By Philip Straffin (editor)

ISBN-10: 0883850850

ISBN-13: 9780883850855

Scholars see how calculus can clarify the constitution of a rainbow, consultant a robotic arm, or examine the unfold of AIDS. every one module begins with a concrete challenge and strikes directly to offer an answer. The discussions are precise, sensible, and pay cautious realization to the method of mathematical modeling. workouts, options, and references are supplied.

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**Extra info for Applications of Calculus (Resources for Calculus Collection)**

**Sample text**

The multiples a1, where a ∈ R, form a subfield of C isomorphic to the real field R. By identifying the real number a with the complex number a1, we may regard R itself as contained in C. Thus we will now stop using matrices and use only the fact that C is a field containing R such that every z ∈ C can be represented in the form z = x + i y, where x, y ∈ R and i ∈ C satisfies i 2 = −1. The representation is necessarily unique, since i ∈ / R. We call x and y the real and imaginary parts of z and denote them by Rz 5 Complex Numbers 41 and I z respectively.

It is easily verified that if x n > 1 and x n2 > a, then x n+1 > 1, x n+1 Since the inequalities hold for n = 1, it follows that they hold for all n. Thus the sequence {x n } is nonincreasing and bounded, and therefore convergent. If x n → b, then a/x n → a/b and x n+1 → b. Hence b = (b + a/b)/2, which simplifies to b2 = a. We consider now sequences of real numbers which are not necessarily monotonic. Lemma 23 Any sequence {an } of real numbers has a monotonic subsequence. Proof Let M be the set of all positive integers m such that am ≥ an for every n > m.

This completes the proof that A + B is a cut. It follows at once from the corresponding properties of rational numbers that addition of cuts satisfies the commutative law (A2) and the associative law (A3). We consider next the connection between addition and order. Lemma 15 For any cut A and any c ∈ P, there exists a ∈ A such that a + c ∈ / A. Proof If c ∈ / A, then a + c ∈ / A for every a ∈ A, since c < a + c. Thus we may suppose c ∈ A. Choose b ∈ P\A. For some positive integer n we have b < nc and hence nc ∈ / A.

### Applications of Calculus (Resources for Calculus Collection) by Philip Straffin (editor)

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