Science Mathematics

Discrete Hilbert-Type Inequalities by Bicheng Yang

By Bicheng Yang

In 1908, H. Wely released the well-known Hilbert’s inequality. In 1925, G. H. Hardy gave an extension of it by means of introducing one pair of conjugate exponents. The Hilbert-type inequalities are a extra broad type of research inequalities that are together with Hardy-Hilbert’s inequality because the specific case. via creating a nice attempt of mathematicians at approximately 100 years, the idea of Hilbert-type vital and discrete inequalities has now come into being. This ebook is a monograph concerning the concept of a number of half-discrete Hilbert-type inequalities. utilizing the tools of genuine research, sensible research and Operator conception, the writer introduces a number of self sufficient parameters to set up sorts of a number of half-discrete Hilbert-type inequalities with the absolute best consistent elements. The identical kinds and the reverses also are thought of. As functions, the writer additionally considers a few double instances of a number of half-discrete Hilbert-type inequalities and lots of examples. For studying and figuring out this booklet, readers may still carry the fundamental wisdom of actual research and sensible research.

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2. Theory and application of infinite series. Londen: Blackie & Son Limited,1928. 3. Qu WL. Combination mathematics. Beijing: Beijing University Press, 1989. 4. Yang BC. A new formula for evaluating the sum of d-th powers of the first n terms of an arithmetic sequence. Journal of South China Normal University, 1996(1):129-137. 5. Yang BC. The formula about the sum of powers of natural numbers relating Bernoulli numbers. Mathematical Practice and Cognition, 1994(4):5256. 6. Cheng QX. Basic on real variable functions and functional analysis.

For m  3 , since ( 12  121m  127m3 )(1  21m )  12  m1 ( 16  241m  2 m1 2  4 m1 3 )  12 , we have 1 2  1  1/ m 0 1 1 1/ r 1 t t ( ) dt , P1 (t )G (t , m)dt . 17) 1/ s G (t , m)  m .  1  121  41m  121m2  161m3  0 , where we define  1  n 1/ s , s  1, m  N, we obtain f m ( s)  1  121s2  (1 s1)2 m  12 s12m2  (13 s1)2 m3 g m ( s) :  121sm  2(1 21s ) m2  127  21m  121m2  127m3 . 16) where, and B (m)   m1 (1  m1 ) 2  m1 (1  m2  m32 ) . It is obvious that the above inequalities are valid for m  1 .

6), both the above inequalities still keep the strict-sign inequalities. 8). 24) p m 1   m 1 mq / r 1 [1 k (1r ) O ( m 3 )]q1   k q (r ) n r bnq . 26) )    ( r , m)   1 x y  rs   m    0  l (r )  s   ( s, n)   max{1m ,n} ( mn ) s  rs . 27), we have the following equivalent inequalities:     1 ( ) r du   sin( / r ) n 1 m 1  1 r p 1 s  q m 1  )  (rs ) p  m    m 1 p 1 s amp . ln( x / y ) x y 1 u r du  [ sin( / r ) ]2  0 . 1 r 1 1  ( r , m)    1 ln( m / y ) m m y y 1  l (r )   1 s p 1 s , then it follows ( ) r du  [ sin( / r ) ]2 ,  ln u 1 0 1 u u { m m 1 1 ( ) r dy  ( )  sin( / r ) .

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