Truncation Error Bounds for Branched Continued Fraction Expansions of Some Appell’s Hypergeometric Functions F₂

Abstract

This paper considers the problem of approximating some Appell’s hypergeometric functions $F_2$ by their branched continued fraction expansions. Using the formula for the difference of two approximants of a branched continued fraction, we established the truncation error bounds for such expansions. In addition, we provided another proof of the convergence of branched continued fraction expansions to the ratio of Appell’s hypergeometric functions $F_2$. Finally, we also provide examples to demonstrate the effectiveness of branched continued fractions as a tool for approximating special functions.

1. Introduction

The Gaussian or ordinary hypergeometric function and its various generalizations, including Appell’s hypergeometric functions, appear in various fields of mathematics and its applications [1,2].

This paper considers the Appell’s hypergeometric function $F_2,$ which is defined as follows (see [3,4]):

$$F_2(a, b_1, b_2; c_1, c_2; \mathbf{z})=\sum _{p,q=0}^{\infty }{\displaystyle \frac{{\left(a\right)}_{p+q}{\left(b_1\right)}_p{\left(b_2\right)}_q}{{\left(c_1\right)}_p{\left(c_2\right)}_q}}{\displaystyle \frac{z_1^p}{p!}}{\displaystyle \frac{z_2^q}{q!}},$$

where $a, b_1, b_2, c_1, c_2\in \mathbb{C};$ $c_1, c_2\notin \{0, -1, -2, \dots \};$

$${\left(\alpha \right)}_k={\displaystyle \frac{\mathsf{\Gamma }(\alpha +k)}{\mathsf{\Gamma }\left(\alpha \right)}}=\left\{\begin{array}{c}1,\mathrm{if}k=0,\hfill \\ \alpha (\alpha +1)(\alpha +2)\dots (\alpha +k-1),\mathrm{if}k\ge 1;\hfill \end{array}\right.$$

$\mathbf{z}=(z_1, z_2)\in P,$

$$\begin{array}{c}\hfill P=\{\mathbf{z}\in {\mathbb{C}}^2:|z_1|+|z_2|<1\}.\end{array}$$

(1)

Currently, there are several papers that deal with this function (see, for example, [5,6]), primarily because of its diverse applications [7,8].

Among the variety of problems related to $F_2$, one of the most intriguing is constructing its representation in terms of a branched continued fraction, which would be analogous to constructing a Gaussian continued fraction for an ordinary hypergeometric function [9]. Such representations are used to compute special functions (see, for example, [10,11]).

In [12], D. Bodnar obtained a formal expansion of the ratio of the functions

$${\displaystyle \frac{F_2(a, b_1, b_2; c_1, c_2; \mathbf{z})}{F_2(a+1, b_1, b_2; c_1+1, c_2; \mathbf{z})}}$$

into a branched continued fraction

$$\begin{array}{c}\hfill 1-\sum _{i_1=1}^2{\displaystyle \frac{p_{i_1}{z}_{i_1}}{1-{\displaystyle \frac{q_{i_1}{z}_{i_1}}{{\displaystyle 1-\sum _{i_2=1}^2{\displaystyle \frac{p_{i_1,i_2}{z}_{i_2}}{1-{\displaystyle \frac{q_{i_1,i_2}{z}_{i_2}}{{1 - }_{\ddots }}}}}}}}}},\end{array}$$

where $p_{i_1},$ $q_{i_1},$ $p_{i_1,i_2},$ $q_{i_1,i_2},$ … are the coefficients expressed in terms of the parameters of Appell’s hypergeometric function $F_2.$ However, there are still no convergence criteria for this branched continued fraction. The problem of convergence of branched continued fractions can be found in [13,14]. The structures of branched continued fraction expansions are considered in [15].

In [16], the authors considered the case when $b_1=c_1,$ from which it follows that the ratio

$$\begin{array}{c}\hfill {\displaystyle \frac{F_2(a, b_1, b_2; b_1, c_2; \mathbf{z})}{F_2(a+1, b_1, b_2; b_1, c_2+1; \mathbf{z})}}\end{array}$$

(2)

has a formal branched continued fraction expansion of the following form:

$$\begin{array}{c}\hfill 1-z_1-{\displaystyle \frac{{\tau }_1{z}_2}{1-{\displaystyle \frac{{\tau }_2{z}_2}{1-z_1-{\displaystyle \frac{{\tau }_3{z}_2}{1-{\displaystyle \frac{{\tau }_4{z}_2}{1-z_1-{\displaystyle \frac{{\tau }_5{z}_2}{{1 - }_{\ddots }}}}}}}}}}},\end{array}$$

(3)

where

$$\begin{array}{c}\hfill {\tau }_{2k-1}={\displaystyle \frac{(b_2+k-1)(c_2-a+k-1)}{(c_2+2k-2)(c_2+2k-1)}},{\tau }_{2k}={\displaystyle \frac{(a+k)(c_2-b_2+k)}{(c_2+2k-1)(c_2+2k)}},k\ge 1.\end{array}$$

Provided that $a,$ $b_2,$ and $c_2$ are the real constants such that

$$\begin{array}{c}\hfill 0<{\tau }_k\le \tau ,k\ge 1,\end{array}$$

(4)

where $\tau$ is a positive number, it is proven that (3) converges uniformly on every compact subset of the domain

$$\begin{array}{c}\hfill D_{\kappa ,\tau }=\left\{\mathbf{z}\in {\mathbb{C}}^2:z_1\notin [1-\kappa , +\infty ),z_2\notin \left[{\displaystyle \frac{\kappa }{4\tau }}, +\infty \right)\right\},0<\kappa <1,\end{array}$$

to a function $f\left(\mathbf{z}\right)$ that is holomorphic in $D_{\kappa ,\tau },$; furthermore, $f\left(\mathbf{z}\right)$ is an analytic continuation of the function (2) in $D_{\kappa ,\tau }.$ In [17] in (2) and (3), the variable $z_2$ is replaced by $-z_2$, and a similar result is shown in some domain of ${\mathbb{C}}^2$, provided that $a,$ $b_2,$ and $c_2$ are complex constants from some parabolic region. Here, the region refers to a domain (an open connected set) which may include all, part, or none of its boundary.

Mainly motivated by the aforementioned recent developments, we establish truncation errors bounds for branched continued fractions (3) in the case of non-negative elements. Such bounds are of great importance for the computation of special functions using branched continued fraction representations [18,19]. In addition, we also provide another proof of the convergence of (3) to function (2).

2. Truncation Error Bounds

Let $g_k\left(\mathbf{z}\right)$ be the kth approximant of the branched continued fraction

$$\begin{array}{c}\hfill {\beta }_0\left(\mathbf{z}\right)+\sum _{i_1=1}^2{\displaystyle \frac{{\alpha }_{i_1}\left(\mathbf{z}\right)}{{\displaystyle {\beta }_{i_1}\left(\mathbf{z}\right)+\sum _{i_2=1}^2{\displaystyle \frac{{\alpha }_{i_1,i_2}\left(\mathbf{z}\right)}{{\beta }_{i_1,i_2}\left(\mathbf{z}\right){ + }_{\ddots }}}}}},\end{array}$$

(5)

where ${\beta }_0\left(\mathbf{z}\right),$ ${\alpha }_{i_1}\left(\mathbf{z}\right),$ ${\beta }_{i_1}\left(\mathbf{z}\right),$ ${\alpha }_{i_1,i_2}\left(\mathbf{z}\right),$ ${\beta }_{i_1,i_2}\left(\mathbf{z}\right),$ $\dots ,$ are functions of $\mathbf{z}$ defined in the region E.

If for each $\mathbf{z}\in E$, the branched continued fraction (5) converges to the finite value $g\left(\mathbf{z}\right)$, then $g\left(\mathbf{z}\right) - g_k\left(\mathbf{z}\right)$ is called the truncation error of the kth approximant. The estimate of the form

$$|g\left(\mathbf{z}\right)-g_k\left(\mathbf{z}\right)|\le C_k\left(\mathbf{z}\right),$$

where $C_k\left(\mathbf{z}\right)\ge 0$ and $C_k\left(\mathbf{z}\right)\to 0$ as $k\to +\infty$ for all $\mathbf{z}\in E$, is called the a priori bound (or truncation error bound) (see [20,21,22]).

Theorem 1.

Suppose that $a,$ $b_2,$ and $c_2$ are real constants that satisfy the inequalities in (4). Then, the following apply:

(i) 

The branched continued fraction (3) converges to a finite value $f\left(\mathbf{z}\right)$ for each $\mathbf{z}\in H_{\kappa },$ where

$$\begin{array}{c}\hfill H_{\kappa }=\{\mathbf{z}\in {\mathbb{R}}^2:z_1\le \kappa ,z_2\le 0\},0<\kappa <1;\end{array}$$

(6)

(ii) 

The convergence is uniform on every compact subset of the domain $Int\left(H_{\kappa }\right),$ and $f\left(\mathbf{z}\right)$ is analytic on $Int\left(H_{\kappa }\right)$;

(iii) 

If $f_k\left(\mathbf{z}\right)$ denotes the kth approximant of (3), then for each $\mathbf{z}\in H_{\kappa }$

$$\begin{array}{c}\hfill |f\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)|\le {\displaystyle \frac{({\delta }_1^{{(-1)}^k}|z_1|(1-{\delta }_1^{{(-1)}^{k+1}}{z}_1)+\tau |z_2\left|\right)\left(\tau \right|z_2{\left|\right)}^k}{(1-z_1)(1-z_1+\tau |z_2{\left|\right)}^{k-1}}},k\ge 2;\end{array}$$

(iv) 

The function $f\left(\mathbf{z}\right)$ is an analytic continuation of (2) in (6).

Proof. 

First we will prove (i). Let

$$\begin{array}{c}\hfill H_k^{\left(k\right)}\left(\mathbf{z}\right)=1,k\ge 1,\end{array}$$

(7)

and

$$\begin{array}{cc}\hfill H_p^{\left(k\right)}\left(\mathbf{z}\right)& =1-{\delta }_1^{{(-1)}^p}{z}_1-{\displaystyle \frac{{\tau }_{p+1}{z}_2}{1-{\delta }_1^{{(-1)}^{p+1}}{z}_1-{\displaystyle \frac{{\tau }_{p+2}{z}_2}{{1 -}_{ \ddots {\displaystyle -{\delta }_1^{{(-1)}^{k-2}}{z}_1-{\displaystyle \frac{{\tau }_{k-1}{z}_2}{1-{\delta }_1^{{(-1)}^{k-1}}{z}_1-{\tau }_k{x}_2}}}}}}}},\hfill \end{array}$$

where $k\ge 2,$ $1\le p\le k-1.$ Then it is easy to see that

$$\begin{array}{c}\hfill H_p^{\left(k\right)}\left(\mathbf{z}\right)=1-{\delta }_1^{{(-1)}^p}{z}_1-{\displaystyle \frac{{\tau }_{p+1}{z}_2}{H_{p+1}^{\left(k\right)}\left(\mathbf{z}\right)}},1\le p\le k-1,k\ge 2.\end{array}$$

(8)

Additionally, we can write the following:

$$\begin{array}{c}\hfill f_k\left(\mathbf{z}\right)=1-z_1-{\displaystyle \frac{{\tau }_1{z}_2}{H_1^{\left(k\right)}\left(\mathbf{z}\right)}},k\ge 1.\end{array}$$

(9)

In the following, we will estimate $|f_{k+p}\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)|,$ where $k\ge 1$ and $p\ge 1$.

Let $\mathbf{z}$ be an arbitrary fixed point in (6). Then, by condition (4), it is easy to see that the elements of the branched continued fraction (3) are non-negative, and thus, from (8) for arbitraries k and p such that $1\le p\le k-1$ and $k\ge 2$, we obtain

$$\begin{array}{c}\hfill H_p^{\left(k\right)}\left(\mathbf{z}\right)\ge 1-{\delta }_1^{{(-1)}^p}{z}_1\ge 1-{\delta }_1^{{(-1)}^p}\kappa >0.\end{array}$$

(10)

Since $H_p^{\left(k\right)}\left(\mathbf{z}\right)\ne 0$ for all indexes and $\mathbf{z}\in H_{\kappa }$, the following formula holds (see [21]):

$$\begin{array}{c}\hfill f_{k+p}\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)=-z_2^k\left({\delta }_1^{{(-1)}^k}{z}_1+{\displaystyle \frac{{\tau }_{k+1}{z}_2}{H_{k+1}^{(k+p)}\left(\mathbf{z}\right)}}\right)\prod _{r=1}^k{\displaystyle \frac{{\tau }_r}{H_r^{(k+p)}\left(\mathbf{z}\right)H_r^{\left(k\right)}\left(\mathbf{z}\right)}},k\ge 1p\ge 1.\end{array}$$

(11)

Indeed, using (9), for $k\ge 1$ and $p\ge 1$, we have that

$$\begin{array}{cc}\hfill f_{k+p}\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)& =1-z_1-{\displaystyle \frac{{\tau }_1{z}_2}{H_1^{(k+p)}\left(\mathbf{z}\right)}}-\left(1-z_1-{\displaystyle \frac{{\tau }_1{z}_2}{H_1^{\left(k\right)}\left(\mathbf{z}\right)}}\right)\hfill \\ \hfill & ={\displaystyle \frac{{\tau }_1{z}_2}{H_1^{(k+p)}\left(\mathbf{z}\right)H_1^{\left(k\right)}\left(\mathbf{z}\right)}}(H_1^{(k+p)}\left(\mathbf{z}\right)-H_1^{\left(k\right)}\left(\mathbf{z}\right))\hfill \end{array}$$

By (8) for $1\le r\le k-1$ and $k\ge 2$, we obtain the following:

$$\begin{array}{cc}\hfill H_r^{(k+p)}\left(\mathbf{z}\right)-H_r^{\left(k\right)}\left(\mathbf{z}\right)& =1-{\delta }_1^{{(-1)}^r}{z}_1-{\displaystyle \frac{{\tau }_{r+1}{z}_2}{H_{r+1}^{(k+p)}\left(\mathbf{z}\right)}}-\left(1-{\delta }_1^{{(-1)}^r}{z}_1-{\displaystyle \frac{{\tau }_{r+1}{z}_2}{H_{r+1}^{\left(k\right)}\left(\mathbf{z}\right)}}\right)\hfill \\ \hfill & ={\displaystyle \frac{{\tau }_{r+1}{z}_2}{H_{r+1}^{(k+p)}\left(\mathbf{z}\right)H_{r+1}^{\left(k\right)}\left(\mathbf{z}\right)}}(H_{r+1}^{(k+p)}\left(\mathbf{z}\right)-H_{r+1}^{\left(k\right)}\left(\mathbf{z}\right))\hfill \end{array}$$

Then, taking (7) into account, we have that

$$\begin{array}{cc}\hfill H_k^{(k+p)}\left(\mathbf{z}\right)-H_k^{\left(k\right)}\left(\mathbf{z}\right))& =1-{\delta }_1^{{(-1)}^k}{z}_1-{\displaystyle \frac{{\tau }_{k+1}{z}_2}{H_{k+1}^{(k+p)}\left(\mathbf{z}\right)}}-1\hfill \\ \hfill & =-\left({\delta }_1^{{(-1)}^k}{z}_1+{\displaystyle \frac{{\tau }_{k+1}{z}_2}{H_{k+1}^{(k+p)}\left(\mathbf{z}\right)}}\right).\hfill \end{array}$$

Now, due to the above, it is easy to obtain Formula (11). For convenience, we write this formula in the following form:

$$\begin{array}{cc}\hfill f_{k+p}\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)& =-{\displaystyle \frac{{\tau }_1{z}_2^k}{H_1^{\left(q\right)}\left(\mathbf{z}\right)H_k^{(k+p)}\left(\mathbf{z}\right)}}\left({\delta }_1^{{(-1)}^k}{z}_1+{\displaystyle \frac{{\tau }_{k+1}{z}_2}{H_{k+1}^{(k+p)}\left(\mathbf{z}\right)}}\right)\hfill \\ & \times \prod _{r=1}^{[k/2]}{\displaystyle \frac{{\tau }_{2r}}{H_{2r-1}^{\left(d\right)}\left(\mathbf{z}\right)H_{2r}^{\left(d\right)}\left(\mathbf{z}\right)}}\prod _{r=1}^{\left[\right(k-1)/2]}{\displaystyle \frac{{\tau }_{2r+1}}{H_{2r}^{\left(q\right)}\left(\mathbf{z}\right)H_{2r+1}^{\left(q\right)}\left(\mathbf{z}\right)}},\hfill \end{array}$$

where $q=k+p$ and $d=k$ if $k=2s,$ and $q=k$ and $d=k+p$ if $k=2s+1,$ $s\ge 1.$

Using (4), (6), and (10), for any $k\ge 2$ and $p\ge 2$, we have that

$${\displaystyle \frac{{\tau }_1{z}_2}{H_1^{\left(k\right)}\left(\mathbf{z}\right)}}\le \tau \left|z_2\right|$$

and

$${\displaystyle \frac{1}{H_k^{(k+p)}\left(\mathbf{z}\right)}}\left({\delta }_1^{{(-1)}^k}{z}_1+{\displaystyle \frac{{\tau }_{k+1}{z}_2}{H_{k+1}^{(k+p)}\left(\mathbf{z}\right)}}\right)\le {\displaystyle \frac{1}{1-{\delta }_1^{{(-1)}^k}{z}_1}}\left({\delta }_1^{{(-1)}^k}\left|z_1\right|+{\displaystyle \frac{\tau |z_2|}{1-{\delta }_1^{{(-1)}^{k+1}}{z}_1}}\right).$$

Next, additionally using (8) and for $1\le p\le k-1$ and $k\ge 2$, we get

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\tau }_{p+1}{z}_2}{H_p^{(k+1)}\left(\mathbf{z}\right)H_{p+1}^{(k+1)}\left(\mathbf{z}\right)}}& ={\displaystyle \frac{{\displaystyle \frac{{\tau }_{p+1}{z}_2}{H_{p+1}^{(k+1)}\left(\mathbf{z}\right)}}}{1-{\delta }_1^{{(-1)}^p}{z}_1-{\displaystyle \frac{{\tau }_{p+1}{z}_2}{H_{p+1}^{(k+1)}\left(\mathbf{z}\right)}}}}\hfill \\ & \le {\displaystyle \frac{{\displaystyle \frac{{\tau }_{p+1}\left|z_2\right|}{H_{p+1}^{(k+1)}\left(\mathbf{z}\right)}}}{1-{\delta }_1^{{(-1)}^p}{z}_1+{\displaystyle \frac{{\tau }_{p+1}\left|z_2\right|}{H_{p+1}^{(k+1)}\left(\mathbf{z}\right)}}}}\hfill \\ & \le {\displaystyle \frac{\tau |z_2|}{1-z_1+\tau \left|z_2\right|}}\hfill \end{array}$$

and, furthermore,

$$\begin{array}{cc}\hfill {\displaystyle \frac{{\tau }_{k+1}{z}_2}{H_k^{(k+1)}\left(\mathbf{z}\right)H_{k+1}^{(k+1)}\left(\mathbf{z}\right)}}& ={\displaystyle \frac{{\tau }_{k+1}{z}_2}{1-{\delta }_1^{{(-1)}^k}{z}_1-{\tau }_{p+1}{z}_2}}\hfill \\ & \le {\displaystyle \frac{\tau |z_2|}{1-{\delta }_1^{{(-1)}^k}{z}_1+\tau \left|z_2\right|}}.\hfill \end{array}$$

Now, due to the above, for $k\ge 2$ and $p\ge 2,$, we obtain the following:

$$\begin{array}{cc}\hfill |f_{k+p}\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)|& \le {\displaystyle \frac{({\delta }_1^{{(-1)}^k}|z_1|(1-{\delta }_1^{{(-1)}^{k+1}}{z}_1)+\tau |z_2\left|\right)\left(\tau \right|z_2{\left|\right)}^k}{(1-{\delta }_1^{{(-1)}^k}{z}_1)(1-{\delta }_1^{{(-1)}^{k+1}}{z}_1)(1-{\delta }_1^{{(-1)}^k}{z}_1+\tau |z_2\left|\right)(1-z_1+\tau |z_2{\left|\right)}^{k-1}}}\hfill \\ \hfill & ={\displaystyle \frac{({\delta }_1^{{(-1)}^k}|z_1|(1-{\delta }_1^{{(-1)}^{k+1}}{z}_1)+\tau |z_2\left|\right)\left(\tau \right|z_2{\left|\right)}^k}{(1-z_1)(1-{\delta }_1^{{(-1)}^k}{z}_1+\tau |z_2\left|\right)(1-z_1+\tau |z_2{\left|\right)}^{k-1}}}.\hfill \end{array}$$

(12)

Obviously, for an arbitrary fixed $\mathbf{z}\in H_{\kappa }$,

$${\displaystyle \frac{({\delta }_1^{{(-1)}^k}|z_1|(1-{\delta }_1^{{(-1)}^{k+1}}{z}_1)+\tau |z_2\left|\right)\left(\tau \right|z_2{\left|\right)}^k}{(1-z_1)(1-{\delta }_1^{{(-1)}^k}{z}_1+\tau |z_2\left|\right)(1-z_1+\tau |z_2{\left|\right)}^{k-1}}}\to 0\mathrm{as}k\to +\infty .$$

Then, due to the arbitrariness of p, we obtain (i).

In what follows, we will prove (iii). Let K be an arbitrary compact subset of $Int\left(H_{\kappa }\right)$; then there exists a constant $M>0$ such that

$$\begin{array}{cc}\hfill |f_{k+p}\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)|& <{\displaystyle \frac{({\delta }_1^{{(-1)}^k}M(1-{\delta }_1^{{(-1)}^{k+1}}\kappa )+\tau M){\left(\tau M\right)}^k}{(1-\kappa )(1-{\delta }_1^{{(-1)}^k}\kappa +\tau M){(1-\kappa +\tau M)}^{k-1}}},k\ge 2p\ge 2,\mathbf{z}\in K.\hfill \end{array}$$

Furthermore, if m and n are arbitrary natural numbers such that $m\ge 2$ and $n\ge k\ge 2,$ then

$$|f_{n+m}\left(\mathbf{z}\right)-f_n\left(\mathbf{z}\right)|\le |f_{n+m}\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)|+|f_n\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)|,\mathbf{z}\in K.$$

Due to

$${\displaystyle \frac{({\delta }_1^{{(-1)}^k}M(1-{\delta }_1^{{(-1)}^{k+1}}\kappa )+\tau M){\left(\tau M\right)}^k}{(1-\kappa )(1-{\delta }_1^{{(-1)}^k}\kappa +\tau M){(1-\kappa +\tau M)}^{k-1}}}\to 0\mathrm{as}k\to +\infty ,$$

we get (ii).

It is clear that (iii) follows from (12) when we pass to the limit as $p\to +\infty$.

Finally, we will prove (iv) by the PF method (based on the so-called ‘fork property’ for a branched continued fraction with positive elements). It is obvious that

$${\displaystyle \frac{F_2(a, b_1, b_2; b_1, c_2; \mathbf{0})}{F_2(a+1, b_1, b_2; b_1, c_2+1; \mathbf{0})}}=1.$$

Then, there exists $0<\epsilon <1/2$ such that function (2) is analytic in domain

$$\begin{array}{c}{P}_{\epsilon }=\left\{\mathbf{z}\in {\mathbb{R}}^2:-\epsilon <z_1<0,-\epsilon <z_2<0,\right\},\end{array}$$

and $P_{\epsilon }\subset (P\cap Int\left(H_{\kappa }\right)),$ where P is defined by (1). In particular, $P_{1/4}\subset (P\cap Int\left(H_{\kappa }\right)).$

In what follows, we will show that function (2) and the elements of the branched continued fraction (3) are positive-valued functions in $P_{\epsilon }$.

Let $\mathbf{z}$ be an arbitrary fixed point in $P_{\epsilon }.$ Then, by condition (4), it is easy to see that the elements of the branched continued fraction (3) are positive, which means that the approximants of (3) satisfy the following inequalities:

$$\begin{array}{c}\hfill f_{2k}\left(\mathbf{z}\right)<f_{2k+2}\left(\mathbf{z}\right)<f_{2k+1}\left(\mathbf{z}\right)<f_{2k-1}\left(\mathbf{z}\right),k\ge 1,\end{array}$$

i.e., the ‘fork property’ holds (see [21]). This, together with (i), for each $\mathbf{z}\in P_{\epsilon }$ ensures the convergence of the sequences of even and odd approximants of (3) to a finite value $f\left(\mathbf{z}\right).$

Next, we will consider

$$\begin{array}{c}\hfill {\displaystyle \frac{F_2(a, b_1, b_2; b_1, c_2; \mathbf{z})}{F_2(a+1, b_1, b_2; b_1, c_2+1; \mathbf{z})}}-f_k\left(\mathbf{z}\right),k\ge 1,\end{array}$$

where

$$\begin{array}{c}{\displaystyle \frac{F_2(a, b_1, b_2; b_1, c_2; \mathbf{z})}{F_2(a+1, b_1, b_2; b_1, c_2+1; \mathbf{z})}}=1-z_1-{\displaystyle \frac{{\tau }_1{z}_2}{1-{\displaystyle \frac{{\tau }_2{z}_2}{{1 - }_{\ddots {\displaystyle -{\delta }_1^{{(-1)}^k}{z}_1-{\displaystyle \frac{{\tau }_{k+1}{z}_2}{E_{k+1}^{(k+1)}\left(\mathbf{z}\right)}}}}}}}},\end{array}$$

and

$$\begin{array}{c}\hfill E_{k+1}^{(k+1)}\left(\mathbf{z}\right)={\displaystyle \frac{{\displaystyle F_2\left(a+\sum _{r=0}^k{\delta }_1^{{(-1)}^r},b,b_2+\sum _{r=0}^{k-1}{\delta }_1^{{(-1)}^r};b,c_2+k+1;\mathbf{z}\right)}}{{\displaystyle F_2\left(a+\sum _{r=0}^{k+1}{\delta }_1^{{(-1)}^r},b,b_2+\sum _{r=0}^k{\delta }_1^{{(-1)}^r};b,c_2+k+2;\mathbf{z}\right)}}},k\ge 1.\end{array}$$

Let

$$\begin{array}{c}\hfill E_p^{(k+1)}\left(\mathbf{z}\right)=1-{\delta }_1^{{(-1)}^p}{z}_1-{\displaystyle \frac{{\tau }_{p+1}{z}_2}{1-{\delta }_1^{{(-1)}^{p+2}}{z}_1-{\displaystyle \frac{{\tau }_{p+2}{z}_2}{{1 - }_{\ddots {\displaystyle -{\delta }_1^{{(-1)}^k}{z}_1-{\displaystyle \frac{{\tau }_{k+1}{z}_2}{E_{k+1}^{(k+1)}\left(\mathbf{z}\right)}}}}}}}},\end{array}$$

where $1\le p\le k,$ $k\ge 1.$ Then, similarly to (8), we have that

$$\begin{array}{c}\hfill E_p^{(k+1)}\left(\mathbf{z}\right)=1-{\delta }_1^{{(-1)}^p}{z}_1-{\displaystyle \frac{{\tau }_{p+1}{z}_2}{E_{p+1}^{(k+1)}\left(\mathbf{z}\right)}},1\le p\le k,k\ge 1.\end{array}$$

It is obvious that $H_p^{\left(k\right)}\left(\mathbf{z}\right)\ne 0$ and $E_p^{\left(k\right)}\left(\mathbf{z}\right)\ne 0,$ $k\ge 1,$ $1\le p\le k,$ and $\mathbf{z}\in P_{\epsilon }$. Taking (11) into account for $k\ge 1$, we obtain the following:

$$\begin{array}{cc}\hfill {\displaystyle \frac{F_2(a, b_1, b_2; b_1, c_2; \mathbf{z})}{F_2(a+1, b_1, b_2; b_1, c_2+1; \mathbf{z})}}-f_k\left(\mathbf{z}\right)& =-z_2^k\left({\delta }_1^{{(-1)}^k}{z}_1+{\displaystyle \frac{{\tau }_{k+1}{z}_2}{E_{k+1}^{(k+1)}\left(\mathbf{z}\right)}}\right)\prod _{r=1}^k{\displaystyle \frac{{\tau }_r}{E_r^{(k+1)}\left(\mathbf{z}\right)H_r^{\left(k\right)}\left(\mathbf{z}\right)}}.\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill f_{2k}\left(\mathbf{z}\right)<{\displaystyle \frac{F_2(a, b_1, b_2; b_1, c_2; \mathbf{z})}{F_2(a+1, b_1, b_2; b_1, c_2+1; \mathbf{z})}}<f_{2k-1}\left(\mathbf{z}\right),k\ge 1,\mathbf{z}\in P_{\epsilon }.\end{array}$$

Now, due to the above, including (i), we have that

$$\begin{array}{c}\hfill \underset{k\to +\infty }{lim}{f}_{2k}\left(\mathbf{z}\right)=\underset{k\to +\infty }{lim}{f}_{2k-1}\left(\mathbf{z}\right)=f\left(\mathbf{z}\right),\mathbf{z}\in P_{\epsilon },\end{array}$$

and, therefore,

$$\begin{array}{c}\hfill f\left(\mathbf{z}\right)={\displaystyle \frac{F_2(a, b_1, b_2; b_1, c_2; \mathbf{z})}{F_2(a+1, b_1, b_2; b_1, c_2+1; \mathbf{z})}},\mathbf{z}\in P_{\epsilon },\end{array}$$

i.e., the branched continued fraction (3) converges to function (2) in the domain $P_{\epsilon }.$

Finally, applying Theorem 5 from [17], we complete the proof of (iv). □

In what follows, setting $a=0$ and replacing $c_2$ with $c_2-1$ in Theorem 1, we obtain the truncation error bound for a branched continued fraction of the form

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{1-z_1-{\displaystyle \frac{{\varrho }_1{z}_2}{1-{\displaystyle \frac{{\varrho }_2{z}_2}{1-z_1-{\displaystyle \frac{{\varrho }_3{z}_2}{1-{\displaystyle \frac{{\varrho }_4{z}_2}{{1 - }_{\ddots }}}}}}}}}}},\end{array}$$

(13)

where

$$\begin{array}{c}\hfill {\varrho }_1={\displaystyle \frac{b_2}{c_2}},{\varrho }_{2k}={\displaystyle \frac{k(c_2-b_2+k-1)}{(c_2+2k-2)(c_2+2k-1)}},{\varrho }_{2k+1}={\displaystyle \frac{(b_2+k)(c_2+k-1)}{(c_2+2k-1)(c_2+2k)}},k\ge 1.\end{array}$$

(14)

Corollary 1.

Suppose that $b_2$ and $c_2$ are real constants that satisfy the following inequality:

$$\begin{array}{c}\hfill 0<{\varrho }_k\le \varrho ,k\ge 1,\end{array}$$

(15)

where ${\varrho }_k,$ $k\ge 1,$ are defined by (14), and ϱ is a positive number. Then, the following apply:

(i) 

The branched continued fraction (13) converges to a finite value $g\left(\mathbf{z}\right)$ for each $\mathbf{z}\in H_{\kappa },$ where $H_{\kappa }$ is defined by (6);

(ii) 

The convergence is uniform on every compact subset of the domain $Int\left(H_{\kappa }\right),$ and $g\left(\mathbf{z}\right)$ is analytic on $Int\left(H_{\kappa }\right)$;

(iii) 

If $g_k\left(\mathbf{z}\right)$ denotes the kth approximant of (13), then for each $\mathbf{z}\in H_{\kappa }$,

$$\begin{array}{c}\hfill |g\left(\mathbf{z}\right) - g_k\left(\mathbf{z}\right)|\le {\displaystyle \frac{({\delta }_1^{{(-1)}^k}|z_1|(1-{\delta }_1^{{(-1)}^{k+1}}{z}_1)+\varrho |z_2\left|\right)\left(\varrho \right|z_2{\left|\right)}^{k-1}}{{(1-z_1)}^3(1 - z_1+\varrho |z_2{\left|\right)}^{k-3}}},k\ge 3;\end{array}$$

(iv) 

The function $g\left(\mathbf{z}\right)$ is an analytic continuation of $F_2(1, b, b_2; b, c_2; \mathbf{z})$ in (6).

Here it is sufficient to note that if $g_k\left(\mathbf{z}\right)$ denotes the kth approximant of (13), then

$$g_1\left(\mathbf{z}\right)=T_0^{\left(0\right)}\left(\mathbf{z}\right)=1$$

and

$$g_k\left(\mathbf{z}\right)={\displaystyle \frac{1}{T_0^{(k-1)}\left(\mathbf{z}\right)}}={\displaystyle \frac{1}{1-z_1-{\displaystyle \frac{{\varrho }_1{z}_2}{T_1^{(k-1)}\left(\mathbf{z}\right)}}}},k\ge 2,$$

where $T_1^{\left(1\right)}\left(\mathbf{z}\right)$ and $T_1^{(k-1)}\left(\mathbf{z}\right),$ $k\ge 3,$ are defined by analogy with (7) and (8), respectively.

Similarly to the proof of (10), it can be shown that

$$T_p^{(k-1)}\left(\mathbf{z}\right)\ge 1-{\delta }_1^{{(-1)}^p}{z}_1\ge 1-{\delta }_1^{{(-1)}^p}\kappa >0,0\le p\le k-1,k\ge 2,\mathbf{z}\in H_{\kappa },$$

where $H_{\kappa }$ is defined by (6). Then the following formula holds (see [21]):

$$\begin{array}{cc}\hfill g_{k+p}\left(\mathbf{z}\right)-g_k\left(\mathbf{z}\right)& ={\displaystyle \frac{z_2^{k-1}}{T_0^{(k+p-1)}\left(\mathbf{z}\right)T_0^{(k-1)}\left(\mathbf{z}\right)}}\left({\delta }_1^{{(-1)}^{k-1}}{z}_1+{\displaystyle \frac{{\varrho }_k{z}_2}{T_k^{(k+p-1)}\left(\mathbf{z}\right)}}\right)\hfill \\ & \times \prod _{r=1}^{k-1}{\displaystyle \frac{{\varrho }_r}{T_r^{(k+p-1)}\left(\mathbf{z}\right)T_r^{(k-1)}\left(\mathbf{z}\right)}},k\ge 2p\ge 1,\mathbf{z}\in H_{\kappa },\hfill \end{array}$$

or

$$\begin{array}{cc}\hfill g_{k+p}\left(\mathbf{z}\right)-g_k\left(\mathbf{z}\right)& ={\displaystyle \frac{{\varrho }_1{z}_2^{k-1}}{T_0^{(k+p-1)}\left(\mathbf{z}\right)T_0^{(k-1)}\left(\mathbf{z}\right)T_1^{\left(q\right)}\left(\mathbf{z}\right)T_{k-1}^{(k+p-1)}\left(\mathbf{z}\right)}}\left({\delta }_1^{{(-1)}^{k-1}}{z}_1+{\displaystyle \frac{{\varrho }_k{z}_2}{T_k^{(k+p-1)}\left(\mathbf{z}\right)}}\right)\hfill \\ & \times \prod _{r=1}^{\left[\right(k-1)/2]}{\displaystyle \frac{{\varrho }_{2r}}{T_{2r-1}^{\left(d\right)}\left(\mathbf{z}\right)T_{2r}^{\left(d\right)}\left(\mathbf{z}\right)}}\prod _{r=1}^{\left[\right(k-2)/2]}{\displaystyle \frac{{\varrho }_{2r+1}}{T_{2r}^{\left(q\right)}\left(\mathbf{z}\right)T_{2r+1}^{\left(q\right)}\left(\mathbf{z}\right)}},\hfill \end{array}$$

where $q=k+p-1$ and $d=k-1$ if $k=2s+1,$ and $q=k-1$ and $d=k+p-1$ if $k=2s+2,$ $s\ge 1,$ $\mathbf{z}\in H_{\kappa }.$ Hence, for each $\mathbf{z}\in H_{\kappa }$, the following ‘fork property’ holds:

$$\begin{array}{c}\hfill g_{2k-1}\left(\mathbf{z}\right)<g_{2k+1}\left(\mathbf{z}\right)<g_{2k+2}\left(\mathbf{z}\right)<g_{2k}\left(\mathbf{z}\right),k\ge 1.\end{array}$$

3. Numerical Experiments

From [23,24], Corollary 1, and Corollary 2.3 from [16], it follows that the function

$$\begin{array}{c}\hfill ln\left(\sqrt{{\displaystyle \frac{z_2}{1+z_1}}}+\sqrt{1+{\displaystyle \frac{z_2}{1+z_1}}}\right)\end{array}$$

(16)

has representations in the series

$$\begin{array}{c}\sqrt{z_2}\sqrt{1+z_1+z_2}{F}_2(1, b_1, 1; b_1, 3/2; -z_1,-z_2)\hfill \\ \hfill =\sqrt{z_2}\sqrt{1+z_1+z_2}\sum _{p,q=0}^{\infty }{\displaystyle \frac{{\left(1\right)}_{p+q}{\left(1\right)}_q}{{(3/2)}_q}}{\displaystyle \frac{{(-z_1)}^p}{p!}}{\displaystyle \frac{{(-z_2)}^q}{q!}}\end{array}$$

(17)

and a branched continued fraction

$$\begin{array}{c}\hfill {\displaystyle \frac{\sqrt{z_2}\sqrt{1+z_1+z_2}}{1+z_1+{\displaystyle \frac{{\displaystyle \frac{2}{3}}{z}_2}{1+{\displaystyle \frac{{\displaystyle \frac{2}{15}}{z}_2}{1+z_1+{\displaystyle \frac{{\displaystyle \frac{12}{35}}{z}_2}{1+{\displaystyle \frac{{\displaystyle \frac{12}{63}}{z}_2}{{1 + }_{\ddots }}}}}}}}}}},\end{array}$$

(18)

which converges and represents a single-valued branch of the function (16) in the domain

$$\begin{array}{c}\hfill D_{\kappa }=\left\{\mathbf{z}\in {\mathbb{C}}^2:z_1\notin (-\infty ,-1+\kappa ],z_2\notin (-\infty ,0]\right\},0<\kappa <1.\end{array}$$

Note that the case $z_2=0$ is trivial.

Plots of the values of the nth approximants of branched continued fraction expansion (18) are shown in Figure 1a,b. On the sets given in color there, all the elements of (18) are positive, so we see the ‘fork property’ for a branched continued fraction with positive elements (see [21]). That is, the plots of the values of even (odd) approximations of expansion (18) approaches from below (above) to the plot of function (16) at a fixed value of $z_2$ (Figure 1a). The plots at fixed values of $z_1$ are similar (see Figure 1b).

Figure 2a,b show 2D plots where the 15th approximant of the branched continued fraction expansion (18) guarantees certain truncation error bounds for (16). Here we see the symmetrical regions, with a cut along the real axis from $-\infty$ to $-1$ (Figure 2a) and a cut along the real axis from $-\infty$ to 0 (Figure 2b).

Figure 3a–d show 2D plots in different planes in ${\mathbb{C}}^2,$ where the 15th approximant of (18) guarantees certain truncation error bounds for function (16). Here we observe symmetrical regions in all cases except Figure 3a.

The results of evaluations (17) and (18) are displayed in

Table 1. Relative error of 20th partial sum and 20th approximants for (16).

z	(16)	(18)	(17)
$(0.1, 0.1)$	$0.2971$	$3.74\times {10}^{-16}$	$7.47\times {10}^{-16}$
$(-0.3, 0.3)$	$0.4636$	$3.59\times {10}^{-16}$	$7.77\times {10}^{-12}$
$(1.5, 0.3)$	$0.3398$	$4.90\times {10}^{-16}$	$2.82\times {10}^3$
$(3, 3)$	$0.2612$	$2.13\times {10}^{-16}$	$3.89\times {10}^{10}$
$(10, 15)$	$0.2116$	$5.25\times {10}^{-16}$	$1.36\times {10}^{25}$
$(100, 100)$	$0.0728$	$2.10\times {10}^{-15}$	$3.60\times {10}^{51}$
$(0.5+0.5i, 0.5+0.5i)$	$0.6165+0.12881i$	$1.97\times {10}^{-16}$	$2.02\times {10}^4$
$(1-3i, 5-2i)$	$0.9556-0.5421i$	$6.52\times {10}^{-12}$	$5.49\times {10}^{35}$
$(10-5i, 4+15i)$	$0.8402+0.6810i$	$1.12\times {10}^{-11}$	$2.43\times {10}^{56}$
$(100+45i, 100+45i)$	$0.8784+0.0013i$	$7.59\times {10}^{-16}$	$1.88\times {10}^{94}$

. The analysis of these computation results shows that the approximation of function (16) is better using the 20th approximant of (18) and then the 20th partial sum of (17). We also see numerical stability of the 20th approximant of the branched continued fraction (18) at extreme inputs (e.g., $\mathbf{z}=(100,100)$, $\mathbf{z}=(100+45i,100+45i)$). Note that this computational property is inherent to continued and branched continued fractions [10,11,22].

The numerical stability problem of (3) and (13) will be studied in our next paper.

Plots of the values of the relative errors of the nth approximants of expansion (18) for function (16) are shown in Figure 4a,b. Here we observe that the relative error values tend to 0 as $k\to +\infty .$

In another example, from [23,24], Corollary 1, and Corollary 2.3 from [16], it follows that the function

$$\begin{array}{c}\hfill ln{\displaystyle \frac{1+\sqrt{{\displaystyle \frac{z_2}{1-z_1}}}}{1-\sqrt{{\displaystyle \frac{z_2}{1-z_1}}}}}\end{array}$$

(19)

has representations in the series

$$\begin{array}{c}\hfill 2\sqrt{z_2(1-z_1)}{F}_2(1, b_1, 1/2; b_1, 3/2; z_1, z_2)=2\sqrt{z_2(1-z_1)}\sum _{p,q=0}^{\infty }{\displaystyle \frac{{\left(1\right)}_{p+q}{(1/2)}_q}{{(3/2)}_q}}{\displaystyle \frac{z_1^p}{p!}}{\displaystyle \frac{z_2^q}{q!}}\end{array}$$

(20)

and a branched continued fraction

$$\begin{array}{c}\hfill {\displaystyle \frac{2\sqrt{z_2(1-z_1)}}{1-z_1-{\displaystyle \frac{{\displaystyle \frac{1}{3}}{z}_2}{1-{\displaystyle \frac{{\displaystyle \frac{4}{15}}{z}_2}{1-z_1-{\displaystyle \frac{{\displaystyle \frac{9}{35}}{z}_2}{1-{\displaystyle \frac{{\displaystyle \frac{16}{63}}{z}_2}{{1 - }_{\ddots }}}}}}}}}}},\end{array}$$

(21)

which converges and represents a single-valued branch of function (19) in the domain

$$\begin{array}{c}\hfill D_{\kappa }=\left\{\mathbf{z}\in {\mathbb{C}}^2:z_1\notin [1-\kappa , +\infty ),z_2\notin (-\infty , 0]\cup \left[{\displaystyle \frac{3\kappa }{4}}, +\infty \right)\right\},0<\kappa <1.\end{array}$$

We note that the case $z_2=0$ is trivial.

In Figure 5a,b, the sets where the elements of (18) are not positive are shown, so we cannot see the ‘fork property’ as in the previous example.

The graphical illustrations of (19) and (21) are given in Figure 6a,b and Figure 7a–d.

The numerical illustrations of (20) and (21) are given in

Table 2. Relative errors of 20th partial sum and 20th approximants for (19).

z	(19)	(21)	(20)
$(-0.1, 0.1)$	$0.6224$	$3, 57\times {10}^{-16}$	$8.92\times {10}^{-16}$
$(-0.5, 0.3)$	$0.9624$	$2.31\times {10}^{-16}$	$6.05\times {10}^{-6}$
$(-0.8, 0.3)$	$0.8670$	$1.28\times {10}^{-16}$	$1.1448$
$(-3, 0.3)$	$0.5621$	$1.98\times {10}^{-16}$	$1.20\times {10}^{12}$
$(-11, 0.5)$	$0.4141$	$1.34\times {10}^{-16}$	$2.55\times {10}^{26}$
$(-131, 0.7)$	$0.1459$	$5.71\times {10}^{-16}$	$2.70\times {10}^{51}$
$(0.5+0.5i, 0.5+0.5i)$	$0.8814+1.5708i$	$8.52\times {10}^{-14}$	$5.16\times {10}^3$
$(1+3i, 5-2i)$	$1.0818+2.0576i$	$6.74\times {10}^{-9}$	$5.59\times {10}^{34}$
$(10-5i, 4+15i)$	$1.02611-1.8952i$	$2.16\times {10}^{-10}$	$2.55\times {10}^{55}$
$(100+45i, 100+45i)$	$0.0019+1.5750i$	$1.57\times {10}^{-15}$	$1.88\times {10}^{93}$

.

Finally, plots of the values of the relative errors of the nth approximants of (21) for function (19) are shown in Figure 8a,b.

Here we have similar results to those in the previous example.

4. Conclusions

Compared to previous works (see [16,17,21]), the novelty of this paper is that under the conditions of (4) and for each $\mathbf{z}\in H_{\kappa },$ where $H_{\kappa }$ is defined by (6), we have established that the branched continued fraction (3) converges to the function $f\left(\mathbf{z}\right)$ at least as fast as geometric series with the following ratio:

$$\varrho \left(\mathbf{z}\right)={\displaystyle \frac{\tau |z_2|}{1-z_1+\tau \left|z_2\right|}}.$$

Thus,

$$\underset{k\to +\infty }{lim\; sup}{|f\left(\mathbf{z}\right)-f_k\left(\mathbf{z}\right)|}^{1/k}\le \varrho \left(\mathbf{z}\right),$$

where $f_k\left(\mathbf{z}\right)$ is the kth approximant of (3). Using the PF method (based on the ‘fork property’ of a branched continued fraction with positive elements), it is also established that $f\left(\mathbf{z}\right)$ is an analytic extension of function (2) in (6). Numerical experiments on new examples demonstrate the effectiveness and feasibility of using branched continued fractions as a tool for approximating special functions in comparison with the corresponding series.

A further research direction is the study of numerical stability (see [25,26]) of branched continued fractions (3) and (13) under conditions (4) and (15), respectively.