Download e-book for iPad: Analisi 1 by Gianni Gilardi

By Gianni Gilardi

ISBN-10: 8838606536

ISBN-13: 9788838606533

Show description

Read Online or Download Analisi 1 PDF

Best mathematical analysis books

New PDF release: Matrix Algorithms, Volume II: Eigensystems

This can be the second one quantity in a projected five-volume survey of numerical linear algebra and matrix algorithms. It treats the numerical answer of dense and large-scale eigenvalue issues of an emphasis on algorithms and the theoretical historical past required to appreciate them. The notes and reference sections comprise tips to different tools besides old reviews.

The Analysis of Linear Partial Differential Operators IV: by Lars Hörmander PDF

From the reviews:"Volumes III and IV entire L. Hörmander's treatise on linear partial differential equations. They represent the main whole and updated account of this topic, via the writer who has ruled it and made the main major contributions within the final a long time. .. .. it's a brilliant publication, which needs to be found in each mathematical library, and an crucial instrument for all - old and young - attracted to the speculation of partial differential operators.

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 .

Download PDF sample

Analisi 1 by Gianni Gilardi


by William
4.4

Rated 4.49 of 5 – based on 31 votes