By Gianni Gilardi

ISBN-10: 8838606536

ISBN-13: 9788838606533

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Extra resources for Analisi 1

Example text

B a h(t)dt = (b − a)h(x) − b a E1 (x,t)h(1) (t)dt. 3. 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)!

4. 11). 4. 55) b−a a b−a a for x ∈ [a, b]. 55) by g(x) and f (x) respectively A[g(x)] − and adding the resulting identities, we have 1 g(x) g(x)A[ f (x)] + f (x)A[g(x)] − b−a b b f (t)dt + f (x) a g(t)dt a b b (−1)n+1 g(x) En (x,t) f (n) (t)dt + f (x) En (x,t)g(n) (t)dt . 55), we get b b 1 A[g(x)] A[ f (x)]A[g(x)] − f (t)dt + A[ f (x)] g(t)dt b−a a a = + 1 (b − a)2 b b f (t)dt g(t)dt a a b b (−1)2n+2 En (x,t) f (n) (t)dt En (x,t)g(n) (t)dt . 49). 59) b−a a n(b − a) a for x ∈ [a, b]. 3, we leave the details to the reader.

I=k+1 By direct computation it is easy to observe that the following discrete identity Proof. 29) n. 27). 5. If we take wi = 1 for i = 1, . . 27) reduces to the discrete Montgomery identity, n−1 1 n xk = ∑ xi + ∑ Dn (k, i)Δxi , n i=1 i=1 where ⎧ ⎪ ⎨ i, 1 i k − 1, Dn (k, i) = i n ⎪ ⎩ − 1, k i n. n Finally, we present the Gr¨uss-type discrete inequalities given in [133]. 5. 31) Let {uk }, {vk } for k = 1, . . 31). 5): Proof. 37) for k = 1, . . , n. 37) by vk and uk respectively, adding the resulting identities and rewriting, we get uk vk − n n n−1 n−1 1 1 vk ∑ ui + uk ∑ vi = vk ∑ Dn (k, i)Δui + uk ∑ Dn (k, i)Δvi .

### Analisi 1 by Gianni Gilardi

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