Application of a Product of Binomials for Computing the Kummer Function
Abstract Generated abstract
This paper develops approximations to the confluent hypergeometric, or Kummer, function for positive parameters satisfying \(2\alpha > \gamma > 0\). It represents the logarithmic derivative of the function by a continued fraction, converts its convergents into polynomial ratios, and derives a product of binomial factors with an exponential multiplier. The construction includes recurrence relations for the required polynomials and explicit bounds for the remainder term, giving a computable approximation scheme. Special cases are shown to recover exact identities and to yield approximations for the probability integral and Fresnel integrals, with sample numerical errors reported.
Full Text
Application of the Product of Binomials to the Computation of the Kummer Function
V. E. Kornilov
(Presented by the Department of Higher Mathematics of the Tomsk Polytechnic Institute)
The article examines the question of obtaining approximations of the Kummer function in the form of a product of binomials. These approximations are obtained on the basis of continued fractions and may be applied to the computation of this function.
The Kummer function contains the parameters \(\alpha\) and \(\gamma\), which in this article are studied for positive values \(2\alpha > \gamma > 0\).
- The degenerate hypergeometric function (Kummer function)
\[ \Phi(\alpha;\gamma;x)=\sum_{k=0}^{\infty}\frac{(\alpha)_k}{(\gamma)_k\,k!}x^k,\quad |x|<N, \]
\[ \gamma \ne 0,-1,\ldots;\quad (\alpha)_k=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)},\quad (\alpha)_0=1 \tag{1} \]
may be represented as the quotient of two functions in the following way:
\[ \Phi(\alpha;\gamma;x)=e^{\ln \Phi(\alpha;\gamma;x)} = \exp\left\{ \int_0^x \frac{d[\Phi(\alpha;\gamma;t)]}{\Phi(\alpha;\gamma;t)} \right\}, \]
i.e.,
\[ \Phi(\alpha;\gamma;x)= \exp\left[ \alpha\int_0^x \frac{\Phi(\alpha+1;\gamma+1;t)} {\gamma\Phi(\alpha;\gamma;t)}\,dt \right]. \tag{2} \]
Next we replace the ratio of two series under the integral sign in (2) by its known expansion into a continued fraction ([1], p. 134) and represent the function (1) in the form of the following product:
\[ \Phi(\alpha;\gamma;x)= \exp\left[ \alpha\int_0^x \frac{P_{2n+1}(t)}{Q_{2n+1}(t)}\,dt \right] \exp\left[ -\alpha\int_0^x R_{2n+1}(t)\,dt \right], \tag{3} \]
where
\[ \rho_{2n+1}(x)=-\alpha\int_0^x R_{2n+1}(t)\,dt, \tag{4} \]
\[ \frac{P_k(t)}{Q_k(t)} \quad \frac{1}{\alpha_1+\dfrac{1}{\alpha_2+\cdots+\dfrac{1}{\alpha_k}}}, \]
\[ \alpha_1=\gamma,\quad \alpha_{2m}=(-1)^m\frac{(\gamma+2m-1)(\alpha+1)_{m-1}}{(\gamma-\alpha)_m}\cdot\frac{1}{t}, \]
\[ \alpha_{2m+1}=(-1)^m\frac{(\gamma+2m)(\gamma-\alpha)_m}{(\alpha+1)_m}, \qquad m=1,2,\ldots; \tag{5} \]
\[ R_{2n+1}(t)=\sum_{k=n+1}^{\infty} \frac{\alpha_{2k+1}}{Q_{2k-1}(t)\,Q_{2k+1}(t)} \quad ([2], \text{ p. }34). \tag{6} \]
For the numerator and denominator of the convergent fraction (5), the following equalities and relation are known ([2], pp. 13, 14)
\[ P_{2n+1}(t)=1+\cdots+ \frac{(\gamma+1)_{2n}}{(\alpha+1)_n}\cdot\frac{1}{t^n} = \frac{(\gamma+1)_{2n}}{(\alpha+1)_n}\cdot\frac{1}{t^n}\,p_{2n+1}(t), \tag{7} \]
\[ Q_{2n+1}(t)=\sum_{m=0}^{n}\alpha_{2m+1}+\cdots+ \frac{(\gamma)_{2n+1}}{(\alpha+1)_n}\cdot\frac{1}{t^n} = \frac{(\gamma)_{2n+1}}{(\alpha+1)_n}\cdot\frac{1}{t^n}\,q_{2n+1}(t), \tag{8} \]
where
\[ D_n=\sum_{m=0}^{n}\alpha_{2m+1} =\sum_{m=0}^{n}(-1)^m \frac{(\gamma+2n)(\gamma-\alpha)_m}{(\alpha+1)_m}. \tag{9} \]
\[ \varphi_{2n+1}(x)= \left[ 1+\frac{2\alpha-\gamma}{(\gamma+2n-2)(\gamma+2n)}\,x \right]\varphi_{2n-1}(x)+ \]
\[ +\frac{(\alpha+n-1)(\gamma-\alpha+n-1)}{(\gamma+2n-2)(\gamma+2n-3)_3}\, x^2\varphi_{2n-3}(x),\qquad n=2,3,\ldots \tag{10} \]
where \(p_{2n+1}(x)\) and \(q_{2n+1}(x)=\varphi_{2n+1}(x)\).
Since the right-hand side of equality (9) can be represented as the difference of two generalized hypergeometric series and, applying the following formula ([3], p. 191) in the special case \((c=1)\):
\[ \sum_{m=0}^{\infty}\alpha_{2m+1} = \gamma\,{}_3F_2\left(1+\frac{\gamma}{2},\ \gamma-\alpha,\ 1;\ \frac{\gamma}{2},\ \alpha+1;\ -1\right) =\alpha, \tag{11} \]
we obtain
\[ D_n=\sum_{m=0}^{n}\alpha_{2m+1} = \frac{(\alpha)_{n+1}+(-1)^n(\gamma-\alpha)_{n+1}}{(\alpha+1)_n}. \tag{12} \]
- For the following values of the parameters \(2\alpha>\gamma>0\) and of the independent variable \(x>0\), we have:
1) for \(\alpha=\gamma+k+\varepsilon\) \((0<\varepsilon<1)\), all \(\alpha_1,\alpha_2,\ldots,\alpha_{2k+3}\) in equalities (5) \((m=1,\ldots,k+1)\) are positive, and the roots of \(Q_{2k+3}(x)\) are negative ([2], p. 14) and are separated by the roots of \(Q_{2k+1}(x)\); therefore, in view of equality (8),
\[ q_1(x)<q_3(x)<\cdots<q_{2n+3}(x); \tag{13} \]
2) for \(\dfrac{\gamma}{2}<\alpha<\gamma+1\) and for \(x=\gamma+k+\varepsilon\) \((m=k+2,\ldots)\), in view of relations (10) and inequalities (13), we obtain:
\[ q_{2n+1}(x)<q_{2n-3}(x)<\cdots;\qquad n=0,1,\ldots \tag{14} \]
Starting from equalities (5), (8), and (14), it is not difficult to obtain an estimate for the functional series (6):
\[ \left|R_{2n+1}(t)\right| < \frac{|(\gamma-\alpha)_{n+1}|\,|(\alpha+1)_n|\,t^{2n+1}} {(\gamma)_{2n+1}(\gamma)_{2n+2}\,[q_{2n+1}(t)]^2}, \qquad t>0 \tag{15} \]
For the modulus of integral (4), in view of inequalities (15), one can establish the upper bound:
\[ |\rho_{2n+1}(x)| < \frac{x^{2n+2}|(\gamma-\alpha)_{n+1}|\,|(\alpha)_{n+1}|} {(\gamma)_{2n+1}(\gamma)_{2n+2}[q_{2n+1}(x)]^2}, \qquad x>0. \tag{16} \]
Equality (3), taking into account (7) and (8), can be represented in the form of the following sum:
\[ \Phi(\alpha;\gamma;x)\exp\left[ \frac{\alpha}{\gamma}\int_0^x \frac{p_{2n+1}(t)}{q_{2n+1}(t)}\,dt \right] +r_{2n+1}(x) = \]
\[ =\Phi_{2n+1}(x)+r_{2n+1}(x), \tag{17} \]
where
\[ |r_{2n+1}(x)| < e^{\frac{\alpha}{\gamma}x} \exp|\rho_1(x)|\,[\exp|\rho_{2n+1}(x)|-1]. \tag{18} \]
3. Theorem. The confluent hypergeometric function (1) is represented by the following product of binomials:
\[ \Phi(\alpha;\gamma;x) = e^{b_0 x}\prod_{m=1}^{n} \left(1+\frac{x}{a_m}\right)^{b_m} + r_{2n+1}(x), \tag{19} \]
where
\[ \prod_{m=1}^{n}\left(1+\frac{x}{a_m}\right)=q_{2n+1}(x), \qquad b_0= \frac{(\alpha)_{n+1}} {(\alpha)_{n+1}+(-1)^n(\gamma-\alpha)_{n+1}}, \]
\[ b_m= \frac{\alpha(\alpha-\gamma)a_m A_{2n+1}(-a_m)} {\gamma^2(\gamma+1)\dfrac{d}{dx}\,[q_{2n+1}(-a_m)]}, \qquad n=1,2,\ldots \tag{20} \]
The polynomials \(A_{2n+1}(x)\), \(q_{2n+1}(x)\), and \(p_{2n+1}(x)\) are the following:
\[ A_1(x)=0,\qquad A_3(x)=q_1(x)=p_1(x)=1, \]
\[ q_3(x)=1+\frac{2\alpha-\gamma}{\gamma(\gamma+2)}x, \qquad p_3(x)=1+\frac{(\alpha+1)}{(\gamma+1)_2}x; \tag{21} \]
for \(n=2,3,\ldots\) they are determined successively by means of relation (10). The estimate of the remainder term \(|r_{2n+1}(x)|\) is established by means of inequalities (18) and (16).
Proof. In view of the following equality
\[ (\gamma-\alpha)xA_{2n+1}(x) = (\gamma)_2[p_{2n+1}(x)-q_{2n+1}(x)], \qquad n=1,2,\ldots; \tag{22} \]
and also
\[ M_{2n+1}(x) = \frac{\alpha}{\gamma}p_{2n+1}(x) - \frac{\alpha}{D_n}q_{2n+1}(x), \tag{23} \]
the polynomials \(A_{2n+1}(x)\), \(M_{2n+1}(x)\), together with \(q_{2n+1}(x)\) and \(p_{2n+1}(x)\), satisfy relation (10).
Applying equalities (7)—(9), (20), and (23), we transform equality (17):
\[ \Phi(\alpha;\gamma;x)\exp\left[\frac{\alpha}{D_n}\int_0^x dt+\int_0^x \frac{M_{2n+1}(t)}{q_{2n+1}(t)}\,dt\right]+r_{2n-1}(x)= \]
\[ =\exp\left[b_0\int_0^x dt+\sum_{m=1}^n \frac{b_m}{a_m}\int_0^x \frac{dt}{1+\dfrac{t}{a_m}}\right]+r_{2n+1}(x), \]
where \(b_0=\dfrac{\alpha}{D_n}\), and, in view of equality (12),
\[ b_0=\frac{(\gamma)_{n+1}}{(\alpha)_{n+1}+(-1)^n(\gamma-\alpha)_{n+1}}, \]
and also
\[ \Phi(\alpha;\gamma;x)=e^{b_0x}\prod_{m=1}^n\left(1+\frac{x}{a_m}\right)^{b_m}+r_{2n+1}(x),\quad b_m= \]
\[ =\frac{M_{2n+1}(-a_m)} {\dfrac{d}{dx}\,[q_{2n+1}(-a_m)]}, \]
moreover, on the basis of (19) and (17), the estimate of \(|r_{2n+1}(x)|\) is found according to inequalities (18) and (16).
In view of the first equality (20), \(q_{2n+1}(-a_m)=0\); therefore, applying equality (23), we obtain
\[ b_m= \frac{ M_{2n+1}(-a_m)+\left(\dfrac{\alpha}{D_n}-\dfrac{\alpha}{\gamma}\right)q_{2n+1}(-a_m) }{ \dfrac{d}{dx}\,[q_{2n+1}(-a_m)] } = \]
\[ = \frac{ \dfrac{\alpha}{\gamma}\,[p_{2n+1}(-a_m)-q_{2n+1}(-a_m)] }{ \dfrac{d}{dx}\,[q_{2n+1}(-a_m)] } \]
and, on the basis of equality (22),
\[ b_m=\frac{\alpha(\alpha-\gamma)a_m A_{2n+1}(-a_m)} {\gamma^2(\gamma+1)\dfrac{d}{dx}\,[q_{2n+1}(-a_m)]}, \]
thus the theorem is completely proved.
4. In the case of complex roots of the polynomial \(q_{2n+1}(x)\), these roots \(a_m\) and the coefficients \(b_m\), respectively, will be pairwise conjugate, and each such two factors can be transformed into elementary functions with real coefficients. For example, after elementary transformations we obtain \((i=\sqrt{-1})\)
\[ \left(1+\frac{x}{a+bi}\right)^{\sigma+\tau i} \left(1+\frac{x}{a-bi}\right)^{\sigma-\tau i} = \]
\[ = \left[ \frac{b^2x^2}{(a^2+b^2)^2} + \left(\frac{ax}{a^2+b^2}+1\right)^2 \right]^\sigma e^{\,2\tau\arctg\frac{bx}{ax+a^2+b^2}}. \tag{24} \]
Applying formula (19) in the case \(n=1\), we obtain:
\[ \Phi(\alpha;\gamma;x)= e^{\frac{\alpha(\alpha+1)x}{(2\alpha-\gamma)(\gamma+1)}} \left(1+\frac{2\alpha-\gamma}{\gamma^2+2\gamma}x\right)^{ \frac{\alpha(\alpha-\gamma)(\gamma+2)^2}{(2\alpha-\gamma)^2(\gamma+1)} } +r_3(x). \tag{25} \]
From equality (25) one can obtain both exact equalities
\[ \Phi(\gamma+1;\gamma;x)=e^x\left(1+\frac{1}{\gamma}x\right); \qquad \Phi(-1;\gamma;x)=1-\frac{x}{\gamma}; \]
and approximate equalities for the probability integral and the Fresnel integrals ([3], p. 254):
\[ \operatorname{Erf}(x)\equiv \int_0^x e^{-t^2}\,dt =x\Phi\left(\frac12;\frac32;-x^2\right)= \]
\[ =xe^{0.6x^2}\left(1+\frac{2}{21}x^2\right)^{-9.8} +xr_3(-x^2); \tag{26} \]
\[ C(x)\equiv \frac{1}{\sqrt{2\pi}}\int_0^x \frac{\cos t}{\sqrt{t}}\,dt = \sqrt{-\frac{x}{2\pi}} \left[ \Phi\left(\frac12;\frac32;-ix\right) + \right. \]
\[ \left. +\Phi\left(\frac12;\frac32;ix\right) \right] \approx \sqrt{\frac{2x}{\pi}} \left(1+\frac{4x^2}{441}\right)^{-4.9} \times \]
\[ \times \cos\left(9.8\arctan\frac{2x}{21}-0.6x\right), \tag{27} \]
\[ S(x)\equiv \frac{1}{\sqrt{2\pi}}\int_0^x t^{-\frac12}\sin t\,dt \approx \]
\[ \approx \sqrt{\frac{2x}{\pi}} \left(1+\frac{4x^2}{441}\right)^{-4.9} \sin\left(9.8\arctan\frac{2x}{21}-0.6x\right), \tag{28} \]
where for \(x=1\) we obtain:
\[ C(1)\approx 0.72203\;(\Delta=-0.00033),\qquad S(1)\approx 0.24774\;(\Delta=-0.00014), \]
here \(\Delta\) is the absolute error.
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