On the Analytic Continuation of Lauricella–Saran Hypergeometric Function F_K(a₁,a₂,b₁,b₂;a₁,b₂,c₃;z)

Abstract

The paper establishes an analytical extension of two ratios of Lauricella–Saran hypergeometric functions $F_K$ with some parameter values to the corresponding branched continued fractions in their domain of convergence. The PC method used here is based on the correspondence between a formal triple power series and a branched continued fraction. As additional results, analytical extensions of the Lauricella–Saran hypergeometric functions $F_K(a_1,a_2,1,b_2;a_1,b_2,c_3;\mathbf{z})$ and $F_K(a_1,1,b_1,b_2;a_1,b_2,c_3;\mathbf{z})$ to the corresponding branched continued fractions were obtained. To illustrate this, we provide some numerical experiments at the end.

1. Introduction

Hypergeometric functions of one and several variables occur naturally in a variety of applied mathematics, statistics and other decision sciences, chemistry and biology, mathematical physics, and engineering sciences. Their investigation has a very long history and a large bibliography (see, for example, [1,2,3,4,5]).

In 1893, G. Lauricella defined and studied four hypergeometric series $F_A,$$F_B,$$F_C,$ and $F_D$ of three variables [6]. He also indicated the existence of ten other hypergeometric functions of three variables $F_E,$$F_F,$ $\dots ,$$F_T,$ which were studied by Sh. Saran in 1954 [7].

Lauricella–Saran hypergeometric function $F_K$ is defined by triple power series

$$\begin{array}{cc}\hfill F_K(a_1,a_2,b_1,b_2;c_1,c_2,c_3;\mathbf{z})& =\sum _{p,q,r=0}^{+\infty }\frac{{\left(a_1\right)}_p{\left(a_2\right)}_{q+r}{\left(b_1\right)}_{p+r}{\left(b_2\right)}_q}{{\left(c_1\right)}_p{\left(c_2\right)}_q{\left(c_3\right)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!},\hfill \end{array}$$

(1)

where $a_1,$$a_2,$$b_1,$$b_2,$$c_1,$$c_2,$ and $c_3$ are complex constants, $c_1,c_2,c_3\notin \{0,-1,-2,\dots \},$ $\mathbf{z}=(z_1,z_2,z_3)\in {\mathrm{D}}_{F_K},$

$${\mathrm{D}}_{F_K}=\{\mathbf{z}\in {\mathbb{C}}^3:|z_k|<1,k=1,2,|z_3|<(1-|z_1\left|\right)(1-|z_2\left|\right)\};$$

${(·)}_k$ is the Pochhammer symbol, defined as follows: ${\left(\alpha \right)}_0=1,$${\left(\alpha \right)}_k=\alpha {(\alpha +1)}_{k-1},$$k\ge 1.$ Applications and recent studies of these functions can be found, for instance, in [8,9,10,11] (see also [12,13,14,15,16,17]).

In this paper, we study the analytic continuation of the Lauricella–Saran hypergeometric function $F_K$ with some parameter values into a branched continued fraction of the form

$$\begin{array}{c}\hfill v_0\left(\mathbf{z}\right)+\sum _{i_1=1}^3\frac{u_{i\left(1\right)}\left(\mathbf{z}\right)}{{\displaystyle v_{i\left(1\right)}\left(\mathbf{z}\right)+\sum _{i_2=1}^3\frac{u_{i\left(2\right)}\left(\mathbf{z}\right)}{{\displaystyle v_{i\left(2\right)}\left(\mathbf{z}\right)+{}_{\ddots }}}}},\end{array}$$

(2)

where the $v_0\left(\mathbf{z}\right)$ and the elements $u_{i\left(k\right)}\left(\mathbf{z}\right)$ and $v_{i\left(k\right)}\left(\mathbf{z}\right),$ $i\left(k\right)\in \mathcal{I},$

$$\mathcal{I}=\{i\left(k\right)=(i_1,i_2,\dots ,i_k):1\le i_r\le 3,1\le r\le k,k\ge 1\},$$

are functions of three variables in the certain domain $\mathrm{D},$$\mathrm{D}\subset {\mathbb{C}}^3,$ (for more details on the branched continued fractions, see, for example, [18]).

The problem of the analytical continuation of the ratio of the Lauricella hypergeometric functions $F_D$ with some real parameters to its branched continued fraction expansion were considered in [19,20]. In particular, it was proved in [19] that the expansion of the ratio is its analytic continuation in the domain

$${\mathrm{K}}_{F_D}=\left\{\mathbf{z}\in {\mathbb{C}}^3:\left|z_k\right|<1,Re\left(z_k\right)<\frac{1}{2},1\le k\le 3\right\}.$$

In [21], it was established that the branched continued fraction expansion of the ratio of the Lauricella–Saran hypergeometric functions $F_S$ with some real parameters is its analytic continuation in the domain

$${\mathrm{K}}_{F_S}=\{\mathbf{z}\in {\mathbb{C}}^3:|z_k|+Re\left(z_k\right)<1,1\le k\le 3\}.$$

The paper is organized as follows. In Section 2, we give two methods for analytically extending a hypergeometric function (or ratio of hypergeometric functions) to a branched continued fraction in its domain of convergence. In Section 3, we derive two three-term recurrence relations for Lauricella–Saran hypergeometric functions $F_K$ and construct the formal branched continued fraction expansions for two ratios of Lauricella–Saran hypergeometric functions $F_K.$ Here, it is also proved that the branched continued fraction, which is an expansion of each ratio, uniformly converges to a holomorphic function of three variables on every compact subset of some domain of ${\mathbb{C}}^3,$ and that this function is an analytic continuation of such a ratio in this domain.

2. Methods of Analytic Continuation

In the analytical theory of branched continued fractions, two methods are used to prove that the branched continued fraction expansion is an analytic continuation of a hypergeometric function (or ratio of hypergeometric functions) in some domain.

2.1. PC Method

The first method—let us call it the “PC method”—uses the so-called “principle of correspondence” (see, [22,23]). Its application requires that the branched continued fraction expansion corresponds at $\mathbf{z}=\mathbf{0}$ to a hypergeometric function (or ratio of hypergeometric functions) and that the sequence of its approximants converges uniformly on each compact subset of some neighborhood of the origin $(\mathbf{z}=\mathbf{0})$ to a function that is holomorphic in this neighborhood. Then, it remains to consistently apply the well-known Weierstrass’ theorem ([24], p. 23) and the principle of analytic continuation ([25], p. 39).

Let us recall the necessary concepts.

An expression of the form

$$\begin{array}{c}\hfill f_n\left(\mathbf{z}\right)=v_0\left(\mathbf{z}\right)+\sum _{i_1=1}^3\frac{u_{i\left(1\right)}\left(\mathbf{z}\right)}{{\displaystyle v_{i\left(1\right)}\left(\mathbf{z}\right)+\sum _{i_2=1}^3\frac{u_{i\left(2\right)}\left(\mathbf{z}\right)}{{\displaystyle v_{i\left(2\right)}\left(\mathbf{z}\right)+{}_{\ddots }{}_{\genfrac{}{}{0pt}{}{}{{\displaystyle +\sum _{i_n=1}^3\frac{u_{i\left(n\right)}\left(\mathbf{z}\right)}{v_{i\left(n\right)}\left(\mathbf{z}\right)}}}}}}}}\end{array}$$

is called an nth approximant of (2) ([18], pp. 15–16).

A branched continued fraction (2) is called convergent at the point $\mathbf{z}={\mathbf{z}}^0$, if at most a finite number of its approximants do not make sense, and if the limit of its sequence of approximants

$$\underset{n\to +\infty }{lim}{f}_n\left({\mathbf{z}}^0\right)$$

exists and is finite (see, [26] and ([27], p. 16)).

A branched continued fraction (2) is called uniformly convergent on subset $\mathrm{E}$ of $\mathrm{D}$ if its sequence $\left\{f_n\left(\mathbf{z}\right)\right\}$ converges uniformly on $\mathrm{E}.$ If, moreover, this occurs for an arbitrary subset $\mathrm{E}$ such that $\overline{E}\subset D$ (here, $\overline{E}$ is the closure of the subset E), then (2) converges uniformly on each compact subset in $\mathrm{D}$ (see, [26] and ([27], p. 16)).

The concept of correspondence at $\mathbf{z}=\mathbf{0}$ (see, [28] and ([29], pp. 30–32)). Let $\mathbb{L}$ be a set of all formal triple power series of the form

$$\begin{array}{c}\hfill L\left(\mathbf{z}\right)=\sum _{p,q,r=0}^{+\infty }{d}_{p,q,r}{z}_1^p{z}_2^q{z}_3^r,\end{array}$$

(3)

where $d_{p,q,r}\in \mathbb{C},$$p\ge 0,$$q\ge 0,$$r\ge 0,$ $\mathbf{z}\in {\mathbb{C}}^3.$ Let $f\left(\mathbf{z}\right)$ be a function of three variables holomorphic in a neighborhood of the origin and let $\Lambda :f\left(\mathbf{z}\right)\to \Lambda \left(f\right)$ be a mapping associate with $f\left(\mathbf{z}\right)$ its Taylor expansion in a neighborhood of the point $\mathbf{0}.$

A sequence $\left\{f_n\left(\mathbf{z}\right)\right\}$ of the functions of three variables holomorphic at the origin is said to correspond at $\mathbf{z}=\mathbf{0}$ to a formal triple power series (3) if

$$\underset{n\to +\infty }{lim}\lambda \left(L-\Lambda \left(f_n\right)\right)=+\infty ,$$

where $\lambda$ is defined to be: $\lambda :\mathbb{L}\to {\mathbb{Z}}_{\ge 0}\cup \{+\infty \}$; if $L\left(\mathbf{z}\right)\equiv 0$, then $\lambda \left(L\right)=+\infty ;$ if $L\left(\mathbf{z}\right)\not\equiv 0$ then $\lambda \left(L\right)=k,$ where k is the smallest degree of homogeneous terms for which $d_{p,q,r}\ne 0,$ that is, $k=p+q+r.$

A branched continued fraction (2) is said to correspond at $\mathbf{z}=\mathbf{0}$ to a formal triple power series (3) (or a function $f\left(\mathbf{z}\right)$ holomorphic at the origin) if its sequence $\left\{f_n\left(\mathbf{z}\right)\right\}$ corresponds to $L\left(\mathbf{z}\right)$ (or a formal triple power series $\Lambda \left(f\right)$).

Theorem 1 

(Weierstrass’ Theorem). Let a sequence $\left\{g_n\left(\mathbf{z}\right)\right\}$ of holomorphic functions in a domain $\mathrm{D},$ $\mathrm{D}\subset {\mathbb{C}}^3,$ converge to a function $g\left(\mathbf{z}\right)$ uniformly on each compact subset in $\mathrm{D},$ then $f\left(\mathbf{z}\right)$ is holomorphic in $\mathrm{D},$ and for any $p\ge 0,$$q\ge 0,$$r\ge 0,$

$$\frac{{\partial }^{p+q+r}{g}_n\left(\mathbf{z}\right)}{\partial z_1^p\partial z_2^q\partial z_3^r}\to \frac{{\partial }^{p+q+r}g\left(\mathbf{z}\right)}{\partial z_1^p\partial z_2^q\partial z_3^r}asn\to +\infty$$

on each compact subset in $\mathrm{D}.$

Theorem 2 

(The Principle of Analytic Continuation). Let the functions $g_1\left(\mathbf{z}\right)$ and $g_2\left(\mathbf{z}\right)$ be holomorphic in the domains ${\mathrm{D}}_1,$ ${\mathrm{D}}_1\subset {\mathbb{C}}^3,$ and ${\mathrm{D}}_2,$ ${\mathrm{D}}_2\subset {\mathbb{C}}^3,$ respectively, and let ${\mathrm{D}}_1\cap {\mathrm{D}}_2$ be the domain. Let, further, in a real neighborhood of the point ${\mathbf{z}}^0$ from ${\mathrm{D}}_1\cap {\mathrm{D}}_2$ the functions $g_1\left(\mathbf{z}\right)$ and $g_2\left(\mathbf{z}\right)$ coincide. Then these functions are an analytic continuation of one another, i.e., there is a unique function $g\left(\mathbf{z}\right)$ that is holomorphic in ${\mathrm{D}}_1\cup {\mathrm{D}}_2$ and coincides with $g_1\left(\mathbf{z}\right)$ in ${\mathrm{D}}_1$ and with $g_2\left(\mathbf{z}\right)$ in ${\mathrm{D}}_2$.

2.2. PF Method

The second method, let us call it the “PF method”, uses the so-called “property of fork” (see, [21,30,31]). This method is used when the hypergeometric function (or the ratio of hypergeometric functions) and the elements of the branched continued fraction expansion are positive-valued functions in some domain $\mathrm{D}.$ If it holds, then its approximant satisfies the “property of fork”: the sequence of even (odd) approximants increases (decreases) and is no greater (no less) than any odd (even) approximant. If, in addition, the branched continued fraction expansion converges, then it converges to the hypergeometric function (or the ratio of hypergeometric functions) in $\mathrm{D}.$ Finally, for the same restrictions on the parameters of the hypergeometric function, it remains to prove the convergence of the branched continued fraction expansion in a wider domain than $\mathrm{D}$ and to apply Theorem 2.

3. Lauricella–Saran Hypergeometric Function F_K(a₁,a₂,b₁,b₂;a₁,b₂,c₃;z)

We set $c_1=a_1$ and $c_2=b_2.$ Then, from (1), it follows

$$\begin{array}{c}\hfill F_K(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})=\sum _{p,q,r=0}^{+\infty }\frac{{\left(a_2\right)}_{q+r}{\left(b_1\right)}_{p+r}}{{\left(c_3\right)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}.\end{array}$$

(4)

3.1. Recurrence Relations

Remark 1. 

In the process of constructing a branched continued fraction expansion of the ratio of hypergeometric functions, recurrent relations (for instance, three-term and/or four-term) play an important role. The problem is not only in the direct construction of such an expansion, but also in obtaining a branched continued fraction of the simplest structure. This, in turn, can provide more opportunities to investigate the convergence of the constructed expansion.

Let us prove the three-term recurrence relations for Lauricella–Saran hypergeometric function (4).

Lemma 1. 

The following relations hold true:

$$\begin{array}{cc}\hfill & F_K(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})=(1-z_1)F_K(a_1,a_2,b_1+1,b_2;a_1,b_2,c_3+1;\mathbf{z})\hfill \\ \hfill & -\frac{a_2(c_3-b_1)}{c_3(c_3+1)}{z}_3{F}_K(a_1,a_2+1,b_1+1,b_2;a_1,b_2,c_3+2;\mathbf{z}),\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill & F_K(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})=(1-z_2)F_K(a_1,a_2+1,b_1,b_2;a_1,b_2,c_3+1;\mathbf{z})\hfill \\ \hfill & -\frac{b_1(c_3-a_2)}{c_3(c_3+1)}{z}_3{F}_K(a_1,a_2+1,b_1+1,b_2;a_1,b_2,c_3+2;\mathbf{z}).\hfill \end{array}$$

(6)

Proof. 

We have

$$\begin{array}{cc}\hfill & F_K(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})-F_K(a_1,a_2,b_1+1,b_2;a_1,b_2,c_3+1;\mathbf{z})\hfill \\ \hfill & =\sum _{p,q,r\ge 0}\frac{{\left(a_2\right)}_{q+r}{\left(b_1\right)}_{p+r}}{{\left(c_3\right)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}-\sum _{p,q,r\ge 0}\frac{{\left(a_2\right)}_{q+r}{(b_1+1)}_{p+r}}{{(c_3+1)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & =\sum _{p+q+r\ge 1}\frac{{\left(a_2\right)}_{q+r}{\left(b_1\right)}_{p+r}}{{\left(c_3\right)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}-\sum _{p+q+r\ge 1}\frac{{\left(a_2\right)}_{q+r}{(b_1+1)}_{p+r}}{{(c_3+1)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & =\sum _{q\ge 0,p+r\ge 1}{\left(a_2\right)}_{q+r}\left(\frac{{\left(b_1\right)}_{p+r}}{{\left(c_3\right)}_r}-\frac{{(b_1+1)}_{p+r}}{{(c_3+1)}_r}\right)\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & =\sum _{q\ge 0,p=0,r\ge 1}{\left(a_2\right)}_{q+r}\frac{{(b_1+1)}_{r-1}}{{(c_3+1)}_{r-1}}\left(\frac{b_1}{c_3}-\frac{b_1+r}{c_3+r}\right)\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & +\sum _{q\ge 0,p\ge 1,r=0}{\left(a_2\right)}_{q+r}\left(b_1-b_1-p\right)\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & +\sum _{q\ge 0,p\ge 1,r\ge 1}\frac{{\left(a_2\right)}_{q+r}{(b_1+1)}_{p+r-1}}{{(c_3+1)}_{r-1}}\left(\frac{b_1}{c_3}-\frac{b_1+p+r}{c_3+r}\right)\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & =-\sum _{q\ge 0,p=0,r\ge 1}\frac{(c_3-b_1){\left(a_2\right)}_{q+r}{(b_1+1)}_{p+r-1}}{c_3(c_3+1){(c_3+2)}_{r-1}}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!(r-1)!}\hfill \\ \hfill & -\sum _{q\ge 0,p\ge 1,r=0}\frac{{\left(a_2\right)}_{q+r}{(b_1+1)}_{p+r-1}}{{\left(c_3\right)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{(p-1)!q!r!}\hfill \\ \hfill & -\sum _{q\ge 0,p\ge 1,r\ge 1}\frac{{\left(a_2\right)}_{q+r}{(b_1+1)}_{p+r-1}(c_3-b_1)}{c_3(c_3+1){(c_3+2)}_{r-1}}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!(r-1)!}\hfill \\ \hfill & -\sum _{q\ge 0,p\ge 1,r\ge 1}\frac{{\left(a_2\right)}_{q+r}{(b_1+1)}_{p+r-1}}{{(c_3+1)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{(p-1)!q!r!}\hfill \\ \hfill & =-\frac{a_2(c_3-b_1)}{c_3(c_3+1)}{z}_3\sum _{q\ge 0,p\ge 0,r\ge 1}\frac{{(a_2+1)}_{q+r-1}{(b_1+1)}_{p+r-1}}{{(c_3+2)}_{r-1}}\frac{z_1^p{z}_2^q{z}_3^{r-1}}{p!q!(r-1)!}\hfill \\ \hfill & -z_1\sum _{q\ge 0,p\ge 1,r\ge 0}\frac{{\left(a_2\right)}_{q+r}{(b_1+1)}_{p+r-1}}{{(c_3+1)}_r}\frac{z_1^{p-1}{z}_2^q{z}_3^r}{(p-1)!q!r!},\hfill \\ \hfill & =-\frac{a_2(c_3-b_1)}{c_3(c_3+1)}{z}_3{F}_K(a_1,a_2+1,b_1+1,b_2;a_1,b_2,c_3+2;\mathbf{z})\hfill \\ \hfill & -z_1{F}_K(a_1,a_2,b_1+1,b_2;a_1,b_2,c_3+1;\mathbf{z}),\hfill \end{array}$$

from which follows the correctness of relation (5).

Similarly, we will prove the relation (6). By definition (4), we get

$$\begin{array}{cc}\hfill & F_K(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})-F_K(a_1,a_2+1,b_1,b_2;a_1,b_2,c_3+1;\mathbf{z})\hfill \\ \hfill & =\sum _{p,q,r\ge 0}\frac{{\left(a_2\right)}_{q+r}{\left(b_1\right)}_{p+r}}{{\left(c_3\right)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}-\sum _{p,q,r\ge 0}\frac{{(a_2+1)}_{q+r}{\left(b_1\right)}_{p+r}}{{(c_3+1)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & =\sum _{p+q+r\ge 1}\frac{{\left(a_2\right)}_{q+r}{\left(b_1\right)}_{p+r}}{{\left(c_3\right)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}-\sum _{p+q+r\ge 1}\frac{{(a_2+1)}_{q+r}{\left(b_1\right)}_{p+r}}{{(c_3+1)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & =\sum _{p\ge 0,q+r\ge 1}{\left(b_1\right)}_{p+r}\left(\frac{{\left(a_2\right)}_{q+r}}{{\left(c_3\right)}_r}-\frac{{(a_2+1)}_{q+r}}{{(c_3+1)}_r}\right)\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & =\sum _{p\ge 0,q=0,r\ge 1}{\left(b_1\right)}_{p+r}\frac{{(a_2+1)}_{r-1}}{{(c_3+1)}_{r-1}}\left(\frac{a_2}{c_3}-\frac{a_2+r}{c_3+r}\right)\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & +\sum _{p\ge 0,q\ge 1,r=0}{\left(b_1\right)}_{p+r}(a_2-a_2-q)\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & +\sum _{p\ge 0,q\ge 1,r\ge 1}\frac{{\left(b_1\right)}_{p+r}{(a_2+1)}_{q+r-1}}{{(c_3+1)}_{r-1}}\left(\frac{a_2}{c_3}-\frac{a_2+q+r}{c_3+r}\right)\frac{z_1^p{z}_2^q{z}_3^r}{p!q!r!}\hfill \\ \hfill & =-\sum _{p\ge 0,q=0,r\ge 1}\frac{(c_3-a_2){\left(b_1\right)}_{p+r}{(a_2+1)}_{q+r-1}}{c_3(c_3+1){(c_3+2)}_{r-1}}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!(r-1)!}\hfill \\ \hfill & -\sum _{p\ge 0,q\ge 1,r=0}\frac{{\left(b_1\right)}_{p+r}{(a_2+1)}_{q+r-1}}{{\left(c_3\right)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!(q-1)!r!}\hfill \\ \hfill & -\sum _{p\ge 0,q\ge 1,r\ge 1}\frac{{(a_2+1)}_{q+r-1}{\left(b_1\right)}_{p+r}(c_3-a_2)}{c_3(c_3+1){(c_3+2)}_{r-1}}\frac{z_1^p{z}_2^q{z}_3^r}{p!q!(r-1)!}\hfill \\ \hfill & -\sum _{p\ge 0,q\ge 1,r\ge 1}\frac{{\left(a_2\right)}_{q+r-1}{\left(b_1\right)}_{p+r}}{{(c_3+1)}_r}\frac{z_1^p{z}_2^q{z}_3^r}{p!(q-1)!r!}.\hfill \\ \hfill & =-\frac{b_1(c_3-a_2)}{c_3(c_3+1)}{z}_3\sum _{p\ge 0,q\ge 0,r\ge 1}\frac{{(a_2+1)}_{q+r-1}{(b_1+1)}_{p+r-1}}{{(c_3+2)}_{r-1}}\frac{z_1^p{z}_2^q{z}_3^{r-1}}{p!q!(r-1)!}\hfill \\ \hfill & -z_2\sum _{p\ge 0,q\ge 1,r\ge 0}\frac{{(a_2+1)}_{q+r-1}{\left(b_1\right)}_{p+r}}{{(c_3+1)}_r}\frac{z_1^p{z}_2^{q-1}{z}_3^r}{p!(q-1)!r!}\hfill \\ \hfill & =-\frac{b_1(c_3-a_2)}{c_3(c_3+1)}{z}_3{F}_K(a_1,a_2+1,b_1+1,b_2;a_1,b_2,c_3+2;\mathbf{z})\hfill \\ \hfill & -z_2{F}_K(a_1,a_2+1,b_1,b_2;a_1,b_2,c_3+1;\mathbf{z}),\hfill \end{array}$$

which had to be proved. □

3.2. Expansions

We set

$$\begin{array}{cc}\hfill R_K^{\left(1\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})& =\frac{F_K(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})}{F_K(a_1,a_2,b_1+1,b_2;a_1,b_2,c_3+1;\mathbf{z})},\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill R_K^{\left(2\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})& =\frac{F_K(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})}{F_K(a_1,a_2+1,b_1,b_2;a_1,b_2,c_3+1;\mathbf{z})}.\hfill \end{array}$$

(8)

The following theorem is true.

Theorem 3. 

A ratio (7) has a formal branched continued fraction of the form

$$\begin{array}{c}\hfill 1-z_1-{\displaystyle \frac{d_1{z}_3}{1-z_2-{\displaystyle \frac{d_2{z}_3}{1-z_1-{\displaystyle \frac{d_3{z}_3}{1-z_2-{\displaystyle \frac{d_4{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }}}}}}}}},\end{array}$$

(9)

where, for all $k\ge 1,$

$$\begin{array}{c}\hfill d_{2k-1}=\frac{(a_2+k-1)(c_3+k-1-b_1)}{(c_3+2k-2)(c_3+2k-1)},d_{2k}=\frac{(b_1+k)(c_3+k-a_2)}{(c_3+2k-1)(c_3+2k)}.\end{array}$$

(10)

Proof. 

Dividing (5) and (6) by

$$F_K(a_1,a_2,b_1+1,b_2;a_1,b_2,c_3+1;\mathbf{z})and{F}_K(a_1,a_2+1,b_1,b_2;a_1,b_2,c_3+1;\mathbf{z}),$$

respectively, we obtain

$$\begin{array}{c}\hfill R_K^{\left(1\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})=1-z_1-{\displaystyle \frac{{\displaystyle \frac{a_2(c_3-b_1)}{c_3(c_3+1)}}{z}_3}{R_K^{\left(2\right)}(a_1,a_2,b_1+1,b_2;a_1,b_2,c_3+1;\mathbf{z})}},\end{array}$$

(11)

$$\begin{array}{c}\hfill R_K^{\left(2\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})=1-z_2-{\displaystyle \frac{{\displaystyle \frac{b_1(c_3-a_2)}{c_3(c_3+1)}}{z}_3}{R_K^{\left(1\right)}(a_1,a_2+1,b_1,b_2;a_1,b_2,c_3+1;\mathbf{z})}}.\end{array}$$

(12)

In fact, in (11), we have Step 1.1 of constructing a branched continued fraction. At Step 1.2, replacing $b_1,$$c_3$ by $b_1+1$ and $c_3+1,$ respectively, in (12), we get

$$\begin{array}{cc}\hfill R_K^{\left(1\right)}& (a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})\hfill \end{array}$$

$$\begin{array}{c}\hfill =1-z_1-{\displaystyle \frac{{\displaystyle \frac{a_2(c_3-b_1)}{c_3(c_3+1)}}{z}_3}{1-z_2-{\displaystyle \frac{{\displaystyle \frac{(b_1+1)(c_3+1-a_2)}{(c_3+1)(c_3+2)}}{z}_3}{R_K^{\left(1\right)}(a_1,a_2+1,b_1+1,b_2;a_1,b_2,c_3+2;\mathbf{z})}}}}.\end{array}$$

(13)

Let us continue the next construction of the branched continued fraction in the same way as in steps 1.1–1.2. It is clear that the following relation holds, for all $k\ge 1,$

$$\begin{array}{cc}\hfill & R_K^{\left(1\right)}(a_1,a_2+k-1,b_1+k-1,b_2;a_1,b_2,c_3+2k-2;\mathbf{z})\hfill \\ \hfill & =1-z_1-{\displaystyle \frac{{\displaystyle \frac{(a_2+k-1)(c_3+k-1-b_1)}{(c_3+2k-2)(c_3+2k-1)}}{z}_3}{1-z_2-{\displaystyle \frac{{\displaystyle \frac{(b_1+k)(c_3+k-a_2)}{(c_3+2k-1)(c_3+2k)}}{z}_3}{R_K^{\left(1\right)}(a_1,a_2+k,b_1+k,b_2;a_1,b_2,c_3+2k;\mathbf{z})}}}}.\hfill \end{array}$$

(14)

At Steps 2.1–2.2, substituting (14) when $k=2$ in (13), we obtain

$$\begin{array}{cc}\hfill & R_K^{\left(1\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})\hfill \\ & =1-z_1-{\displaystyle \frac{{\displaystyle \frac{a_2(c_3-b_1)}{c_3(c_3+1)}}{z}_3}{1-z_2-{\displaystyle \frac{{\displaystyle \frac{(b_1+1)(c_3+1-a_2)}{(c_3+1)(c_3+2)}}{z}_3}{1-z_1-{\displaystyle \frac{{\displaystyle \frac{(a_2+1)(c_3+1-b_1)}{(c_3+2)(c_3+3)}}{z}_3}{1-z_2-{\displaystyle \frac{{\displaystyle \frac{(b_1+2)(c_3+2-a_2)}{(c_3+3)(c_3+4)}}{z}_3}{R_K^{\left(1\right)}(a_1,a_2+2,b_1+2,b_2;a_1,b_2,c_3+4;\mathbf{z})}}}}}}}}.\hfill \end{array}$$

Next, by (14) after the Steps n.1–n.2, we have

$$\begin{array}{cc}\hfill & R_K^{\left(1\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})=1-z_1\hfill \\ & -{\displaystyle \frac{{\displaystyle \frac{a_2(c_3-b_1)}{c_3(c_3+1)}}{z}_3}{1-z_2-{\displaystyle \frac{{\displaystyle \frac{(b_1+1)(c_3+1-a_2)}{(c_3+1)(c_3+2)}}{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }\genfrac{}{}{0pt}{}{{\displaystyle }}{{\displaystyle -z_1-{\displaystyle \frac{{\displaystyle \frac{(a_2+n-1)(c_3+n-1-b_1)}{(c_3+2n-2)(c_3+2n-1)}}{z}_3}{1-z_2-{\displaystyle \frac{{\displaystyle \frac{(b_1+n)(c_3+n-a_2)}{(c_3+2n-1)(c_3+2n)}}{z}_3}{R_K^{\left(1\right)}(a_1,a_2+n,b_1+n,b_2;a_1,b_2,c_3+2n;\mathbf{z})}}}}}}}}}}.\hfill \end{array}$$

Finally, as $n\to +\infty$, we obtain the formal expansion of (7) into branched continued fraction (9). □

The following theorem can be proved in much the same way as Theorem 3.

Theorem 4. 

A ratio (8) has a formal branched continued fraction of the form

$$\begin{array}{c}\hfill 1-z_2-{\displaystyle \frac{h_1{z}_3}{1-z_1-{\displaystyle \frac{h_2{z}_3}{1-z_2-{\displaystyle \frac{h_3{z}_3}{1-z_1-{\displaystyle \frac{h_4{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }}}}}}}}},\end{array}$$

(15)

where, for all $k\ge 1,$

$$\begin{array}{c}\hfill h_{2k-1}=\frac{(b_1+k-1)(c_3+k-1-a_2)}{(c_3+2k-2)(c_3+2k-1)},h_{2k}=\frac{(a_2+k)(c_3+k-b_1)}{(c_3+2k-1)(c_3+2k)}.\end{array}$$

(16)

3.3. Analytic Continuation

We will apply the PC method to prove that expansion (9) is an analytic continuation of ratio (7) in some domain.

The following corollary follows directly from Theorem 1 [26].

Corollary 1. 

Let $g_{0,0,k},$$k\ge 1,$ be real numbers such that, for all $k\ge 1,$

$$\begin{array}{c}\hfill 0<g_{0,0,k}\le 1.\end{array}$$

Then, the branched continued fraction,

$$\begin{array}{c}\hfill 1-z_{1,0,0}-{\displaystyle \frac{g_{0,0,1}{z}_{0,0,1}}{1-(1-g_{0,0,1})z_{0,1,1}-{\displaystyle \frac{g_{0,0,2}(1-g_{0,0,1})z_{0,0,2}}{1-(1-g_{0,0,2})z_{1,0,2}-{\displaystyle \frac{g_{0,0,3}(1-g_{0,0,2})z_{0,0,3}}{1-\genfrac{}{}{0pt}{}{}{\ddots }}}}}}},\end{array}$$

converges if, for all $k\ge 0,$

$$\begin{array}{c}\hfill |z_{1,0,2k}|\le \frac{1}{2},|z_{0,1,2k+1}|\le \frac{1}{2},\left|z_{0,0,k+1}\right|\le \frac{1}{2}.\end{array}$$

From the proof of Lemma 4.41 [32], we have following corollary.

Corollary 2. 

If $x\ge c>0$ and $v^2\le 4u+4,$ where $u,v\in \mathbb{R},$ then

$$\underset{-\infty <y<+\infty }{min}Re\left(\frac{u+iv}{x+iy}\right)=-\frac{\sqrt{u^2+v^2}-u}{2x}.$$

Moreover, the following theorem clearly follows from Theorem 2.17 [18] (see also ([27], Theorem 24.2)).

Theorem 5. 

Let a sequence of holomorphic functions $\left\{g_n\left(\mathbf{z}\right)\right\}$ on the domain $\mathrm{D},$$\mathrm{D}\subset {\mathbb{C}}^3,$ is uniform bounded on every compact subset of $\mathrm{D}.$ If, moreover, the sequence $\left\{g_n\left(\mathbf{z}\right)\right\}$ converges at each point of the set $\mathrm{E},$$\mathrm{E}\subset \mathrm{D},$ which is the real neighborhood of the point ${\mathbf{z}}^0$ in $\mathrm{D},$ then its converges uniformly on every compact subset of $\mathrm{D}$ to a holomorphic function in $\mathrm{D}.$

We will prove the following theorem.

Theorem 6. 

Let $a_2,$$b_1,$ and $c_3$ be constants such that, for all $k\ge 1,$

$$\begin{array}{c}\hfill 0<d_k\le r,\end{array}$$

(17)

where $d_k,$$k\ge 1,$ are defined by (10), r is a positive number. Then:

(A) 

The branched continued fraction (9) converges uniformly on every compact subset of

$$\begin{array}{c}\hfill {\mathrm{H}}_{r,r^{\ast }}=\bigcup _{-\pi /2<\alpha <\pi /2}{\mathrm{H}}_{r,r^{\ast },\alpha },\end{array}$$

(18)

where $0<r^{\ast }<1$ and

$$\begin{array}{cc}\hfill & {\mathrm{H}}_{r,r^{\ast },\alpha }\hfill \end{array}$$

$$\begin{array}{c}\hfill =\left\{\mathbf{z}\in {\mathbb{C}}^3:\frac{|z_k|+Re\left(z_k{e}^{-2i\alpha }\right)}{2(1-r^{\ast }){cos}^2\alpha }<1,k=1,2,\frac{|z_3|+Re\left(z_3{e}^{-2i\alpha }\right)}{r^{\ast }{cos}^2\alpha }<\frac{1}{2r}\right\},\end{array}$$

(19)

to a holomorphic function $f\left(\mathbf{z}\right)$ in ${\mathrm{H}}_{r,r^{\ast }};$

(B) 

The function $f\left(\mathbf{z}\right)$ is an analytic continuation of (7) in the domain (18).

Proof. 

We set, for $n\ge 1,$

$$\begin{array}{c}\hfill G_n^{\left(n\right)}\left(\mathbf{z}\right)=1\end{array}$$

(20)

and, for $n\ge 1$ and $1\le k\le n,$

$$\begin{array}{cc}\hfill G_{2k-1}^{\left(2n\right)}\left(\mathbf{z}\right)& =1-z_2-{\displaystyle \frac{d_{2k}{z}_3}{1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }{}_{{\displaystyle -z_1-{\displaystyle \frac{d_{2n-1}{z}_3}{1-z_2-d_{2n}{z}_3}}}}}}}},\hfill \\ \hfill G_{2k}^{\left(2n\right)}\left(\mathbf{z}\right)& =1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{1-z_2-{\displaystyle \frac{d_{2k+2}{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }{}_{{\displaystyle -z_1-{\displaystyle \frac{d_{2n-1}{z}_3}{1-z_2-d_{2n}{z}_3}}}}}}}},\hfill \\ \hfill G_{2k-1}^{(2n+1)}\left(\mathbf{z}\right)& =1-z_2-{\displaystyle \frac{d_{2k}{z}_3}{1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }{}_{{\displaystyle -z_2-{\displaystyle \frac{d_{2n}{z}_3}{1-z_1-d_{2n+1}{z}_3}}}}}}}},\hfill \end{array}$$

$$\begin{array}{cc}\hfill G_{2k}^{(2n+1)}\left(\mathbf{z}\right)& =1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{1-z_2-{\displaystyle \frac{d_{2k+2}{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }{}_{{\displaystyle -z_2-{\displaystyle \frac{d_{2n}{z}_3}{1-z_1-d_{2n+1}{z}_3}}}}}}}},\hfill \end{array}$$

which gives us, for $n\ge 1$ and $1\le k\le n,$

$$\begin{array}{c}\hfill G_{2k-1}^{\left(2n\right)}\left(\mathbf{z}\right)=1-z_2-{\displaystyle \frac{d_{2k}{z}_3}{G_{2k}^{\left(2n\right)}\left(\mathbf{z}\right)}},G_{2k}^{\left(2n\right)}\left(\mathbf{z}\right)=1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{G_{2k+1}^{\left(2n\right)}\left(\mathbf{z}\right)}},\end{array}$$

(21)

and

$$\begin{array}{c}\hfill G_{2k-1}^{(2n+1)}\left(\mathbf{z}\right)=1-z_2-{\displaystyle \frac{d_{2k}{z}_3}{G_{2k}^{(2n+1)}\left(\mathbf{z}\right)}},G_{2k}^{(2n+1)}\left(\mathbf{z}\right)=1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{G_{2k+1}^{(2n+1)}\left(\mathbf{z}\right)}}.\end{array}$$

(22)

Thus, we write the nth approximants of (9) in the form

$$\begin{array}{c}\hfill f_n\left(\mathbf{z}\right)=1-z_1-{\displaystyle \frac{d_1{z}_3}{G_1^{\left(n\right)}\left(\mathbf{z}\right)}}.\end{array}$$

(23)

Let n be an arbitrary natural number, let $\alpha$ be an arbitrary real from $(-\pi /2,\pi /2),$ and let $\mathbf{z}$ be an arbitrary fixed point from (19). Then, the following inequalities are held, for all $1\le k\le n,$

$$\begin{array}{cc}\hfill Re\left(G_{2k-1}^{\left(2n\right)}\left(\mathbf{z}\right)e^{-i\alpha }\right)& >\frac{r^{\ast }cos\alpha }{2}>0\hfill \end{array}$$

(24)

and

$$\begin{array}{cc}\hfill Re\left(G_{2k-1}^{(2n+1)}\left(\mathbf{z}\right)e^{-i\alpha }\right)& >\frac{r^{\ast }cos\alpha }{2}>0.\hfill \end{array}$$

(25)

Let us prove that (24) is true. In view of (20), it is obvious that (24) holds for $k=n.$ Assuming, by the induction, that (24) holds for $k=p+1,$$p+1\le n,$ from (21) one obtains, for $k=p,$

$$\begin{array}{c}\hfill G_{2p-1}^{\left(2n\right)}\left(\mathbf{z}\right)e^{-i\alpha }=e^{-i\alpha }-\frac{z_2{e}^{-2i\alpha }}{e^{-i\alpha }}-\frac{d_{2p}{z}_3{e}^{-2i\alpha }}{G_{2p}^{\left(2n\right)}\left(\mathbf{z}\right)e^{-i\alpha }}\end{array}$$

and

$$\begin{array}{c}\hfill G_{2p}^{\left(2n\right)}\left(\mathbf{z}\right)e^{-i\alpha }=e^{-i\alpha }-\frac{z_1{e}^{-2i\alpha }}{e^{-i\alpha }}-\frac{d_{2p+1}{z}_3{e}^{-2i\alpha }}{G_{2p+1}^{\left(2n\right)}\left(\mathbf{z}\right)e^{-i\alpha }}.\end{array}$$

Then, using (17), (19), Corollary 2, and the induction hypothesis, we have

$$\begin{array}{cc}\hfill Re\left(G_{2p}^{\left(2n\right)}\left(\mathbf{z}\right)e^{-i\alpha }\right)& \ge cos\alpha -\frac{|z_1{e}^{-2i\alpha }|+Re\left(z_1{e}^{-2i\alpha }\right)}{2Re\left(e^{-i\alpha }\right)}-\frac{d_{2p+1}\left(\right|z_3{e}^{-2i\alpha }|+Re\left(z_3{e}^{-2i\alpha }\right))}{2Re\left(G_{2p+1}^{\left(2n\right)}\left(\mathbf{z}\right)e^{-i\alpha }\right)}\hfill \\ \hfill & >cos\alpha -(1-r^{\ast })cos\alpha -\frac{r^{\ast }cos\alpha }{2}\hfill \\ \hfill & =\frac{r^{\ast }cos\alpha }{2}>0\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill Re\left(G_{2p-1}^{\left(2n\right)}\left(\mathbf{z}\right)e^{-i\alpha }\right)& \ge cos\alpha -\frac{|z_2{e}^{-2i\alpha }|+Re\left(z_2{e}^{-2i\alpha }\right)}{2Re\left(e^{-i\alpha }\right)}-\frac{d_{2p}\left(\right|z_3{e}^{-2i\alpha }|+Re\left(z_3{e}^{-2i\alpha }\right))}{2Re\left(G_{2p}^{\left(2n\right)}\left(\mathbf{z}\right)e^{-i\alpha }\right)}\hfill \\ \hfill & >cos\alpha -(1-r^{\ast })cos\alpha -\frac{r^{\ast }cos\alpha }{2}\hfill \\ \hfill & =\frac{r^{\ast }cos\alpha }{2}>0.\hfill \end{array}$$

In the same way, we obtain the inequalities (25).

Thus, for all $n\ge 1$ and $\mathbf{z}\in {\mathrm{H}}_{r,r^{\ast },\alpha },$

$$\begin{array}{c}\hfill G_1^{\left(n\right)}\left(\mathbf{z}\right)\ne 0.\end{array}$$

This means that the sequence $\left\{f_n\left(\mathbf{z}\right)\right\}$ is a sequence of holomorphic functions in (19), and, therefore, in domain ${\mathrm{H}}_{r,r^{\ast }}$ due to the arbitrariness $\alpha .$

Let $\mathrm{K}$ be an arbitrary compact subset of ${\mathrm{H}}_{r,r^{\ast }}.$ Then, there exists an open triple-disk

$$\begin{array}{c}\hfill {\mathrm{H}}_l=\{\mathbf{z}\in {\mathbb{C}}^3:|z_k|<l,k=1,2.3\},l>0,\end{array}$$

such that $\mathrm{K}\subset {\mathrm{H}}_l.$ Now, cover $\mathrm{K}$ by domains of the form

$$\begin{array}{c}\hfill {\mathrm{H}}_{r,r^{\ast },l,\alpha }={\mathrm{H}}_{r,r^{\ast },\alpha }\bigcap {\mathrm{H}}_l\end{array}$$

and choose from this cover a finite subcover,

$$\begin{array}{c}\hfill {\mathrm{H}}_{r,r^{\ast },l,{\alpha }_1},{\mathrm{H}}_{r,r^{\ast },l,{\alpha }_2},\dots ,{\mathrm{H}}_{r,r^{\ast },l,{\alpha }_k}.\end{array}$$

Using (23)–(25), for any $n\ge 1,$ $p\in \{1,2,\dots ,k\}$ and $\mathbf{z}\in {\mathrm{H}}_{r,r^{\ast },l,{\alpha }_p}$, we have

$$\begin{array}{cc}\hfill |f_n\left(\mathbf{z}\right)|& \le 1+|z_1|+\frac{d_1\left|z_3\right|}{\mathrm{Re}\left(G_1^{\left(n\right)}\left(\mathbf{z}\right)e^{-i{\alpha }_p}\right)}\hfill \\ & <1+l+\frac{2rl}{cos{\alpha }_p}\hfill \\ & =C\left({\mathrm{H}}_{r,r^{\ast },l,{\alpha }_p}\right).\hfill \end{array}$$

Setting

$$C\left(\mathrm{K}\right)=\underset{1\le p\le k}{max}C\left({\mathrm{H}}_{r,r^{\ast },l,{\alpha }_p}\right),$$

for any $n\ge 1$ and $\mathbf{z}\in K$, we obtain

$$\begin{array}{c}\hfill |f_n\left(\mathbf{z}\right)|\le C\left(\mathrm{K}\right).\end{array}$$

This means that the sequence $\left\{f_n\left(\mathbf{z}\right)\right\}$ is uniformly bounded on every compact subset of the domain ${\mathrm{H}}_{r,r^{\ast }}.$

It is clear that, for each real $l^{\ast }$ such that

$$0<l^{\ast }<min\left\{\frac{1}{4},\frac{1}{8r}\right\},$$

the domain

$$\begin{array}{c}\hfill {\mathrm{H}}_{l^{\ast }}=\left\{\mathbf{z}\in {\mathbb{R}}^3:-l^{\ast }<z_k<0,1\le k\le 3\right\}\end{array}$$

is contained in ${\mathrm{H}}_{r,r^{\ast }},$ in particular, ${\mathrm{H}}_{l^{\ast }/2}\subset {\mathrm{H}}_{r,r^{\ast }}.$

Taking into account (17), it is easy to show that, for any $\mathbf{z}\in {\mathrm{H}}_{l^{\ast }},$ ${\mathrm{H}}_{l^{\ast }}\subset {\mathrm{H}}_{r,r^{\ast }},$ the following inequalities hold, for all $k\ge 1,$

$$\begin{array}{c}\hfill |z_k| <\frac{1}{4}k=1,2,\left|d_k{z}_3\right|<\frac{1}{8}.\end{array}$$

This means that the elements of branched continued fraction (9) satisfy the conditions of Corollary 1, with $g_{0,0,k}=1/2$ for all $k\ge 1.$ By this corollary the branched continued fraction (9) converges in ${\mathrm{H}}_{l^{\ast }},$ ${\mathrm{H}}_{l^{\ast }}\subset {\mathrm{H}}_{r,r^{\ast }}.$ It follows from Theorem 5 that the convergence is uniform on compact subsets of ${\mathrm{H}}_{r,r^{\ast }}$ to a holomorphic function $f\left(\mathbf{z}\right)$ in ${\mathrm{H}}_{r,r^{\ast }}.$ This proves (A).

Now, we prove (B). Setting, for $n\ge 1,$

$$\begin{array}{cc}\hfill F_{2n}^{\left(2n\right)}\left(\mathbf{z}\right)& =R_K^{\left(1\right)}(a_1,a_2+n,b_1+n,b_2;a_1,b_2,c_3+2n;\mathbf{z}),\hfill \\ \hfill F_{2n+1}^{(2n+1)}\left(\mathbf{z}\right)& =R_K^{\left(2\right)}(a_1,a_2+n,b_1+n+1,b_2;a_1,b_2,c_3+2n+1;\mathbf{z}),\hfill \end{array}$$

and, for $n\ge 1$ and $1\le k\le n,$

$$\begin{array}{cc}\hfill F_{2k-1}^{\left(2n\right)}\left(\mathbf{z}\right)& =1-z_2-{\displaystyle \frac{d_{2k}{z}_3}{1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }{}_{{\displaystyle -z_1-{\displaystyle \frac{d_{2n-1}{z}_3}{1-z_2-{\displaystyle \frac{d_{2n}{z}_3}{F_{i\left(2n\right)}^{\left(2n\right)}\left(\mathbf{z}\right)}}}}}}}}}},\hfill \\ \hfill F_{2k}^{\left(2n\right)}\left(\mathbf{z}\right)& =1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{1-z_2-{\displaystyle \frac{d_{2k+2}{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }{}_{{\displaystyle -z_1-{\displaystyle \frac{d_{2n-1}{z}_3}{1-z_2-{\displaystyle \frac{d_{2n}{z}_3}{F_{i\left(2n\right)}^{\left(2n\right)}\left(\mathbf{z}\right)}}}}}}}}}},\hfill \\ \hfill F_{2k-1}^{(2n+1)}\left(\mathbf{z}\right)& =1-z_2-{\displaystyle \frac{d_{2k}{z}_3}{1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }{}_{{\displaystyle -z_2-{\displaystyle \frac{d_{2n}{z}_3}{1-z_1-{\displaystyle \frac{d_{2n+1}{z}_3}{F_{i(2n+1)}^{(2n+1)}\left(\mathbf{z}\right)}}}}}}}}}},\hfill \\ \hfill F_{2k}^{(2n+1)}\left(\mathbf{z}\right)& =1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{1-z_2-{\displaystyle \frac{d_{2k+2}{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }{}_{{\displaystyle -z_2-{\displaystyle \frac{d_{2n}{z}_3}{1-z_1-{\displaystyle \frac{d_{2n+1}{z}_3}{F_{i(2n+1)}^{(2n+1)}\left(\mathbf{z}\right)}}}}}}}}}},\hfill \end{array}$$

we have, for $n\ge 1$ and $1\le k\le n,$

$$\begin{array}{c}\hfill F_{2k-1}^{\left(2n\right)}\left(\mathbf{z}\right)=1-z_2-{\displaystyle \frac{d_{2k}{z}_3}{F_{2k}^{\left(2n\right)}\left(\mathbf{z}\right)}},F_{2k}^{\left(2n\right)}\left(\mathbf{z}\right)=1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{F_{2k+1}^{\left(2n\right)}\left(\mathbf{z}\right)}},\end{array}$$

(26)

and

$$\begin{array}{c}\hfill F_{2k-1}^{(2n+1)}\left(\mathbf{z}\right)=1-z_2-{\displaystyle \frac{d_{2k}{z}_3}{F_{2k}^{(2n+1)}\left(\mathbf{z}\right)}},F_{2k}^{(2n+1)}\left(\mathbf{z}\right)=1-z_1-{\displaystyle \frac{d_{2k+1}{z}_3}{F_{2k+1}^{(2n+1)}\left(\mathbf{z}\right)}}.\end{array}$$

(27)

Hence, and from the proof of Theorem 3, it follows that for each $n\ge 1$,

$$\begin{array}{cc}\hfill R_K^{\left(1\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})& =1-z_1-{\displaystyle \frac{d_1{z}_3}{1-z_2-{\displaystyle \frac{d_2{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }\genfrac{}{}{0pt}{}{{\displaystyle }}{{\displaystyle -z_1-{\displaystyle \frac{d_{2n-1}{z}_3}{1-z_2-{\displaystyle \frac{d_{2n}{z}_3}{F_{2n}^{\left(2n\right)}\left(\mathbf{z}\right)}}}}}}}}}}\hfill \\ \hfill & =1-z_1-{\displaystyle \frac{d_1{z}_3}{F_1^{\left(2n\right)}\left(\mathbf{z}\right)}}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill R_K^{\left(1\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})& =1-z_1-{\displaystyle \frac{d_1{z}_3}{1-z_2-{\displaystyle \frac{d_2{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }\genfrac{}{}{0pt}{}{{\displaystyle }}{{\displaystyle -z_2-{\displaystyle \frac{d_{2n}{z}_3}{1-z_1-{\displaystyle \frac{d_{2n+1}{z}_3}{F_{2n+1}^{(2n+1)}\left(\mathbf{z}\right)}}}}}}}}}}\hfill \\ \hfill & =1-z_1-{\displaystyle \frac{d_1{z}_3}{F_1^{(2n+1)}\left(\mathbf{z}\right)}}.\hfill \end{array}$$

Since $F_k^{\left(n\right)}\left(\mathbf{0}\right)=1$ and $G_k^{\left(n\right)}\left(\mathbf{0}\right)=1$ for any $1\le k\le n,$$n\ge 1,$ then there exist $\Lambda (1/F_k^{\left(n\right)})$ and $\Lambda (1/G_k^{\left(n\right)}),$ i.e., the $1/F_k^{\left(n\right)}$ and $1/G_k^{\left(n\right)}$ have Taylor expansions in a neighborhood of the origin. It is clear that $F_k^{\left(n\right)}\left(\mathbf{z}\right)\not\equiv 0$ and $G_k^{\left(n\right)}\left(\mathbf{z}\right)\not\equiv 0$ for all indices. Applying the method suggested in ([18], p. 28) and (20)–(22), (26), and (27), for each $n\ge 1$ one obtains

$$\begin{array}{cc}\hfill & R_K^{\left(1\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})-f_{2n-1}\left(\mathbf{z}\right)\hfill \\ \hfill & =\frac{d_1{z}_3}{F_1^{\left(2n\right)}\left(\mathbf{z}\right)G_1^{(2n-1)}\left(\mathbf{z}\right)}\dots \frac{d_{2n-1}{z}_3}{F_{2n-1}^{\left(2n\right)}\left(\mathbf{z}\right)G_{2n-1}^{(2n-1)}\left(\mathbf{z}\right)}\left(-z_2-\frac{d_{2n}{z}_3}{F_{2n}^{\left(2n\right)}\left(\mathbf{z}\right)}\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & R_K^{\left(1\right)}(a_1,a_2,b_1,b_2;a_1,b_2,c_3;\mathbf{z})-f_{2n}\left(\mathbf{z}\right)\hfill \\ \hfill & =\frac{d_1{z}_3}{F_1^{(2n+1)}\left(\mathbf{z}\right)G_1^{\left(2n\right)}\left(\mathbf{z}\right)}\dots \frac{d_{2n}{z}_3}{F_{2n}^{(2n+1)}\left(\mathbf{z}\right)G_{2n}^{\left(2n\right)}\left(\mathbf{z}\right)}\left(-z_1-\frac{d_{2n+1}{z}_3}{F_{2n+1}^{(2n+1)}\left(\mathbf{z}\right)}\right).\hfill \end{array}$$

Hence, in a neighborhood of origin for any $n\ge 1$, we have

$$\Lambda \left(R_K^{\left(1\right)}\right)-\Lambda \left(f_n\right)=\sum _{\begin{array}{c}p+q+r\ge n\\ p\ge 0,q\ge 0,r\ge 0\end{array}}{d}_{p,q,r}^{\left(n\right)}{z}_1^p{z}_2^q{z}_3^r,$$

where $d_{p,q,r}^{\left(n\right)},$$p\ge 0,$$q\ge 0,$$r\ge 0,$$p+q+r\ge n,$ are some coefficients. It follows that

$$\lambda (\Lambda \left(R_1\right)-\Lambda \left(f_n\right))=n+1$$

tends monotonically to $+\infty$ as $n\to +\infty .$

Thus, the branched continued fraction (9) corresponds at $\mathbf{z}=\mathbf{0}$ to a formal triple power series $\Lambda \left(R_K^{\left(1\right)}\right).$

Let $\Delta$ be the neighborhood of the origin which contained (18), and in which

$$\Lambda \left(R_1\right)=\sum _{p,q,r=0}^{+\infty }{d}_{p,q,r}{z}_1^p{z}_2^q{z}_3^r.$$

(28)

From part (A), it follows that the sequence $\left\{f_n\left(\mathbf{z}\right)\right\}$ converges uniformly on each compact subset of the domain $\Delta$ to function $f\left(\mathbf{z}\right),$ which is holomorphic in $\Delta .$ Then, according to Theorem 1 for arbitrary $k+l,k\ge 0,l\ge 0,$ we have

$$\frac{{\partial }^{p+q+r}{f}_n\left(\mathbf{z}\right)}{\partial z_1^p\partial z_2^q\partial z_3^r}\to \frac{{\partial }^{p+q+r}f\left(\mathbf{z}\right)}{\partial z_1^p\partial z_2^q\partial z_3^r}asn\to +\infty$$

on each compact subset of the domain $\Delta .$ And now, according to the above proven, the expansion of each approximant $f_n\left(\mathbf{z}\right),n\ge 1,$ into formal triple power series and series (28) agree for all homogeneous terms up to and including degree $(n-1).$ Then, for arbitrary $p+q+r,$$p\ge 0,$ $q\ge 0,$$r\ge 0$, we obtain

$$\underset{n\to +\infty }{lim}\left(\frac{{\partial }^{p+q+r}{f}_n}{\partial z_1^p\partial z_2^q\partial z_3^r}\left(\mathbf{0}\right)\right)=\frac{{\partial }^{p+q+r}f}{\partial z_1^p\partial z_2^q\partial z_3^r}\left(\mathbf{0}\right)=p!q!r!d_{p,q,r}.$$

Hence,

$$f\left(\mathbf{z}\right)=\sum _{p,q,r=0}^{+\infty }\frac{1}{p!q!r!}\left(\frac{{\partial }^{p+q+r}f}{\partial z_1^p\partial z_2^q\partial z_3^r}\left(\mathbf{0}\right)\right)z_1^p{z}_2^q{z}_3^r=\sum _{p,q,r=0}^{+\infty }{\alpha }_{p,q,r}{z}_1^p{z}_2^q{z}_3^r$$

for all $\mathbf{z}\in \Delta .$

Finally, Theorem 2 follows part (B). □

Setting $b_1=0$ and replacing $c_3$ by $c_3-1$ in Theorem 6, we have the following result.

Corollary 3. 

Let $a_2$ and $c_3$ be constants such that, for all $k\ge 1,$

$$\begin{array}{c}\hfill 0<\frac{(a_2+k-1)(c_3+k-2)}{(c_3+2k-3)(c_3+2k-2)}\le r,0<\frac{k(c_3+k-1-a_2)}{(c_3+2k-2)(c_3+2k-1)}\le r,\end{array}$$

where r is a positive number. Then:

(A) 

The branched continued fraction

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{1-z_1-{\displaystyle \frac{d_1{z}_3}{1-z_2-{\displaystyle \frac{d_2{z}_3}{1-z_1-{\displaystyle \frac{d_3{z}_3}{1-z_2-{\displaystyle \frac{d_4{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }}}}}}}}}}},\end{array}$$

(29)

where, for all $k\ge 1,$

$$\begin{array}{c}\hfill d_{2k-1}=\frac{(a_2+k-1)(c_3+k-2)}{(c_3+2k-3)(c_3+2k-2)},d_{2k}=\frac{k(c_3+k-1-a_2)}{(c_3+2k-2)(c_3+2k-1)},\end{array}$$

converges uniformly on every compact subset of the domain (18) to a function $f\left(\mathbf{z}\right)$ holomorphic in ${\mathrm{H}}_{r,r^{\ast }};$

(B) 

The function $f\left(\mathbf{z}\right)$ is an analytic continuation of $F_K(a_1,a_2,1,b_2;a_1,b_2,c_3;\mathbf{z})$ in ${\mathrm{H}}_{r,r^{\ast }}.$

The following theorem can be proved in much the same way as Theorem 6 using Theorems 1, 2, and 5 and Corollaries 1 and 2.

Theorem 7. 

Let $a_2,$$b_1,$ and $c_3$ be constants such that, for all $k\ge 1,$

$$\begin{array}{c}\hfill 0<h_k\le r,\end{array}$$

where $h_k,$$k\ge 1,$ are defined by (16), r is a positive number. Then:

(A) 

The branched continued fraction (15) converges uniformly on every compact subset of the domain (18) to a function $f\left(\mathbf{z}\right)$ holomorphic in ${\mathrm{H}}_{r,r^{\ast }};$

(B) 

The function $f\left(\mathbf{z}\right)$ is an analytic continuation of (8) in the domain (18).

Corollary 4. 

Let $b_1,$ and $c_3$ be constants such that, for all $k\ge 1,$

$$\begin{array}{c}\hfill 0<\frac{(b_1+k-1)(c_3+k-2)}{(c_3+2k-3)(c_3+2k-2)}\le r,0<\frac{k(c_3+k-1-b_1)}{(c_3+2k-2)(c_3+2k-1)}\le r,\end{array}$$

where r is a positive number. Then:

(A) 

The branched continued fraction

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{1-z_2-{\displaystyle \frac{d_1{z}_3}{1-z_1-{\displaystyle \frac{d_2{z}_3}{1-z_2-{\displaystyle \frac{d_3{z}_3}{1-z_1-{\displaystyle \frac{d_4{z}_3}{1-\genfrac{}{}{0pt}{}{}{\ddots }}}}}}}}}}},\end{array}$$

(30)

where, for all $k\ge 1,$

$$\begin{array}{c}\hfill d_{2k-1}=\frac{(b_1+k-1)(c_3+k-2)}{(c_3+2k-3)(c_3+2k-2)},d_{2k}=\frac{k(c_3+k-1-b_1)}{(c_3+2k-2)(c_3+2k-1)},\end{array}$$

converges uniformly on every compact subset of the domain (18) to a holomorphic function $f\left(\mathbf{z}\right)$ in ${\mathrm{H}}_{r,r^{\ast }};$

(B) 

The function $f\left(\mathbf{z}\right)$ is an analytic continuation of $F_K(a_1,1,b_1,b_2;a_1,b_2,c_3;\mathbf{z})$ in ${\mathrm{H}}_{r,r^{\ast }}.$

Remark 2. 

Theorems 6 and 7, as and Corollaries 3 and 4, establish the convergence criteria for the constructed branched continued fraction expansions for real parameter values of the Lauricella–Saran hypergeometric function (4). The method used for this also allows us to obtain the convergence criteria for complex parameter values, sacrificing the domain for variable $z_3.$

Remark 3. 

Estimates of the rate of convergence for the branched continued fractions (9), (15), (29), and (30) can be established in the same way as in [33].

4. Numerical Experiments

By Corollary 3, we have

$$\begin{array}{cc}\hfill ln\left(1+\frac{z_3}{(1+z_1)(1+z_2)}\right)& =z_3{F}_K(a_1,1,1,b_2;a_1,b_2,2;-z_1,-z_2,-z_3)\hfill \\ \hfill & ={\displaystyle \frac{z_3}{1+z_1+{\displaystyle \frac{d_1{z}_3}{1+z_2+{\displaystyle \frac{d_2{z}_3}{1+z_1+{\displaystyle \frac{d_3{z}_3}{1+z_2+{\displaystyle \frac{d_4{z}_3}{1+\genfrac{}{}{0pt}{}{}{\ddots }}}}}}}}}}},\hfill \end{array}$$

(31)

where, for $k\ge 1,$

$$d_{2k-1}={\displaystyle \frac{k}{2(2k-1)}},d_{2k}={\displaystyle \frac{k}{2(2k+1)}}.$$

The branched continued fraction in (31) converges and represents a single-valued branch of the function,

$$\begin{array}{c}\hfill \left(1+\frac{z_3}{(1+z_1)(1+z_2)}\right),\end{array}$$

(32)

in the domain

$${\mathrm{H}}_{r^{\ast }}=\left\{\mathbf{z}\in {\mathbb{C}}^3:|arg(z_k+1-r^{\ast })|<\pi ,k=1,2,\left|arg\left(z_3+\frac{r^{\ast }}{2}\right)\right|<\pi \right\},0<r^{\ast }<1.$$

The numerical illustration of triple power series

$$\begin{array}{cc}\hfill ln\left(1+\frac{z_3}{(1+z_1)(1+z_2)}\right)& =z_3{F}_K(a_1,1,1,b_2;a_1,b_2,2;-z_1,-z_2,-z_3)\hfill \\ \hfill & =z_3-z_1{z}_3-z_2{z}_3-z_3^2+\dots ,\hfill \end{array}$$

(33)

and the branched continued fraction (31) is given in Table 1.

Table 1. Relative error of 5th partial sum and 5th approximant for (32).

z	(32)	(33)	(31)
$(0.1,0.1,-0.4)$	$-0.40134$	$2.96\times {10}^{-1}$	$1.00\times {10}^{-1}$
$(0.4,0.4,-0.4)$	$-0.22826$	$4.92\times {10}^{-1}$	$4.00\times {10}^{-1}$
$(0.5,0.5,0.5)$	$0.20067$	$8.22\times {10}^{-1}$	$5.00\times {10}^{-1}$
$(0.9,0.9,0.9)$	$0.22259$	$2.22\times {10}^1$	$9.00\times {10}^{-1}$
$(0.1,0.1,10)$	$2.22619$	$9.10\times {10}^3$	$1.89\times {10}^{-1}$
$(-0.01,-0.01,10)$	$2.41619$	$7.54\times {10}^3$	$9.03\times {10}^{-2}$
$(0.1,0.1,50)$	$3.74531$	$1.66\times {10}^7$	$9.32\times {10}^{-1}$
$(-0.9,-0.9,9)$	$6.80351$	$3.32\times {10}^2$	$7.14\times {10}^{-1}$

Calculations were performed using Wolfram Mathematica software 13.1.0.0 for Linux.

5. Conclusions

In this paper, we constructed two formal branched continued fraction expansions for Lauricella–Saran hypergeometric function ratios defined by (7) and (8). Our method is based on the classical method of constructing a Gaussian continued fraction [34], which can be applied to other Lauricella–Saran functions. To prove the convergence of expansions to ratios, we used the PC method, which is described in Section 2.1. These branched continued fractions are fascinating in their forms and have good approximate properties (in particular, compared with triple power series under certain conditions, they have wider convergence domains and are endowed with the property of numerical stability). They can bring new insights into the study of the hypergeometric functions of several variables. Their potential wide domain convergence and estimates of the rate of convergence are an interesting direction worth exploring in the future. Along this path, ideas implemented in [35,36,37,38,39] can be used.