Download PDF by George A. Anastassiou: Advanced Inequalities

By George A. Anastassiou

ISBN-10: 9814317632

ISBN-13: 9789814317634

This monograph provides univariate and multivariate classical analyses of complex inequalities. This treatise is a end result of the author's final 13 years of analysis paintings. The chapters are self-contained and a number of other complicated classes could be taught out of this ebook. huge historical past and motivations are given in every one bankruptcy with a accomplished checklist of references given on the finish.

the subjects lined are wide-ranging and numerous. fresh advances on Ostrowski variety inequalities, Opial sort inequalities, Poincare and Sobolev variety inequalities, and Hardy-Opial sort inequalities are tested. Works on traditional and distributional Taylor formulae with estimates for his or her remainders and purposes in addition to Chebyshev-Gruss, Gruss and comparability of capacity inequalities are studied.

the consequences provided are commonly optimum, that's the inequalities are sharp and attained. functions in lots of parts of natural and utilized arithmetic, resembling mathematical research, likelihood, usual and partial differential equations, numerical research, details concept, etc., are explored intimately, as such this monograph is acceptable for researchers and graduate scholars. it will likely be an invaluable educating fabric at seminars in addition to a useful reference resource in all technology libraries.

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Example text

N j [ai , bi ] , i=1 Then (bj − aj )m−1 j−1 j=1 i=1 ∂mf (. . , xj+1 , . . , xn ) ∂xm j (bi − ai ) j 1, [ai ,bi ] i=1 Bm (t) − Bm xj − a j bj − a j . 5in Book˙Adv˙Ineq ADVANCED INEQUALITIES 52 f |E2r (x1 , . . , xn )| ≤ 1 (2r)! n (bj − aj )2r−1 ∂ 2r f (. . , xj+1 , . . , xn ) ∂x2r j j−1 j=1 i=1 (bi − ai ) j 1, [ai ,bi ] i=1 xj − a j bj − a j × (1 − 2−2r )|B2r | + 2−2r B2r − B2r . 81) 2) When m = 2r + 1, r ∈ N, then f |E2r+1 (x1 , . . , xn )| ≤ × 1 (2r + 1)! n (bj j−1 − aj )2r i=1 (bi − ai ) j=1 ∂ 2r+1 f (.

Xn ) ds1 ds2 · · · dsj . 43) is zero. Proof. 13. 43). Then we prove it for n+1, n ∈ N. We have f (x1 , x2 , . . , xn , xn+1 ) = 1 n i=1 n (bi − ai ) + [ai ,bi ] i=1 Tj , for j = 1, . . , n, j=1 where f (s1 , . . , sn , xn+1 )ds1 , . . 5in Book˙Adv˙Ineq ADVANCED INEQUALITIES 40 Tj := Tj xj , xj+1 , . . , xn , xn+1 = (j−1) i=1 xj − a j bj − a j Bk j−1 [ai ,bi ] i=1 m−1 1 k=1 (bi − ai ) ∂ k−1 f (s1 , . . , sj−1 , bj , xj+1 , . . , xn , xn+1 ) ∂xjk−1 ∂ k−1 f s1 , . . , sj−1 , aj , xj+1 , .

18. 8. Additionally suppose that f (n) ∞ < +∞. Then n−2 b b 1 1 f (k+1) (s1 )dg(s1 ) f (s1 )dg(s1 ) − f (x) − · (g(b) − g(a)) a (g(b) − g(a)) a k=0 b · ··· a k b ≤ f (n) a P (g(x), g(s1 )) · b ∞ · a i=1 P (g(si ), g(si+1 ))ds1 · · · dsk+1 n−1 b ··· |P (g(x), g(s1 ))| · a i=1 |P (g(si ), g(si+1 ))|ds1 · · · dsn . 19. 10. Additionally assume that ∂2f < +∞, for all i, j = 1, . . , k. γij := ∂xi ∂xj ∞ Then k 1 1 1 ∂f (t1 t2 x)dt1 dt2 f (x) − f (t1 x)dt1 − xj t1 ∂x j 0 0 0 j=1   k k 1 |xi | |xj | · γij  .

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Advanced Inequalities by George A. Anastassiou


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