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**Unformatted text preview: **University of Illinois Spring 2010 ECE 313: Problem Set 0: Problems and Solutions Calculus Review Due: Friday January 22 at 4 p.m. Reading: Ross Chapter 1.1–1.4, Chapter 2.1–2.5. Noncredit Exercises: Chap. 1: Probs. 1-5,7,9. Theoretical exercises 4,8,13. Self-test probs. 1-15. Chap. 2: Probs. 3,4,9,10, 11-14. Theoretical exerecises 1-3,6,7,10,11,12,16,19,20. Self-test probs. 1-8. About the problems below: Calculus, a prerequisite of this course, will be used mainly in the second half of the semester. The parts of calculus primarily needed are integration and differentiation, Taylor series, L’Hopital’s Rule, integration by parts, and double integrals (setting up limits of integration and change of variables such as in rectangular to polar coordinates). This problem set will help you review these topics, and identify areas in which you may need additional review. 1. [Geometric, MacLaurin, and Taylor series; L’Hopital’s rule] (a) Prove that 1 + x + x 2 + ··· + x n- 1 = 1 − x n 1 − x for all x negationslash = 1 and all integers n ≥ 1. Solution: Let n denote a positive integer. Then, (1 − x ) bracketleftbig 1 + x + x 2 + ··· + x n- 1 bracketrightbig = bracketleftbig 1 + x + x 2 + ··· + x n- 1 bracketrightbig − bracketleftbig x + x 2 + ··· + x n bracketrightbig = 1 − x n . If x negationslash = 1, divide both sides by (1 − x ) to get 1 + x + x 2 + ··· + x n- 1 = 1 − x n 1 − x . (b) Continuing to assume that n is a positive integer, what is the value of the sum 1+ x + x 2 + ··· + x n- 1 when x = 1? Does your answer equal lim x → 1 1 − x n 1 − x ? Solution: At x = 1, the sum is 1+1 1 +1 2 + ··· +1 n- 1 = n . Now, since 1 − x n 1 − x is the indeterminate form at x = 1, we apply L’Hˆ opital’s rule to get lim x → 1 1 − x n 1 − x = lim x → 1 − nx n- 1 − 1 = n . Therefore, the value of 1 + x + x 2 + ··· + x n- 1 at x = 1 does equal lim x → 1 1 − x n 1 − x . (c) Assuming that | x | < 1, find the sum of the series 1 + x + x 2 + ··· . Hint: it is the limit of the finite sum 1 + x + x 2 + ··· + x n- 1 as n → ∞ . Solution: Since | x | < 1, 1 + x + x 2 + ··· = lim n →∞ 1 + x + x 2 + ··· + x n- 1 = lim n →∞ 1 − x n 1 − x = 1 1 − x . (d) Prove that lim x → sin( x ) x = 1 using L’Hˆ opital’s rule. Solution: L’Hˆ opital’s rule addresses a fraction whose numerator and denominator both go to zero at the same place, for example, sin( x ) and x both go to zero at x = 0. Therefore lim x → sin( x ) x = lim x → d sin( x ) /dx dx/dx = lim x → cos( x ) 1 = 1 (e) Prove that lim x → sin( x ) x = 1 without using L’Hˆ opital’s rule; instead, use the MacLaurin series (Taylor series in the vicinity of...

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