Limit of FD of PBCA using the GKZ hypergeometric function
Definition of the GKZ hypergeometric function and its properties
In this section, we define the GKZ hypergeometric function and its properties. First, let us define \(B_{k}^{\gamma }(x)\) by
$$\begin{aligned} B_{k}^{\gamma }(x)=\frac{x^{\gamma +k}}{\varGamma (\gamma +k+1)}. \end{aligned}$$
It satisfies the following contiguity relations.
$$\begin{aligned} \frac{d}{dx}B_{k}^{\gamma }(x)& = B_{k-1}^{\gamma }(x)=B_{k}^{\gamma -1}(x),\\ x B_{k}^{\gamma }(x)& = (\gamma +k+1)B_{k+1}^{\gamma }(x)=(\gamma +k+1)B_{k}^{\gamma +1}(x). \end{aligned}$$
For an \(m\times n\) matrix \(A=(a_{ij})\) and an n-dimensional vector \(\mathbf {\beta }\), general solution to
$$\begin{aligned} A\mathbf {\gamma }=\mathbf {\beta }, \end{aligned}$$
is expressed by
$$\begin{aligned} \mathbf {\gamma }=\mathbf {\gamma _0}+\mathbf {k}, \quad \mathbf {k}\in \ker A, \end{aligned}$$
where \(\mathbf {\gamma _0}\) is a special solution to the equation. Using A, \(\mathbf {\beta }\), \(\mathbf {\gamma _0}\) and \(\mathbf {k}\), the GKZ hypergeometric function \(\varphi (\mathbf {\beta };\mathbf {x})\) is defined by
$$\begin{aligned} \varphi (\mathbf {\beta };\mathbf {x})=\sum _{\mathbf {k}\in \ker A \cap \mathbb {Z}^n} \prod _{j=1}^{n} B_{k_j}^{\gamma _{0,j}}(x_j), \end{aligned}$$
(10)
where \(\mathbf {x}\) is an n-dimensional vector and \(\gamma _{0,j}\), \(k_j\) and \(x_j\) are jth element of \(\mathbf {\gamma _0}\) , \(\mathbf {k}\), and \(\mathbf {x}\), respectively. Although \(\varphi \) also depends on the matrix A, A is fixed below and we omit the dependency. Using contiguity relations, the following two properties for GKZ hypergeometric function are derived.
First, for a given vector
$$\begin{aligned} \mathbf {b}= \left( \begin{array}{c} b_1\\ b_2\\ \vdots \\ b_n \end{array} \right) \in \ker A \cap \mathbb {Z}^n, \end{aligned}$$
\(J_{+}(\mathbf {b})\) and \(J_{-}(\mathbf {b})\) are defined by subsets of \(\{0,1,\ldots ,n \}\) as
$$\begin{aligned} J_{+}(\mathbf {b})=\{j|b_j > 0 \} ,\qquad J_{-}(\mathbf {b})= \{j|b_j < 0\}. \end{aligned}$$
Then, we obtain
$$\begin{aligned} \prod _{j \in J_{-}(\mathbf {b})} \left( \frac{\partial }{\partial x_j}\right) ^{-b_j}\varphi& = \sum _{\mathbf {k}\in \ker A \cap \mathbb {Z}^n} \prod _{j \in J_{-}(\mathbf {b})} \left( \frac{\partial }{\partial x_j}\right) ^{-b_j} \prod _{j=1}^n B_{k_j}^{\gamma _{0,j}}(x_j) \\& = \sum _{\mathbf {k}\in \ker A \cap \mathbb {Z}^n} \prod _{j \in J_{+}(\mathbf {b})} B_{k_j}^{\gamma _{0,j}}(x_j) \prod _{j \in J_{-}(\mathbf {b})} B_{k_j+b_j}^{\gamma _{0,j}}(x_j) \\& = \sum _{\mathbf {k}\in \ker A \cap \mathbb {Z}^n} \prod _{j \in J_{+}(\mathbf {b})} \left( \frac{\partial }{\partial x_j}\right) ^{b_j} \prod _{j=1}^n B_{k_j}^{\gamma _{0,j}}(x_j) \\& = \prod _{j \in J_{+}(\mathbf {b})} \left( \frac{\partial }{\partial x_j}\right) ^{b_j}\varphi . \end{aligned}$$
Therefore, \(\varphi \) satisfies a differential equation
$$\begin{aligned} \left\{ \prod _{j \in J_{+}(\mathbf {b})} \left( \frac{\partial }{\partial x_j}\right) ^{b_j}-\prod _{j \in J_{-}(\mathbf {b})} \left( \frac{\partial }{\partial x_j}\right) ^{-b_j} \right\} \varphi =0. \end{aligned}$$
(11)
Second, for an operator \(\theta _j=x_j\frac{\partial }{\partial x_j}\) and \(\mathbf {\theta }\) defined by
$$\begin{aligned} \mathbf {\theta }= \left( \begin{array}{c} \theta _1\\ \theta _2\\ \vdots \\ \theta _n \end{array} \right) , \end{aligned}$$
\(\varphi \) satisfies
$$\begin{aligned} A\mathbf {\theta }\varphi (\mathbf {\beta };\mathbf {x}) =A(\mathbf {\gamma }+\mathbf {k})\varphi =\mathbf {\beta }\varphi . \end{aligned}$$
Using \(\alpha \in \mathbb {C}^{\times }\) and the ith row vector \(\mathbf {a_i}=(a_{i1},\ldots ,a_{in})\) of A, define \(\alpha ^{D(\mathbf {a_i})}\) by
$$\begin{aligned} \alpha ^{D(\mathbf {a_i})}= \hbox {diag}(\alpha ^{a_{i1}},\ldots ,\alpha ^{a_{in}}). \end{aligned}$$
Since
$$\begin{aligned} B_{k_j}^{\gamma _j} (\alpha ^{a_{ij}} x_j) = \alpha ^{a_{ij} (\gamma _j + k_j)} B_{k_j}^{\gamma _j}(x_j), \end{aligned}$$
\(B_{k_j}^{\gamma _j} (x_j)\) satisfies
$$\begin{aligned} \prod _{j=1}^{n}B_{k_j}^{\gamma _j} (\alpha ^{a_{ij}} x_j) = \alpha ^{\sum _{j=1}^{n}a_{ij} (\gamma _j + k_j)} \prod _{j=1}^{n} B_{k_j}^{\gamma _j}(x_j) = \alpha ^{\beta _i} \prod _{j=1}^{n} B_{k_j}^{\gamma _j}(x_j). \end{aligned}$$
Thus,
$$\begin{aligned} \varphi (\mathbf {\beta };\alpha ^{D(\mathbf {a_i})} \mathbf {x}) = \alpha ^{\beta _i}\varphi (\mathbf {\beta };\mathbf {x}), \end{aligned}$$
(12)
is obtained.
FD of PBCA expressed by the GKZ hypergeometric function
In this subsection, we express FD of PBCA by the GKZ hypergeometric function. The FD of PBCA (3) for the infinite space size is
$$\begin{aligned} Q=\lim _{\begin{array}{c} L\rightarrow \infty \\ m=\rho L \end{array}}Q_{L,\alpha }(m)=\lim _{\begin{array}{c} L\rightarrow \infty \\ m=\rho L \end{array}}\frac{\alpha }{L}\frac{\sum _{k=1}^{m} k \left( \frac{1}{1-\alpha }\right) ^{k-1} N_{L,m}(k)}{\sum _{k=1}^{m} \left( \frac{1}{1-\alpha }\right) ^{k-1} N_{L,m}(k)}, \end{aligned}$$
where density \(\rho \) is constant and
$$\begin{aligned} N_{L,m}(k)=\frac{(m-1)!L(L-m-1)! }{(m-k)!(k-1)!k! (L-m-k)!}. \end{aligned}$$
To evaluate the above limit, we choose the following matrix A and vector \(\mathbf {\beta _1}\) in the definition of the GKZ hypergeometric function in A.1.
$$\begin{aligned} A = \left( \begin{array}{cccc} 0 &{} -1 &{} 1 &{} 0 \\ 1 &{} 1 &{} 1 &{} 1 \\ 1 &{} 0 &{} 1 &{} 0 \end{array} \right) , \qquad \mathbf {\beta _1}= \left( \begin{array}{c} 1\\ L-1\\ m \end{array} \right) . \end{aligned}$$
The general solution to \(A \mathbf {\gamma }=\mathbf {\beta _1}\) is
$$\begin{aligned} \mathbf {\gamma }=\mathbf {\gamma _0}+\mathbf {k}\qquad (\mathbf {k}\in \ker A \cap \mathbb {Z}^4 ), \end{aligned}$$
where
$$\begin{aligned} \mathbf {\gamma _0}= \left( \begin{array}{c} \gamma _{0,1}\\ \gamma _{0,2}\\ \gamma _{0,3}\\ \gamma _{0,4}\\ \end{array} \right) = \left( \begin{array}{c} m\\ -1\\ 0 \\ L-m \end{array} \right) , \qquad \mathbf {k}= \left( \begin{array}{c} k_{1}\\ k_{2}\\ k_{3}\\ k_{4}\\ \end{array} \right) = k \left( \begin{array}{c} -1\\ 1\\ 1 \\ -1 \end{array} \right) \quad (k\in \mathbb {Z}). \end{aligned}$$
(13)
The form of A is chosen to give \(\mathbf {\gamma }\) of which elements corresponding to the variables of factorials in the denominator of \(N_{L,m}(k)\). Introducing a new notation \(\varphi _{\beta _1, \beta _2, \beta _3}(x_1, x_2, x_3, x_4)\) defined by \(\varphi (\mathbf {k}, \mathbf {x})\) for \(\mathbf {\beta }=(\beta _1, \beta _2, \beta _3)\) and \(\mathbf {x}=(x_1, x_2, x_3, x_4)\) and \(\lambda \) by \(1/(1-\alpha )\), we obtain
$$\begin{aligned} \varphi _{1,L-1,m}(1,\lambda ,1,1)& = \sum _{\mathbf {k}\in \ker A \cap \mathbb {Z}^4} B_{k_1}^{\gamma _{0,1}}(1) B_{k_2}^{\gamma _{0,2}}(\lambda ) B_{k_3}^{\gamma _{0,3}}(1) B_{k_4}^{\gamma _{0,4}}(1) \\& = \sum _{k=1}^{m} \frac{\lambda ^{k-1}}{(m-k)! (k-1)! k! (L-m-k)!}. \end{aligned}$$
Considering \(\varphi _{\beta _1, \beta _2, \beta _3}(1,\lambda ,1,1)\) as the function on \(\lambda \), let us introduce a notation \(F_{\beta _1, \beta _2, \beta _3}(\lambda )=\varphi _{\beta _1, \beta _2, \beta _3}(1,\lambda ,1,1)\).
On the other hand, the general solution to \(A \mathbf {\gamma '}=\mathbf {\beta _2}\) for
$$\begin{aligned} \mathbf {\beta _2}= \left( \begin{array}{c} 0\\ L-2\\ m-1 \end{array} \right) , \end{aligned}$$
is
$$\begin{aligned} \mathbf {\gamma '}=\mathbf {\gamma '_0}+\mathbf {k}\qquad (\mathbf {k}\in \ker A \cap \mathbb {Z}^4 ), \end{aligned}$$
where
$$\begin{aligned} \mathbf {\gamma '_0}= \left( \begin{array}{c} \gamma '_{0,1}\\ \gamma '_{0,2}\\ \gamma '_{0,3}\\ \gamma '_{0,4}\\ \end{array} \right) = \left( \begin{array}{c} m\\ -1\\ -1 \\ L-m \end{array} \right) . \end{aligned}$$
(14)
Therefore,
$$\begin{aligned} \varphi _{0,L-2,m-1}(1,\lambda ,1,1)& = \sum _{\mathbf {k}\in \ker A \cap \mathbb {Z}^4} B_{k_1}^{\gamma '_{0,1}}(1) B_{k_2}^{\gamma '_{0,2}}(\lambda ) B_{k_3}^{\gamma '_{0,3}}(1) B_{k_4}^{\gamma '_{0,4}}(1) \\& = \sum _{k=1}^{m} \frac{\left( \frac{1}{1-\alpha }\right) ^{k-1}}{(m-k)! (k-1)! (k-1)! (L-m-k)!} \\& = F_{0,L-2,m-1}(\lambda ), \end{aligned}$$
is obtained. Thus, we have
$$\begin{aligned} Q=\lim _{\begin{array}{c} L\rightarrow \infty \\ m=\rho L \end{array}}\frac{\alpha }{L}\frac{F_{0,L-2,m-1}(\lambda )}{F_{1,L-1,m}(\lambda )}. \end{aligned}$$
(15)
Differential equation on \(F_{1,L-1,m}\)
In this section, we derive a differential equation on \(F_{1,L-1,m}\). Since the relations
$$\begin{aligned} \varphi _{1,L-1,m}\bigg (x_1, \frac{x_2}{a}, a x_3, x_4\bigg )& = a \varphi _{1,L-1,m}(x_1,x_2,x_3,x_4), \\ \varphi _{1,L-1,m}(b x_1, b x_2, b x_3, b x_4)& = b^{L-1}\varphi _{1,L-1,m}(x_1,x_2,x_3,x_4), \\ \varphi _{1,L-1,m}(c x_1, x_2, c x_3, x_4)& = c^m \varphi _{1,L-1,m}(x_1,x_2,x_3,x_4), \end{aligned}$$
for \(\varphi _{1,L-1,m}(x_1,x_2,x_3,x_4)\) and a, b ,\(c \in \mathbb {C}^{\times }\) are obtained from (12), the relation
$$\begin{aligned} \varphi _{1,L-1,m}\bigg (bc x_1,\frac{b}{a}x_2,abc x_3,b x_4\bigg )=ab^{L-1} c^{m} \varphi _{1,L-1,m}(x_1,x_2,x_3,x_4), \end{aligned}$$
holds. If we assume
$$\begin{aligned} \bigg (bc x_1,\frac{b}{a}x_2,abc x_3,b x_4\bigg )=(1,\lambda ,1,1), \end{aligned}$$
that is,
$$\begin{aligned} a=\frac{x_1}{x_3},\quad b=\frac{1}{x_4},\quad c=\frac{x_4}{x_1},\quad \lambda =\frac{x_2 x_3}{x_1 x_4}, \end{aligned}$$
(16)
the relation between \(\varphi _{1,L-1,m}\) and \(F_{1,L-1,m}\) is
$$\begin{aligned} \varphi _{1,L-1,m}(x_1,x_2,x_3,x_4)=x_1^{m-1} x_3 x_4^{L-m-1} F_{1,L-1,m}(\lambda ). \end{aligned}$$
(17)
Moreover, we can derive
$$\begin{aligned} \frac{\partial ^2}{\partial x_2 \partial x_3} \varphi _{1,L-1,m}& = 2x_1^{m-2} x_3 x_4^{L-m-2} F'_{1,L-1,m}(\lambda )+x_1^{m-3} x_2 x_3^2 x_4^{L-m-3} F''_{1,L-1,m}(\lambda ), \\ \frac{\partial ^2}{\partial x_1 \partial x_4} \varphi _{1,L-1,m}& = (m-1)(L-m-1)x_1^{m-2} x_3 x_4^{L-m-2}F_{1,L-1,m}(\lambda ) \\&-(L-m-1)x_1^{m-3} x_2 x_3^2 x_4^{L-m-3} F'_{1,L-1,m}(\lambda ) \\&-(m-2)x_1^{m-3} x_2 x_3^2 x_4^{L-m-3} F'_{1,L-1,m}(\lambda ) \\&+x_1^{m-4} x_2^2 x_3^3 x_4^{L-m-4} F''_{1,L-1,m}(\lambda ). \end{aligned}$$
Substituting these equations into the differential equation
$$\begin{aligned} \left( \frac{\partial ^2}{\partial x_2 \partial x_3} - \frac{\partial ^2}{\partial x_1 \partial x_4}\right) \varphi _{1,L-1,m}=0, \end{aligned}$$
which is derived from (11), the following differential equation on \(F_{1,L-1,m}\) is obtained.
$$\begin{aligned}&\lambda (1-\lambda ) F''_{1,L-1,m}(\lambda )+\{ (L-3)\lambda +2 \} F'_{1,L-1,m}(\lambda ) \nonumber \\&\qquad \qquad -(m-1)(L-m-1) F_{1,L-1,m}(\lambda )=0. \end{aligned}$$
(18)
Contiguity relation between \(F_{1,L-1,m}\) and \(F_{0,L-2,m-1}\)
From (13) and (14), we have
$$\begin{aligned} \mathbf {\gamma '}-\mathbf {\gamma }= \left( \begin{array}{c} 0\\ 0\\ -1 \\ 0 \end{array} \right) +k \left( \begin{array}{c} -1\\ 1\\ 1 \\ -1 \end{array} \right) \qquad (k\in \mathbb {Z}). \end{aligned}$$
Therefore,
$$\begin{aligned} \varphi _{0,L-2,m-1}(x_1,x_2,x_3,x_4)& = \sum _{\mathbf {k}\in \ker A \cap \mathbb {Z}^4} B_{k_1}^{\gamma '_{0,1}}(x_1) B_{k_2}^{\gamma '_{0,2}}(x_2) B_{k_3}^{\gamma '_{0,3}}(x_3) B_{k_4}^{\gamma '_{0,4}}(x_4) \nonumber \\& = \sum _{\mathbf {k}\in \ker A \cap \mathbb {Z}^4} B_{k_1}^{\gamma _{0,1}}(x_1) B_{k_2}^{\gamma _{0,2}}(x_2) B_{k_3}^{\gamma _{0,3}-1}(x_3) B_{k_4}^{\gamma _{0,4}}(x_4) \nonumber \\& = \frac{\partial }{\partial x_3} \varphi _{1,L-1,m}(x_1,x_2,x_3,x_4), \end{aligned}$$
(19)
is derived. From (12), since the relations
$$\begin{aligned} \varphi _{0,L-2,m-1}\left( x_1,\frac{1}{a}x_2,a x_3, x_4\right)& = \varphi _{0,L-2,m-1}(x_1,x_2,x_3,x_4), \\ \varphi _{0,L-2,m-1}(bx_1,bx_2,bx_3,bx_4)& = b^{L-2}\varphi _{0,L-2,m-1}(x_1,x_2,x_3,x_4), \\ \varphi _{0,L-2,m-1}(c x_1,x_2,c x_3,x_4)& = c^{m-1} \varphi _{0,L-2,m-1}(x_1,x_2,x_3,x_4), \end{aligned}$$
are obtained, the relation
$$\begin{aligned} \varphi _{0,L-2,m-1}\left( bc x_1,\frac{b}{a}x_2,abc x_3,b x_4\right) = b^{L-2} c^{m-1} \varphi _{0,L-2,m-1}(x_1,x_2,x_3,x_4), \end{aligned}$$
holds. Substituting (16) into this equation, we have
$$\begin{aligned} \varphi _{0,L-2,m-1}(x_1,x_2,x_3,x_4)=x_1^{m-1} x_4^{L-m-1} F_{0,L-2,m-1}(\lambda ). \end{aligned}$$
(20)
Thus, we obtain
$$\begin{aligned} x_1^{m-1} x_4^{L-m-1} F_{0,L-2,m-1}(\lambda )=\frac{\partial }{\partial x_3} \varphi _{1,L-1,m}(x_1,x_2,x_3,x_4), \end{aligned}$$
(21)
from (19) and (20).
On the other hand, derivative of (17) with respect to \(x_3\) is
$$\begin{aligned}&\frac{\partial }{\partial x_3} \varphi _{1,L-1,m}(x_1,x_2,x_3,x_4) \nonumber \\&\quad =x_1^{m-1}x_4^{L-m-1} F_{1,L-1,m}(\lambda )+x_1^{m-2} x_2 x_3 x_4^{L-m-2} F'_{1,L-1,m}(\lambda ). \end{aligned}$$
(22)
Therefore, from (21) and (22),
$$\begin{aligned} F_{0,L-2,m-1}(\lambda )=F_{1,L-1,m}(\lambda )+\lambda F'_{1,L-1,m}(\lambda ), \end{aligned}$$
(23)
is obtained. Derivative of (23) with respect to \(\lambda \) is
$$\begin{aligned} F'_{0,L-2,m-1}(\lambda )= 2F'_{1,L-1,m}(\lambda ) + \lambda F''_{1,L-1,m}(\lambda ). \end{aligned}$$
(24)
Substituting (18) into (24) gives
$$\begin{aligned} F'_{0,L-2,m-1}(\lambda )=\frac{(m-1)(L-m-1)}{1-\lambda } F_{1,L-1,m}(\lambda )-\frac{(L-1)\lambda }{1-\lambda }F'_{1,L-1,m}(\lambda ). \end{aligned}$$
(25)
From (23) and (25), contiguity relation between \(F_{1,L-1,m}\) and \(F_{0,L-2,m-1}\) is
$$\begin{aligned} \left( \begin{array}{c} F_{0,L-2,m-1}(\lambda )\\ F_{0,L-2,m-1}'(\lambda ) \end{array} \right) = \left( \begin{array}{cc} 1 &{} \lambda \\ \frac{(m-1)(L-m-1)}{1-\lambda } &{} \frac{(L-1)\lambda }{1-\lambda } \end{array} \right) \left( \begin{array}{c} F_{1,L-1,m}(\lambda )\\ F_{1,L-1,m}'(\lambda ) \end{array} \right) . \end{aligned}$$
(26)
Limit of contiguity relation
From the contiguity relation (26), we have
$$\begin{aligned} \left( \begin{array}{c} F_{0,L-2,m-1}(\lambda )\\ \frac{F'_{0,L-2,m-1}(\lambda )}{m} \end{array} \right) = \left( \begin{array}{cc} 1 &{} \lambda m \\ \frac{(m-1)(L-m-1)}{(1-\lambda ) m} &{} \frac{(L-1)\lambda }{1-\lambda } \end{array} \right) \left( \begin{array}{c} F_{1,L-1,m}(\lambda )\\ \frac{F'_{1,L-1,m}(\lambda )}{m} \end{array} \right) . \end{aligned}$$
(27)
Dividing both side of (27) by \(LF_{1,L-1,m}(\lambda )\),
$$\begin{aligned} \left( \begin{array}{c} \frac{F_{0,L-2,m-1}(\lambda )}{L F_{1,L-1,m}(\lambda )}\\ \frac{F'_{0,L-2,m-1}(\lambda )}{m L F_{1,L-1,m}(\lambda )} \end{array} \right) = \left( \begin{array}{cc} \frac{1}{L} &{} \frac{\lambda m}{L} \\ \frac{(m-1)(L-m-1)}{(1-\lambda ) m L} &{} \frac{(L-1)\lambda }{(1-\lambda ) L} \end{array} \right) \left( \begin{array}{c} 1\\ \frac{F'_{1,L-1,m}(\lambda )}{m F_{1,L-1,m}(\lambda )} \end{array} \right) , \end{aligned}$$
(28)
is obtained. If we define
$$\begin{aligned} g=\frac{F'_{1,L-1,m}(\lambda )}{F_{1,L-1,m}(\lambda )}, \end{aligned}$$
and substitute g into (18), we obtain
$$\begin{aligned} \lambda (1-\lambda )(g'+g^2)+\{ (L-3)\lambda +2 \} g-(m-1)(L-m-1)=0. \end{aligned}$$
Since \(\rho =m/L\), we have
$$\begin{aligned} \lambda (1-\lambda )(g'+g^2)+\left\{ \left( \frac{m}{\rho }-3\right) \lambda +2 \right\} g-(m-1)\left( \frac{m}{\rho }-m-1\right) =0. \end{aligned}$$
(29)
We can assume the following expansion of g for \(m=\infty \):
$$\begin{aligned} g=g_1 m+g_0 + g_{-1}m^{-1} +g_{-2}m^{-2} +\cdots . \end{aligned}$$
From the balance of \({\mathcal {O}}(m^2)\) terms of (29), we have
$$\begin{aligned} \lambda (1-\lambda )g_1^2+\frac{\lambda }{\rho }g_1+1-\frac{1}{\rho }=0. \end{aligned}$$
Solving this relation,
$$\begin{aligned} g_1=\frac{-\frac{\lambda }{\rho }-\sqrt{\left( \frac{\lambda }{\rho }\right) ^2-4\lambda (1-\lambda )\left( 1-\frac{1}{\rho }\right) }}{2\lambda (1-\lambda )}, \end{aligned}$$
(30)
is obtained. Therefore, we can derive the limit of the first component of (28) using \(g_1\) as
$$\begin{aligned} \lim _{\begin{array}{c} L\rightarrow \infty \\ m=\rho L \end{array}}\frac{\alpha }{L}\frac{F_{0,L-2,m-1}(\lambda )}{ F_{1,L-1,m}(\lambda )}=\alpha \rho \lambda g_1=\frac{1-\sqrt{1-4\alpha \rho (1-\rho )}}{2}. \end{aligned}$$
(31)