta_{n}=\frac{\left(w;\frac{q}{w};q\right)_{n}}{(q;q)_{2n}}.$

Using this Bailey pair with $(w,q)=(-\zeta q,q^{2})$ in (3.7) we compute

$\displaystyle\sum_{n\geq 0}\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{n}q^{2n+1}$ $\displaystyle\hskip
68.28644pt=\frac{q}{\left(q^{2};q^{2}\right)_{\infty}}\left({\vphantom{\sum_{\begin{subarray}{c}n\geq
0\\\ r\geq 1\end{subarray}}}}\right.\sum_{n,r\geq
0}(-1)^{n}\zeta^{r}q^{3n^{2}+3n+4nr+r^{2}+2r}+\sum_{\begin{subarray}{c}n\geq
0\\\ r\geq
1\end{subarray}}(-1)^{n}\zeta^{-r}q^{3n^{2}+3n+4nr+r^{2}+2r}\left.{\vphantom{\sum_{\begin{subarray}{c}n\geq
0\\\ r\geq 1\end{subarray}}}}\right)$ $\displaystyle\hskip
68.28644pt=\frac{q}{\left(q^{2};q^{2}\right)_{\infty}}\left(\sum_{n,r\geq
0}(-1)^{n}\zeta^{r}q^{3n^{2}+3n+4nr+r^{2}+2r}-\sum_{n,r<0}(-1)^{n}\zeta^{r}q^{3n^{2}+3n+4nr+r^{2}+2r}\right),$

where the last line follows upon replacing $(n,r)$ by $(-n-1,-r)$. This gives
(1.10).

Finally, we treat (1.11). Again we use Bailey pairs. This time we require the
fact that if $(\alpha,\beta_{n})$ is a Bailey pair relative to $(a,q)$, then
[25, Theorem 10.1]

$\sum_{n\geq
0}(b,c;q)_{n}\left(\frac{aq}{bc}\right)^{n}\beta_{n}=\frac{\left(\frac{aq}{b},\frac{aq}{c};q\right)_{\infty}}{\left(aq,\frac{aq}{bc};q\right)_{\infty}}\sum_{n\geq
0}\frac{(b,c;q)_{n}\left(\frac{aq}{bc}\right)^{n}}{\left(\frac{aq}{b},\frac{aq}{c};q\right)_{n}}\alpha_{n},$
(3.8)

along with the Bailey pair444We point out to the reader that the $A_{n}$ in
Andrews’ paper are related to the $\alpha_{n}$ via
$\alpha_{n}=a^{n}q^{n^{2}}A_{n}$. relative to $(q,q)$ from [2, equation
(5.11)],

$\alpha_{n}=\frac{q^{2n^{2}+n}\left(1-q^{2n+1}\right)}{1-q}\sum_{j=-n}^{n}(-1)^{j}q^{-\frac{j(3j+1)}{2}},\qquad\beta_{n}=1.$

Using this Bailey pair from (3.8) with $(b,c,q)=(-\zeta q,-\zeta^{-1}q,q^{2})$
and employing (3.3), we compute

$\displaystyle\sum_{n\geq 0}\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{n}q^{2n+1}$ $\displaystyle\hskip
5.69046pt=\frac{q\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{\infty}}{\left(q^{2};q^{2}\right)_{\infty}^{2}}\sum_{n\geq
0}\frac{q^{4n^{2}+4n}\left(1-q^{4n+2}\right)}{\left(1+\zeta
q^{2n+1}\right)\left(1+\zeta^{-1}q^{2n+1}\right)}\sum_{j=-n}^{n}(-1)^{j}q^{-j(3j+1)}$
$\displaystyle\hskip 5.69046pt=\frac{q\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{\infty}}{\left(q^{2};q^{2}\right)_{\infty}^{2}}\left(\sum_{n\geq
0}\sum_{j=-n}^{n}\frac{(-1)^{j}q^{4n^{2}+4n-j(3j+1)}}{1+\zeta
q^{2n+1}}-\zeta^{-1}\sum_{n\geq
0}\sum_{j=-n}^{n}\frac{(-1)^{j}q^{4n^{2}+6n+1-j(3j+1)}}{1+\zeta^{-1}q^{2n+1}}\right)$
$\displaystyle\hskip 5.69046pt=\frac{q\left(-\zeta
q,-\zeta^{-1}q;q^{2}\right)_{\infty}}{\left(q^{2};q^{2}\right)_{\infty}^{2}}\left(\sum_{n\geq
0}\sum_{j=-n}^{n}\frac{(-1)^{j}q^{4n^{2}+4n-j(3j+1)}}{1+\zeta
q^{2n+1}}-\sum_{n\leq
0}\sum_{j=n}^{-n}\frac{(-1)^{j}q^{4n^{2}-4n-j(3j+1)}}{1+\zeta
q^{2n-1}}\right).$

Now letting $(n,j)=(\frac{r+s}{2},\frac{r-s}{2})$ in the first sum on the
right-hand side, letting $(n,j)=(\frac{r+s+2}{2},\frac{r-s}{2})$ in the second
sum, and then simplifying gives (1.11). This completes the proof of Theorem
1.2. ∎

##  4\. Proof of Theorem 1.3

In this section, we prove Theorem 1.3.

###### Proof of Theorem 1.3.

From Theorem 1.1 we have that

$\sum_{n\geq 0}\mathrm{ou}(n)q^{n}=\sum_{n\geq
0}\frac{q^{2n+1}}{\left(q;q^{2}\right)_{n+1}^{2}}=\sum_{n\geq
0}(-1)^{n+1}q^{n(3n+2)}\left(1+q^{2n+1}\right)+\frac{1}{\left(q;q^{2}\right)_{\infty}^{2}}\sum_{n\geq
0}(-1)^{n}q^{n^{2}+n}.$ (4.1)

We apply Proposition 2.1. To begin, it is not hard to see that the ${\rm
ou}(n)$ are monotonic, since

$(1-q)\sum_{n\geq 0}{\rm ou}(n)q^{n}=\sum_{n\geq
0}\frac{q^{2n+1}}{(q^{3};q^{2})_{n}(q;q^{2})_{n+1}},$

and the right-hand side has non-negative coefficients. Alternatively, note
that for any $n\in\mathbb{N}$

$\left(a_{1},\dots,a_{r},\overline{c},b_{1},\dots,b_{s}\right)\mapsto\left(1,a_{1},\dots,a_{r},\overline{c},b_{1},\dots,b_{s}\right)$

is an injective mapping from the set of odd unimodal sequences of weight $n$
to the set of odd unimodal sequences of weight $n+1$.

Next, let $F_{2}(q)$ denote the right-hand side of (4.1). We analyze each of
the terms separately with the goal of showing that as $z\to 0$,

$F_{2}\left(e^{-z}\right)\sim\frac{e^{\frac{\pi^{2}}{6z}}}{4}.$

The modularity of the Dedekind $\eta$-function implies that

$\frac{1}{\left(e^{-z};e^{-z}\right)_{\infty}}\sim\sqrt{\frac{z}{2\pi}}e^{\frac{\pi^{2}}{6z}}\qquad\text{as
}z\to 0.$ (4.2)

Thus, with $q=e^{-z}$, we have

$\frac{1}{\left(q;q^{2}\right)_{\infty}^{2}}=\frac{\left(q^{2};q^{2}\right)_{\infty}^{2}}{(q;q)_{\infty}^{2}}\sim\frac{e^{\frac{\pi^{2}}{6z}}}{2}.$
(4.3)

We apply Proposition 2.2 to the two sums in (4.1) . We start by splitting the
first sum according to the parity of the summation variable in order to
rewrite it as

$\sum_{n\geq
0}(-1)^{n+1}q^{n(3n+2)}\left(1+q^{2n+1}\right)=q^{-\frac{1}{3}}\sum_{n\geq
0}\left(q^{12\left(n+\frac{2}{3}\right)^{2}}+q^{12\left(n+\frac{5}{6}\right)^{2}}-q^{12\left(n+\frac{1}{6}\right)^{2}}-q^{12\left(n+\frac{1}{3}\right)^{2}}\right).$

Now we can apply Proposition 2.2 with $f(z):=e^{-12z^{2}}$ and
$a\in\\{\frac{2}{3},\frac{5}{6},\frac{1}{6},\frac{1}{3}\\}$. The main terms
from Proposition 2.2 cancel and using that $q^{-\frac{1}{3}}=O(1)$ we obtain
that the first sum is $O(1)$.

For the second sum we write, using Proposition 2.2

$\displaystyle\sum_{n\geq 0}(-1)^{n}e^{-\left(n^{2}+n\right)z}$
$\displaystyle=\sum_{n\geq 0}e^{-\left(4n^{2}+2n\right)z}-\sum_{n\geq
0}e^{-\left((2n+1)^{2}+(2n+1)\right)z}$
$\displaystyle=e^{\frac{z}{4}}\left(\sum_{n\geq
0}\left(e^{-4\left(n+\frac{1}{4}\right)^{2}z}-e^{-4\left(n+\frac{3}{4}\right)^{2}z}\right)\right)\sim-
B_{1}\left(\tfrac{1}{4}\right)+B_{1}\left(\tfrac{3}{4}\right)=\tfrac{1}{2}.$

Combining with (4.3), Proposition 2.1 with $\lambda=\frac{1}{4}$,
$\alpha=\beta=0$ gives the claim. ∎

##  5\. Proof of Theorem 1.4

In this section, we prove Theorem 1.4. As in the previous section, we wish to
apply Proposition 2.1, though in this case the details are much more involved.
To begin, we record the monotonicity of the sequence $\mathrm{ou}^{*}(n)$.

######  Lemma 5.1.

For $n\geq 3$ we have that ${\rm ou}^{*}(n)\geq{\rm ou}^{*}(n-1)$.

###### Proof.

We give two proofs, one employing $q$-series and one using a combinatorial
argument. For the $q$-series proof, first observe that

$\displaystyle(1-q)\sum_{n\geq 0}\mathrm{ou}^{*}(n)q^{n}$
$\displaystyle=(1-q)\sum_{n\geq
0}\left(-q;q^{2}\right)_{n}^{2}q^{2n+1}=q(1-q)+(1-q)\sum_{n\geq
1}\left(-q;q^{2}\right)_{n}^{2}q^{2n+1}$
$\displaystyle=q(1-q)+q^{3}\left(1-q^{2}\right)(1+q)\sum_{n\geq
0}\left(-q^{3};q^{2}\right)_{n}^{2}q^{2n}.$ (5.1)

We now require the $q$-binomial theorem [18, Exercise 1.2]

$\sum_{m=0}^{n}\frac{(q;q)_{n}w^{n}q^{\frac{m(m-1)}{2}}}{(q;q)_{m}(q;q)_{n-m}}=(-w;q)_{n}$
(5.2)

and a transformation of Jackson [18, Appendix (III.4)],

$\sum_{n\geq
0}\frac{(a,b;q)_{n}w^{n}}{(c,q;q)_{n}}=\frac{(aw;q)_{\infty}}{(w;q)_{\infty}}\sum_{n\geq
0}\frac{\left(a,\frac{c}{b};q\right)_{n}(-bw)^{n}q^{\frac{n(n-1)}{2}}}{(c,aw,q;q)_{n}}.$
(5.3)

Using these, we rewrite the final sum in (5.1) as follows:

$\displaystyle\sum_{n\geq
0}\left(-q^{3};q^{2}\right)_{n}^{2}q^{2n}=\sum_{n\geq
0}\left(-q^{3};q^{2}\right)_{n}q^{2n}\sum_{m=0}^{n}\frac{\left(q^{2};q^{2}\right)_{n}q^{m^{2}+2m}}{\left(q^{2};q^{2}\right)_{m}\left(q^{2};q^{2}\right)_{n-m}}$
$\displaystyle\hskip 17.07182pt=\sum_{m\geq
0}\frac{q^{m^{2}+2m}}{\left(q^{2};q^{2}\right)_{m}}\sum_{n\geq
m}\frac{\left(-q^{3},q^{2};q^{2}\right)_{n}q^{2n}}{\left(q^{2};q^{2}\right)_{n-m}}=\sum_{m\geq
0}q^{m^{2}+4m}\left(-q^{3};q^{2}\right)_{m}\sum_{n\geq
0}\frac{\left(-q^{2m+3},q^{2m+2};q^{2}\right)_{n}q^{2n}}{\left(q^{2};q^{2}\right)_{n}}$
$\displaystyle\hskip 17.07182pt=\sum_{m\geq
0}\frac{q^{m^{2}+4m}\left(-q^{3};q^{2}\right)_{m}}{\left(q^{2};q^{2}\right)_{m}}\sum_{n\geq
0}\frac{q^{n^{2}+4n+2nm}}{\left(q^{2};q^{2}\right)_{n}\left(1-q^{2n+2m+2}\right)}.$
(5.4)

Here the first equality follows from (5.2) and the final equality implied by
(5.3) with $(a,b,c,w,q)=(q^{2m+2},-q^{2m+3},0,q^{2},q^{2})$. Combining (5.1)
and (5.4) gives

$(1-q)\sum_{n\geq
0}\mathrm{ou}^{*}(n)q^{n}=q(1-q)+q^{2}\left(1-q^{2}\right)(1+q)\sum_{n,m\geq
0}\frac{q^{n^{2}+4n+m^{2}+4m+2nm}\left(-q^{3};q^{2}\right)_{m}}{\left(q^{2};q^{2}\right)_{n}\left(q^{2};q^{2}\right)_{m}\left(1-q^{2n+2m+2}\right)}.$

It is straightforward to see that the coefficient of $q^{n}$ on the right-hand
side is non-negative for $n\geq 3$.

Alternatively, one may deduce the montonicity using a combinatorial argument.
For $n\geq 3$ we define a mapping on odd strongly unimodal sequences of weight
$n$ as follows:

$\left(a_{1},\dots,a_{r},\overline{c},b_{1},\dots,b_{s}\right)\mapsto\begin{cases}\left(1,a_{1},\dots,a_{r},\overline{c},b_{1},\dots,b_{s}\right)&\text{if
}a_{1}\neq 1,\\\
\left(a_{2},\dots,a_{r},\overline{c+2},b_{1},\dots,b_{s}\right)&\text{if
}a_{1}=1.\end{cases}$

It is not hard to see that in either case the image is an odd strongly
unimodal sequence of weight $n+1$ and that the mapping is injective. This
gives the desired inequality ${\rm ou}^{*}(n)\geq{\rm ou}^{*}(n-1)$ for $n\geq
4$, and the case $n=3$ follows from the fact that ${\rm ou}^{*}(3)=1$ and
${\rm ou}^{*}(2)=0$. ∎

We now turn to the proof of Theorem 1.4.

###### Proof of Theorem 1.4.

Lemma 5.1 implies that the ${\rm ou}^{*}(n)$ are monotonic. The rest of the
proof is devoted to showing that the remaining conditions of Proposition 2.1
are satisfied. By (1.10)

$\sum_{n\geq 0}{\rm
ou}^{*}(n)q^{n}=\frac{q}{\left(q^{2};q^{2}\right)_{\infty}}\left(\sum_{r,n\geq
0}-\sum_{r,n<0}\right)(-1)^{n}q^{3n^{2}+4nr+r^{2}+3n+2r}.$

Denoting by $F_{1}(q)$ the right-hand side, we aim to prove that

$F_{1}\left(e^{-t}\right)\sim\frac{e^{\frac{\pi^{2}}{12t}}}{2}\quad\text{as
}t\to 0,\qquad F\left(e^{-z}\right)\ll e^{\frac{\pi^{2}}{12|z|}}\quad\text{as
}z\to 0,$ (5.5)

with $z=x+iy$ ($x,y\in\mathbb{R}$, $x>0$, $|y|\leq\Delta x$, $\Delta>0$). We
first consider the outside factor. By (4.2) with $q=e^{-z}$ we have, as $z\to
0$,

$\frac{q}{\left(q^{2};q^{2}\right)_{\infty}}\sim\sqrt{\frac{z}{\pi}}e^{\frac{\pi^{2}}{12z}}.$

Next define

$G(q):=\frac{1}{2}\sum_{n,r\in\mathbb{Z}}\left(\operatorname{sgn}\left(n+\tfrac{1}{2}\right)+\operatorname{sgn}\left(r+\tfrac{1}{2}\right)\right)(-1)^{n}q^{3n^{2}+4nr+r^{2}+3n+2r},$

so $F_{1}(q)=\frac{G(q)}{(q^{2};q^{2})_{\infty}}$. We realize $G$ as
“holomorphic part” of an indefinite theta function. For this set

$g(\tau):=2q^{\frac{3}{4}}G(q)=i\sum_{\bm{n}\in\mathbb{Z}^{2}+\bm{a}}(\operatorname{sgn}(B(\bm{c_{1}},\bm{n}))-\operatorname{sgn}(B(\bm{c_{2}},\bm{n})))e^{2\pi
iB(\bm{b},\bm{n})}q^{Q(\bm{n})},$

where $Q(\bm{n}):=3n_{1}^{2}+4n_{1}n_{2}+n_{2}^{2}$, $\bm{c_{1}}:=(1,-2)^{T}$,
$\bm{c_{2}}:=(2,-3)^{T}$, $\bm{b}:=(-\frac{1}{4},\frac{1}{2})^{T}$, and
$\bm{a}:=(\frac{1}{2},0)^{T}$.

Using (2.2), we may decompose

$g(\tau)=\Theta(\tau)+\Theta^{-}(\tau),$

where

$\displaystyle\Theta(\tau)$
$\displaystyle:=i\sum_{\bm{n}\in\mathbb{Z}^{2}+\bm{a}}\left(E\left(\tfrac{B(\bm{c_{1}},\bm{n})}{\sqrt{-Q(\bm{c_{1}})}}\sqrt{v}\right)-E\left(\tfrac{B(\bm{c_{2}},\bm{n})}{\sqrt{-Q(\bm{c_{2}})}}\sqrt{v}\right)\right)e^{2\pi
iB(\bm{b},\bm{n})}q^{Q(\bm{n})},$ $\displaystyle\Theta^{-}(\tau)$
$\displaystyle:=\sum_{\bm{n}\in\mathbb{Z}^{2}+\bm{a}}\left(\operatorname{sgn}(n_{1})\beta\left(4n_{1}^{2}v\right)+\operatorname{sgn}(n_{2})\beta\left(\tfrac{4n_{2}^{2}v}{3}\right)\right)(-1)^{n_{1}-\frac{1}{2}}q^{3n_{1}^{2}+4n_{1}n_{2}+n_{2}^{2}}.$

In fact the identity holds termwise and we use that

$\displaystyle 3n_{1}^{2}+4n_{1}n_{2}+n_{2}^{2}$
$\displaystyle=Q(\bm{n}),\quad(-1)^{n_{1}-\frac{1}{2}}=-ie^{2\pi
iB(\bm{b},\bm{n})},\quad n_{1}=-\tfrac{1}{2}B(\bm{c_{1}},\bm{n}),\quad
n_{2}=\tfrac{1}{2}B(\bm{c_{2}},\bm{n}),$ $\displaystyle Q(\bm{c_{1}})$
$\displaystyle=-1,\qquad Q(\bm{c_{2}})=-3.$

We determine the asymptotic behavior of $\Theta$ and $\Theta^{-}$ separately:
For $\Theta$ we use modularity and for $\Theta^{-}$ the Euler–Maclaurin
summation formula.

We start with $\Theta$. We have that (in the notation of Subsection 2.2)

$\Theta(\tau)=i\vartheta_{\bm{a},\bm{b}}(\tau).$

We apply Theorem 2.4 to obtain

$\vartheta_{\bm{a},\bm{b}}(\tau)=-\frac{1}{2\tau}\sum_{\bm{\ell}\in\left\\{\bm{0},\left(\frac{1}{2},0\right),\left(0,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2}\right)\right\\}}\vartheta_{\bm{b}+\bm{\ell},-\bm{a}}\left(-\frac{1}{\tau}\right).$

Now write

$\vartheta_{\bm{b}+\bm{\ell},-\bm{a}}(\tau)=-\sum_{\bm{n}\in\mathbb{Z}^{2}+\bm{b}+\bm{\ell}}\left(E\left(2n_{1}\sqrt{v}\right)+E\left(\tfrac{2n_{2}\sqrt{v}}{\sqrt{3}}\right)\right)e^{2\pi
iB(-\bm{a},\bm{n})}q^{Q(\bm{n})}.$ (5.6)

Using that $E(x)\sim\operatorname{sgn}(x)$ as $|x|\to\infty$ the terms in
(5.6) exponentially decay.

We next investigate $\Theta^{-}$ and write it as

$\displaystyle\Theta^{-}(\tau)$
$\displaystyle=\sum_{\bm{n}\in\mathbb{Z}^{2}+\bm{a}}\left(\operatorname{sgn}(n_{1})\beta\left(4n_{1}^{2}v\right)+\operatorname{sgn}(n_{2})\beta\left(\tfrac{4n_{2}^{2}v}{3}\right)\right)(-1)^{n_{1}-\tfrac{1}{2}}q^{3n_{1}^{2}+4n_{1}n_{2}+n_{2}^{2}}$
$\displaystyle=\sum_{\bm{n}\in\mathbb{Z}^{2}}\left(\operatorname{sgn}\left(n_{1}+\tfrac{1}{2}\right)\beta\left(4\left(n_{1}+\tfrac{1}{2}\right)^{2}v\right)+\operatorname{sgn}(n_{2})\beta\left(\tfrac{4n_{2}^{2}v}{3}\right)\right)(-1)^{n_{1}}$
$\displaystyle\hskip 270.30118pt\times
q^{3\left(n_{1}+\tfrac{1}{2}\right)^{2}+4\left(n_{1}+\tfrac{1}{2}\right)n_{2}+n_{2}^{2}}$
$\displaystyle=\sum_{\begin{subarray}{c}n_{1}\geq 0\\\ n_{2}\geq
1\end{subarray}}(-1)^{n_{1}}\left(\beta\left(4\left(n_{1}+\tfrac{1}{2}\right)^{2}v\right)+\beta\left(\tfrac{4}{3}n_{2}^{2}v\right)\right)q^{3\left(n_{1}+\tfrac{1}{2}\right)^{2}+4\left(n_{1}+\tfrac{1}{2}\right)n_{2}+n_{2}^{2}}$
$\displaystyle+\sum_{\begin{subarray}{c}n_{1}\geq 0\\\ n_{2}\geq
1\end{subarray}}(-1)^{-n_{1}-1}\left(-\beta\left(4\left(-n_{1}-1+\tfrac{1}{2}\right)^{2}v\right)+\beta\left(\tfrac{4}{3}n_{2}^{2}v\right)\right)q^{3\left(-n_{1}-1+\tfrac{1}{2}\right)^{2}+4\left(-n_{1}-1+\tfrac{1}{2}\right)n_{2}+n_{2}^{2}}$
$\displaystyle+\sum_{\begin{subarray}{c}n_{1}\geq 0\\\ n_{2}\geq
1\end{subarray}}(-1)^{n_{1}}\left(\beta\left(4\left(n_{1}+\tfrac{1}{2}\right)^{2}v\right)-\beta\left(\tfrac{4}{3}(-n_{2})^{2}v\right)\right)q^{3\left(n_{1}+\tfrac{1}{2}\right)^{2}+4\left(n_{1}+\tfrac{1}{2}\right)(-n_{2})+(-n_{2})^{2}}$
$\displaystyle+\sum_{n_{1}\geq
0}(-1)^{-n_{1}-1}\left(-\beta\left(4\left(-n_{1}-1+\tfrac{1}{2}\right)^{2}v\right)-\beta\left(\tfrac{4}{3}(-n_{2})^{2}v\right)\right)$
$\displaystyle\hskip 199.16928pt\times
q^{3\left(-n_{1}-1+\tfrac{1}{2}\right)^{2}+4\left(-n_{1}-1+\tfrac{1}{2}\right)(-n_{2})+(-n_{2})^{2}}$
$\displaystyle=2\sum_{\begin{subarray}{c}n_{1}\geq 0\\\ n_{2}\geq
1\end{subarray}}(-1)^{n_{1}}\left(\beta\left(4\left(n_{1}+\tfrac{1}{2}\right)^{2}v\right)+\beta\left(\tfrac{4n_{2}^{2}v}{3}\right)\right)q^{3\left(n_{1}+\tfrac{1}{2}\right)^{2}+4\left(n_{1}+\tfrac{1}{2}\right)n_{2}+n_{2}^{2}}$
$\displaystyle+2\sum_{\begin{subarray}{c}n_{1}\geq 0\\\ n_{2}\geq
1\end{subarray}}(-1)^{n_{1}}\left(\beta\left(4\left(n_{1}+\tfrac{1}{2}\right)^{2}v\right)-\beta\left(\tfrac{4n_{2}^{2}v}{3}\right)\right)q^{3\left(n_{1}+\tfrac{1}{2}\right)^{2}-4\left(n_{1}+\tfrac{1}{2}\right)n_{2}+n_{2}^{2}}$
$\displaystyle=2\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{1},n_{2}\geq
0}\left(\beta\left(16\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4}\right)^{2}v\right)\pm\beta\left(\tfrac{4}{3}(n_{2}+1)^{2}v\right)\right)$
$\displaystyle\hskip 199.16928pt\times
q^{12\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4}\right)^{2}\pm
8\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4}\right)(n_{2}+1)+(n_{2}+1)^{2}}.$

We now first show the first asymptotic in (5.5). For this, let
$\tau=\frac{it}{2\pi}$. Then

$\Theta^{-}\left(\tfrac{it}{2\pi}\right)=2\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{1},n_{2}\geq
0}f_{\pm}\left(\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4},n_{2}+1\right)\sqrt{t}\right),$

where

$f_{\pm}(x_{1},x_{2}):=\left(\beta\left(\tfrac{8x_{1}^{2}}{\pi}\right)\pm\beta\left(\tfrac{2x_{2}^{2}}{3\pi}\right)\right)e^{-12x_{1}^{2}\mp
8x_{1}x_{2}-x_{2}^{2}}.$

We now use Proposition 2.3. The term with the double integral term vanishes
(the two $\delta$-terms cancel). The second term contributes

$-\frac{2}{\sqrt{t}}\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{1}=0}^{N-1}\frac{B_{n_{1}+1}\left(\frac{\delta}{2}+\frac{1}{4}\right)t^{\frac{n_{1}}{2}}}{(n_{1}+1)!}\int_{0}^{\infty}f_{\pm}^{(n_{1},0)}(0,w_{2})dw_{2}.$

By combining the term for $\delta=0$ and $\delta=1$, using properties of
Bernoulli polynomials it is not hard to see that only $n_{1}$ even survive.
The terms from $n_{1}\geq 1$ yield a contribution overall that is $O(t)$.
Using that $\beta(0)=1$, the term $n_{1}=0$ gives

$-\frac{4}{\sqrt{t}}\sum_{\pm}B_{1}\left(\tfrac{1}{4}\right)\int_{0}^{\infty}\left(1\pm\beta\left(\tfrac{2x_{2}^{2}}{3\pi}\right)\right)e^{-x_{2}^{2}}dx_{2}=\frac{2}{\sqrt{t}}\int_{0}^{\infty}e^{-w^{2}_{2}}dw_{2}=\sqrt{\tfrac{\pi}{t}}.$

For the third term, we have

$-\frac{2}{\sqrt{t}}\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{2}=0}^{N-1}\frac{B_{n_{2}+1}(1)}{(n_{2}+1)!}t^{\frac{n_{2}}{2}}\int_{0}^{\infty}f_{\pm}^{(0,n_{2})}(w_{1},0)dw_{1}=0,$

because the $\delta$-terms cancel. The final term in Proposition 2.3 is in
$O(t)$. This gives that the first asymptotic in (5.5) holds.

We next need to show that the second asymptotic in (5.5) holds. For this, we
need to prove that

$\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{1},n_{2}\geq
0}\left(\beta\left(\tfrac{8}{\pi}\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4}\right)^{2}x\right)\pm\beta\left(\tfrac{2}{3\pi}(n_{2}+1)^{2}x\right)\right)\\\
\times e^{-\left(12\left(n_{1}+\frac{\delta}{2}+\frac{1}{4}\right)^{2}\pm
8\left(n_{1}+\frac{\delta}{2}+\frac{1}{4}\right)(n_{2}+1)+(n_{2}+1)^{2}\right)z}\ll|z|^{\frac{1}{2}}.$
(5.7)

The proof follows by a lengthy calculation from the following refinement of
Proposition 2.3 in the one-dimensional case, namely (see (5.8) of [10])

$\sum_{n\geq
0}f((n+a)z)=\frac{1}{z}\int_{0}^{\infty}f(w)dw-\sum_{n=0}^{N-1}\frac{B_{n+1}(a)f^{(n)}(0)}{(n+1)!}z^{n}+\mathcal{E}(a;z),$
(5.8)

where ($C_{R}(0)$ is the circle around $0$ with radius $R$,
$\widetilde{B}_{N}(x):=B_{N}(x-\lfloor x\rfloor)$)

$\mathcal{E}(a;z):=-\sum_{k\geq
N}\frac{f^{(k)}(0)a^{k+1}}{(k+1)!}z^{k}-\frac{z^{N}}{2\pi
i}\sum_{n=0}^{N-1}\frac{B_{n+1}(0)a^{N-n}}{(n+1)!}\int_{C_{R}(0)}\frac{f^{(n)}(w)}{w^{N-n}(w-az)}dw\\\
-\frac{(-1)^{N}z^{N-1}}{N!}\int_{az}^{z\infty}f^{(N)}(w)\widetilde{B}_{N}\left(\frac{w}{z}-a\right)dw.$

For the reader’s convenience we defer the full proof of (5.7) to Appendix A.

Combining and using Proposition 2.1 with $\gamma=\frac{\pi^{2}}{12}$,
$\beta=0$, and $\lambda=\frac{1}{2}$ gives the claim. ∎

##  6\. Congruences for $\mathrm{ou}^{*}(n)$ modulo $4$ and the proof of
Theorem 1.5

In this section we prove Theorem 6.2 below. Note that this reduces to Theorem
1.5 for $k=0$. First, we determine the parity of ${\rm ou}^{*}(n)$.

######  Proposition 6.1.

For $n\in\mathbb{N}$, we have that $\mathrm{ou}^{*}(n)$ is odd if and only if
$6n-2$ is a square.

###### Proof.

We require a classical $q$-series identity from Ramanujan’s lost notebook [5,
Entry 9.5.2],

$\sum_{n\geq 0}\left(q;q^{2}\right)_{n}q^{n}=\sum_{n\geq
0}(-1)^{n}q^{3n^{2}+2n}\left(1+q^{2n+1}\right).$

Using this along with (1.8), we have

$\displaystyle\sum_{n\geq 0}\mathrm{ou}^{*}(n)q^{n}$
$\displaystyle=\sum_{n\geq
0}\left(-q;q^{2}\right)_{n}^{2}q^{2n+1}\equiv\sum_{n\geq
0}\left(q^{2};q^{4}\right)_{n}q^{2n+1}=\sum_{n\geq
0}(-1)^{n}q^{6n^{2}+4n+1}\left(1+q^{4n+2}\right)$
$\displaystyle\equiv\sum_{n\in\mathbb{Z}}q^{6n^{2}+4n+1}\ (\mathrm{mod}\,2).$

Now in the extremes of this string of equations we replace $q$ by $q^{6}$ and
multiply by $q^{-2}$ to obtain

$\sum_{n\geq
0}\mathrm{ou}^{*}(n)q^{6n-2}\equiv\sum_{n\in\mathbb{Z}}q^{(6n+2)^{2}}\
(\mathrm{mod}\,2),$

and the result follows. ∎

We now state the main result of this section.

######  Theorem 6.2.

Let $k\in\mathbb{N}$ and for $r$ with $1\leq r\leq k+1$, let $p_{r}\geq 5$ be
prime. For any $j\not\equiv 0\ (\mathrm{mod}\,p_{k+1})$, if $p_{k+1}\not\equiv
7,13\ (\mathrm{mod}\,24)$ or $p_{k+1}\equiv 7,13\ (\mathrm{mod}\,24)$ and
$(\frac{3j}{p_{k+1}})=-1$, then we have:

  1. (1)

If $j$ is odd, then we have

${\rm ou}^{*}\left(4p_{1}^{2}\cdots p_{k+1}^{2}n+2p_{1}^{2}\cdots
p_{k}^{2}p_{k+1}j+\tfrac{8p_{1}^{2}\cdots p_{k+1}^{2}+1}{3}\right)\equiv 0\
\left(\mathrm{mod}\,4\right).$

  2. (2)

If $j$ is even, then we have

${\rm ou}^{*}\left(4p_{1}^{2}\cdots p_{k+1}^{2}n+2p_{1}^{2}\cdots
p_{k}^{2}p_{k+1}j+\tfrac{2p_{1}^{2}\cdots p_{k+1}^{2}+1}{3}\right)\equiv 0\
\left(\mathrm{mod}\,4\right).$

To make the proof smoother, we first prove a simple lemma.

######  Lemma 6.3.

For $n\in\mathbb{N}$ we have that modulo $4$,

$\left(-q;q^{2}\right)_{n}^{2}-\left(-q^{2};q^{4}\right)_{n}$

is an odd polynomial.

###### Proof.

We prove the claim by induction. The case $n=1$ is clear.

Now assume that the claim holds for $n\in\mathbb{N}$. Then

$\left(-q;q^{2}\right)^{2}_{n+1}-\left(-q^{2};q^{4}\right)_{n+1}=\left(1+q^{4n+2}\right)\left(\left(-q;q^{2}\right)^{2}_{n}-\left(-q^{2};q^{4}\right)_{n}\right)+2q^{2n+1}\left(-q;q^{2}\right)^{2}_{n}.$

By the induction assumption the first term is an odd polynomial $\
\left(\mathrm{mod}\,4\right)$. Thus we are left to show that
$(-q;q^{2})^{2}_{n}$ is an even polynomial $\ \left(\mathrm{mod}\,2\right)$.
Now we have the even polynomial

$\left(-q;q^{2}\right)^{2}_{n}\equiv\left(-q^{2};q^{4}\right)_{n}\
(\mathrm{mod}\,2).\qed$

Next we use a result of Chen and Chen [15], who employed the theory of class
numbers to prove congruences modulo $4$ for $\mathcal{E}\mathcal{O}(n)$,
defined via the infinite product

$\sum_{n\geq
0}\mathcal{E}\mathcal{O}(n)q^{n}:=\frac{\left(q^{4};q^{4}\right)_{\infty}^{3}}{\left(q^{2};q^{2}\right)_{\infty}^{2}}.$
(6.1)

######  Theorem 6.4 ([15]).

Let $k\in\mathbb{N}_{0}$. For $i$ with $1\leq i\leq k+1$, let $p_{i}\geq 5$ be
prime. For $j\not\equiv 0\ (\mathrm{mod}\,p_{k+1})$, if $p_{k+1}\not\equiv
7,13\ (\mathrm{mod}\,24)$ or $p_{k+1}\equiv 7,13\ (\mathrm{mod}\,24)$ and
$(\frac{3j}{p_{k+1}})=-1$, then for $n\in\mathbb{N}_{0}$

$\mathcal{E}\mathcal{O}\left(p_{1}^{2}\cdots p_{k+1}^{2}n+p_{1}^{2}\cdots
p_{k}^{2}p_{k+1}j+\tfrac{p_{1}^{2}\cdots p_{k+1}^{2}-1}{3}\right)\equiv 0\
(\mathrm{mod}\,4).$ (6.2)

We are now ready to prove Theorem 6.2.

###### Proof of Theorem 6.2.

First recall the third order mock theta function,

$\nu(q):=\sum_{n\geq 0}\left(q;q^{2}\right)_{n}(-q)^{n}=:\sum_{n\geq
0}c(n)q^{n}.$

Using Lemma 6.3 we have that

$\sum_{n\geq 0}\mathrm{ou}^{*}(n)q^{n}-\sum_{n\geq
0}(-1)^{n}c(n)q^{2n+1}=\sum_{n\geq
0}\left(-q;q^{2}\right)_{n}^{2}q^{2n+1}-\sum_{n\geq
0}\left(-q^{2};q^{4}\right)_{n}q^{2n+1}$

is supported on even exponents of $q$ modulo $4$. Therefore

${\rm ou}^{*}(2n+1)\equiv(-1)^{n}c(n)\ \left(\mathrm{mod}\,4\right).$ (6.3)

Next we note that by [17, equation (26.88)] we have

$\sum_{n\geq
0}c(2n)q^{2n}=\frac{\left(q^{4};q^{4}\right)_{\infty}^{3}}{\left(q^{2};q^{2}\right)_{\infty}^{2}}.$

This is the same product as in (6.1), and so $c(n)=\mathcal{E}\mathcal{O}(n)$
if $n$ is even. Therefore the congruences for $\mathcal{E}\mathcal{O}(n)$ in
Theorem 6.4 imply congruences for $c(n)$. The argument on the left-hand side
of (6.2) is even if and only if $n\equiv j\ (\mathrm{mod}\,2)$. So for $j$
even, we let $n\mapsto 2n$ and for $j$ odd, we let $n\mapsto 2n+1$ to obtain:

  1. (1)

If $j$ is odd, then

$c\left(2p_{1}^{2}\cdots p_{k+1}^{2}n+p_{1}^{2}\cdots
p_{k}^{2}p_{k+1}j+\tfrac{4p_{1}^{2}\cdots p_{k+1}^{2}-1}{3}\right)\equiv 0\
(\mathrm{mod}\,4).$

  2. (2)

If $j$ is even, then

$c\left(2p_{1}^{2}\cdots p_{k+1}^{2}n+p_{1}^{2}\cdots
p_{k}^{2}p_{k+1}j+\tfrac{p_{1}^{2}\cdots p_{k+1}^{2}-1}{3}\right)\equiv 0\
(\mathrm{mod}\,4).$

By (6.3) the theorem follows. ∎

##  7\. Conclusion

This paper contains a preliminary investigation of odd unimodal sequences,
establishing generating functions, basic asymptotics, and some congruence
properties modulo $4$. While other variants of unimodal sequences have arisen
in the literature (e.g. [8, 9, 21]), odd unimodal sequences are perhaps the
most natural. Below we leave two ideas for further study. The motivated reader
will surely find many more.

First, improve upon the asymptotics in Theorems 1.3 and 1.4 to find
Rademacher-type formulas for ${\rm ou}(n)$ and ${\rm ou}^{*}(n)$. Up to simple
terms the generating function for $\mathrm{ou}(n)$ and is a mixed false theta
function while the generating function for $\mathrm{ou}^{*}(n)$ is a mixed
mock theta function. In both cases the weight is $\frac{1}{2}$. One could now
use the Circle Method to deduce asymptotic formulas for $\mathrm{ou}(n)$ and
$\mathrm{ou}^{*}(n)$. However, an exact formula is out of reach with this
method because the weight is too large. To obtain an exact formula one would
need to find new methods (like Poincaré type series).

Second, it appears that the arithmetic progressions in some of the congruences
in Theorem 6.2 can be enlarged. For example, with $k=0$ (i.e., in the case of
Theorem 1.5) the congruences corresponding to the primes $5,7$, and $11$ are

$\displaystyle{\rm ou}^{*}(100n+r)$ $\displaystyle\equiv 0\
\left(\mathrm{mod}\,4\right)\text{for $r\in\\{37,57,77,97\\}$},$ (7.1)
$\displaystyle{\rm ou}^{*}(196n+r)$ $\displaystyle\equiv 0\
\left(\mathrm{mod}\,4\right)\text{for $r\in\\{61,89,145\\}$},$ (7.2)
$\displaystyle{\rm ou}^{*}(484n+r)$ $\displaystyle\equiv 0\
\left(\mathrm{mod}\,4\right)\text{for
$r\in\\{125,169,213,257,301,345,389,433,477,521\\}$}.$ (7.3)

Computations suggest that the cases $r\in\\{37,97\\}$ of (7.1) are special
cases of the congruences

${\rm ou}^{*}(50n+r)\equiv 0\ \left(\mathrm{mod}\,4\right)\text{for
$r\in\\{37,47\\}$},$

the cases $r\in\\{61,145\\}$ of (7.2) are special cases of the congruences

${\rm ou}^{*}(98n+r)\equiv 0\ \left(\mathrm{mod}\,4\right)\text{for
$r\in\\{47,61\\}$},$

and all of the congruences in (7.3) are special cases of congruences in the
corresponding arithmetic progressions modulo $242$. We leave it as an open
problem to establish exactly which of the congruences in Theorem 6.2 (or
Theorem 1.5) can be strengthened in this way.

## Appendix A

Here we prove the second bound from (5.7). We require the relation
$\beta(x)=\operatorname{erfc}(\sqrt{\pi x})$ and use the bound
$\operatorname{erfc}(x)\ll 1$. To use (5.8), we write

$\displaystyle\Theta^{-}\left(\tfrac{iz}{2\pi}\right)=$
$\displaystyle-2\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}$
$\displaystyle\left(\sum_{n_{2}\geq 0}e^{-(n_{2}+1)^{2}z}\sum_{n_{1}\geq
0}\beta\left(\tfrac{8}{\pi}\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4}\right)^{2}x\right)e^{-12\left(n_{1}+\frac{\delta}{2}+\frac{1}{4}\right)^{2}z}e^{\mp
8\left(n_{1}+\frac{\delta}{2}+\frac{1}{4}\right)(n_{2}+1)z}\right.$
$\displaystyle\qquad\left.\pm\sum_{n_{2}\geq
0}\beta\left(\tfrac{2}{3\pi}(n_{2}+1)^{2}x\right)e^{-(n_{2}+1)^{2}z}\sum_{n_{1}\geq
0}e^{-12\left(n_{1}+\frac{\delta}{2}+\frac{1}{4}\right)^{2}z\mp
8\left(n_{1}+\frac{\delta}{2}+\frac{1}{4}\right)(n_{2}+1)z}\right).$ (A.1)

We write the first term as

$-2\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{2}\geq
0}e^{-(n_{2}+1)^{2}z}\sum_{n_{1}\geq
0}H_{n_{2},\pm}\left(\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4}\right)\sqrt{x}\right),$

where

$H_{n_{2},\pm}(w_{1}):=\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12\frac{z}{x}w_{1}^{2}\mp
8(n_{2}+1)\frac{z}{\sqrt{x}}w_{1}}.$

Now (5.8) with $N=0$, gives that

$\sum_{n_{1}\geq
0}H_{n_{2},\pm}\left(\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4}\right)\sqrt{x}\right)=\frac{1}{\sqrt{x}}\int_{0}^{\infty}H_{n_{2},\pm}(w_{1})dw_{1}+\mathcal{E}\left(\tfrac{\delta}{2}+\tfrac{1}{4};x\right),$

where

$\mathcal{E}(a;x):=-\sum_{k_{1}\geq
0}\frac{H_{n_{2},\pm}^{(k_{1})}(0)}{(k_{1}+1)!}a^{k_{1}+1}x^{\frac{k_{1}}{2}}-\frac{1}{\sqrt{x}}\int_{a\sqrt{x}}^{\infty}H_{n_{2},\pm}(w_{1})\widetilde{B}_{0}\left(\tfrac{w_{1}}{\sqrt{x}}-a\right)dw_{1}.$
(A.2)

Note that $\widetilde{B}_{0}(x)=1$. The contribution from the main term
vanishes (because of the $(-1)^{\delta}$).

The first term in the error contributes

$\displaystyle
2\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{2}\geq
0}e^{-(n_{2}+1)^{2}z}\sum_{k_{1}\geq 0}\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}}H_{n_{2},\pm}(w_{1})\right]_{w_{1}=0}\frac{\left(\frac{\delta}{2}+\frac{1}{4}\right)}{(k_{1}+1)!}^{k_{1}+1}x^{\frac{k_{1}}{2}}$
$\displaystyle=2\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{k_{1}\geq
0}\frac{\left(\frac{\delta}{2}+\frac{1}{4}\right)}{(k_{1}+1)!}^{k_{1}+1}x^{\frac{k_{1}}{2}}\sum_{\begin{subarray}{c}0\leq\ell_{1},\ell_{2}\leq
k_{1}\\\
\ell_{1}+\ell_{2}=k_{1}\end{subarray}}\binom{k_{1}}{\ell_{1}}\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{\ell_{1}}\left(\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12\frac{z}{x}w_{1}^{2}}\right)\right]_{w_{1}=0}$
$\displaystyle\hskip 250.38434pt\times\sum_{n_{2}\geq 0}\left(\mp
8(n_{2}+1)\frac{z}{\sqrt{x}}\right)^{\ell_{2}}e^{-(n_{2}+1)^{2}z}.$ (A.3)

Now only $\ell_{2}$ even survive (otherwise the $\pm$ cancels). We now
determine the asymptotic behaviors of ($\ell_{2}$ even)

$\sum_{n_{2}\geq
0}(n_{2}+1)^{\ell_{2}}e^{-(n_{2}+1)^{2}z}=(-1)^{\frac{\ell_{2}}{2}}\left(\frac{\partial}{\partial
z}\right)^{\frac{\ell_{2}}{2}}\sum_{n\geq 1}e^{-n^{2}z}.$

For this recall the modular theta function

$\vartheta(\tau):=\sum_{n\in\mathbb{Z}}e^{\pi in^{2}\tau}.$

It satisfies

$\vartheta(\tau)=(-i\tau)^{-\tfrac{1}{2}}\vartheta\left(-\tfrac{1}{\tau}\right).$

Thus

$\sum_{n\geq
1}e^{-n^{2}z}=\tfrac{1}{2}\sum_{n\in\mathbb{Z}}e^{-n^{2}z}-\tfrac{1}{2}=\tfrac{1}{2}\vartheta\left(\tfrac{iz}{\pi}\right)-\tfrac{1}{2}=\tfrac{1}{2}\left(\tfrac{z}{\pi}\right)^{-\tfrac{1}{2}}\vartheta\left(\tfrac{\pi
i}{z}\right)-\tfrac{1}{2}.$ (A.4)

The second term contributes to (A.3) (it only survives if $\ell_{2}=0$)

$-\tfrac{1}{2}\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{k_{1}\geq
0}\frac{\left(\frac{\delta}{2}+\frac{1}{4}\right)^{k_{1}+1}}{(k_{1}+1)!}x^{\frac{k_{1}}{2}}\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}}\left(\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12\frac{z}{x}w_{1}^{2}}\right)\right]_{w_{1}=0}\\\
\ll\sum_{k_{1}\geq
0}\frac{x^{\frac{k_{1}}{2}}}{(k_{1}+1)!}\left(1+\left|\frac{z}{x}\right|^{k_{1}}\right)\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}}\left(\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12w_{1}^{2}}\right)\right]_{w_{1}=0}\ll
1.$

The first term from (A.4) contributes to (A.3) (noting that $\ell$, $k_{1}$
need to be even)

$\ll\sum_{k_{1}\geq 0}\frac{x^{{k_{1}}}}{(2k_{1}+1)!}\sum_{0\leq\ell\leq
k_{1}}\binom{2k_{1}}{2\ell}\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{2k_{1}-2\ell}\left(\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12\frac{z}{x}w_{1}^{2}}\right)\right]_{w_{1}=0}\left(\tfrac{8z}{\sqrt{x}}\right)^{2\ell}\frac{\partial^{2\ell}}{\partial
z^{2\ell}}\frac{\vartheta\left(\frac{\pi i}{z}\right)}{z^{\frac{1}{2}}}.$
(A.5)

Now assume that $\frac{x}{|z|^{2}}>1$ (which is true as $x\to 0$). Then

$z^{2\ell}\frac{\partial^{2\ell}}{\partial
z^{2\ell}}\frac{\vartheta\left(\frac{\pi
i}{z}\right)}{z^{\frac{1}{2}}}\ll\left[\frac{\partial^{2\ell}}{\partial
w_{2}^{2\ell}}\frac{\vartheta\left(\frac{\pi
i}{w_{2}}\right)}{{w_{2}}^{\frac{1}{2}}}\right]_{w_{2}=1}.$

Moreover, as above,

$\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{2k_{1}-2\ell}\left(\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12\frac{z}{x}w_{1}^{2}}\right)\right]_{w_{1}=0}\ll\left(1+\Delta^{2k_{1}-2\ell}\right)\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{2k_{1}-2\ell}\left(\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12w_{1}^{2}}\right)\right]_{w_{1}=0}.$

This yields that (A.5) can be bound against $O(1)$. Thus the first term in
(A.2) is overall $O(1)$.

The second term of (A.2) contributes

$\displaystyle\tfrac{2}{\sqrt{x}}\sum_{\pm}\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{2}\geq
0}e^{-(n_{2}+1)^{2}z}\int_{\left(\frac{\delta}{2}+\frac{1}{4}\right)\sqrt{x}}^{\infty}H_{n_{2},\pm}(w_{1})dw_{1}$
$\displaystyle\hskip 85.35826pt=\tfrac{2}{\sqrt{x}}\sum_{\pm}\sum_{n_{2}\geq
0}e^{-(n_{2}+1)^{2}z}\left(\int_{\frac{\sqrt{x}}{4}}^{\infty}-\int_{\frac{3\sqrt{x}}{4}}^{\infty}\right)H_{n_{2},\pm}(w_{1})dw_{1}$
$\displaystyle\hskip
85.35826pt=\tfrac{2}{\sqrt{x}}\sum_{\pm}\int_{\frac{\sqrt{x}}{4}}^{\frac{3\sqrt{x}}{4}}\beta\left(\tfrac{8}{\pi}w_{1}^{2}\right)e^{-12\frac{z}{x}w_{1}^{2}}\sum_{n_{2}\geq
0}e^{-(n_{2}+1)^{2}z\mp 8(n_{2}+1)\frac{z}{\sqrt{x}}w_{1}}dw_{1}.$ (A.6)

The sum on $n_{2}$ may be written as

$\sum_{n_{2}\geq 0}h_{[1]}\left((n_{2}+1)\sqrt{x}\right),$

where $h_{[1]}(w_{2}):=e^{-\frac{z}{x}w_{2}^{2}\mp 8\frac{z}{x}w_{1}w_{2}}$.
Using (5.8), with $N=0$, we have

$\sum_{n_{2}\geq
0}h_{[1]}\left((n_{2}+1)\sqrt{x}\right)=\frac{1}{\sqrt{x}}\int_{0}^{\infty}h_{[1]}(w_{2})dw_{2}+\mathcal{E}^{[1]}(x),$

where

$\mathcal{E}^{[1]}(x):=-\sum_{k_{2}\geq
0}\frac{h_{[1]}^{(k_{2})}(0)}{(k_{2}+1)!}x^{\frac{k_{2}}{2}}-\frac{1}{\sqrt{x}}\int_{\sqrt{x}}^{\infty}h_{[1]}(w_{2})dw_{2}.$

The main term contributes to (A.6) as

$\tfrac{2}{x}\sum_{\pm}\int_{\frac{\sqrt{x}}{4}}^{\frac{3\sqrt{x}}{4}}\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12\frac{z}{x}w_{1}^{2}}\int_{0}^{\infty}h_{[1]}(w_{2})dw_{2}dw_{1}=\tfrac{2}{x}\sum_{\pm}\int_{\frac{\sqrt{x}}{4}}^{\frac{3\sqrt{x}}{4}}\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{4\frac{z}{x}w_{1}^{2}}\int_{0}^{\infty}e^{-\frac{z}{x}(w_{2}\pm
4w_{1})^{2}}dw_{2}.$

Using that
$\int_{4w_{1}}^{\infty}+\int_{-4w_{1}}^{\infty}=2\int_{0}^{\infty}$, the above
is in $O(\frac{1}{\sqrt{z}})$.

We next consider the first term in $\mathcal{E}^{[1]}(x)$ which contributes

$-\tfrac{2}{\sqrt{x}}\sum_{\pm}\int_{\frac{\sqrt{x}}{4}}^{\frac{3\sqrt{x}}{4}}\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12\frac{z}{x}w_{1}^{2}}\sum_{k_{2}\geq
0}\left[\tfrac{\partial^{k^{2}}}{\partial
w_{2}^{k^{2}}}e^{-\frac{z}{x}w_{2}^{2}\mp
8\frac{z}{x}w_{1}w_{2}}\right]_{w_{2}=0}dw_{1}\frac{x^{\frac{k_{2}}{2}}}{(k_{2}+1)!}.$

We have

$\left[\frac{\partial^{k_{2}}}{\partial
w_{2}^{k_{2}}}e^{-\frac{z}{x}w_{2}^{2}\mp
8\frac{z}{x}w_{1}w_{2}}\right]_{w_{2}=0}=\sum_{\ell=0}^{k}\binom{k_{2}}{\ell}\left(\mp
8\frac{z}{x}w_{1}\right)^{\ell}\left(\frac{\sqrt{z}}{\sqrt{x}}\right)^{k_{2}-\ell}\left[\frac{\partial^{k_{2}-\ell}}{\partial
w_{2}^{k_{2}-\ell}}e^{-w_{2}^{2}}\right]_{w_{2}=0}.$

Now the $\pm$ enforces $\ell$, $k_{2}-\ell$ to be even because of the $\pm$
and bound overall against $O(\frac{1}{\sqrt{z}})$.

We next consider the second term in $\mathcal{E}^{[1]}(x)$. This contributes

$\tfrac{1}{x}\int_{\frac{\sqrt{x}}{4}}^{\frac{3\sqrt{x}}{4}}\beta\left(\tfrac{8w_{1}^{2}}{\pi}\right)e^{-12\frac{z}{x}w_{1}^{2}}\int_{\sqrt{x}}^{\infty}e^{-\frac{z}{x}w_{2}^{2}\mp
8\frac{z}{x}w_{1}w_{2}}dw_{2}dw_{1}.$ (A.7)

We write

$e^{-12\frac{z}{x}w_{1}^{2}}\int_{\sqrt{x}}^{\infty}e^{-\frac{z}{x}w_{2}^{2}\mp
8\frac{z}{x}w_{1}w_{2}}dw_{2}\ll
e^{4\frac{z}{x}w_{1}^{2}}\int_{\sqrt{x}}^{\infty}e^{-\frac{z}{x}(w_{2}\pm
4w_{1})^{2}}dw_{2}\ll e^{4w_{1}^{2}}\int_{\mathbb{R}}e^{-w_{2}^{2}}dw_{2}.$

Thus (A.7) may be bound against $O(\frac{1}{\sqrt{z}})$. Combining gives that
the first term is $O(\frac{1}{\sqrt{z}})$.

We next turn to the second term in (A.1) and proceed as before, again first
considering the sum on $n_{1}$. We have

$\sum_{n_{1}\geq
0}e^{-12\left(n_{1}+\frac{\delta}{2}+\frac{1}{4}\right)^{2}z\mp
8\left(n_{1}+\frac{\delta}{2}+\frac{1}{4}\right)(n_{2}+1)z}=\sum_{n_{1}\geq
0}G_{n_{2},\pm}\left(\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4}\right)\sqrt{z}\right),$

where $G_{n_{2},\pm}(w_{1}):=e^{-12w_{1}^{2}\mp 8(n_{2}+1)\sqrt{z}w_{1}}$. Now
(5.8) gives (with $N=0$) that

$\sum_{n_{1}\geq
0}G_{n_{2},\pm}\left(\left(n_{1}+\tfrac{\delta}{2}+\tfrac{1}{4}\right)\sqrt{z}\right)=\frac{1}{\sqrt{z}}\int_{0}^{\infty}G_{n_{2},\pm}(w_{1})dw_{1}+E\left(\tfrac{\delta}{2}+\tfrac{1}{4};z\right),$

where

$E(a;z):=-\sum_{k_{1}\geq
0}\frac{G_{n_{2},\pm}^{(k_{1})}(0)}{(k_{1}+1)!}a^{k_{1}+1}z^{\frac{k_{1}}{2}}-\frac{1}{\sqrt{z}}\int_{a\sqrt{z}}^{\sqrt{z}\infty}G_{n_{2},\pm}(w_{1})\widetilde{B}_{0}\left(\tfrac{w_{1}}{\sqrt{z}}-a\right)dw_{1}.$

The main term contributes overall

$-\tfrac{2}{\sqrt{z}}\sum_{\pm}\pm\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{2}\geq
0}\beta\left(\tfrac{2}{3\pi}(n_{2}+1)^{2}x\right)e^{-(n_{2}+1)^{2}z}\int_{0}^{\infty}G_{n_{2},\pm}(w_{1})dw_{1}=0.$

The first term in the error $E(\frac{\delta}{2}+\frac{1}{4};z)$ contributes

$2\sum_{\pm}\pm\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{2}\geq
0}\beta\left(\tfrac{2}{3\pi}(n_{2}+1)^{2}x\right)e^{-(n_{2}+1)^{2}z}\sum_{k_{1}\geq
0}\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}}G_{n_{2},\pm}(w_{1})\right]_{w_{1}=0}\\\
\times\frac{\left(\frac{\delta}{2}+\frac{1}{4}\right)^{k_{1}+1}}{(k_{1}+1)!}z^{\frac{k_{1}}{2}}=2\sum_{\pm}\pm\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{k_{1}\geq
0}\left(\tfrac{\delta}{2}+\tfrac{1}{4}\right)^{k_{1}+1}\frac{z^{\frac{k_{1}}{2}}}{(k_{1}+1)!}\\\
\times\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}}\left(e^{-12w_{1}^{2}}\sum_{n_{2}\geq
0}\beta\left(\tfrac{2}{3\pi}(n_{2}+1)^{2}x\right)e^{-(n_{2}+1)^{2}z\mp
8(n_{2}+1)\sqrt{z}w_{1}}\right)\right]_{w_{1}=0}.$

We write the sum on $n_{2}$ as

$\sum_{n_{2}\geq 0}g_{[1],\pm}\left((n_{2}+1)\sqrt{x}\right),$

where
$g_{[1],\pm}(w_{2}):=\beta(\tfrac{2w_{2}^{2}}{3\pi})e^{-\frac{z}{x}w_{2}^{2}\mp
8\frac{\sqrt{z}}{\sqrt{x}}w_{1}w_{2}}$. Now (5.8), with $N=0$, gives that

$\sum_{n_{2}\geq
0}g_{[1],\pm}\left((n_{2}+1)\sqrt{x}\right)=\frac{1}{\sqrt{x}}\int_{0}^{\infty}g_{[1],\pm}(w_{2})dw_{2}+E^{[1]}(x),$

where

$E^{[1]}(x):=-\sum_{k_{2}\geq
0}\frac{g_{[1],\pm}^{(k_{2})}(0)}{(k_{2}+1)!}x^{\frac{k_{2}}{2}}-\frac{1}{\sqrt{x}}\int_{\sqrt{x}}^{\infty}g_{[1],\pm}(w_{2})\widetilde{B}_{0}\left(\tfrac{w_{2}}{\sqrt{x}}-1\right)dw_{2}.$

The main term contributes

$\tfrac{2}{\sqrt{x}}\sum_{\pm}\pm\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{k_{1}\geq
0}\left(\tfrac{\delta}{2}+\tfrac{1}{4}\right)^{k_{1}+1}\frac{z^{\frac{k_{1}}{2}}}{(k_{1}+1)!}\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}}\left(e^{-12w_{1}^{2}}\int_{0}^{\infty}g_{[1],\pm}(w_{2})dw_{2}\right)\right]_{w_{1}=0}.$
(A.8)

We rewrite

$\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}}\left(e^{-12w_{1}^{2}}\int_{0}^{\infty}g_{[1],\pm}(w_{2})dw_{2}\right)\right]_{w_{1}=0}\\\
=\sum_{j=0}^{k_{1}}\left(\begin{smallmatrix}k_{1}\\\
j\end{smallmatrix}\right)\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}-j}e^{-12w_{1}^{2}}\right]_{w_{1}=0}\int_{0}^{\infty}\beta\left(\tfrac{2w_{2}^{2}}{3\pi}\right)e^{-\frac{z}{x}w_{2}^{2}}\left(\mp
8\sqrt{\tfrac{z}{x}}w_{2}\right)^{j}dt_{2}.$

The $\pm$ forces $j$ to be even. Also $k_{1}$ is even. Thus (A.8) is
$O(\frac{1}{\sqrt{z}})$.

The first term from $E^{[1]}$ contributes

$2\sum_{\pm}\pm\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{k_{1}\geq
0}\left(\tfrac{\delta}{2}+\tfrac{1}{4}\right)^{k_{1}+1}\Bigg{[}\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}}\\\ \Bigg{(}e^{-12w_{1}^{2}}\sum_{k_{2}\geq
0}\left[\left(\tfrac{\partial}{\partial
w_{2}}\right)^{k_{2}}\left(g_{[1],\pm}(w_{2})\right)\right]_{w_{2}=0}\Bigg{)}\Bigg{]}_{w_{1}=0}\frac{z^{\frac{k_{1}}{2}}}{(k_{1}+1)!}\frac{x^{\frac{k_{2}}{2}}}{(k_{2}+1)!}\ll
1.$

The second term from $E^{[1]}$ contributes

$\tfrac{2}{\sqrt{x}}\sum_{\pm}\pm\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{k_{1}\geq
0}\left(\tfrac{\delta}{2}+\tfrac{1}{4}\right)^{k_{1}+1}\frac{z^{\frac{k_{1}}{2}}}{(k_{1}+1)!}{\vphantom{\left(\frac{\partial}{\partial
w_{2}}\right)^{k_{2}}}}\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}}\left(e^{-12w_{1}^{2}}\int_{\sqrt{x}}^{\infty}g_{[1],\pm}(w_{2})dw_{2}\right)\right]_{w_{1}=0}.$

We compute

$\left[\left(\tfrac{\partial}{\partial w_{1}}\right)^{k_{1}}e^{-12w_{1}^{2}\mp
8\frac{\sqrt{z}}{\sqrt{x}}w_{1}w_{2}}\right]_{w_{1}=0}=\sum_{\ell=0}^{k_{1}}\binom{k_{1}}{\ell}\left[\left(\tfrac{\partial}{\partial
w_{1}}\right)^{k_{1}-\ell}e^{-12w_{1}^{2}}\right]_{w_{1}=0}\left(\mp
8\tfrac{\sqrt{z}}{\sqrt{x}}w_{2}\right)^{\ell}.$

Arguing as above we need $\ell$, $k_{1}$ to be even and obtain overall
$O(\frac{1}{\sqrt{z}})$.

The second term in the error $E(\frac{\delta}{2}+\frac{1}{4};z)$ contributes

$\tfrac{2}{\sqrt{z}}\sum_{\pm}\pm\sum_{\delta\in\\{0,1\\}}(-1)^{\delta}\sum_{n_{2}\geq
0}\beta\left(\tfrac{2}{3\pi}(n_{2}+1)^{2}x\right)e^{-(n_{2}+1)^{2}z}\int_{\left(\frac{\delta}{2}+\frac{1}{4}\right)\sqrt{z}}^{\sqrt{z}\infty}e^{-12w_{1}^{2}\mp
8(n_{2}+1)\sqrt{z}w_{1}}dw_{1}\\\
=\tfrac{2}{\sqrt{z}}\sum_{\pm}\pm\sum_{n_{2}\geq
0}\beta\left(\tfrac{2}{3\pi}(n_{2}+1)^{2}x\right)e^{-(n_{2}+1)^{2}z}\int_{\frac{\sqrt{z}}{4}}^{\frac{3\sqrt{z}}{4}}e^{-12w_{1}^{2}\mp
8(n_{2}+1)\sqrt{z}w_{1}}dw_{1}.$

Now the sum on $n_{2}$ is

$\sum_{n_{2}\geq 0}g_{[2]}\left((n_{2}+1)\sqrt{x}\right),$

where

$g_{[2]}(w_{2}):=\beta\left(\tfrac{2w_{2}^{2}}{3\pi}\right)e^{-\frac{z}{x}w_{2}^{2}\mp
8\frac{\sqrt{z}}{\sqrt{x}}w_{1}w_{2}}.$

Thus (5.8), with $N=0$, gives

$\sum_{n_{2}\geq
0}g_{[2]}\left((n_{2}+1)\sqrt{x}\right)=\frac{1}{\sqrt{x}}\int_{0}^{\infty}g_{[2]}(t_{2})dt_{2}+E^{[4]}(x),$

where

$E^{[4]}(x):=-\sum_{k_{2}\geq
0}\frac{g_{[2]}^{(k_{2})}(0)}{(k_{2}+1)!}x^{\frac{k_{2}}{2}}-\frac{1}{\sqrt{x}}\int_{\sqrt{x}}^{\infty}g_{[2]}(w_{2})\widetilde{B}_{0}\left(\tfrac{w_{2}}{\sqrt{x}}-1\right)dw_{2}.$

The main term contributes

$\tfrac{2}{\sqrt{z}\sqrt{x}}\sum_{\pm}\pm\int_{\frac{\sqrt{z}}{4}}^{\frac{3\sqrt{z}}{4}}e^{-12w_{1}^{2}}\int_{0}^{\infty}g_{[2]}(w_{2})dw_{2}dw_{1}.$

We write

$e^{-12w_{1}^{2}}\int_{0}^{\infty}g_{[2]}(w_{2})dw_{2}=e^{4w_{1}^{2}}\int_{0}^{\infty}\beta\left(\tfrac{2w_{2}^{2}}{3\pi}\right)e^{-\frac{z}{x}\left(w_{2}\mp
4\frac{\sqrt{x}}{\sqrt{z}}\right)^{2}}dw_{2}\ll e^{4w_{1}^{2}}.$

Thus we have overall $O(\frac{1}{\sqrt{z}})$.

The first term in the error $E^{[4]}$ contributes

$-\tfrac{2}{\sqrt{z}}\sum_{\pm}\pm\int_{\frac{\sqrt{z}}{4}}^{\frac{3\sqrt{z}}{4}}e^{-12w_{1}^{2}}\sum_{k_{2}\geq
0}\left[\left(\tfrac{\partial}{\partial
w_{2}}\right)^{k_{2}}\beta\left(\tfrac{2w_{2}^{2}}{3\pi}\right)e^{-\frac{z}{x}w_{2}^{2}\mp
8\frac{\sqrt{z}}{\sqrt{x}}w_{1}w_{2}}\right]_{w_{2}=0}\frac{x^{\frac{k_{2}}{2}}}{(k_{2}+1)!}.$

We bound, since $|z|\leq(1+\Delta)x$,

$\left[\left(\tfrac{\partial}{\partial
w_{2}}\right)^{k_{2}}\left(\beta\left(\tfrac{2w_{2}^{2}}{3\pi}\right)e^{-\frac{z}{x}w_{2}^{2}\mp
8\frac{\sqrt{z}}{\sqrt{x}}w_{1}w_{2}}\right)\right]_{w_{2}=0}\ll\left(1+\Delta^{\frac{k_{2}}{2}}\right)\left[\left(\tfrac{\partial}{\partial
w_{2}}\right)^{k_{2}}\beta\left(\tfrac{2w_{2}^{2}}{3\pi}\right)e^{-w_{2}^{2}\mp
8w_{1}w_{2}}\right]_{w_{2}=0}.$

Then overall we have

$\ll\tfrac{1}{\sqrt{z}}\sum_{k_{2}\geq
0}\left(1+\Delta^{\frac{k_{2}}{2}}\right)\left[\left(\tfrac{\partial}{\partial
w_{2}}\right)^{k_{2}}\left(\beta\left(\tfrac{2w_{2}^{2}}{3\pi}\right)e^{-w_{2}^{2}}\int_{\frac{\sqrt{z}}{4}}^{\frac{3\sqrt{z}}{4}}e^{-12w_{1}^{2}\mp
8w_{1}w_{2}}dw_{1}\right)\right]_{w_{2}=0}\frac{x^{\frac{k_{2}}{2}}}{(k_{2}+1)!}\ll
1.$

The second term in the error $E^{[4]}$ can be estimated as before as
$O(\frac{1}{\sqrt{z}})$.

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\tfrac{1}{2!}\mathcal{U}^{[2]}(Q_{1})(q_{1}-Q_{1})^{2}+\cdots$ (191)

with a similar expression for $\mathcal{V}_{1}$. Then, equating coefficients
of $(q_{1}-Q_{1})^{0}$, $(q_{1}-Q_{1})^{1}$, and $(q_{1}-Q_{1})^{2}$ on the
left- and right-hand sides of the resulting expression, we obtain
_approximate_ ordinary differential equations for $\gamma$, $\Sigma$, and
$P_{1}$. Along with equation (188a), this gives a complete dynamical system
for $\\{Q_{1},P_{1},\gamma,\Sigma\\}$:

$\displaystyle\frac{dQ_{1}}{dt}$ $\displaystyle=\frac{P_{1}}{m},$ (192a)
$\displaystyle\frac{dP_{1}}{dt}$
$\displaystyle=-\mathcal{U}^{[1]}(Q_{1})-\lambda\langle
V_{2}\rangle\mathcal{V}_{1}^{[1]}(Q_{1}),$ (192b)
$\displaystyle\frac{d\gamma}{dt}$
$\displaystyle=\frac{P_{1}^{2}}{2m}+\frac{i\Sigma}{m}-\mathcal{U}(Q_{1})-\lambda\mathcal{V}_{1}(Q_{1})\langle
V_{2}\rangle,$ (192c) $\displaystyle\frac{d\Sigma}{dt}$
$\displaystyle=-\frac{2\Sigma^{2}}{m}-\frac{\mathcal{U}^{[2]}(Q_{1})+\lambda\mathcal{V}_{1}^{[2]}(Q_{1})\langle
V_{2}\rangle}{2}.$ (192d)

If $\\{Q_{1},P_{1},\gamma,\Sigma\\}$ satisfy these equations and the
wavepacket width is small, then we expect that (183) is a good approximate
solution to (182). To confirm this, we can calculate the norm of the
difference between the left- and right-hand sides of (182) when (192) is
enforced:

$\int\limits_{-\infty}^{\infty}\left|\left\\{i\partial_{t}+\frac{\partial_{q_{1}}^{2}}{2m}-\mathcal{U}(q_{1})-\lambda\mathcal{V}_{1}(q_{1})\langle
V_{2}\rangle\right\\}\varphi_{1}\right|^{2}dq_{1}=\\\
\frac{5}{768}\frac{[\mathcal{U}^{[3]}(Q_{1})+\lambda\langle
V_{2}\rangle\mathcal{V}_{1}^{[3]}(Q_{1})]^{2}}{(\text{Im}\,\Sigma)^{3}}+\cdots.$
(193)

Here, the “$\cdots$” on the right-hand side indicates terms involving higher
derivatives of $U$ and $V_{1}$ as well as higher inverse powers of
$\text{Im}\,\Sigma$. Hence, if $\imaginary\,\Sigma$ is sufficiently large,
$\varphi_{1}$ will be a good approximate solution to the wave equation (182).
In this case, the evolution of the width of the wavepacket will be given by
(192d). For particular choices of $\mathcal{U}$ and $\mathcal{V}$, one would
need to solve the dynamical system 192) for $\Sigma(t)$ to determine if and
when the conditions (189) break down. If the conditions do indeed fail over a
characteristic timescale when the initial wavepacket width is small, that
timescale is known as the “Ehrenfest time.”

We now turn our attention to subsystem 2. In the product state approximation,
its effective Hamiltonian is

$\hat{H}^{\text{\tiny{eff}}}_{2}=\hat{H}_{2}+\lambda\langle
V_{1}(q_{1})\rangle\hat{V}_{2}.$ (194)

Making use of (186), we can rewrite this as

$\hat{H}^{\text{\tiny{eff}}}_{2}=\hat{H}_{2}+\frac{\lambda\mathcal{V}_{1}(Q_{1})\hat{V}_{2}}{1-\varepsilon},$
(195)

where, as in section §III.2, $\varepsilon$ is given by

$\displaystyle\varepsilon$ $\displaystyle=\frac{\langle
V_{1}(q_{1})\rangle-\mathcal{V}_{1}(Q_{1})}{\langle V_{1}(q_{1})\rangle}$
$\displaystyle=\frac{\mathcal{V}_{1}^{[2]}(Q_{1})}{8\mathcal{V}_{1}(Q_{1})\,\text{Im}\,\Sigma}+\mathcal{O}\left[\frac{\mathcal{V}_{1}^{[4]}(Q_{1})}{\mathcal{V}_{1}(Q_{1})\,(\text{Im}\,\Sigma)^{2}}\right].$
(196)

Hence, if

$\left|\frac{\mathcal{V}_{1}^{[2k]}(Q_{1})}{\mathcal{V}_{1}(Q_{1})}\right|\ll(\text{Im}\,\Sigma)^{k},\quad
k=1,2,3\ldots,$ (197)

then the $\varepsilon$ term in (194) can be neglected and subsystem 2 will be
governed by the effective Hamiltonian

$\displaystyle\hat{H}_{\text{\tiny{CQ}}}^{\text{\tiny{eff}}}=\hat{H}_{2}+\lambda\mathcal{V}_{1}(Q_{1})\hat{V}_{2}.$
(198)

We have hence recovered the subsystem 2 classical-quantum Hamiltonian operator
(48) of §III.2 specialized to our choice of $\hat{V}_{1}$.

To summarize, for the particular choices (181) the sharply-peaked wave-
function _ansatz_ (183) leads us to the conclusion that subsystems 1 and 2 are
well described by the coupled classical and quantum Hamiltonians

$\displaystyle\mathcal{H}_{\text{\tiny{CQ}}}^{\text{\tiny{eff}}}$
$\displaystyle=P_{1}^{2}/(2m)+\mathcal{U}(Q_{1})+\lambda\mathcal{V}_{1}(Q_{1})\langle\hat{V}_{2}\rangle,$
(199a) $\displaystyle\hat{H}_{\text{\tiny{CQ}}}^{\text{\tiny{eff}}}$
$\displaystyle=\hat{H}_{2}+\lambda\mathcal{V}_{1}(Q_{1})\hat{V}_{2}$ (199b)

provided that the conditions (189) and (197) hold. Furthermore, the evolution
of the width of subsystem 1’s wavefunction will be governed by (192) as long
as the width $(\imaginary\,\Sigma)^{-1/2}$ is sufficiently small.

###  B.2 Quadratic potentials

In this subsection, we specialize to the case when both the main $\mathcal{U}$
and interaction $\mathcal{V_{1}}$ potentials are quadratic monomial functions:

$\mathcal{U}(q_{1})=\tfrac{1}{2}\Omega_{1}q_{1}^{2},\quad\mathcal{V}_{1}(q_{1})=\sqrt{\bar{\Omega}}q_{1}^{2},\quad
m=\frac{1}{\Omega_{1}}.$ (200)

This is similar to the $\nu=2$ coupled oscillator case considered in §V except
that we do not necessarily assume that subsystem 2 is also an oscillator.

This case is interesting because several of the exact formulae from the
previous subsection simplify considerably.555We should stress that in this
appendix, the phrase “exact” is meant to be applied to for formulae that hold
in the product state approximation without any additional assumptions. For
example, the Ehrenfest equations of motion (188) become exactly

$\displaystyle\partial_{t}Q_{1}$ $\displaystyle=\Omega_{1}P_{1},$ (201a)
$\displaystyle\partial_{t}P_{1}$
$\displaystyle=-\left(\Omega_{1}+\lambda\sqrt{\bar{\Omega}}\langle\hat{V}_{2}\rangle\right)Q_{1}.$
(201b)

That is, the exact expectation values of $\hat{q}_{1}$ and $\hat{p}_{1}$
evolve according to the CQ Hamiltonian

$\displaystyle\mathcal{H}_{\text{\tiny{CQ}}}^{\text{\tiny{eff}}}$
$\displaystyle=\tfrac{1}{2}\Omega_{1}(Q_{1}^{2}+P_{1}^{2})+\lambda\sqrt{\bar{\Omega}}Q_{1}^{2}\langle\hat{V}_{2}\rangle.$
(202)

Also, the right-hand side of (193) is identically equal to zero, implying that
(183) is an exact solution to the wave equation provided that (188) are
satisfied in addition to

$\displaystyle\partial_{t}\gamma$
$\displaystyle=\tfrac{1}{2}\Omega_{1}P_{1}^{2}+i\Omega_{1}\Sigma-\tfrac{1}{2}\left(\Omega_{1}+\lambda\sqrt{\bar{\Omega}}\langle
V_{2}\rangle\right)Q_{1}^{2},$ $\displaystyle\partial_{t}\Sigma$
$\displaystyle=-2\Omega_{1}\Sigma^{2}-\tfrac{1}{2}\left(\Omega_{1}+\lambda\sqrt{\bar{\Omega}}\langle
V_{2}\rangle\right).$ (203)

Another exact relation is

$\langle
V_{1}(q_{1})\rangle=\frac{\sqrt{\bar{\Omega}}}{2}\left(Q_{1}^{2}+\frac{1}{4\imaginary\,\Sigma}\right),$
(204)

which gives that the effective Hamiltonian of subsystem 2 is

$\hat{H}^{\text{\tiny{eff}}}_{2}=\hat{H}_{2}+\frac{\lambda\sqrt{\bar{\Omega}}}{2}\left(Q_{1}^{2}+\frac{1}{4\imaginary\,\Sigma}\right)\hat{V}_{2}.$
(205)

In this scenario, the classical-quantum approximation will be realized if one
assumes $Q_{1}^{2}\gg|4\,\text{Im}\,\Sigma|^{-1}$ so that

$\hat{H}^{\text{\tiny{eff}}}_{2}\approx\hat{H}_{2}+\frac{1}{2}\lambda\sqrt{\bar{\Omega}}Q_{1}^{2}\hat{V}_{2},$
(206)

which then implies that equations (203) need not be explicitly solved for
$\Sigma$ and $\gamma$ in order to solve for $Q_{1}$, $P_{1}$, and
$|\varphi_{2}\rangle$.

Of course, to actually determine if $Q_{1}^{2}\gg|4\,\text{Im}\,\Sigma|^{-1}$
is a reasonable assumption, one does need to solve (203) for $\Sigma$. When
the interaction is small, $\lambda\ll 1$, we can easily obtain

$\displaystyle Q_{1}(t)$
$\displaystyle=\sqrt{2}|\alpha|\cos(\Omega_{1}t+\phi_{1})+\mathcal{O}(\lambda),$
(207a) $\displaystyle\Sigma(t)$
$\displaystyle=\frac{2\Sigma_{0}-\tan\Omega_{1}t}{2(1+2\Sigma_{0}\tan\Omega_{1}t)}+\mathcal{O}(\lambda),$
(207b)

where $\Sigma_{0}=\Sigma(0)$ with $\text{Im}\,\Sigma_{0}>0$ and we have
parametrized the solution for $Q_{1}(t)$ in a manner consistent with §V.4 and
§D. Notice that if $\Sigma_{0}=i/2$, we obtain that
$\Sigma(t)=i/2+\mathcal{O}(\lambda)$ for all time; i.e., the wavepacket has a
nearly constant width. Other than the $\mathcal{O}(\lambda)$ correction, these
are the well-known coherent state solutions of the simple harmonic oscillator.
When $\Sigma_{0}\neq i/2$, we recover the oscillator squeezed state solutions
subject to $\mathcal{O}(\lambda)$ corrections.

From (207), we see that the condition
$Q_{1}^{2}\gg|4\,\text{Im}\,\Sigma|^{-1}$ can be satisfied for almost all $t$
if $|\alpha|^{2}\gg|\Sigma_{0}|^{-1}$. This is essentially the same as
demanding that subsystem 1 be initially prepared in a high energy state.
However, it should be noted that because there are times for which
$Q_{1}(t)=0$, it is impossible to satisfy
$Q_{1}^{2}\gg|4\,\text{Im}\,\Sigma|^{-1}$ for all $t$. However, the time
intervals over which the condition fails to hold will be very short if
$|\alpha|^{2}\gg|\Sigma_{0}|^{-1}$. Outside of these intervals, the effective
Hamiltonian of subsystem 2 will reduce down to the classical-quantum form

$\hat{H}^{\text{\tiny{eff}}}_{2}\approx\hat{H}^{\text{\tiny{eff}}}_{\text{\tiny{CQ}}}=\hat{H}_{2}+\tfrac{1}{2}\lambda\sqrt{\bar{\Omega}}Q_{1}^{2}\hat{V}_{2}.$
(208)

##  Appendix C Born-Oppenheimer approximation for one-dimensional potential
problems

In this appendix, we consider the Born-Oppenheimer approximation to our
bipartite system when subsystems 1 and 2 each represent particles moving in
one-dimensional potentials. Our treatment is somewhat similar to a calculation
presented in Singh and Padmanabhan (1989), but there are some key differences.
While Singh and Padmanabhan (1989) analyzes both time-independent (via the WKB
approximation) and time-dependent Schrödinger equations (via sharply peaked
states), we will only work on the latter. There are also some technical
assumptions in Singh and Padmanabhan (1989) that we prefer to avoid; these are
discussed in further detail below.

For this appendix, we assume that

$\displaystyle\hat{H}_{1}$
$\displaystyle={\hat{p}_{1}^{2}}/{2m_{1}}+m_{1}\mathcal{U}_{1}(\hat{q}_{1}),$
$\displaystyle\hat{V}_{1}$ $\displaystyle=\mathcal{V}_{1}(\hat{q}_{1}),$
$\displaystyle\hat{H}_{2}$
$\displaystyle={\hat{p}_{2}^{2}}/{2m_{2}}+m_{2}\mathcal{U}_{2}(\hat{q}_{2}),$
$\displaystyle\hat{V}_{2}$ $\displaystyle=\mathcal{V}_{2}(\hat{q}_{2}),$ (209)

where $[\hat{q}_{i},\hat{p}_{j}]=i\,\delta_{ij}$ and the $\mathcal{U}_{1,2}$
and $\mathcal{V}_{1,2}$ potential functions are real. Notice that the
$\mathcal{U}_{1,2}$ potentials each have a prefactor of the particle masses
$m_{1,2}$ for later convenience. As elsewhere in this paper, subsystem 1 is
expected to behave classically, which, in the context of the Born-Oppenheimer
approximation, means that the mass of particle 1 should be large. That is, we
will work in the $m_{1}\rightarrow\infty$ limit.

The complete time dependent Schrödinger equation for the wavefunction
$\psi=\psi(t,q_{1},q_{2})$ is

$i\frac{\partial\psi}{\partial
t}=-\frac{1}{2m_{1}}\frac{\partial^{2}\psi}{\partial
q_{1}^{2}}+m_{1}\mathcal{U}_{1}(q_{1})\psi-\frac{1}{2m_{2}}\frac{\partial^{2}\psi}{\partial
q_{2}^{2}}\\\
+m_{2}\mathcal{U}_{2}(q_{2})\psi+\lambda\mathcal{V}_{1}(q_{1})\mathcal{V}_{2}(q_{2}).$
(210)

We will make an explicit product state _ansatz_ for $\psi$:

$\psi(t,q_{1},q_{2})=\psi_{1}(t,q_{1})\psi_{2}(t,q_{2}).$ (211)

Since subsystem 1 is meant to exhibit classical properties, we assume that its
wavefunction is a Gaussian sharply peaked about $q_{1}=Q_{1}(t)$:

$\psi_{1}=\left({2}/{\pi}\right)^{1/4}e^{im_{1}\Gamma(t)}e^{im_{1}v_{1}(t)[q_{1}-Q_{1}(t)]}\\\
\times e^{im_{1}^{2}\varsigma(t)[q_{1}-Q_{1}(t)]^{2}}.$ (212)

This is the same _ansatz_ (183) we made use of in the previous appendix to
derive the classical-quantum approximation for a similar potential problem
with a few changes of notation:

$\gamma=m_{1}\Gamma,\quad P_{1}=m_{1}v_{1},\quad\Sigma=m_{1}^{2}\varsigma.$
(213)

The $m_{1}$ scalings appearing on the right-hand sides of these definitions
are required to make the $m\to\infty$ Born-Oppenheimer limit work out
correctly.

The normalization of $\psi$ requires that

$1=\int_{-\infty}^{\infty}|\psi_{1}|^{2}dq_{1}=\int_{-\infty}^{\infty}|\psi_{2}|^{2}dq_{2},$
(214)

which yields that

$\displaystyle
0<m_{1}^{2}\,\text{Im}\,\varsigma=\exp(-4m_{1}\,\text{Im}\,\Gamma)$
$\displaystyle\langle q_{1}\rangle=Q_{1}(t),\quad\langle
p_{1}\rangle=m_{1}v_{1}(t).$ (215)

Using these and our assumed Hamiltonian, we can use the Ehrenfest theorem to
easily show that

$v_{1}=\frac{dQ_{1}}{dt}.$ (216)

As in the previous appendix, we assume that $\imaginary\,\Sigma$ is large such
that the width of the wavefunction in the $q_{1}$ direction is much smaller
than the characteristic variation scale of the potentials $\mathcal{U}_{1}$
and $\mathcal{V}_{1}$. We therefore seek a solution of (210) valid in some
neighbourhood of $q_{1}=Q_{1}(t)$. More specifically, we substitute (211) into
(210) and expand the result in powers of $\delta q_{1}=q_{1}-Q_{1}(t)$. We
obtain

$0=c_{0}+c_{1}\delta q_{1}+c_{2}\delta q_{1}^{2}+\mathcal{O}(\delta
q_{1}^{3}).$ (217)

Here,

$\displaystyle c_{0}=$
$\displaystyle\,\,m_{1}\left[\frac{d\Gamma}{dt}-i\varsigma-\frac{v_{1}^{2}}{2}+\mathcal{U}_{1}(Q_{1})\right]-i\frac{1}{\psi_{2}}\frac{\partial\psi_{2}}{\partial
t}$
$\displaystyle\,\,-\frac{1}{2m_{2}}\frac{1}{\psi_{2}}\frac{\partial^{2}\psi_{2}}{dq_{2}^{2}}+m_{2}\mathcal{U}_{2}(q_{2})+\lambda\mathcal{V}_{1}(Q_{1})\mathcal{V}_{2}(q_{2}),$
$\displaystyle c_{1}=$
$\displaystyle\,\,m_{1}\frac{dv_{1}}{dt}+\frac{\partial}{\partial
q_{1}}\left[m_{1}\mathcal{U}_{1}(q_{1})+\lambda\mathcal{V}_{1}(q_{1})\mathcal{V}_{2}(q_{2})\right]_{q_{1}=Q_{1}},$
$\displaystyle c_{2}=$
$\displaystyle\,\,m_{1}^{2}\frac{d\varsigma}{dt}+2m_{1}^{3}\varsigma^{2}$
$\displaystyle\,\,+\frac{\partial^{2}}{\partial
q_{1}^{2}}\left[m_{1}\mathcal{U}_{1}(q_{1})+\lambda\mathcal{V}_{1}(q_{1})\mathcal{V}_{2}(q_{2})\right]_{q_{1}=Q_{1}}.$
(218)

As in the previous appendix, we demand that each of the coefficents of $\delta
q_{1}^{0}$, $\delta q_{1}^{1}$, and $\delta q_{1}^{2}$ vanish independently in
(217).666At this stage, our approximation deviates from that of Singh and
Padmanabhan (1989). In that paper, the authors assume that $\delta
q_{1}=m_{1}^{-1/2}y$ and then demand that the coefficient of each power of
$m_{1}$ vanish. This results in approximate equations of motion involving an
arbitrary function $y=y(t,q_{1})$; i.e., an underdetermined system of
equations. Furthermore, let us make the approximation that in the
$m_{1}\rightarrow\infty$ limit

$m_{1}\mathcal{U}_{1}(q_{1})+\lambda\mathcal{V}_{1}(q_{1})\mathcal{V}_{2}(q_{2})\approx
m_{1}\mathcal{U}_{1}(q_{1}).$ (219)

Finally, we demand that the coefficients of $m_{1}^{0}$ and $m_{1}^{1}$ in the
above expression for $c_{0}$ individually vanish. This gives us a complete set
of equations for all of the unknown functions:

$\displaystyle\frac{dQ_{1}}{dt}$ $\displaystyle=v_{1},$ (220a)
$\displaystyle\frac{dv_{1}}{dt}$
$\displaystyle=-\mathcal{U}_{1}^{[1]}(Q_{1}),$ (220b)
$\displaystyle\frac{d\Gamma}{dt}$
$\displaystyle=i\varsigma+\frac{v_{1}^{2}}{2}-\mathcal{U}_{1}(Q_{1}),$ (220c)
$\displaystyle\frac{d\varsigma}{dt}$
$\displaystyle=-2m_{1}\varsigma^{2}-\frac{1}{m_{1}}\mathcal{U}^{[2]}_{1}(Q_{1}),$
(220d) $\displaystyle i\frac{\partial\psi_{2}}{\partial t}$
$\displaystyle=-\frac{1}{2m_{2}}\frac{\partial^{2}\psi_{2}}{dq_{2}^{2}}+$
$\displaystyle\quad\quad[m_{2}\mathcal{U}_{2}(q_{2})+\lambda\mathcal{V}_{1}(Q_{1})\mathcal{V}_{2}(q_{2})]\psi_{2}.$
(220e)

We see that (220a) and (220b) are the classical equations of motion for
subsystem 1 neglecting subsystem 2 (or, in other words, in the $\lambda\to 0$
limit). Apart from a change in notation, equations (220c) and (220d) are the
same as evolution equations for the phase (192c) and width (192d) of a
Gaussian wavepacket in the $\lambda\to 0$ limit of the classical-quantum
approximation. Equation (220) is exactly the time-dependent Schrödinger
equation for subsystem 2 in the classical-quantum approximation for the system
studied in this appendix. Therefore, we have essentially recovered the
classical-background approximation (54) for the system described by the
choices (209).

Before moving on, we make note of something interesting. Suppose that we do
not make the approximation (219). Rather, let us replace the equations
$c_{1}=0$ and $c_{2}=0$ with their expectation values in the state (211);
i.e.,

$\langle\psi|c_{1}|\psi\rangle=\langle\psi|c_{2}|\psi\rangle=0,$ (221)

or, more explicitly,

$\int_{-\infty}^{\infty}dq_{2}\,\psi_{2}^{*}c_{1}\psi_{2}=\int_{-\infty}^{\infty}dq_{2}\,\psi_{2}^{*}c_{2}\psi_{2}=0.$
(222)

We then find that equations (220b) and (220d) get modified to

$\displaystyle\frac{dv_{1}}{dt}$
$\displaystyle=-\frac{m_{1}\mathcal{U}_{1}^{[1]}(Q_{1})+\lambda\mathcal{V}_{1}^{[1]}(Q_{1})\langle
V_{2}\rangle}{m_{1}},$ (223a) $\displaystyle\frac{d\varsigma}{dt}$
$\displaystyle=-2m_{1}\varsigma^{2}-\frac{m_{1}\mathcal{U}^{[2]}_{1}(Q_{1})+\lambda\mathcal{V}_{1}^{[2]}(Q_{1})\langle
V_{2}\rangle}{m_{1}^{2}}.$ (223b)

These are the analogues of the classical-quantum equations of motion in the
previous appendix. Perhaps this is not overly surprising as the strategy of
replacing equations like $c_{1}=c_{2}=0$ with their expectation values is not
dissimilar to the leap made from the classical Einstein equations to the
semiclassical system (2). However, this discussion does provide an intriguing
counterpoint to our derivation of the classical-quantum equations of motion
using density matrices in §III.2.

##  Appendix D Properties of coherent states

In this appendix, we review some basic properties of coherent states
$|\alpha\rangle$ of the simple harmonic oscillator. The Hamiltonian of the
system is, as usual,

$\hat{H}=\Omega(\hat{a}^{\dagger}\hat{a}+\tfrac{1}{2}),\quad[\hat{a},\hat{a}^{\dagger}]=1.$
(224)

In the Heisenberg representation, coherent states are eigenstates of the
$\hat{a}$ operator with eigenvalue $\alpha$,

$\hat{a}|\alpha\rangle=\alpha|\alpha\rangle,$ (225)

where $\alpha$ is a complex constant. Energy eigenstates of the Hamiltonian
are, as usual,

$\displaystyle\hat{H}|n\rangle=\Omega(n+\tfrac{1}{2})|n\rangle,\quad
n=0,1,2\ldots.$ (226)

The expansion of an arbitrary coherent state in the energy eigenbasis reads

$|\alpha\rangle=\sum_{n}\langle
n|\alpha\rangle|n\rangle=\sum_{n}\frac{\alpha^{n}e^{-|\alpha|^{2}/2}}{\sqrt{n!}}|n\rangle.$
(227)

The position and momentum operators are

$\hat{q}=\frac{\hat{a}+\hat{a}^{\dagger}}{\sqrt{2}},\quad\hat{p}=\frac{\hat{a}-\hat{a}^{\dagger}}{\sqrt{2}i}.$
(228)

We have the expectation values

$\displaystyle\langle\hat{q}\rangle$
$\displaystyle=+\sqrt{2}|\alpha|\cos(\Omega t-\phi),$
$\displaystyle\langle\hat{q}^{2}\rangle$
$\displaystyle=\langle\hat{q}\rangle^{2}+\tfrac{1}{2},$ (229a)
$\displaystyle\langle\hat{p}\rangle$
$\displaystyle=-\sqrt{2}|\alpha|\sin(\Omega t-\phi),$
$\displaystyle\langle\hat{p}^{2}\rangle$
$\displaystyle=\langle\hat{p}\rangle^{2}+\tfrac{1}{2},$ (229b)

where $\phi=\text{arg}(\alpha)$. Expectation values of higher powers of
$\hat{q}$ can be obtained by making use of the identity

$[\hat{A}^{n},\hat{B}]=n\hat{A}^{n-1}[\hat{A},\hat{B}],$ (230)

which holds if $[\hat{A},[\hat{A},\hat{B}]]=[\hat{B},[\hat{A},\hat{B}]]=0$. We
can use this to derive the following recursion relation for the epxection
values of $\hat{q}^{n}$ for coherent states:

$\langle\hat{q}^{n}\rangle=\langle\hat{q}\rangle\langle\hat{q}^{n-1}\rangle+\tfrac{1}{2}(n-1)\langle\hat{q}^{n-2}\rangle.$
(231)

This yields

$\displaystyle\langle\hat{q}^{3}\rangle$
$\displaystyle=\langle\hat{q}\rangle^{3}+\tfrac{3}{2}\langle\hat{q}\rangle,$
(232a) $\displaystyle\langle\hat{q}^{4}\rangle$
$\displaystyle=\langle\hat{q}\rangle^{4}+3\langle\hat{q}\rangle^{2}+\tfrac{3}{4},$
(232b) $\displaystyle\langle\hat{q}^{5}\rangle$
$\displaystyle=\langle\hat{q}\rangle^{5}+5\langle\hat{q}\rangle^{3}+\tfrac{15}{4}\langle\hat{q}\rangle,$
(232c)

and so on. A very similar derivation results in a recursion relation for the
the central moments of $\hat{q}$ for coherent states:

$\langle(\hat{q}-\langle\hat{q}\rangle)^{n}\rangle=\tfrac{1}{2}(n-1)\langle(\hat{q}-\langle\hat{q}\rangle)^{n-2}\rangle.$
(233)

Making note of $\langle(\hat{q}-\langle\hat{q}\rangle)^{0}\rangle=1$ and
$\langle(\hat{q}-\langle\hat{q}\rangle)^{1}\rangle=0$, this yields

$\langle(\hat{q}-\langle\hat{q}\rangle)^{n}\rangle=\begin{cases}0,&n\text{
odd},\\\ 2^{-n/2}(n-1)!!,&n\text{ even}.\end{cases}$ (234)

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