Qiao Chen Journal of Inequalities Applications (2018 2018:141 https://doiorg/101186/s13660-018-1734-7 R E S E A R C H Open Access Approimations to inverse tangent function Quan-Xi Qiao 1 Chao-Ping Chen 1* * Correspondence: chenchaoping@sohucom 1 School of Mathematics Informatics, Henan Polytechnic University, Jiaozuo, China Abstract In this paper, we present a sharp Shafer-type inequality for the inverse tangent function Based on the Padé approimation method, we give approimations to the inverse tangent function Based on the obtained result, we establish new bounds for arctan MSC: 26D05 Keywords: Inverse trigonometric function; Inequality; Approimation 1 Introduction In 1966, Shafer [1] posed, as a problem, the following inequality: 3 1+2 < arctan, > 0 (11 1+2 Three proofs of it were later given in [2] Shafer s inequality (11 was sharpened generalized by Qi et al in [3] A survey epository of some old new inequalities associated with trigonometric functions can be found in [4] Chen et al [5]presentedanew method to sharpen bounds of both sinc arcsin functions, the inequalities in eponential form as well For each a > 0, Chen Cheung [6]determinedthelargestnumberb the smallest number c such that the inequalities b 1+a 1+ arctan c 2 1+a (12 1+ 2 are valid for all 0 More precisely, these author proved that the largest number b the smallest number c required by inequality (12are when 0 < a 2, b a, c 1+a; 2 when 2 < a 2 2, b 4(a2 1, c 1+a; a 2 2 when 2 < a <2, b 4(a2 1, c a 2 2 a; when 2 a <, b 1+a, c 2 a The Author(s 2018 This article is distributed under the terms of the Creative Commons Attribution 40 International License (http://creativecommonsorg/licenses/by/40/, which permits unrestricted use, distribution, reproduction in any medium, provided you give appropriate credit to the original author(s the source, provide a link to the Creative Commons license, indicate if changes were made
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 2 of 14 In 1974, Shafer [7] indicated several elementary quadratic approimations of selected functions without proof Subsequently, Shafer [8] established these results as analytic inequalities For eample, Shafer [8]provedthat,for >0, 8 < arctan (13 3+ 25 + 803 2 The inequality (13 canalsobe foundin [9] The inequality (13isanimprovementofthe inequality (11 Zhu [10]developed(13 to produce a symmetric double inequality More precisely, the author proved that, for >0, 8 < arctan < 3+ 25 + 803 2 where the constants 80/3 256/ 2 are the best possible 8, (14 3+ 25 + 256 2 2 Remark 11 For > 0, the following symmetric double inequality holds: 8 < arctan < 3+ 25 + 803 2 3+ 2 15 3 25 + 80 3 2, (15 where the constants 8 2 15 3 are the best possible We here point out that, for >0, the upper bound in (14 is better than the upper bound in (15 Based on the following power series epansion: ( arctan 3+ 25 + 803 2 8 + 32 4725 7 64 4725 9 + 25,376 1,299,375 11, Sun Chen [11] presented a new upper bound proved that, for >0, 8 + 32 4725 arctan < 7 (16 3+ 25 + 80 3 2 Moreover, these authors pointed out that, for 0 < < 0 14243, the upper bound in (16 is better than the upper bound in (14 In fact, we have the following approimation formulas near the origin: 8 ( arctan O 3, 3+ 25 + 256 2 2 3 arctan 1+2 1+ O( 5, 2 8 arctan O 3+ 25 ( 7, + 803 2
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 3 of 14 32 8 + 4725 arctan 7 O 3+ 25 ( 9 + 803 2 Nishizawa [12]provedthat,for >0, 2 4+ ( 2 4 2 +(2 2 < arctan < 2 4+ 32 + (2 2, (17 where the constants ( 2 4 2 32 are the best possible Using the Maple software, we derive the following asymptotic formulas in the Appendi: arctan arctan 2 4+ 32 + (2 12 2 ( 1 2 3 2 + O, (18 4 5 3 2 24 2 + 432 24 2 + 4 12(12 2 +(6 2 + 4 ( 72 1 18 3 + O 5 6, (19 ( 2 arctan 2 + 4 ( 15 1 2 + 3 + O 5 6 (110 as In this paper, motivated by (19, we establish a symmetric double inequality for arctan Based on the Padé approimation method, we develop the approimation formula (110to produce a general result More precisely, we determine the coefficients a j b j (1 j k such that ( 2 arctan 2k + a 1 2(k 1 ( + + a k 1 + O,, 2k + b 1 2(k 1 + + b k 4k+2 where k 1 is any given integer Based on the obtained result, we establish new bounds for arctan Some computations in this paper were performed using Maple software 2 Lemma It is well known that 2n+1 ( 1 k 2k+1 2n (2k +1! < sin < ( 1 k 2k+1 (2k +1! k0 k0 (21 2n+1 ( 1 k 2k k0 (2k! < cos < 2n k0 ( 1 k 2k (2k! (22 for >0n N 0 : N {0},whereN denotes the set of positive integers
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 4 of 14 The following lemma will be used in our present investigation Lemma 21 For 0< u < /2, cos u sin 2 u > u 2 5 6 u4 + 91 360 u6 41 1008 u8 (23 sin 3 u > u 3 1 2 u5 + 13 120 u7 41 3024 u9 (24 Proof We find that where cos u sin 2 u 1 4 ( cos u cos(3u u 2 5 6 u4 + 91 360 u6 41 1008 u8 + sin 3 u 1 ( 3 sin u sin(3u 4 u 3 1 2 u5 + 13 120 u7 41 3024 u9 + ( 1 n 1 w n (u (25 n5 ( 1 n 1 W n (u, (26 n5 w n (u 9n 1 (2n! u2n W n (u 3(9n 1 4 (2n +1! u2n+1 Elementary calculations reveal that, for 0 < u < /2 n 5, w n+1 (u w n (u W n+1 (u W n (u < u 2 (9 n+1 1 2(2n +1(n + 1(9 n 1 < (/2 2 (9 n+1 1 2(2n +1(n + 1(9 n 1 3 9 n+1 2(2n +1(n + 1(9 n 1 27 27 2(2n +1(n +1 { 1+ 1 9 5 1 2(2n +1(n +1 } u 2 (9 n+1 1 2(2n +3(n + 1(9 n 1 < w n+1(u <1 w n (u { 1+ 1 9 n 1 } 1,594,323 118,096(2n +1(n +1 <1 Therefore, for fied u (0, /2, the sequences n w n (u n W n (u areboth strictly decreasing for n 5 From (25 (26, we obtain the desired results (23 (24 The proof of Theorem 31 makes use of the inequalities (21 (24
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 5 of 14 3 Sharp Shafer-type inequality Equation (19 motivated us to establish a symmetric double inequality for arctan Theorem 31 For >0,we have 3 2 24 2 + α 12(12 2 +36 2 2 < arctan < 3 2 24 2 + β 12(12 2 +36 2 2, (31 with the best possible constants α 432 24 2 + 4 292538 β 576 192 2 +16 4 239581 (32 Proof The inequality (31 can be written for >0 as ( 3 2 2 β < arctan ( 24 2 2 +12 ( 12 2 36 2 2 < α (33 Bytheelementarychangeofvariablet arctan ( >0,(33becomes β < ϑ(t<α, 0<t < 2, (34 where ( 3 2 tan 2 t ϑ(t ( 24 2 2 t +12 ( 12 2 tan t 36 2 tan 2 t Elementary calculations reveal that lim ϑ(t 576 192 2 +16 4 t 0 + lim ϑ(t 432 24 2 + 4 t /2 Inordertoprove(34, it suffices to show that ϑ(t is strictly increasing for 0 < t < /2 Differentiation yields t 3 cos 3 tϑ (t ( 24t 3 t sin t cos 2 t + ( 3 3 t 12t 3 sin t ( 3 3 + ( 24 3 t 2 ( 24 2 2 t 3 cos t +3 3 cos 3 t : λ(t We now consider two cases to prove λ(t> 0 for 0 < t < /2
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 6 of 14 Case 1: 0 < t 06 Using (21(22, we have, for 0 < t 06, λ(t (6 14 3 t sin(3t+ 3 4 3 cos(3t+ {(6 + 114 } t 3 12t 3 sin t > ( 3 4 3 ( 3 24 t 2 ( 24 2 2 t 3 cos t ( 3t 9 2 t3 + 81 40 t5 243 560 t7 (6 14 3 t + 3 4 3 ( 1 9 2 t2 + 27 8 t4 81 80 t6 + {(6 + 114 t 3 12t }(t 3 16 t3 ( 3 4 3 ( 3 24 t 2 ( 24 2 2 t (1 3 12 t2 + 124 t4 t {24 3 2 2 (28 83 t 3 ( 12 2 } t 2 { 263 + t 6 20 235 192 3 + (1 112 ( 729 2 t 280 243 } t 2240 3 2 Each function in curly braces is positive for t (0, 06] Thus, λ(t >0fort (0, 06] Case 2: 06 < t < /2 We now prove λ(t >0for06<t < /2 Replacing t by u leads to an equivalent 2 inequality: μ(u>0, 0<u < 2 06, where μ(u ( 24 3( 2 u { 3 3 + ( 24 3( 2 u Using (21 (24, we have, for 0 < u < 2 06, ( ( 2 cos u sin 2 u + {3 3 u 12 3 } 2 u cos u 2 ( 24 2 2( 3 } 2 u sin u +3 3 sin 3 u μ(u > ( 24 3( ( 2 u u 2 5 6 u4 + 91 360 u6 41 1008 u8 ( 2 3 }( + {3 3 u 12 ( 2 u { 3 3 + ( 24 3( 2 u +3 3 ( u 3 1 2 u5 + 13 120 u7 41 u 4 { 1 3 4 24+ 1 1 2 u2 + 1 24 u4 1 720 u6 2 ( 24 2 2( 3 }( 2 u 3024 u9 (12 95 ( 3 u + 2 2 11 90 4 +4 u 2 u 1 6 u3 + 1 120 u5
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 7 of 14 ( + 82 360 3 15 + 199 u 3 ( + 1 5 25 56 2 + 41 2016 4 u 4 + ( 403 420 41 } u 504 3 5 >0 We then obtain λ(t>0ϑ (t > 0 for all 0 < t < /2 Hence, ϑ(t isstrictlyincreasing for 0 < t < /2 The proof is complete From (17(31, we obtain the following approimation formulas: arctan n n 2 4+ 32 + (2n : a 2 n (35 arctan n n 3 2 24 2 + 432 24 2 + 4 12(12 2 n +(6n 2 : b n, (36 as n The following numerical computations (see Table 1 would show that, for n N,Eq(36 is sharper than Eq (35 In fact, we have, as n, arctan n n ( 1 a n + O n 4 arctan n n ( 1 b n + O n 5 4 Approimations to arctan For later use, we introduce the Padé approimant (see [13 16] Let f be a formal power series, f (tc 0 + c 1 t + c 2 t 2 + (41 ThePadéapproimationoforder(p, q of the function f is the rational function, denoted by [p/q] f (t p j0 a jt j 1+ q j1 b jt j, (42 Table 1 Comparison between approimation formulas (35 (36 n a n arctan n n arctan n n 1 7055 10 3 5259 10 3 10 595 10 6 3939 10 7 100 7066 10 10 4492 10 12 1000 7182 10 14 4546 10 17 10,000 7193 10 18 4552 10 22 b n
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 8 of 14 where p 0q 1 are two given integers, the coefficients a j b j are given by (see [13 15] a 0 c 0, a 1 c 0 b 1 + c 1, a 2 c 0 b 2 + c 1 b 1 + c 2, a p c 0 b p + + c p 1 b 1 + c p, 0c p+1 + c p b 1 + + c p q+1 b q, 0c p+q + c p+q 1 b 1 + + c p b q, (43 the following holds: [p/q] f (t f (to ( t p+q+1 (44 Thus, the first p + q + 1 coefficients of the series epansion of [p/q] f are identical to those of f From the epansion (see [17,p81] we obtain where arctan 2 + j1 ( 2 arctan ( 1 j, >1, (2j 12j 1 c j 1 1 2j 3 + 1 2 5 1 +, (45 4 76 j0 c j ( 1j 2j +1 for j 0 (46 Let f (t j0 c j t j, (47 with the coefficients c j givenin(46 Then we have f ( 2 j0 ( c j 2j 2 arctan (48 In what follows, the function f is given in(47
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 9 of 14 Based on the Padé approimation method, we now give a derivation of Eq (110 To this end, we consider [1/1] f (t 1 j0 a jt j 1+ 1 j1 b jt j Noting that c 0 1, c 1 1 3, c 2 1 5, (49 holds, we have, by (43, a 0 1, a 1 b 1 1 3, 0 1 5 1 3 b 1, that is, a 0 1, a 1 4 15, b 1 3 5 We thus obtain [1/1] f (t 1+ 4 15t 1+ 3 5t 15t +4 3(5t +3, (410 we have, by (44, f (t 15t +4 3(5t +3 + O ( 1 t 3, t (411 Replacing t by 2 in (411yields(110 From the Padé approimation method the epansion (47, we now present a general result Theorem 41 ThePadéapproimationoforder(p, q of the function f (t c j j0 (at the t j point t is the following rational function: [p/q] f (t 1+ p j1 a jt j 1+ q j1 b jt j t q p ( t p + a 1 t p 1 + + a p t q + b 1 t q 1 + + b q, (412
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 10 of 14 where p 1 q 1 are any given integers, the coefficients a j b j are given by a 1 b 1 + c 1, a 2 b 2 + c 1 b 1 + c 2, a p b p + + c p 1 b 1 + c p, 0c p+1 + c p b 1 + + c p q+1 b q, 0c p+q + c p+q 1 b 1 + + c p b q, (413 c j is given in (46, the following holds: ( 1 f (t [p/q] f (to, t (414 t p+q+1 In particular, replacing t by 2 in (414 yields ( 2 arctan 2(q p ( 2p + a 1 2(p 1 + + a p 2q + b 1 2(q 1 + + b q with the coefficients a j b j given by (413 ( + O 1 2(p+q+1,, (415 Setting (p, q(k, kin(415, we obtain the following corollary Corollary 41 As, ( 2 arctan 2k + a 1 2(k 1 ( + + a k 1 + O, (416 2k + b 1 2(k 1 + + b k 4k+2 where k 1 is any given integer, the coefficients a j b j (1 j k are given by a 1 b 1 + c 1, a 2 b 2 + c 1 b 1 + c 2, a k b k + + c k 1 b 1 + c k, 0c k+1 + c k b 1 + + c 1 b k, 0c 2k + c 2k 1 b 1 + + c k b k, (417 c j is given in (46
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 11 of 14 Setting k 2in(416yields,as, ( 2 arctan 9454 + 735 2 ( +64 1 15(63 4 +70 2 + 15 + O, (418 10 which gives arctan 2 9454 + 735 2 ( +64 1 15(63 4 +70 2 + 15 + O 11 Using the Maple software, we find, as, arctan 2 9454 + 735 2 +64 15(63 4 +70 2 + 15 + 64 ( 1856 1 464,373 + O 13 15 43,659 11 (419 Equation (419motivated usto establish new bounds for arctan Theorem 42 For >0,we have 2 9454 + 735 2 +64 15(63 4 +70 2 + 15 + 64 43,659 11 1856 464,373 13 < arctan < 2 9454 + 735 2 +64 15(63 4 +70 2 + 15 + 64 (420 43,65911 Proof For >0,let ( I(arctan 2 9454 + 735 2 +64 15(63 4 +70 2 + 15 + 64 43,659 1856 11 464,373 13 ( J(arctan 2 9454 + 735 2 +64 15(63 4 +70 2 + 15 + 64 43,659 11 Differentiation yields I ( 64(230,3918 + 372,680 6 + 236,885 4 + 65,400 2 + 6525 35,721 14 (1 + 2 (63 4 +70 2 + 15 2 <0 J ( 64(12,7898 + 15,610 6 + 8890 4 + 2325 2 + 225 3969 12 (1 + 2 (63 4 +70 2 + 15 2 >0 Hence, I( is strictly decreasing J( is strictly increasing for >0, we have I(> lim t I(t0 J(< lim t J(t0 for >0 The proof is complete
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 12 of 14 Remark 41 We point out that, for > 10213, the lower bound in (420isbetterthan the one in (17 For > 0854439, the upper bound in (420 is better than the one in (17 For > 0947273, the lower bound in (420 is better than the one in (31 For > 0792793, the upper bound in (420 is better than the one in (31 5 Conclusions In this paper, we establish a symmetric double inequality for arctan (Theorem 31 We determine the coefficients a j b j (1 j ksuchthat ( 2 arctan 2k + a 1 2(k 1 ( + + a k 1 + O,, 2k + b 1 2(k 1 + + b k 4k+2 where k 1isanygiveninteger(seeCorollary41 Based on the obtained result, we establish new bounds for arctan (Theorem 42 Appendi: A derivation of (18, (19, (110 Define the function F(by F( arctan 1 a + b + c 2 We are interested in finding the values of the parameters a, b, c such that F( converges as fast as possible to zero, as This provides the best approimations of the form arctan 1 a + b + c 2, Using the Maple software, we find, as, F( c 2 2 c + a c c + b 2a2 2 2c 3/2 3 ( 1 + 3a3 + c 2 3ab 3c 2 4 + O 5 The three parameters a, b, c, which produce the fastest convergence of the function F(, are given by c 20, a c 0, b 2a 2 0, namely, if a 4 2, b 32 4, c 4 2
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 13 of 14 We then obtain, as, arctan 1 12 2 4 32 + + 4 2 4 2 2 3 2 4 2 4+ 32 + (2 12 2 2 3 2 + O 4 Define the function G(by G( arctan 1 p + q + r + s 2 Using the Maple software, we find, as, For ( 1 + O 5 ( 1 G( s 2 2 s + r 2s3/2 +2p s + 4qs 3r2 8sp 2 8 spr 2s 3/2 2 8s 5/2 3 + 48s3/2 pq +48 spr 2 36rqs +15r 3 +48s 3/2 p 3 +72sp 2 r +16s 7/2 48s 7/2 4 + ( 120sqr 2 128 spr 3 128s 2 p 4 240sp 2 r 2 + 256rs 3/2 pq + 192s 2 p 2 q 256s 3/2 p 3 r 35r 4 48q 2 s 2 / ( 128s 9/2 5 ( 1 + O 6 p 24 2 3 2, q 432 24 2 + 4 9 4, r 4(12 2 3 3, s 4 2, we obtain, as, arctan 3 2 24 2 + 432 24 2 + 4 12(12 2 +(6 2 + 4 ( 72 1 18 3 + O 5 6 Define the function H(by ( H( 2 arctan 2 + a 1 + a 2 2 + b 1 + b 2 Using the Maple software, we find, as, H( b 1 a 1 5 3a 2 3b 2 3a 1 b 1 +3b 2 1 +1 + a 1b 2 2b 1 b 2 + a 2 b 1 a 1 b 2 1 + b3 1 3 2 3 1 5a 2b 2 +5b 2 2 +10a 1b 1 b 2 15b 2 1 b 2 +5a 2 b 2 1 5a 1b 3 1 +5b4 1 5 4 + a 1b 2 2 +3b 1b 2 2 2a 2b 1 b 2 +3a 1 b 2 1 b 2 4b 3 1 b 2 + a 2 b 3 1 a 1b 4 1 + b5 1 5 ( 1 + O 6
Qiao Chen Journal of Inequalities Applications (2018 2018:141 Page 14 of 14 For a 1 0, b 1 0, a 2 4 15, b 2 3 5, we obtain, as, ( 2 arctan 2 + 4 ( 15 1 2 + 3 + O 5 6 Acknowledgements We thank the editor referees for their careful reading valuable suggestions to make the article more easily readable Funding Not applicable Competing interests The authors declare that they have no competing interests Authors contributions Both authors contributed equally to this work Both authors read approved the final manuscript Publisher s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps institutional affiliations Received: 26 January 2018 Accepted: 13 June 2018 References 1 Shafer, RE: Problem E1867 Am Math Mon 73, 309 (1966 2 Shafer, RE: Problems E1867 (solution Am Math Mon 74, 726 727 (1967 3 Qi, F, Zhang, SQ, Guo, BN: Sharpening generalizations of Shafer s inequality for the arc tangent function J Inequal Appl 2009, Article ID 930294 (2009 https://doiorg/101155/2009/930294 4 Qi, F, Niu, DW, Guo, BN: Refinements, generalizations, applications of Jordan s inequality related problems J Inequal Appl 2009, Article ID 271923 (2009 https://doiorg/101155/2009/271923 5 Chen, XD, Shi, J, Wang, Y, Pan, X: A new method for sharpening the bounds of several special functions Results Math 72, 695 702 (2017 https://doiorg/101007/s00025-017-0700-6 Chen, CP, Cheung, WS, Wang, W: On Shafer Carlson inequalities J Inequal Appl 2011, Article ID840206 (2011 https://doiorg/101155/2011/840206 7 Shafer, RE: On quadratic approimation SIAM J Numer Anal 11, 447 460 (1974https://doiorg/101137/0711037 8 Shafer, RE: Analytic inequalities obtained by quadratic approimation Publ Elektroteh Fak Univ Beogr, Ser Mat Fiz 57(598, 96 97 (1977 9 Shafer, RE: On quadratic approimation, II Publ Elektroteh Fak Univ Beogr, Ser Mat Fiz 602(633,163 170 (1978 10 Zhu, L: On a quadratic estimate of Shafer J Math Inequal 2, 571 574 (2008 http://filesele-mathcom/articles/jmi-02-51pdf 11 Sun, J-L, Chen, C-P: Shafer-type inequalities for inverse trigonometric functions Gauss lemniscate functions J Inequal Appl 2016, 212 (2016 https://doiorg/101186/s13660-016-1157-2 12 Nishizawa, Y: Refined quadratic estimations of Shafer s inequality J Inequal Appl 2017, 40 (2017 https://doiorg/101186/s13660-017-1312-4 13 Bercu, G: Padé approimant related to remarkable inequalities involving trigonometric functions J Inequal Appl 2016, 99 (2016 https://doiorg/101186/s13660-016-1044-14 Bercu, G: The natural approach of trigonometric inequalities Padé approimant J Math Inequal 11, 181 191 (2017 https://doiorg/107153/jmi-11-18 15 Bercu, G, Wu, S: Refinements of certain hyperbolic inequalities via the Padé approimation method J Nonlinear Sci Appl 9,5011 5020 (2016https://doiorg/1022436/jnsa0090705 16 Brezinski, C, Redivo-Zaglia, M: New representations of Padé, Padé-type, partial Padé approimants J Comput Appl Math 284, 69 77 (2015 https://doiorg/101016/jcam201407007 17 Abramowitz, M, Stegun, IA (eds: Hbook of Mathematical Functions with Formulas, Graphs, Mathematical Tables, 9th printing edn Applied Mathematics Series, vol 55 National Bureau of Stards, Washington (1972