Application of the Product of Binomials to the Computation of the Gauss Function
Abstract Generated abstract
This paper develops approximations for the Gauss hypergeometric function in the form of finite products of binomials, under specified restrictions on the parameters and for complex arguments in certain regions. Using recurrence relations and the continued fraction expansion of a ratio of contiguous hypergeometric functions, the function is transformed into an exponential integral and then into a product representation involving the roots of an associated polynomial. The article derives recurrence formulas for the required polynomials and exponents, proves the product formula, and gives explicit remainder estimates for the approximation in domains defined by conditions on the complex variable.
Full Text
APPLICATION OF THE PRODUCT OF BINOMIALS TO THE COMPUTATION OF THE GAUSS FUNCTION
V. E. KORNILOV
(Presented by the Department of Higher Mathematics of the Tomsk Polytechnic Institute)
The article discusses the problem of obtaining approximations to the Gauss function in the form of a product of binomials. These approximations are obtained on the basis of the theory of continued fractions and can be used for computing the function under study.
The Gauss function contains the parameters \(\alpha\), \(\beta\), and \(\gamma\), which in this article are considered as varying in the following intervals: \(\gamma>\alpha>-1\), \(\gamma>\beta>0\), \(\gamma\ge 2\beta-1\); \(z\) is a complex variable.
- The hypergeometric function—the Gauss function
\[ F(\alpha,\beta;\gamma;-z)=\sum_{k=0}^{\infty} \frac{(\alpha)_k(\beta)_k}{(\gamma)_k k!}(-z)^k, \]
\[ \gamma\ne 0,-1,\ldots;\quad |\arg(1+z)|<\pi, \tag{1} \]
where
\[ (a)_k=a(a+1)\ldots(a+k-1)=\frac{\Gamma(a+k)}{\Gamma(a)},\qquad (a)_0=1, \]
is transformed as follows:
\[ F(\alpha,\beta;\gamma;-z)=e^{\ln F(\alpha,\beta;\gamma;-z)}= \]
\[ =\exp\left\{\int_0^z [F(\alpha,\beta;\gamma;-z)]^{-1}\,d[F(\alpha,\beta;\gamma;-z)]\right\}, \]
i.e.,
\[ F(\alpha,\beta;\gamma;-z)= \exp\left[ -\int_0^z \frac{\alpha\beta F(\alpha+1,\beta+1;\gamma+1;-t)} {\gamma F(\alpha,\beta;\gamma;-t)} \,dt \right]. \tag{2} \]
Next, to the numerator of the integrand in (2) we apply the recurrence relation ([1], p. 313)
\[ \frac{\beta}{\gamma}F(\alpha+1,\beta+1;\gamma+1;-t) = \frac{1}{1+t}F(\alpha,\beta;\gamma;-t) - \]
\[ -\frac{\gamma-\beta}{\gamma}\cdot\frac{1}{1+t} F(\alpha+1,\beta;\gamma+1;-t) \tag{3} \]
and replace the ratio of the two series by its known expansion into a continued fraction ([2], p. 132); then, after some simplifications, we represent the function (1) in the form of the following product:
\[ F(\alpha,\beta;\gamma;-z)=(1+z)^{-\alpha}\exp\left[\alpha(\gamma-\beta)\int_0^z \frac{P_{2n+1}(t)}{Q_{2n-1}(t)}\cdot \frac{dt}{1+t}\right]\times \]
\[ \times \exp\left[-\alpha(\gamma-\beta)\int_0^z R_{2n+1}(t)\frac{dt}{1+t}\right], \tag{4} \]
where
\[ \rho_{2n+1}(z)=-\alpha(\gamma-\beta)\int_0^z R_{2n+1}(t)\frac{dt}{1+t}; \tag{5} \]
\[ \frac{P_k(t)}{Q_k(t)} = \cfrac{1}{a_1+\cfrac{1}{a_2+\cdots+\cfrac{1}{a_k}}}, \]
\[ a_1=\gamma,\qquad a_{2m}= \frac{(\gamma+2m-1)(\alpha+1)_{m-1}(\gamma-\beta+1)_{m-1}} {(\beta)_m(\gamma-\alpha)_m}\cdot \frac{1}{t}, \]
\[ a_{2m+1}= \frac{(\gamma+2m)(\beta)_m(\gamma-\alpha)_m} {(\alpha+1)_m(\gamma-\beta+1)_m}, \qquad m=1,2,\ldots; \tag{6} \]
\[ R_{2n+1}(t)= \sum_{k=n+1}^{\infty} \frac{a_{2k+1}}{Q_{2k-1}(t)Q_{2k+1}(t)}. \tag{7} \]
For the numerator and denominator of the convergent of the fraction (6) the following equalities and relation are known ([3], pp. 13, 14):
\[ P_{2n+1}(t)=1+\ldots+ \frac{(\gamma+1)_{2n}}{(\alpha+1)_n(\gamma-\beta+1)_n}\cdot \frac{1}{t^n} = \frac{(\gamma+1)_{2n}}{(\alpha+1)_n(\gamma-\beta+1)_n}\cdot \frac{1}{t^n}p_{2n+1}(t), \tag{8} \]
\[ Q_{2n+1}(t)= \sum_{m=0}^{n} a_{2m+1}+\ldots+ \frac{(\gamma)_{2n+1}}{(\alpha+1)_n(\gamma-\beta+1)_n}\cdot \frac{1}{t^n} = \frac{(\gamma)_{2n+1}}{(\alpha+1)_n(\gamma-\beta+1)_n}\cdot \frac{1}{t^n}q_{2n+1}(t). \tag{9} \]
\[ Q_{2n+1}(z)= \left(a_{2n}a_{2n+1}+1+\frac{a_{2n+1}}{a_{2n-1}}\right)Q_{2n-1}(z) - \frac{a_{2n+1}}{a_{2n-1}}Q_{2n-3}(z). \tag{10} \]
- Since
\[ \lim_{k\to\infty}\frac{1}{t a_k a_{k+1}}=\frac14, \]
it follows, in view of ([3], p. 26), that all roots of the polynomial \(Q_{2n+1}(t)\) are real and lie in the interval \((-\infty,-1)\); therefore, also in view of equality (9),
\[ Q_{2n+1}(t)= \frac{(\gamma)_{2n+1}}{(\alpha+1)_n(\gamma-\beta+1)_n} \prod_{m=1}^{n}\left(\frac{1}{t}+\frac{1}{a_m}\right), \]
\[ a_1>a_2>\ldots>a_n>1. \tag{11} \]
Hence it is not difficult to conclude that, as \(t\) varies along the rectilinear segment joining the points \(0\) and \(z\):
1) in the regions
\[ \operatorname{Re}(t)\geq 0 \quad \text{and} \quad \operatorname{Re}\left(\frac{1}{t}\right)<-1:\quad \left|\operatorname{Re}\left(\frac{1}{t}\right)\right|>1>\frac{1}{a_m}, \]
the minimum of
\[ \left|\frac{1}{t}+\frac{1}{a_m}\right| \]
will occur at \(t=z\);
2) the maximum of the quotient
\[ \frac{1}{|Q_{2k-1}(t)|} \]
in the indicated regions of variation of the complex variable \(t\) will also occur at \(t=z\). Therefore one can obtain the following inequality for the modulus of the integral (5):
\[ |\rho_{2n+1}(z)|<|\alpha|(\gamma-\beta)|R_{2n+1}(z)| \int_0^z \frac{|dt|}{|1+t|}. \tag{12} \]
When integrating along the radius of the point \(z\), we obtain ([4], p. 430):
1) if \(\operatorname{Re}(z)\geq 0\), then
\[ \int_0^z \frac{|dt|}{|1+t|} < \int_0^{|z|} d|t|=|z|; \]
2) if \(\operatorname{Re}\left(\frac{1}{z}\right)<-1\), then \(|z|<1\) and
\[ \int_0^z \frac{|dt|}{|1+t|} \leq \int_0^{|z|} \frac{d|t|}{1-|t|} = |\ln(1-|z|)|. \]
In view of what has been stated, introduce the function
\[ \tau(z)= \begin{cases} |z|, & \text{for } \operatorname{Re}(z)\geq 0,\\ |\ln(1-|z|)|, & \text{for } \operatorname{Re}\left(\dfrac{1}{z}\right)<-1 \end{cases} \tag{13} \]
and inequality (12) is simplified:
\[ |\rho_{2n+1}(z)|<\tau(z)|\alpha|(\gamma-\beta)R_{2n+1}(z). \tag{14} \]
- The infinite functional series (7) is estimated in modulus analogously to the way this was done for the continued fraction (6) for \(\beta=1\) in the paper ([5], p. 24), namely: in view of (7), (9), (14), and \(\gamma\geq 2\beta-1\), after some simplifications we obtain:
\[ |\rho_{2n+1}(z)| < \frac{\tau(z)}{2} \sum_{k=n+1}^{\infty} \frac{|(\alpha)_k|\,|(\beta)_{k-1}(\gamma-\beta)_k(\gamma-\alpha)_k|\,|z|^{2k-1}} {|(\gamma)_{2k-1}|^2\,|q_{2n+1}(z)|^2\,|\sigma(z)|^{2k-2n-1}}, \tag{15} \]
where
\[ \sigma(z)= \begin{cases} 1, & \text{for } \operatorname{Re}(z)\geq 0,\\ |1+z|, & \text{for } \operatorname{Re}\left(\dfrac{1}{z}\right)<-1. \end{cases} \tag{16} \]
It is not difficult to verify that the ratio of the subsequent term to the preceding one in the sum (15) is less than \(|4\sigma(z)\cdot z^{-1}|^{-2}\), and also
\[ (\alpha+1)_n(\beta)_n(\gamma-\beta+1)_n(\gamma-\alpha+1)_n[(\gamma+1)_{2n}]^{-2}<2^{-4n}, \]
therefore, after simplifications, we finally obtain the following estimate in modulus for integral (5):
\[ |\rho_{2n+1}(z)|< \frac{\tau(z)\sigma(z)\,|\alpha z^{2n+1}|}{|4\sigma(z)|^2-|z|^2}\cdot \frac{(\gamma-\beta)(\gamma-\alpha)}{2^{4n-3}\gamma\,|q_{2n+1}(z)|^2}, \]
\[ 4\sigma(z)>|z|, \tag{17} \]
where \(\tau(z)\) and \(\sigma(z)\) take values according to equalities (13) and (16).
Equality (4), taking into account (8) and (9), may be written in the form of the following sum:
\[ F(\alpha,\beta;\gamma;-z) =(1+z)^{-\alpha}\exp\left[ \frac{\alpha(\gamma-\beta)}{\gamma} \int_0^z \frac{p_{2n+1}(t)}{q_{2n+1}(t)}\cdot\frac{dt}{1+t} \right]+ \]
\[ +r_{2n+1}(z)=F_{2n+1}(z)+r_{2n+1}(z), \tag{18} \]
where
\[ r_{2n+1}(z)=F_{2n+1}(z)\{\exp[\rho_{2n+1}(z)]-1\}. \]
It is not difficult to compute that
\[ |F_{2n+1}(z)|<|(1+z)^{-\frac{\alpha\beta}{\gamma}}|\,\exp|\rho_1(z)|, \]
therefore
\[ |r_{2n+1}(z)|<|(1+z)^{-\frac{\alpha\beta}{\gamma}}|\,\exp|\rho_1(z)|\,|\exp|\rho_{2n+1}(z)|-1|. \tag{19} \]
4. Theorem. The hypergeometric function (1) is represented by the following product of binomials:
\[ F(\alpha,\beta;\gamma;-z)=(1+z)^{b_0}\prod_{m=1}^{n}\left(1+\frac{z}{a_m}\right)^{b_m} +r_{2n+1}(z), \tag{20} \]
where
\[ \prod_{m=1}^{n}\left(1+\frac{z}{a_m}\right)=q_{2n+1}(z), \qquad b_0=-\frac{(\alpha)_{n+1}(\beta)_{n+1}}{(\gamma)_{2n+1}q_{2n+1}(-1)}, \]
\[ b_m= \frac{\alpha\beta(\gamma-\alpha)(\gamma-\beta)a_m A_{2n+1}(-a_m)} {\gamma^2(\gamma+1)(1-a_m)\dfrac{d}{dz}[q_{2n+1}(-a_m)]}; \qquad n=1,2,\ldots . \tag{21} \]
The polynomials \(A_{2n+1}(z)\), \(q_{2n+1}(z)\), \(p_{2n+1}(z)\) are as follows:
\[ A_1(z)=0;\qquad A_3(z)=q_1(z)=p_1(z)=1, \]
\[ q_3(z)=1+\frac{(\alpha+\beta+1)\gamma-2\alpha\beta}{\gamma(\gamma+2)}\,z, \qquad p_3(z)=1+\frac{(\alpha+1)(\gamma-\beta+1)}{(\gamma+1)(\gamma+2)}\,z; \tag{22} \]
for \(n=2,3,\ldots\) they are determined successively with the aid of relation (25).
An upper bound for \(|r_{2n+1}(z)|\) in the regions of the complex variable \( \operatorname{Re}(z)\ge 0\) and \(\operatorname{Re}\left(\frac{1}{z}\right)<-1\), bounded by \(4\sigma(z)>|z|\), is established by inequalities (19) and (17).
Proof. If we introduce the equality
\[ \beta(\gamma-\alpha)zA_{2n+1}(z) =\gamma(\gamma+1)[q_{2n+1}(z)-p_{2n+1}(z)],\quad n=1,2,\ldots; \tag{23} \]
then, in view of equalities (4), (6), (8)--(10), (22), the polynomials \(A_{2n+1}(z)\), \(q_{2n+1}(z)\), \(p_{2n+1}(z)\), and also
\[ M_{2n+1}(z)=\frac{\alpha(\gamma-\beta)}{\gamma}\,p_{2n+1}(z)-\alpha q_{2n+1}(z) \tag{24} \]
satisfy the recurrence relation
\[ \varphi_{2n+1}(z)= \left[ 1+\frac{(\alpha+\beta+2n-1)\gamma+2n(n-1)-2\alpha\beta} {(\gamma+2n-2)(\gamma+2n)}\,z \right]\varphi_{2n-1}(z)- \]
\[ -\frac{(\alpha+n-1)(\beta+n-1)(\gamma-\alpha+n-1)(\gamma-\beta+n-1)} {(\gamma+2n-3)_3(\gamma+2n-2)}\,z^2\varphi_{2n-3}(z). \tag{25} \]
\[ \tag{25} \]
The equality
\[ M_{2n+1}(-1)=-\frac{(\alpha)_{n+1}(\beta)_{n-1}}{(\gamma)_{2n+1}} \tag{26} \]
is easily proved by mathematical induction with the aid of relation (25).
Applying equalities (21), (24), (26), we transform equality (18):
\[ F(\alpha,\beta;\gamma;-z)= \exp\left[ \int_0^z \frac{M_{2n+1}(t)\,dt}{(1+t)q_{2n+1}(t)} \right]+r_{2n+1}(z)= \]
\[ =\exp\left[ b_0\int_0^z\frac{dt}{1+t} +\sum_{m=1}^n \frac{b_m}{a_m}\int_0^z\frac{dt}{1+\frac{t}{a_m}} \right]+r_{2n+1}(z), \]
i.e.
\[ F(\alpha,\beta;\gamma;-z)=(1+z)^{b_0}\prod_{m=1}^n \left(1+\frac{z}{a_m}\right)^{b_m}+r_{2n+1}(z), \tag{20} \]
where, in view of equality (26),
\[ b_0=-\frac{(\alpha)_{n+1}(\beta)_{n+1}} {(\gamma)_{2n+1}q_{2n+1}(-1)},\qquad b_m=\frac{M_{2n+1}(-a_m)} {(1-a_m)\dfrac{d}{dz}\,[q_{2n+1}(-a_m)]}, \]
moreover, on the basis of (20) and (18), the upper bound for \(|r_{2n+1}(z)|\) is estimated by inequalities (19) and (17).
In view of equalities (9) and (11), \(q_{2n+1}(-a_m)=0\), \(m=1,\ldots,n\); therefore, applying equality (24), we obtain
\[ b_m= \frac{M_{2n+1}(-a_m)+\dfrac{\alpha\beta}{\gamma}q_{2n+1}(-a_m)} {(1-a_m)\dfrac{d}{dz}\,[q_{2n+1}(-a_m)]} = \]
\[ =\frac{\alpha(\gamma-\beta)\,[p_{2n+1}(-a_m)-q_{2n+1}(-a_m)]} {\gamma(1-a_m)\dfrac{d}{dz}\,[q_{2n+1}(-a_m)]}, \]
and on the basis of equality (23)
\[ b_m= \frac{\alpha\beta(\gamma-\alpha)(\gamma-\beta)a_m A_{2n+1}(-a_m)} {\gamma^2(\gamma+1)(1-a_m)\dfrac{d}{dz}\,[q_{2n+1}(-a_m)]}, \]
which proves the theorem completely.
References
-
N. N. Lebedev. Special Functions and Their Applications. Moscow, Gostekhizdat, 1953.
-
A. N. Khovanskii. Application of Continued Fractions and Their Generalizations to Problems of Approximate Analysis. Moscow, GITTL, 1956.
-
T. I. Stieltjes. Investigations on Continued Fractions. Moscow, ONTI, 1936.
-
I. I. Privalov. Introduction to the Theory of Functions of a Complex Variable. Moscow–Leningrad, Gostekhizdat, 1948.
-
V. E. Kornilov. Application of continued fractions to the computation of integrals of binomial differentials. Izv. TPI, vol. 131, pp. 21–25, 1965.