APPLICATION OF CONTINUED FRACTIONS TO THE COMPUTATION OF FUNCTIONS OF TWO VARIABLES

V. E. Kornilov

(Presented by the Department of Higher Mathematics)

In the present article, the hypergeometric function of two variables (1) is represented in the form of a sum of suitable fractions. For the remainder term an estimate in modulus is given in a certain complex domain of the complex variables \(x\) and \(y\) (7).

In the article the following abbreviated notation for the product is introduced:

\[ (a)_n=a(a+1)\ldots(a+n-1),\quad (a)_0=1. \]

1. The function of two variables [1], p. 219, (7)

\[ F_2(\alpha,\beta,b,\gamma,c,x,y) = \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \frac{(\alpha)_{m+n}(\beta)_m(b)_n}{(\gamma)_m(c)_n\,m!\,n!}\,x^m y^n \tag{1} \]

is represented by the multiple integral [1], p. 224, (2)

\[ F_2(\alpha,\beta,b,\gamma,c,x,y) = F_2(x,y) = \frac{\Gamma(\gamma)\Gamma(c)} {\Gamma(\beta)\Gamma(b)\Gamma(\gamma-\beta)\Gamma(c-b)} \times \]

\[ \times \int_0^1\int_0^1 u^{\beta-1}v^{b-1}(1-u)^{\gamma-\beta-1} (1-v)^{c-b-1}(1-ux-vy)^{-\alpha}\,du\,dv, \tag{2} \]

\[ \beta>0,\quad b>0,\quad \gamma-\beta>0,\quad c-b>0. \]

With respect to the function (1) we shall prove a theorem.

Theorem. The function (1) with real parameters

\[ 0<2\alpha\leq 1,\quad 0<2\beta\leq\gamma,\quad 0<2b\leq c \tag{3} \]

is represented in the form of a sum of suitable fractions and a remainder term:

\[ F_2(x,y) = R_{nn}(x,y) + \sum_{\lambda=1}^{n}\sum_{\mu=1}^{n} \frac{a_\lambda b_\lambda c_\mu d_\mu}{a_\lambda c_\mu-x} \times \]

\[ \times \frac{ \displaystyle \sum_{m=1}^{n} \left( \frac{x-a_\lambda c_\mu}{c_\mu y} \right)^m \sum_{k=0}^{\,n-m} C_n^{m+k}(-1)^k \frac{(c+n-1)_{m+k}}{(b+k)_m(c)_k} }{ \displaystyle \sum_{m=0}^{n} C_n^m \frac{(c+n-1)_m}{(b)_m} \left( \frac{x-a_\lambda c_\mu}{c_\mu y} \right)^m }. \tag{4} \]

The coefficients \(a_\lambda,\ b_\lambda,\ c_\mu,\ d_\mu\) in equality (4) have positive values [2], pp. 5, 24, and, in view of [3], p. 20, (1) and [4], p. 23 (5), are determined according to the equalities

\[ \left\{ \begin{aligned} &\prod_{\lambda=1}^{n}\left(1-\frac{x}{a_\lambda}\right) = \sum_{m=0}^{n} C_n^m \frac{(1-\alpha-n)_m}{(1-2n)_m}(-x)^m = q_{2n}(x);\\ &\prod_{\mu=1}^{n}\left(1-\frac{x}{c_\mu}\right) = \sum_{m=0}^{n} C_n^m \frac{(1-\beta-n)_m}{(2-\gamma-2n)_m}(-x)^m = Q_{2n}(x);\\ &1<a_1<a_2<\cdots<a_n,\qquad 1<c_1<c_2<\cdots<c_n . \end{aligned} \right. \tag{5} \]

\[ \left\{ \begin{aligned} b_\lambda &=-p_{2n}(a_\lambda):\left\{a_\lambda\frac{d}{dx}\,[q_{2n}(a_\lambda)]\right\},\\[4pt] p_{2n}(x) &=\sum_{m=1}^{n-1} C_{n-1}^{m}\, \frac{(\beta-n)_m}{(1-2n)_m}(-x)^m,\\[4pt] d_\mu &=-P_{2n}(c_\mu):\left\{c_\mu\frac{d}{dx}\,[Q_{2n}(c_\mu)]\right\},\\[4pt] P_{2n}(x) &=\sum_{m=1}^{n}(-x)^{\,n-m}\sum_{\kappa=0}^{\,n-m} C_n^{m+\kappa}(-1)^\kappa \frac{(\gamma+n-1)_{m+\kappa}(\beta)_n} {(\beta+\kappa)_m(\gamma)_\kappa(\gamma+n-1)_n}. \end{aligned} \right. \tag{6} \]

The absolute value of the remainder term \(R_{nn}(x,y)\) in equality (4) is less than the following sum:

\[ \left\{ \begin{aligned} |R_{nn}(x,y)|<&\, \frac{(\alpha)_n(1-\alpha)_n\,|1+\alpha_1(1-x-y)|^2\,\sigma^{2n}:[(n)_n(n+1)_n]} {|q_{2n}(x+y)q_{2n+2}(x+y)|\,[\,|1+\alpha_1(1-x-y)|^2-\sigma^2\,]}\\ &+\sum_{\lambda=1}^{n} \frac{a_\lambda b_\lambda(\beta)_n(\gamma-\beta)_n\,n!\,|1+\beta_1(1-x_1)|^2\,|x:\tau|^{2n+1}:[\,|x|(\gamma)_{2n}\,]} {(\gamma+n-1)_n\,|Q_{2n}(x_1)Q_{2n+2}(x_1)|\,[\,|1+\beta_1(1+x_1)|^2-|x:\tau|^2\,]}\\ &+\sum_{\lambda=1}^{n}\sum_{\mu=1}^{n} \frac{a_\lambda b_\lambda d_\mu(b)_n(c-b)_n\,n!\,|y_1|^{2n+1}|1+b_1(1-y_1)|^2:[\,|y|(c)_{2n}\,]} {(c+n-1)_n\,|Q_{2n}(y_1)Q_{2n+2}(y_1)|\,[\,|1+b_1(1-y_1)|^2-|y_1|^2\,]};\\[4pt] &\sigma=\max|ux+vy|,\quad |\tau|=\min|a_\lambda-vy|\quad \text{for }0\le u\le 1,\ 0\le v\le 1;\\ &\sigma<|1+\alpha_1(1-x-y)|,\quad |x:\tau|<|1+\beta_1(1-x_1)|;\\[4pt] &\alpha_1=\frac{\alpha+a}{1-\alpha+a},\qquad b_1=\frac{b+n}{c-b+n},\qquad \beta_1=\frac{\beta+n}{\gamma-\beta+n};\\[6pt] &x_1=\frac{x}{a_\lambda-y},\qquad y_1=\frac{yc_\mu}{a_\lambda c_\mu-x};\\[6pt] &0\le \operatorname{Re}(x+y)<1,\qquad 0\le \operatorname{Re}\frac{x}{1-y}\le 1,\\[6pt] &0\le \operatorname{Re}\frac{y}{1-x}\le 1,\qquad n=1,2,\ldots . \end{aligned} \right. \tag{7} \]

Proof. On the basis of the formula [1], p. 116, (1) and the second equality (5), we have

\[ Q_{2n}(x)=F(-n,1-\beta-n;\,2-\gamma-2n;\,x)= \]

\[ =\frac{(\gamma-\beta)_n}{(\gamma+n-1)_n} F(-n,1-\beta-n;\,\gamma-\beta;\,1-x) =\frac{(\gamma-\beta)_n}{(\gamma+n-1)_n}T_{2n}(x). \tag{8} \]

In view of formulas (1)—(7), [4], pp. 23–24 and [5], p. 312, (9), we transform the double integral (2) in the following way (below the abbreviated notation
\[ A_1=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)},\qquad A=\frac{\Gamma(\gamma)\Gamma(c)}{\Gamma(\beta)\Gamma(b)\Gamma(\gamma-\beta)\Gamma(c-b)} \]
is used):
\[ \left\{ \begin{aligned} F_2(x,y)={}&A\sum_{\lambda=1}^{n} b_\lambda\int_{0}^{1}v^{b-1}(1-v)^{c-b-1} \left(\frac{a_\lambda-vy}{a_\lambda}\right)^{-1}dv \times \\ &\times \int_{0}^{1}u^{\beta-1}(1-u)^{\gamma-\beta-1} \left(1-\frac{ux}{a_\lambda-vy}\right)^{-1}du \\ ={}&A\int_{0}^{1}v^{b-1}(1-v)^{c-b-1}dv \times \\ &\times \int_{0}^{1}u^{\beta-1}(1-u)^{\gamma-\beta-1} r_{2n}(ux+vy)\,du +A_1\sum_{\lambda=1}^{n}\sum_{\mu=1}^{n} \frac{a_\lambda b_\lambda c_\mu d_\mu}{a_\lambda c_\mu-x}\times \\ &\times \int_{0}^{1}v^{b-1}(1-v)^{c-b-1}(1-vy_1)^{-1}dv +\rho_{2n}(ux+vy)+ \\ &+A_1\sum_{\lambda=1}^{n}a_\lambda b_\lambda\int_{0}^{1} \frac{v^{b-1}(1-v)^{c-b-1}}{a_\lambda-vy} R_{2n}\left(\frac{x}{a_\lambda-vy}\right)dv \\ ={}&\sum_{\lambda=1}^{n}\sum_{\mu=1}^{n} \frac{a_\lambda b_\lambda c_\mu d_\mu}{a_\lambda c_\mu-x} \left[\frac{P_{2n}(y_1)}{Q_{2n}(y_1)}+\sigma_{2n}(y_1)\right]+ \\ &+\rho_{2n}(ux+vy)+s_{2n}\left(\frac{x}{a_\lambda-vy}\right). \end{aligned} \right. \tag{9} \]

The remainder term \(s_{2n}\), according to equalities (7)—(9), [4] p. 21, (1) and inequality (15), is transformed as follows:
\[ \left(T_n\left(\frac{x}{a_\lambda-vy}\right)\equiv T_n\right) \]
\[ \left\{ \begin{aligned} \left|s_{2n}\left(\frac{x}{a_\lambda-vy}\right)\right| &<\sum_{\lambda=1}^{n}\frac{b_\lambda a_\lambda}{|x|} \left|\sum_{\kappa=n}^{\infty} \frac{(\beta)_\kappa\,\kappa!\,(\gamma+2\kappa)\,|x:\tau|^{2\kappa+1}} {(\gamma)_\kappa(\gamma-\beta)_{\kappa+1}T_{2\kappa}T_{2\kappa+2}} \right| \\ &<\sum_{\lambda=1}^{n} \frac{a_\lambda b_\lambda(\beta)_n(\gamma-\beta)_n n!\,|1+\beta_1(1-x_1)|^2\,|x:\tau|^{2n+1}} {(\gamma+n-1)_n\,|Q_{2n}(x_1)Q_{2n+2}(x_1)|\,[\,|1+\beta_1(1-x_1)|^2-|x:\tau|^2\,]}, \end{aligned} \right. \tag{10} \]
where
\[ |\tau|=\min |a_\lambda-vy|,\qquad 0\le v\le 1, \]
\[ \left|\frac{x}{\tau}\right|<|1+\beta_1(1-x_1)|,\qquad 0<2\beta\le\gamma. \]
\[ \beta_1=\frac{\beta+n}{\gamma-\beta+n};\qquad x_1=\frac{x}{a_\lambda-y}. \]

The right-hand side of inequality (10) coincides with the second term on the right-hand side of inequality (7). Estimates for the moduli of the remainder terms \(\rho_{2n}\) and \(\sigma_{2n}\) in equality (9) are obtained analogously. In view of equality (9) and inequality (10), the theorem is proved.

Let us prove the theorem on the inequality of polynomials.

Theorem. If the roots of the polynomial \(Q_{2k+2}(z)\) are separated by the roots of the polynomial \(Q_{2k}(z)\), i.e.,

\[ \begin{gathered} Q_{2k}(z)=(z+\beta_1)\ldots(z+\beta_k),\\ Q_{2k+2}(z)=(z+\alpha_1)\ldots(z+\alpha_{k+1}),\\ \beta_0=\alpha_1>\beta_1>\alpha_2>\beta_2>\ldots>\alpha_k>\beta_k>\beta_{k+1}=\alpha_{k+1}>0, \end{gathered} \tag{11} \]

then, with respect to these polynomials, the inequality

\[ |Q_{2k+2}(z)|\ge |Q_{2k}(z)|\,|z+\alpha|,\quad k=1,2,\ldots, \]

where \(\operatorname{Re} z\ge 0,\ \alpha_1\alpha_2\ldots\alpha_{k+1}=\beta_1\ldots\beta_k\alpha,\)

\[ \alpha_1>\alpha>\alpha_{k+1}. \tag{12} \]

Proof. By direct calculation it is not difficult to prove that inequality (12) is valid for two factors, i.e.,

\[ |z+c|\,|z+d|\ge |z+a|\,|z+b|,\quad \operatorname{Re}z\ge 0,\quad ab=cd, \]

\[ 0<a\le b<d. \tag{13} \]

Let \(\beta_{l-1}>\alpha>\beta_l,\quad l=1,\ldots,k+1\); then, on the basis of inequalities (11), (13), and the equalities (11), (12), we obtain the following inequalities:

\[ \left\{ \begin{array}{l} |z+\alpha_1|\,|z+\alpha_2|\ge |z+\beta_1|\left|z+\dfrac{\alpha_1\alpha_2}{\beta_1}\right|,\\[1.2em] \left(\alpha_1>\dfrac{\alpha_1\alpha_2}{\beta_1}>\alpha_2\right);\\ \cdots\\[0.6em] \left|z+\dfrac{\alpha_1\ldots\alpha_l}{\beta_1\ldots\beta_{l-1}}\right|\,|z+\alpha_{l+1}| \ge |z+\alpha|\left|z+\dfrac{\beta_l\ldots\beta_k}{\alpha_{l+2}\ldots\alpha_{k+1}}\right|,\\[1.2em] \left(\dfrac{\alpha_1\ldots\alpha_l}{\beta_1\ldots\beta_{l-1}}> \dfrac{\alpha_1\ldots\alpha_{l+1}}{\beta_1\ldots\beta_{l-1}\alpha} = \dfrac{\beta_l\ldots\beta_k}{\alpha_{l+2}\ldots\alpha_{k+1}} >\beta_l\right);\\ \cdots\\[0.6em] \left|z+\dfrac{\beta_{k-1}\beta_k}{\alpha_{k+1}}\right|\,|z+\alpha_{k+1}| \ge |z+\beta_{k-1}|\,|z+\beta_k|,\\[1.2em] \left(\dfrac{\beta_{k-1}\beta_k}{\alpha_{k+1}}>\beta_{k-1}>\beta_k>\alpha_{k+1}\right). \end{array} \right. \tag{14} \]

Multiplying the left- and right-hand sides of inequalities (14) and canceling both sides of the resulting inequality by equal factors, we obtain inequality (12), as was required to prove.

In view of equality (8) and inequality (12), we have:

\[ |T_{2n+2m}(z)|<|T_{2n}(z)|\left|1+\frac{\beta+n}{\gamma-\beta+n}(1-z)\right|^m, \]

\[ \gamma>2\beta>0,\quad m=1,2,\ldots \tag{15} \]

REFERENCES

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