Advanced Calculus: Theory and Practice - download pdf or read online

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 direction, Advanced Calculus: idea 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 abilities, familiarizes them with the historic improvement of calculus recommendations, and is helping them comprehend the connections between diversified topics.

The publication takes a motivating method that makes rules much less summary to scholars. It explains how a number of issues in calculus could appear unrelated yet in fact have universal roots. Emphasizing old views, the textual content supplies scholars a glimpse into the advance of calculus and its principles 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 boost the mandatory abilities to heavily research the theorems. Proofs also are awarded in an obtainable solution to students.

By strengthening talents won via hassle-free calculus, this textbook leads scholars towards getting to know calculus strategies. it's going to support them achieve their destiny mathematical or engineering studies.

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Example text

Proof. We will first prove the case a = 1. Let b be a positive real number and let B = {n ∈ Z : n ≤ b}. 2, the set B is bounded above and sup B = ⌊b⌋. If N = ⌊b⌋ + 1 then N ∈ N and N > b. The general case a = 1 follows easily now. Since a, b > 0, the number b/a is defined and positive. By the first part of the proof, there exists N ∈ N such that N > b/a. It follows that aN > b and the proof is complete. 5. Earlier, when we talked about Hilbert’s axioms of R, we only specified two groups of axioms (Field and Order Axioms).

Then an + bn < lim sup an + lim sup bn + ε, so any accumulation point of {an + bn } (including the largest) cannot exceed lim sup an + lim sup bn + ε. Consequently, lim sup(an + bn ) ≤ lim sup an + lim sup bn + ε. Since ε is arbitrary, (c) is proved. Parts (a) and (b) show that the rules for the limit superior and limit inferior are not the same as for the limits. Part (c) is even worse: where we were hoping for an equality, we have an inequality. Now, is it because proving the equality might be too hard, or could it be that the left and the right sides are not always equal?

In order to show that it is bounded above, we will consider a sequence bn = (1+1/n)n+1 . An argument, similar to the one above, can be used to establish that bn+1 /bn ≤ 1 so {bn } is a decreasing sequence. Further, an ≤ bn ≤ b1 for any n ∈ N, so the sequence {an } is bounded above by b1 = 4. Therefore, it is convergent, and the proof is complete. 3. The limit of the sequence {an } lies between a1 = 2 and b1 = 4. This number is the well-known constant e, and it can be calculated to any number of decimals.

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Advanced Calculus: Theory and Practice by John Srdjan Petrovic


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