$\displaystyle\mathsf{N}^{(1|2)}_{\emptyset,\lambda^{(1)}}(qt^{-1/2}|q,t^{-1/2}),$

which is non vanishing, if and only if $\lambda^{(1)}=\emptyset$. Hence the
non vanishing contributions for $\mathcal{F}^{(2)}$ arise if and only if we
have $(\lambda^{(1)},\lambda^{(2)})=(\emptyset,(n+m,n))$ $(m,n\geq 0)$. Making
explicit calculation, we have (B.11).

∎

We have seen how both partition functions $\Psi(t\Lambda,x)$ and
$\mathcal{F}^{(1)}$ are simplified to the asymptotically free Macdonald
functions in the limit (B.8). If Conjectures 2.4 and 5.4 are true, they are
related by the following gauge transformation.

######  Conjecture B.6.

$\displaystyle\mathcal{F}^{(1)}$
$\displaystyle={1\over\Phi(t^{2}T_{2}T_{3}\Lambda)\Phi(qtT_{1}T_{4}\Lambda)}\varphi(q^{1/2}t^{1/2}T_{1}x)\varphi(q^{1/2}t^{1/2}T_{4}\Lambda/x)\psi(\Lambda,x)$
$\displaystyle={\Phi(qtT_{2}T_{3}\Lambda)\Phi(t^{2}T_{1}T_{4}\Lambda)\over\Phi(t^{2}T_{2}T_{3}\Lambda)\Phi(qtT_{1}T_{4}\Lambda)}\varphi(q^{1/2}t^{1/2}T_{1}x)\varphi(q^{1/2}t^{1/2}T_{4}\Lambda/x)\mathcal{A}_{3}(t\Lambda,tqQx)\Psi(t\Lambda,x).$

We can prove this is indeed the case.

######  Lemma B.7.

We have

$\displaystyle
f^{\mathfrak{gl}_{2}}(1/\Lambda,1/x|1/Q,1|q,t)={(t\Lambda;q)_{\infty}\over(q\Lambda;q)_{\infty}}{(tx;q)_{\infty}\over(qx;q)_{\infty}}f^{\mathfrak{gl}_{3}}(1/\Lambda,1/x,1|1/Q,1,0|q,t).$
(B.12)

###### Proof.

We find that the Macdonald operators
$\mathcal{D}^{\mathfrak{gl}_{3}}(1/Q,1,0|q,q/t)$ and
$\mathcal{D}^{\mathfrak{gl}_{2}}(1/Q,1|q,q/t)$ are gauge equivalent

$\displaystyle{(t\Lambda;q)_{\infty}\over(q\Lambda;q)_{\infty}}{(tx;q)_{\infty}\over(qx;q)_{\infty}}\mathcal{D}^{\mathfrak{gl}_{3}}(1/Q,1,0|q,q/t){(q\Lambda;q)_{\infty}\over(t\Lambda;q)_{\infty}}{(qx;q)_{\infty}\over(tx;q)_{\infty}}=\mathcal{D}^{\mathfrak{gl}_{2}}(1/Q,1|q,q/t).$

Then the equality (B.12) follows from the uniqueness of the normalized
asymptotic eigenfunctions of $\mathcal{D}^{\mathfrak{gl}_{3}}(1/Q,1,0|q,q/t)$
and $\mathcal{D}^{\mathfrak{gl}_{2}}(1/Q,1|q,q/t)$. ∎

######  Proposition B.8.

Conjecture B.6 holds in the Macdonald limit (B.8).

###### Proof.

Under the condition (B.8), we have

$\displaystyle{\Phi(qtT_{2}T_{3}\Lambda)\Phi(t^{2}T_{1}T_{4}\Lambda)\over\Phi(t^{2}T_{2}T_{3}\Lambda)\Phi(qtT_{1}T_{4}\Lambda)}\varphi(q^{1/2}t^{1/2}T_{1}x)\varphi(q^{1/2}t^{1/2}T_{4}\Lambda/x)\mathcal{A}_{3}(t\Lambda,tqQx)={(t\Lambda;q)_{\infty}\over(q\Lambda;q)_{\infty}}{(tx;q)_{\infty}\over(qx;q)_{\infty}}.$

∎

Now we are ready to prove Conjecture 5.4 in the Macdonald limit. We need the
following formulas for the proof.

######  Lemma B.9 ([44]).

For $n\in\mathbb{Z}$, we have

$\displaystyle\mathcal{B}\cdot{1\over\varphi(\alpha
x)\varphi(\beta\Lambda/x)}x^{n}={\varphi(-q^{1+n}\alpha
x)\varphi(-q^{-n}\beta\Lambda/x)\over\varphi(\alpha\beta\Lambda)}q^{n(n+1)/2}x^{n},$
(B.13) $\displaystyle\mathcal{B}^{-1}\cdot{\varphi(\alpha
x)\varphi(\beta\Lambda/x)}x^{n}={\varphi(q^{-1}\alpha\beta\Lambda)\over\varphi(-q^{-1-n}\alpha
x)\varphi(-q^{n}\beta\Lambda/x)}q^{-n(n+1)/2}x^{n}.$ (B.14)

######  Proposition B.10 ($q$-Chu-Vandermonde sums [22]).

For $n\in\mathbb{Z}_{\geq 0}$, we have

$\displaystyle{}_{2}\phi_{1}\left[\left.{a,q^{-n}\atop
c}\right|q;q\right]={(c/a;q)_{n}\over(c;q)_{n}}a^{n},$ (B.15)
$\displaystyle{}_{2}\phi_{1}\left[\left.{a,q^{-n}\atop c}\right|q;{cq^{n}\over
a}\right]={(c/a;q)_{n}\over(c;q)_{n}}.$ (B.16)

######  Proposition B.11.

Conjecture 5.4 holds in the Macdonald limit (B.8).

$\displaystyle
T_{(qt^{1/2}Q)^{1/2},x}\mathcal{F}^{(2)}=\varphi(q\Lambda){1\over\varphi(-qtx)\varphi(-\Lambda/x)}\cdot\mathcal{B}\cdot{1\over\varphi(tx)\varphi(q\Lambda/tx)}\mathcal{F}^{(1)},$
(B.17)
$\displaystyle\mathcal{F}^{(1)}=\varphi(t\Lambda){1\over\varphi(qx)\varphi(t\Lambda/x)}\cdot\mathcal{B}\cdot{1\over\varphi(-x)\varphi(-q\Lambda/x)}(T_{t^{1/2},x}^{-1}T_{t,\Lambda}^{-1})T_{(qt^{1/2}Q)^{-1/2},x}\mathcal{F}^{(2)}.$
(B.18)

###### Proof.

First, we show (B.17). By using (B.13) and the $q$-binomial formula, we have

$\displaystyle\varphi(q\Lambda){1\over\varphi(-qtx)\varphi(-\Lambda/x)}\cdot\mathcal{B}\cdot{1\over\varphi(tx)\varphi(q\Lambda/tx)}\mathcal{F}^{(1)}$
$\displaystyle={}$
$\displaystyle\varphi(q\Lambda){1\over\varphi(-qtx)\varphi(-\Lambda/x)}\cdot\mathcal{B}\cdot{1\over\varphi(tx)\varphi(q\Lambda/tx)}\sum_{n\geq
0}{(q/t;q)_{n}(qQ/t;q)_{n}\over(q;q)_{n}(qQ;q)_{n}}t^{n}(\Lambda/x)^{n}$
$\displaystyle={}$ $\displaystyle\sum_{n\geq
0}{\varphi(-q^{1-n}tx)\varphi(-q^{1+n}\Lambda/tx)\over\varphi(-qtx)\varphi(-\Lambda/x)}{(q/t;q)_{n}(qQ/t;q)_{n}\over(q;q)_{n}(qQ;q)_{n}}t^{n}q^{n(n-1)/2}(\Lambda/x)^{n}$
$\displaystyle={}$ $\displaystyle\sum_{k,l,n\geq
0}{(q^{-n};q)_{k}\over(q;q)_{k}}(-qtx)^{k}{(q^{n+1}/t;q)_{l}\over(q;q)_{l}}(-\Lambda/x)^{l}{(q/t;q)_{n}(qQ/t;q)_{n}\over(q;q)_{n}(qQ;q)_{n}}t^{n}q^{n(n-1)/2}(\Lambda/x)^{n}.$

Note that we have the truncation of the summation as $\sum_{k,l,n\geq
0}=\sum_{k,l\geq 0}\sum_{n=k}^{\infty}$ because of the factor
$(q^{-n};q)_{k}$. Shifting the running index $n$ by $k$ as $n\rightarrow n+k$
(hence changing the summation range as $\sum_{k,l\geq
0}\sum_{n=k}^{\infty}\rightarrow\sum_{k,l\geq 0}\sum_{n=0}^{\infty}$
accordingly), we have

$\displaystyle={}$ $\displaystyle\sum_{k,l,n\geq
0}{(q^{-k-n};q)_{k}\over(q;q)_{k}}(-qt)^{k}{(q^{k+n+1}/t;q)_{l}\over(q;q)_{l}}(-1)^{l}$
$\displaystyle\cdot{(q/t;q)_{k+n}(qQ/t;q)_{k+n}\over(q;q)_{k+n}(qQ;q)_{k+n}}t^{k+n}q^{(k+n)(k+n-1)/2}\cdot(\Lambda/x)^{l+n}\Lambda^{k}.$

Then we change the running index $l$ to $m$ (where $m=l+n$) as $\sum_{l,n\geq
0}=\sum_{m\geq 0}\sum_{n=0}^{m}$. Simplifying the factors, we have

$\displaystyle={}$ $\displaystyle\sum_{k\geq 0}\sum_{m\geq
0}{(q^{k+1}/t;q)_{m}\over(q;q)_{m}}(-\Lambda/x)^{m}{(q/t;q)_{k}\over(q;q)_{k}}{(qQ/t;q)_{k}\over(qQ;q)_{k}}(t^{2}\Lambda)^{k}\sum_{n=0}^{m}{(q^{-m};q)_{n}\over(q;q)_{n}}{(q^{k+1}Q/t;q)_{n}\over(q^{k+1}Q;q)_{n}}q^{mn}t^{n}$
$\displaystyle={}$ $\displaystyle\sum_{k\geq 0}\sum_{m\geq
0}{(q^{k+1}/t;q)_{m}\over(q;q)_{m}}{(t;q)_{m}\over(q^{k+1}Q;q)_{m}}(-\Lambda/x)^{m}{(q/t;q)_{k}\over(q;q)_{k}}{(qQ/t;q)_{k}\over(qQ;q)_{k}}(t^{2}\Lambda)^{k}$
$\displaystyle={}$ $\displaystyle T_{(qt^{1/2}Q)^{1/2},x}\mathcal{F}^{(2)}.$

Here, we have used the $q$-Chu-Vandermonde summation formula (B.16) in the
last step.

The second equation (B.18) can be shown in exactly the same manner as above.
Note that (B.18) is equivalent to

$\displaystyle(T_{t^{1/2},x}^{-1}T_{t,\Lambda}^{-1})T_{(qt^{1/2}Q)^{-1/2},x}\mathcal{F}^{(2)}=$
$\displaystyle{}\,{1\over\varphi(t\Lambda)}{\varphi(-x)\varphi(-q\Lambda/x)}\cdot\mathcal{B}^{-1}\cdot{\varphi(qx)\varphi(t\Lambda/x)}\mathcal{F}^{(1)}.$

By using (B.14) and the $q$-binomial formula, we have

$\displaystyle{1\over\varphi(t\Lambda)}{\varphi(-x)\varphi(-q\Lambda/x)}\cdot\mathcal{B}^{-1}\cdot{\varphi(qx)\varphi(t\Lambda/x)}\mathcal{F}^{(1)}$
$\displaystyle={}$
$\displaystyle{1\over\varphi(t\Lambda)}{\varphi(-x)\varphi(-q\Lambda/x)}\cdot\mathcal{B}^{-1}\cdot{\varphi(qx)\varphi(t\Lambda/x)}\sum_{n\geq
0}{(q/t;q)_{n}(qQ/t;q)_{n}\over(q;q)_{n}(qQ;q)_{n}}t^{n}(\Lambda/x)^{n}$
$\displaystyle={}$ $\displaystyle\sum_{n\geq
0}{\varphi(-x)\varphi(-q\Lambda/x)\over\varphi(-q^{n}x)\varphi(-q^{-n}t\Lambda/x)}{(q/t;q)_{n}(qQ/t;q)_{n}\over(q;q)_{n}(qQ;q)_{n}}t^{n}q^{-n(n-1)/2}(\Lambda/x)^{n}$
$\displaystyle={}$ $\displaystyle\sum_{k,l,n\geq
0}{(q^{-n};q)_{k}\over(q;q)_{k}}(-q^{n}x)^{k}{(q^{n+1}/t;q)_{l}\over(q;q)_{l}}(-q^{-n}t\Lambda/x)^{l}{(q/t;q)_{n}(qQ/t;q)_{n}\over(q;q)_{n}(qQ;q)_{n}}t^{n}q^{-n(n-1)/2}(\Lambda/x)^{n}.$

Note that we have $\sum_{k,l,n\geq 0}=\sum_{k,l\geq 0}\sum_{n=k}^{\infty}$
because of the factor $(q^{-n};q)_{k}$. Shifting the running index $n$ by $k$
as $n\rightarrow n+k$ (and changing the summation range as $\sum_{k,l\geq
0}\sum_{n=k}^{\infty}\rightarrow\sum_{k,l\geq 0}\sum_{n=0}^{\infty}$
accordingly), we have

$\displaystyle={}$ $\displaystyle\sum_{k,l,n\geq
0}{(q^{-k-n};q)_{k}\over(q;q)_{k}}(-q^{n})^{k}{(q^{k+n+1}/t;q)_{l}\over(q;q)_{l}}(-q^{-n}t)^{l}$
$\displaystyle\cdot{(q/t;q)_{k+n}(qQ/t;q)_{k+n}\over(q;q)_{k+n}(qQ;q)_{k+n}}t^{k+n}q^{-(k+n)(k+n-1)/2}\cdot(\Lambda/x)^{l+n}\Lambda^{k}.$

Changing the running index $l$ to $m$ (where $m=l+n$) as $\sum_{l,n\geq
0}=\sum_{m\geq 0}\sum_{n=0}^{m}$, and simplifying the factors, we have

$\displaystyle={}$ $\displaystyle\sum_{k\geq 0}\sum_{m\geq
0}{(q^{k+1}/t;q)_{m}\over(q;q)_{m}}(-\Lambda/x)^{m}{(q/t;q)_{k}\over(q;q)_{k}}{(qQ/t;q)_{k}\over(qQ;q)_{k}}\Lambda^{k}q^{-km}t^{k+m}\sum_{n=0}^{m}{(q^{-m};q)_{n}\over(q;q)_{n}}{(q^{k+1}Q/t;q)_{n}\over(q^{k+1}Q;q)_{n}}q^{n}$
$\displaystyle={}$ $\displaystyle\sum_{k\geq 0}\sum_{m\geq
0}{(q^{k+1}/t;q)_{m}\over(q;q)_{m}}{(t;q)_{m}\over(q^{k+1}Q;q)_{m}}(-qQ\Lambda/x)^{m}{(q/t;q)_{k}\over(q;q)_{k}}{(qQ/t;q)_{k}\over(qQ;q)_{k}}(t\Lambda)^{k}$
$\displaystyle={}$
$\displaystyle(T_{t^{1/2},x}^{-1}T_{t,\Lambda}^{-1})T_{(qt^{1/2}Q)^{-1/2},x}\mathcal{F}^{(2)}.$

Here, we have used the $q$-Chu-Vandermonde summation formula (B.15) in the
last step. ∎

####  B.2.3. Macdonald limit in [44]

For the sake of readers’ convenience, we recollect the facts concerning the
choice of the parameters investigated in [44];

$\displaystyle T_{1}=vt^{-1},\qquad T_{2}=v^{-1},\qquad T_{3}=v^{-1},\qquad
T_{4}=vt^{-1}.$ (B.19)

Note that in the limit (B.19) we have (See Conjecture 2.4);

$\displaystyle\mathcal{A}_{1}(\Lambda,x)={1\over\varphi(qt^{-1}x)\varphi(t\Lambda/x)},$
$\displaystyle\mathcal{A}_{2}(\Lambda,x)={\varphi(t\Lambda)\varphi(qt^{-2}\Lambda)\over\varphi(-t^{-1}x)\varphi(-Q^{-1}x)\varphi(-qQ\Lambda/x)\varphi(-q\Lambda/x)},$
$\displaystyle\mathcal{A}_{3}(\Lambda,x)={1\over\varphi(q^{-1}Q^{-1}x)\varphi(q^{2}t^{-1}Q\Lambda/x)}.$

Set

$\displaystyle U(\Lambda,x)=\sum_{k,l\geq
0}{(q/t;q)_{k}(q^{l}qQ/t;q)_{k}\over(q;q)_{k}(q^{l}qQ;q)_{k}}(\Lambda/x)^{k}{(t;q)_{l}(tQ;q)_{l}\over(q;q)_{l}(qQ;q)_{l}}(q\Lambda/t^{2})^{l},$
$\displaystyle V(\Lambda,x)=\sum_{k,l\geq
0}{(q/t;q)_{k}(t;q)_{k}\over(q;q)_{k}(qQ;q)_{k}}(-qQ\Lambda/x)^{k}{(q^{k}t^{2};q)_{l}(tQ;q)_{l}\over(q;q)_{l}(q^{k}qQ;q)_{l}}(q\Lambda/t^{2})^{l}.$

######  Proposition B.12 ([44]).

When (B.19) is satisfied, we have

$\displaystyle\Psi(\Lambda,x)=U(\Lambda,x),$

and

$\displaystyle
V(\Lambda,x)={\varphi(t\Lambda)\over\varphi(-Q^{-1}x)\varphi(-qQ\Lambda/x)}\cdot\mathcal{B}\cdot{1\over\varphi(q^{-1}Q^{-1}x)\varphi(q^{2}t^{-1}Q\Lambda/x)}U\left(\Lambda,{x\over
tqQ}\right),$ $\displaystyle
U(t\Lambda,x)={\varphi(qt^{-2}\Lambda)\over\varphi(qt^{-1}x)\varphi(t\Lambda/x)}\cdot\mathcal{B}\cdot{1\over\varphi(-t^{-1}x)\varphi(-q\Lambda/x)}V\left(\Lambda,x\right).$

######  Proposition B.13.

In the limit (B.19), we have

$\displaystyle{\varphi(t\Lambda/x)\over\varphi(q\Lambda/tx)}\Psi(t\Lambda,x)$
$\displaystyle=\sum_{k,l\geq
0}{(t;q)_{k}(q^{l}tQ;q)_{k}\over(q;q)_{k}(q^{l}qQ;q)_{k}}(q\Lambda/tx)^{k}{(t;q)_{l}(tQ;q)_{l}\over(q;q)_{l}(qQ;q)_{l}}(q\Lambda/t)^{l}$
$\displaystyle=f^{\mathfrak{gl}_{3}}(1/\Lambda,1/x,1|tQ^{-1},t,1|q,t),$
$\displaystyle\mathcal{D}^{\mathfrak{gl}_{3}}(tQ^{-1},t,1|q,t){\varphi(t\Lambda/x)\over\varphi(q\Lambda/tx)}\Psi(t\Lambda,x)=(tQ^{-1}+t+1){\varphi(t\Lambda/x)\over\varphi(q\Lambda/tx)}\Psi(t\Lambda,x),$

where

$\displaystyle\mathcal{D}^{\mathfrak{gl}_{3}}(tQ^{-1},t,1|q,t)$
$\displaystyle=tQ^{-1}{1-\Lambda/tx\over 1-\Lambda/x}{1-\Lambda/t\over
1-\Lambda}T_{q,\Lambda}^{-1}$ $\displaystyle+t{1-t\Lambda/x\over
1-\Lambda/x}{1-x/t\over 1-x}T_{q,x}^{-1}+{1-t\Lambda/x\over
1-\Lambda/x}{1-tx\over 1-x}T_{q,\Lambda}T_{q,x}.$

######  Remark B.14.

In the limit (B.19), the set of partitions giving rise to the non vanishing
contributions for $\mathcal{F}^{(1)}$ remains rather big. This seems to be
consistent with the appearance of double infinite products in the gauge
transformation;

$\displaystyle{\Phi(q^{-1}t^{3}\Lambda)\Phi(q^{2}t^{-2}\Lambda)\over\Phi(qt^{-1}\Lambda)\Phi(t^{2}\Lambda)}{\varphi(tx)\over\varphi(qt^{-1}x)}\mathcal{F}^{(1)}=\Psi(t\Lambda,x).$

In four dimensional theory, the set of partitions giving rise to the non
vanishing contributions is the same. The four dimensional limit of the ratio
of double infinite products is
$(1-\Lambda)^{-\frac{2(\epsilon_{1}+\epsilon_{2})^{2}}{\epsilon_{1}\epsilon_{2}}}$.
On the other hand the corresponding set of partitions for $\mathcal{F}^{(2)}$
becomes small and gives us

$T_{(qt^{1/2}Q)^{1/2},x}\mathcal{F}^{(2)}=\sum_{m,n\geq
0}{(t;1)_{m}(q^{n}q/t;1)_{m}\over(q;q)_{m}(q^{n}qQ;q)_{m}}(-\Lambda/x)^{m}{(q/t;1)_{n}(qQ/t;1)_{n}\over(q;q)_{n}(qQ;q)_{n}}(t^{2}\Lambda)^{n}.$

##  Appendix C Four dimensional limit in a factorized form

In section 4, we computed the four dimensional limit of the operator $\SS$.
Here we will consider the limit of each factor

$\begin{array}[]{l}\displaystyle
K_{1}=\frac{1}{\varphi(\Lambda)}\varphi(-d_{1}x)\varphi(-d_{3}\frac{\Lambda}{x})\mathcal{B}^{-1}\varphi(qx)\varphi(\frac{\Lambda}{x}),\\\
\displaystyle
K_{2}=\frac{1}{\varphi(q^{-1}d_{2}d_{4}\Lambda)}\varphi(q^{-1}d_{1}d_{2}x)\varphi(d_{3}d_{4}\frac{\Lambda}{x})\mathcal{B}^{-1}\varphi(-d_{2}x)\varphi(-d_{4}\frac{\Lambda}{x}),\\\
\displaystyle
N=\frac{\varphi(q^{-1}d_{2}d_{4}\Lambda)}{\varphi(q^{-1}d_{1}d_{2}d_{3}d_{4}\Lambda)},\end{array}$
(C.1)

in the decomposition

$\SS^{-1}=NK_{2}K_{1},$ (C.2)

separately. We show that each factor $K_{i}$ has also a well defined four
dimensional limit, though the result is not usual differencial operator any
more. To describe the results in a simple form, it is convenient to use the
normal ordering $:\ :$ which is a $\mathbb{C}$-linear map from a commutative
ring $\mathbb{C}(x,\vartheta_{x})_{/\vartheta_{x}x=x\vartheta_{x}}$ to the
corresponding non-commutative ring
$\mathbb{C}(x,\vartheta_{x})_{/\vartheta_{x}x=x(\vartheta_{x}+1)}$ defined by

$:x^{n}F(x,\vartheta_{x}):\
=x^{n}:F(x,\vartheta_{x}):,\quad:F(x,\vartheta_{x})\vartheta_{x}^{n}:\ =\
:F(x,\vartheta_{x}):{\vartheta_{x}}^{n},\quad:1:=1.$ (C.3)

######  Theorem C.1.

For $q=e^{h},d_{i}=q^{m_{i}}$, we have

$\displaystyle K_{1}$
$\displaystyle=~{}:(1+x)^{-m_{1}-\vartheta_{x}}(1+\frac{\Lambda}{x})^{-m_{3}+\vartheta_{x}}\left(1+\frac{h}{2}A_{1}\right)+{\mathcal{O}}(h^{2}):,$
$\displaystyle K_{2}$
$\displaystyle=~{}:(1-x)^{-m_{1}-\vartheta_{x}}(1-\frac{\Lambda}{x})^{-m_{3}+\vartheta_{x}}\left(1+\frac{h}{2}A_{2}\right)+{\mathcal{O}}(h^{2}):,$
$\displaystyle N$
$\displaystyle=(1-\Lambda)^{-(m_{1}+m_{3})}\left(1+\frac{h}{2}\frac{(m_{1}+m_{3})(m_{1}+2m_{2}+m_{3}+2m_{4}-3)}{(1-\Lambda)}\right)+{\mathcal{O}}(h^{2}),$

where

$\displaystyle A_{1}$
$\displaystyle=\frac{x(\vartheta_{x}-m_{1}+1)(\vartheta_{x}+m_{1})}{1+x}+\frac{\frac{\Lambda}{x}(\vartheta_{x}+m_{3}-1)(\vartheta_{x}-m_{3})}{1+\frac{\Lambda}{x}}-\vartheta_{x}(\vartheta_{x}+1),$
$\displaystyle A_{2}$
$\displaystyle=-\frac{x(\vartheta_{x}-m_{1}-2m_{2}+3)(\vartheta_{x}+m_{1})}{1-x}-\frac{\frac{\Lambda}{x}(\vartheta_{x}+m_{3}+2m_{4}-1)(\vartheta_{x}-m_{3})}{1-\frac{\Lambda}{x}}-\vartheta_{x}(\vartheta_{x}+1).$

###### Proof.

From (B.14) in Appendix B, we have

$\mathcal{B}^{-1}\varphi(qx)\varphi(\frac{\Lambda}{x})x^{n}=\frac{\varphi(\Lambda)}{\varphi(-q^{-n}x)\varphi(-q^{n}\frac{\Lambda}{x})}q^{-n(n+1)}x^{n},$
(C.4)

and hence

$K_{1}x^{n}=\frac{\varphi(-d_{1}x)\varphi(-d_{3}\frac{\Lambda}{x})}{\varphi(-q^{-n}x)\varphi(-q^{n}\frac{\Lambda}{x})}q^{-n(n+1)}x^{n}.$
(C.5)

Then, using the limiting formula of the $q$-binomial theorem

$\frac{\varphi(q^{j}x)}{\varphi(x)}=\sum_{k=1}^{\infty}\frac{(q^{j})_{k}}{(q)_{k}}x^{k}=(1-x)^{-j}\big{\\{}1+\frac{hj(j-1)}{2}\frac{x}{1-x}\big{\\}}+{\mathcal{O}}(h^{2}),$
(C.6)

we obtain

$\begin{array}[]{l}\displaystyle
K_{1}x^{n}=(1+x)^{-m_{1}-n}(1+\frac{\Lambda}{x})^{-m_{3}+n}(1+\frac{h}{2}A_{1})+{\mathcal{O}}(h^{2})\\\
\displaystyle\qquad
A_{1}=\frac{x(n-m_{1}+1)(n+m_{1})}{1+x}+\frac{\frac{\Lambda}{x}(n+m_{3}-1)(n-m_{3})}{1+\frac{\Lambda}{x}}-n(n+1),\end{array}$
(C.7)

as desired.

The expression for $K_{2}$ follows from that for $K_{1}$ by the substitution

$x\to-q^{-1}d_{2}x,\quad\Lambda\to q^{-1}d_{2}d_{4}\Lambda.$ (C.8)

The limit of $N$ follows directly as

$\begin{array}[]{l}\displaystyle
N=\exp\sum_{n=1}^{\infty}\frac{(q^{-1}d_{2}d_{4}\Lambda)^{n}-(q^{-1}d_{1}d_{2}d_{3}d_{4}\Lambda)^{n}}{n(1-q^{n})}=\exp\sum_{n=1}^{\infty}\frac{(q^{m_{2}+m_{4}-1}\Lambda)^{n}(1-q^{m_{1}+m_{3}+n})}{n(1-q^{n})}\\\
\displaystyle=\exp\\{\sum_{n=1}^{\infty}\frac{1}{n}(1+nh(m_{2}+m_{4}-1))\Lambda^{n}(m_{1}+m_{3})(1+\frac{nh}{2}(m_{1}+m_{3}-1))\\}+{\mathcal{O}}(h^{2}),\\\
\displaystyle=\exp\\{(m_{1}+m_{3})\sum_{n=1}^{\infty}\frac{\Lambda^{n}}{n}(1+\frac{nh(m_{1}+2m_{2}+m_{3}+2m_{4}-3)}{2})\\}+{\mathcal{O}}(h^{2}),\\\
\displaystyle=(1+\frac{h(m_{1}+2m_{2}+m_{3}+2m_{4}-3)}{2}\vartheta_{\Lambda})\exp\\{(m_{1}+m_{3})\sum_{n=1}^{\infty}\frac{\Lambda^{n}}{n}\\}+{\mathcal{O}}(h^{2}),\\\
\displaystyle=(1-\Lambda)^{-(m_{1}+m_{3})}(1+\frac{h}{2}\frac{(m_{1}+m_{3})(m_{1}+2m_{2}+m_{3}+2m_{4}-3)}{(1-\Lambda)})+{\mathcal{O}}(h^{2}).\end{array}$
(C.9)

∎

######  Remark C.2.

The result of Theorem C.1 is consistent with the result in section 4. For
instance, the consistency in the leading order is given by the identity

$:(1-x)^{-m_{1}-\vartheta_{x}}(1-\frac{\Lambda}{x})^{-m_{3}+\vartheta_{x}}::(1+x)^{-m_{1}-\vartheta_{x}}(1+\frac{\Lambda}{x})^{-m_{3}+\vartheta_{x}}:=(1-\Lambda)^{m_{1}+m_{3}},$
(C.10)

which follows from formal computations such as

$:(1+x)^{-m_{1}-\vartheta_{x}}(1+\frac{\Lambda}{x})^{-m_{3}+\vartheta_{x}}:f(x)=(1+x)^{-m_{1}}(1+\frac{\Lambda}{x})^{-m_{3}}f(x\frac{1+\frac{\Lambda}{x}}{1+x}).$
(C.11)

In relation to (C.11), the formula

$:(1-x)^{-b-\vartheta_{x}}:F(a,b,c;-x)=(1-x)^{-b}F(a,b,c,\frac{x}{x-1})=F(c-a,b,c;x).$
(C.12)

for the Gauss Hypergeometric series $F(a,b,c;x)$ will be instructive.

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