By Arieh Iserles

Acta Numerica anually surveys crucial advancements in numerical arithmetic and medical computing. the themes and authors of the substantial articles are selected by way of a wonderful overseas editorial board, with a view to file crucial and well timed advancements in a fashion available to the broader group of execs with an curiosity in medical computing. Acta Numerica volumes are a important instrument not just for researchers and execs wishing to boost their realizing of numerical suggestions and algorithms and stick to new advancements. also they are used as complex educating aids at schools and universities (many of the unique articles are used because the leading source for graduate courses).

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47 First we observe that 1 n 1 ∑ pi fi − Pn Pn i=1 = = 1 n ∑ pi Pn i=1 fi gi − fi 1 n ∑ pi fi gi − Pn i=1 1 Pn n ∑ pj fj j=1 n = n 1 Pn ∑ p jg j j=1 n 1 n n j=1 j=1 1 ∑ p j g j −gi Pn ∑ p j f j + Pn2 ∑ p j f j ∑ p j g j j=1 j=1 1 n ∑ pi fi Pn i=1 + gi − n 1 Pn 1 n ∑ pi Pn i=1 1 Pn 1 n ∑ pi fi gi − Pn i=1 ∑ p jg j − j=1 n ∑ pj fj j=1 1 Pn 1 n ∑ pi fi Pn i=1 1 n ∑ pi gi Pn i=1 1 Pn n ∑ pj fj j=1 n ∑ p jg j j=1 1 n ∑ pi gi Pn i=1 = Cn (p, f , g). 10) and the proof is complete. 2. By taking pi = 1 for i = 1, .

22) holds for n and let us prove it for n + 1. That is, we have to prove the equality (b − x)k+1 + (−1)k (x − a)k+1 (k) h (x) (k + 1)! n b h(t)dt = a ∑ k=0 b +(−1)n+1 a En+1 (x,t)h(n+1) (t)dt. 25) It is easy to observe that b En+1 (x,t)h(n+1) (t)dt = a x a (t − a)n+1 (n+1) h (t)dt + (n + 1)! x = 1 (t − a)n+1 (n) h (t) − (n + 1)! n! a x a b + = (t − b)n+1 (n) 1 h (t) − (n + 1)! n! x b x b x (t − b)n+1 (n+1) h (t)dt (n + 1)! (t − a)n h(n) (t)dt (t − b)n h(n) (t)dt (x − a)n+1 + (−1)n+2 (b − x)n+1 (n) h (x) − (n + 1)!

2. By taking pi = 1 for i = 1, . . 10), we get |Cn ( f , g)| 1 n ∑ n i=1 fi − 1 n n ∑ fj j=1 gi − 1 n n ∑ gj . 13) can be considered as the discrete version of the Gr¨usstype integral inequality given by Dragomir and McAndrew in [34]. ˇ The discrete Cebyˇ sev-type inequality established in [128] is given in the following theorem. 3. Let f = ( f1 , . . , fn ), g = (g1 , . . 2) and Δ fk = fk+1 − fk . 14) 48 Analytic Inequalities: Recent Advances Proof. 3) holds. 5). It is easy to observe that the following identity holds: fi − f j = i−1 i−1 k= j k= j ∑ ( fk+1 − fk ) = ∑ Δ fk .