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1、HEBEI UNIVERSITY 密级: 分 类 号: 学校代码: 密级: 分 类 号: 学校代码:10075 学号:学号:20090881 硕士学位论文硕士学位论文 形心平均的最优凸组合界 学位申请人: 郭建玲 指 导 教 师 : 高红亚 教授 学 位 类 别 : 理学硕士 学 科 专 业 : 应 用 数 学 授 予 单 位 : 河北大学 答 辩 日 期 : 二O一二年五月 Classihed Index:CODE: 10075 U.D.C. :NO: 20090881 A Dissertation for the Degree of Master Science Optimal Conve
2、x Combination Bounds for Centroidal Mean Candidate : Guo Jianling Supervisor : Prof. Gao Hongya Academic Degree Applied for : Master of Science Speciality : Applied Mathematics University : Hebei University Date of Accomplishment : May, 2012 ? ? ? ?.? ?,?,?,?, Seifert?. ?1, 1, 2, 2, 3, 3, 4, 4? 1C(a
3、,b) 十 (1 1)Mq(a,b) E(a,b) 1C(a,b) 十 (1 1)Mq(a,b), 2C(a,b) 十 (1 2)G(a,b) E(a,b) 2C(a,b) 十 (1 2)G(a,b), 3C(a,b) 十 (1 3)L(a,b) E(a,b) 3C(a,b) 十 (1 3)L(a,b), 4C(a,b) 十 (1 4)P(a,b) E(a,b) 4C(a,b) 十 (1 4)P(a,b) ?,?a,b 0?a b.?Mq(a,b), C(a,b), G(a,b), L(a,b), P(a,b), E(a,b)? ?a, b?,?,?,?, Seifert? ?. ?Seife
4、rt? ? I Abstract Abstract Various means of two positive numbers a and b and their inequalities have wide applications in the helds of statistics, mathematics and engineering.In this paper, we consider optimal convex combination bounds for the centroidal mean in terms of contraharmonic, geometric, lo
5、garithmic, Seifert and golden section means. We hnd the optimal values for 1, 1, 2, 2, 3, 3, 4and 4such that the inequalities 1C(a,b) 十 (1 1)Mq(a,b) E(a,b) 1C(a,b) 十 (1 1)Mq(a,b), 2C(a,b) 十 (1 2)G(a,b) E(a,b) 2C(a,b) 十 (1 2)G(a,b), 3C(a,b) 十 (1 3)L(a,b) E(a,b) 3C(a,b) 十 (1 3)L(a,b), 4C(a,b) 十 (1 4)P
6、(a,b) E(a,b) 4C(a,b) 十 (1 4)P(a,b) hold for all a,b 0 with a b, where C(a,b), Mq(a,b), G(a,b), L(a,b), P(a,b) and E(a,b) be the contraharmonic mean, golden section mean, geometric mean, logarithmic mean, Seifert,s mean and centroidal mean of two positive numbers a and b, respectively. Keywordsconvex
7、 combination boundgolden section meangeometric meanlog- arithmic meanSeifert meancentroidal meancontraharmonic mean II ? ? ?1?. . 1 ?2?1.1-1.4?6 ?. 16 ?.18 ? 20 ?. 21 III Contents Contents Chapter 1Introduction and Statement of Results1 Chapter 2Proof of Theorems 1.1-1.4 6 Epilogue 16 References18 A
8、cknowledgements.20 Theses of written during postgraduate period21 IV Chapter 1Introduction and Statement of Results Chapter 1Introduction and Statement of Results Let a and b be positive real numbers with a b. The arithmetic mean, geometric mean and logarithmic mean of a and b are dehned by A(a,b) a
9、 十 b 2 , G G(a,b) ) ab and L L(a,b) a b loga logb respectively. It is well-known that for a b one has (see e.g. 1) G(a,b) L(a,b). In 1993, Seifert 2 introduced the mean P P(a,b) a b 4arctan JO b , and proved in 2 that for a b L(a,b) P(a,b). Many remarkable inequalities for the Seifert,s mean can be
10、found in the literature 3- 8. The Seifert,s mean P(a,b) and the second Seifert,s mean can be rewritten as (see 5,(2.4) P(a,b) a b 2arcsin O b Ob and T(a,b) a b 2arctanO b Ob The golden section mean Mq(a,b) is dehned by Mq(a,b) q(Ob)2Ob 2q(Ob) where q 1V5 2 . We refer the readers to 9,10 for general
11、information about the golden section mean. -1- ? The centroidal mean, harmonic mean and contraharmonic mean are dehned as E(a,b) 2 3 (a2 十 b2十 ab a 十 b ) , H(a.b) 2ab a 十 b and C(a,b) a2十 b2 a 十 b respectively. In order to establish our main results we need two double inequalities, which we present
12、in this section. Lemma 1.1 For q 1V5 2 , the double inequality H(a,b) Mq(a,b) E(a,b) holds for all a,b 0 with a b. PToof Firstly, we prove that for a,b 0 with a b, H(a,b) Mq(a,b). Without loss of generality, we assume a b. Let t O b 1. Then 1 b (Mq(a,b) H(a,b) Mq(t,1) H(t,1) q(t 1)2 (2q 十 t 十 1)(t 十
13、 1) 0. Secondly, we prove that for a,b 0 with a b, Mq(a,b) E(a,b) Without loss of generality, we assume a b. Let t O b 1. Then 1 b (Mq(a,b) E(a,b) Mq(t,1) E(t,1) (t 1)2(2t 十 q 十 2) 3(t 十 1)(2q 十 t 十 1) 0. -2- Chapter 1Introduction and Statement of Results Lemma 1.2 The double inequality T(a,b) E(a,b
14、) C(a,b) holds for all a,b 0 with a b. PToof Firstly, we prove that for a,b 0 with a b, E(a,b) C(a,b) Without loss of generality, we assume a b. Let t O b 1. Then 1 b (E(a,b) C(a,b) E(t,1) C(t,1) (t 1)2 3(t 十 1) 0. Secondly, we prove that T(a,b) E(a,b) Let t O b 1. Then 1 b (E(a,b) T(a,b) E(t,1) T(t
15、,1) t2十 t 十 1 6(t 十 1)arctan t 1 t1 h(t), where h(t) 4arctan t 1 t 十 1 3(t2 1) t2十 t 十 1. Since lim tH1十 h(t) 0, hI(t) 3t4十 4t3十 10t2十 4t 十 3 (t2十 1)(t2十 t 十 1)2 0, then by the continuity of h(t), we get h(t) 0, which implies that T(a,b) E(a,b). Combining Lemmas 1.1 and 1.2 and the known results on means, we have mina,b H(a,b) G(a,b) L(a,b) P(a,b) T(a,b) E(a,b)