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A Rational Approximation for the Complete Elliptic Integral of the First Kind †

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A Rational Approximation for the Complete Elliptic Integral of the First Kind † mathematics Article A Rational Approximation for the Complete Elliptic Integral of the First Kind † Zhen-Hang Yang 1,2 , Jing-Feng Tian 3,* and Ya-Ru Zhu 3 1 Engineering Research Center of Intelligent Computing for Complex Energy Systems of Ministry of Education, North China Electric Power University, Yonghua Street 619, Baoding 071003, China; [email protected] 2 Zhejiang Society for Electric Power, Hangzhou 310014, China 3 Department of Mathematics and Physics, North China Electric Power University, Yonghua Street 619, Baoding 071003, China; [email protected] * Correspondence: [email protected] † Dedicated to the 60th anniversary of Zhejiang Electric Power Company Research Institute. Received: 24 March 2020; Accepted: 18 April 2020; Published: 21 April 2020 Abstract: Let K (r) be the complete elliptic integral of the first kind. We present an accurate rational 0 2 0 2 5(r ) +126r +61 lower approximation for K (r). More precisely, we establish the inequality K (r) > 2 p p 61(r0) +110r0+21 for r 2 (0, 1), where r0 = 1 − r2. The lower bound is sharp. Keywords: complete integrals of the first kind; arithmetic-geometric mean; rational approximation MSC: Primary 33E05, 41A20; Secondary 26E60, 40A99 1. Introduction The complete elliptic integral of the first kind K (r) is defined on (0, 1) by Z p/2 1 K (r) = p dt, 0 1 − r2 sin2 t which can be also represented by the Gaussian hypergeometric function 2 1 1 ¥ (1/2)2 K ( ) = 2 = n 2n r F , ; 1; r ∑ 2 r , (1) p 2 2 n=0 (n!) where (a)0 = 1 for a 6= 0, (a)n = a(a + 1) ··· (a + n − 1) = G(a + n)/G(a) is the shifted factorial R ¥ x−1 −t function and G(x) = 0 t e dt (x > 0) is the gamma function [1,2]. The famous Landen identities [3], p. 507 show that p 2 r 1 − r 1 + r K = (1 + r) K (r) and K = K r0 (2) 1 + r 1 + r 2 for all r 2 (0, 1). Moreover, an asymptotic formula for K (r) as r ! 1− is given by 4 K (r) ∼ ln as r ! 1− (3) r0 p (see [3], p. 299), where and in what follows r0 = 1 − r2. Mathematics 2020, 8, 635; doi:10.3390/math8040635 www.mdpi.com/journal/mathematics Mathematics 2020, 8, 635 2 of 9 As we all know, analytic inequality plays a very important role in mathematics and in other parts of science (see for example, [4–6]) since it offers certain computable and accurate bounds for given complicated functions. As far as the complete elliptic integral of the first kind is concerned, there are at least two kinds of bounds for K (r). The first kind of bounds for K (r) are in terms of the inverse hyperbolic tangent function, that is, 1 1 + r artanhr = ln . 2 1 − r In 1992, Anderson, Vamanamurthy and Vuorinen [7] mentioned such bound, and they presented the double inequality artanhr 1/2 2 artanhr < K (r) < (4) r p r for r 2 (0, 1). This was improved by Alzer and Qiu in [8] (Theorem 19) as artanhr a1 2 artanhr b1 < K (r) < (5) r p r for r 2 (0, 1) with the best weights a1 = 3/4 and b1 = 1. In the same paper [8] (Theorem 19), the authors gave another better double inequality for K (r): artanhr 2 artanhr 1 − a + a < K (r) < 1 − b + b (6) 2 2 r p 2 2 r for r 2 (0, 1) with the best weights a2 = 2/p and b2 = 3/4. Recently, Yang, Qian, Chu and Zhang [9] (Theorem 3.2) obtain more accurate bounds: (3p − 7) r0 + 1 artanhr 2 17r0 + 23 artanhr 2 < K (r) < 3 (7) (5p − 12) r0 + p r p 31r0 + 89 r for r 2 (0, 1). The second kind of bounds for K (r) are related to the asymptotic Equation (3). In 1985, Carlson and Guatafson [10] showed that K (r) 4 1 < < (8) ln (4/r0) 3 + r2 for r 2 (0, 1). The first inequality of Equation (8) was improved in [7] as 9 K (r) < (9) 8.5 + r2 ln (4/r0) for r 2 (0, 1). Qiu and Vamanamurthy [11] and Alzer [12] proved the double inequality p 2 K (r) 1 2 1 + − 1 r0 < < 1 + r0 (10) ln 16 ln (4/r0) 4 for r 2 (0, 1), where the constant factors 1/4 and p/ ln 16 − 1 in Equation (10) are the best possible. The recent advance on such inequalities can be found in [13–15]. The aim of this paper is to provide the third kind of bounds (rational bounds) for K (r). More precisely, we will prove the following theorem. Mathematics 2020, 8, 635 3 of 9 Theorem 1. Let q ≥ 2. The inequality 0 2 0 2 1 (16 − 3q)(r ) + (96 + 22q) r + (61q − 112) 0 K (r) > = Lq r (11) p 8 10 (r0)2 + (7q − 4) r0 + (3q − 6) holds for all r 2 (0, 1) if and only if q ≥ q0 = 192/61. In particular, we have 0 2 0 2 5 (r ) + 126r + 61 0 K (r) > = Lq0 r p 61 (r0)2 + 110r0 + 21 for r 2 (0, 1). 2. Proof of Theorem1 To prove Theorem1, we need a sign rule for a type of special power series and polynomials, which has been proven in [9,15,16] by Yang et al. It should be noted that this sign rule plays an important role in the study for certain special functions, see, for example, [17–21]. ¥ ¥ Lemma 1. Let fakgk=0 be a nonnegative real sequence with am > 0 and ∑k=m+1 ak > 0 and let m ¥ k k S (t) = − ∑ akt + ∑ akt k=0 k=m+1 be a convergent power series on the interval (0, r) (r > 0). (i) If S (r−) ≤ 0, then S (t) < 0 for all t 2 (0, r). − (ii) If S (r ) > 0, then there is a unique t0 2 (0, r) such that S (t) < 0 for t 2 (0, t0) and S (t) > 0 for t 2 (t0, r). Remark 1. If r = ¥, then Lemma1 is reduced to [ 22] (Lemma 6.3). Furthermore, if ak = 0 for k ≥ k2 + 1, then Lemma1 yields [16] (Lemma 7). Remark 2. It follows from Lemma1 that if there is a t1 2 (0, r) such that S (t1) < 0, then we have S (t) < 0 for all t 2 (0, t1); if there is a t2 2 (0, r) such that S (t2) > 0, then we have S (t) > 0 for all t 2 (t2, r). For convenience, we introduce a notation: (1/2)n G (n + 1/2) Wn = = . (12) n! G (1/2) G (n + 1) Then (−1/2)n (−1/2) G (n + 1/2) 1 = = − Wn, n! n − 1/2 n!G (1/2) 2n − 1 (−3/2) (−3/2)(−1/2) G (n + 1/2) 3W n = = n . n! (n − 3/2)(n − 1/2) n!G (1/2) (2n − 1)(2n − 3) Clearly, Wn satisfies the recurrence relation: 2n W − = W . (13) n 1 2n − 1 n Using this notation, the Gaussian hypergeometric Equation (1) can be written as ¥ 2 1 1 2 2 2n K (r) = F , ; 1; r = ∑ Wnr . (14) p 2 2 n=0 Mathematics 2020, 8, 635 4 of 9 p 3/2 In addition, the functions r0 = 1 − r2 and (r0)3 = 1 − r2 can be expanded in power sereis p ¥ (−1/2) ¥ W r0 = 1 − r2 = n r2n = − n r2n, (15) ∑ ∑ − n=0 n! n=0 2n 1 ¥ ¥ 3 3/2 (−3/2) 3W r0 = 1 − r2 = n r2n = n r2n. (16) ∑ ∑ ( − )( − ) n=0 n! n=0 2n 1 2n 3 We are now in a position to prove our main result. Proof. (i) Necessity. Expanding in power series yields 2 1 1 1 9 25 1225 K (r) = F , ; 1; r2 = 1 + r2 + r4 + r6 + r8 + O r10 p 2 2 4 64 256 16384 1 9 25 379q + 12 L r0 = 1 + r2 + r4 + r6 + r8 + O r10 , q 4 64 256 5120q which gives 2 61q − 192 K (r) − L r0 = r8 + O r10 . (17) p q 81920q If the inequality seen in Equation (11) holds for all r 2 (0, 1), then 0 (2/p) K (r) − Lq (r ) 61q − 192 lim = ≥ 0, r!0+ r8 81920q which implies q ≥ 192/61 = q0. This proves the necessity. (ii) Sufficiency. Differentiation yields ¶ 15 (1 − r0)4 L r0 = − < 0 ¶q q 4 h i2 10 (r0)2 + (7q − 4) r0 + (3q − 6) 0 for r 2 (0, 1). That is, the lower bound Lq (r ) is decreasing with respect to q on [2, ¥). Thus, to prove the sufficiency, it is enough to prove the inequality in Equation (11) holds for r 2 (0, 1) and q = q0. Let 2 2 2 f (r) = 61 r0 − 110r0 + 21 61 r0 + 110r0 + 21 K (r) − L r0 . 0 p q0 Obviously, f0 (r) has at least one zero on (0, 1) which satisfies 2 61 r0 − 110r0 + 21 = 0. Solving the equation yields s p 440 109 − 1048 r = = 0.97617.... (18) 0 3721 On the other hand, using the power series representations in Equations (14)–(16), f0 (r) can be expressed as h 4 2 i 2 f (r) = 3721 r0 − 9538 r0 + 441 K (r) 0 p h 4 3 2 i − 305 r0 + 7136 r0 − 10034 r0 − 4064r0 + 1281 Mathematics 2020, 8, 635 5 of 9 ¥ 4 2 2 2n h 4 2 = 3721r + 2096r − 5376 ∑ Wnr − 305r + 9424r − 8448 n=0 # ¥ 3W ¥ W +7136 n r2n + 4064 n r2n , ∑ ( − )( − ) ∑ − n=0 2n 1 2n 3 n=0 2n 1 which, by the recurrence relations in Equation (13), is simplified to ¥ 16 441n4 + 7774n3 − 21131n2 + 16128n − 3024 = 2 2n ∑ 2 2 Wnr n=3 (2n − 1) (2n − 3) ¥ 64 (127n + 144) ¥ − W r2n = a r2n, ∑ ( − )( − ) n ∑ n n=3 2n 1 2n 3 n=3 where 64 (127n + 144) W a = n a∗, (19) n (2n − 1)(2n − 3) n 441n4 + 7774n3 − 21131n2 + 16128n − 3024 a∗ = W − 1.
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