Download PDF by John Srdjan Petrovic: Advanced Calculus: Theory and Practice

By John Srdjan Petrovic

ISBN-10: 1351381431

ISBN-13: 9781351381437

ISBN-10: 1466565640

ISBN-13: 9781466565647

ISBN-10: 2592642692

ISBN-13: 9782592642691

Suitable for a one- or two-semester path, Advanced Calculus: concept and Practice expands at the fabric coated in effortless calculus and offers this fabric in a rigorous demeanour. The textual content improves scholars’ problem-solving and proof-writing talents, familiarizes them with the historic improvement of calculus techniques, and is helping them comprehend the connections between various topics.

The publication takes a motivating strategy that makes rules much less summary to scholars. It explains how quite a few themes in calculus could seem unrelated yet actually have universal roots. Emphasizing old views, the textual content supplies scholars a glimpse into the advance of calculus and its rules from the age of Newton and Leibniz to the 20 th century. approximately three hundred examples bring about vital theorems in addition to support scholars enhance the mandatory talents to heavily study the theorems. Proofs also are offered in an available technique to students.

By strengthening abilities received via user-friendly calculus, this textbook leads scholars towards studying calculus ideas. it is going to aid them achieve their destiny mathematical or engineering studies.

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Lim(1 + 3n)1/n . 6. lim 2−1/ n . 7. Given x ≥ 1, show that lim(2 n x − 1)n = x2 . 8. Let {an } be a sequence of positive numbers. Prove that lim sup a1 + an+1 an n ≥ e. 9. Suppose that the terms of the sequence {an } satisfy the inequalities 0 ≤ an+m ≤ an + am . Prove that the sequence {an /n} converges. 6). In this chapter, our goal is to prove these theorems. When Cauchy did that in Cours d’analyse he took some properties of real numbers as self-evident. In the course of the 19th century it became clear that these needed to be proved as well, and for that it was necessary to make a precise definition of real numbers.

Let an = 1 + nπ n cos . What are the convergent subsequences of n+1 2 {an }? It is helpful to consider bn = cos nπ 2 for several values of n. We calculate 3π π = 0, b2 = cos π = −1, b3 = cos = 0, 2 2 5π b5 = cos = 0, b6 = cos 3π = −1, etc. , bn+4 = bn , for all n ∈ N. Trigonometric considerations confirm this: bn+4 = cos nπ + 4π nπ (n + 4)π nπ = cos = cos + 2π = cos = bn . 2 2 2 2 Since b1 = b3 = 0, we have that b2k−1 = 0, for all k ∈ N. It follows that a2k−1 = 1, and it is obvious that the limit of this subsequence is 1.

However, an + bn = 0, for every n ∈ N. Thus, lim sup(an + bn ) = 0. 1. Let {an } be a bounded sequence such that lim sup an = lim inf an . Prove that {an } is a convergent sequence. 2. Let {an } be a convergent sequence and let lim an = L. Prove that every subsequence of {an } converges to L. 3. Let {an } be a sequence of positive numbers. Prove that lim sup 1 an = 1 . 4. Let an , bn > 0 for all n ∈ N. Prove that lim sup(an bn ) ≤ lim sup an lim sup bn . Give an example to show that the inequality may be strict.

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Advanced Calculus: Theory and Practice by John Srdjan Petrovic

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