In the present paper, we answer the question: for 0a what are the greatest value p(a) and the least value q(a) such that the double inequality Jp(a,b)aA(a,b)+ (1-a)G(a,b)Jq(a,b) holds for all a,b>0 with a is not eq...In the present paper, we answer the question: for 0a what are the greatest value p(a) and the least value q(a) such that the double inequality Jp(a,b)aA(a,b)+ (1-a)G(a,b)Jq(a,b) holds for all a,b>0 with a is not equal to?b ?展开更多
In the article,we prove that the double inequalities Gp[λ1a+(1-λ1)b,λ1 b+(1-λ1)a]A1-p(a,b)<T[A(a,b),G(a,b)]<Gp[μ1 a+(1-μ1)b,μ1b+(1-μ1)a]A1-p(a,b),Cs[λ^(2) a+(1-λ2)b,λ2 b+(1-λ2)a]A1-s(a,b)<T[A(a,b)...In the article,we prove that the double inequalities Gp[λ1a+(1-λ1)b,λ1 b+(1-λ1)a]A1-p(a,b)<T[A(a,b),G(a,b)]<Gp[μ1 a+(1-μ1)b,μ1b+(1-μ1)a]A1-p(a,b),Cs[λ^(2) a+(1-λ2)b,λ2 b+(1-λ2)a]A1-s(a,b)<T[A(a,b),Q(a,b)]<Cs[μ2 a+(1-μ2)b,μ2 b+(1-μ2)a]A1-p(a,b)hold for all a,b>0 with a≠b if and only ifλ1≤1/2-(1-(2/π)2/p)1/2/2,μ1≥1/2-(2p)1/2/(4 p),λ2≤1/2+(2(3/(2 s)(E(21/2/2)/π)1/s)-1)1/2/2 andμ2≥1/2+s1/2/(4 s)ifλ1,μ1∈(0,1/2),λ2,μ2∈(1/2,1),p≥1 and s≥1/2,where G(a,b)=(ab)1/2,A(a,b)=(a+b)/2,T(a,b)=∫0π/2(a2 cos2 t+b2 sin2)1/2 tdt/π,Q(a,b)=((a2+b2)/2)1/2,C(a,b)=(a2+b2)/(a+b)and E(r)=∫0π/2(1-r^(2) sin^(2))1/2 tdt.展开更多
In this paper we will extend the well-known chain of inequalities involving the Pythagorean means, namely the harmonic, geometric, and arithmetic means to the more refined chain of inequalities by including the logari...In this paper we will extend the well-known chain of inequalities involving the Pythagorean means, namely the harmonic, geometric, and arithmetic means to the more refined chain of inequalities by including the logarithmic and identric means using nothing more than basic calculus. Of course, these results are all well-known and several proofs of them and their generalizations have been given. See [1-6] for more information. Our goal here is to present a unified approach and give the proofs as corollaries of one basic theorem.展开更多
In the present paper, we answer the question: for 0 what are the greatest value p(a) and the least value q(a) such that the inequality. For more information about abstract,please download the PDF file.
By virtue of Cauchy’s integral formula in the theory of complex functions,the authors establish an integral representation for the weighted geometric mean,apply this newly established integral representation to show ...By virtue of Cauchy’s integral formula in the theory of complex functions,the authors establish an integral representation for the weighted geometric mean,apply this newly established integral representation to show that the weighted geometric mean is a complete Bernstein function,and find a new proof of the well-known weighted arithmetic-geometric mean inequality.展开更多
In this paper, taking the Lorenz system as an example, we compare the influences of the arithmetic mean and the geometric mean on measuring the global and local average error growth. The results show that the geometri...In this paper, taking the Lorenz system as an example, we compare the influences of the arithmetic mean and the geometric mean on measuring the global and local average error growth. The results show that the geometric mean error （GME） has a smoother growth than the arithmetic mean error （AME） for the global average error growth, and the GME is directly related to the maximal Lyapunov exponent, but the AME is not, as already noted by Krishnamurthy in 1993. Besides these, the GME is shown to be more appropriate than the AME in measuring the mean error growth in terms of the probability distribution of errors. The physical meanings of the saturation levels of the AME and the GME are also shown to be different. However, there is no obvious difference between the local average error growth with the arithmetic mean and the geometric mean, indicating that the choices of the AME or the GME have no influence on the measure of local average predictability.展开更多
文摘In the present paper, we answer the question: for 0a what are the greatest value p(a) and the least value q(a) such that the double inequality Jp(a,b)aA(a,b)+ (1-a)G(a,b)Jq(a,b) holds for all a,b>0 with a is not equal to?b ?
基金supported by the Natural Science Foundation of China(61673169,11301127,11701176,11626101,11601485)。
文摘In the article,we prove that the double inequalities Gp[λ1a+(1-λ1)b,λ1 b+(1-λ1)a]A1-p(a,b)<T[A(a,b),G(a,b)]<Gp[μ1 a+(1-μ1)b,μ1b+(1-μ1)a]A1-p(a,b),Cs[λ^(2) a+(1-λ2)b,λ2 b+(1-λ2)a]A1-s(a,b)<T[A(a,b),Q(a,b)]<Cs[μ2 a+(1-μ2)b,μ2 b+(1-μ2)a]A1-p(a,b)hold for all a,b>0 with a≠b if and only ifλ1≤1/2-(1-(2/π)2/p)1/2/2,μ1≥1/2-(2p)1/2/(4 p),λ2≤1/2+(2(3/(2 s)(E(21/2/2)/π)1/s)-1)1/2/2 andμ2≥1/2+s1/2/(4 s)ifλ1,μ1∈(0,1/2),λ2,μ2∈(1/2,1),p≥1 and s≥1/2,where G(a,b)=(ab)1/2,A(a,b)=(a+b)/2,T(a,b)=∫0π/2(a2 cos2 t+b2 sin2)1/2 tdt/π,Q(a,b)=((a2+b2)/2)1/2,C(a,b)=(a2+b2)/(a+b)and E(r)=∫0π/2(1-r^(2) sin^(2))1/2 tdt.
基金国家自然科学基金(61640020)江苏省农业自主创新项目(CX(13)3054+5 种基金CX(16)1006)江苏省重点研发计划(BE2016368-1)江苏省科技重点及面上项目(SBE2018310371)弹总装线***技术研究(JCKY2017***)Postgraduate Research&Practice Innovation Program of Jiangsu Province(SJCX17_0107)北京市教育委员会科技计划面上项目(KM201510028019)
文摘In this paper we will extend the well-known chain of inequalities involving the Pythagorean means, namely the harmonic, geometric, and arithmetic means to the more refined chain of inequalities by including the logarithmic and identric means using nothing more than basic calculus. Of course, these results are all well-known and several proofs of them and their generalizations have been given. See [1-6] for more information. Our goal here is to present a unified approach and give the proofs as corollaries of one basic theorem.
文摘In the present paper, we answer the question: for 0 what are the greatest value p(a) and the least value q(a) such that the inequality. For more information about abstract,please download the PDF file.
文摘By virtue of Cauchy’s integral formula in the theory of complex functions,the authors establish an integral representation for the weighted geometric mean,apply this newly established integral representation to show that the weighted geometric mean is a complete Bernstein function,and find a new proof of the well-known weighted arithmetic-geometric mean inequality.
基金Supported by the National Natural Science Foundation of China （40805022 and 40675046）the National Basic Research and Development （973） Program of China （2010CB950400）
文摘In this paper, taking the Lorenz system as an example, we compare the influences of the arithmetic mean and the geometric mean on measuring the global and local average error growth. The results show that the geometric mean error （GME） has a smoother growth than the arithmetic mean error （AME） for the global average error growth, and the GME is directly related to the maximal Lyapunov exponent, but the AME is not, as already noted by Krishnamurthy in 1993. Besides these, the GME is shown to be more appropriate than the AME in measuring the mean error growth in terms of the probability distribution of errors. The physical meanings of the saturation levels of the AME and the GME are also shown to be different. However, there is no obvious difference between the local average error growth with the arithmetic mean and the geometric mean, indicating that the choices of the AME or the GME have no influence on the measure of local average predictability.
