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Spring 2004 - Bender's Class - Exam 1

Spring 2004 - Bender's Class - Exam 1 - Math 20D First...

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Unformatted text preview: Math 20D First Midterm 48 points April 21, 2004 a Print Name, ID number and Section on your blue book. a BOOKS and OALOULATOQS are NOT allowed. One side of one page of NO. .LS is allowed. a You must show your work to receive credit. 1. [5 points) Let Zane” be the Maclaurin series for fir) = (cosm) sin(m2). Compute a1, a2, a3 a4 and a5. 1. While one could differentiate five times, it is easier to multiply series. In the following, - - - stands for terms of degree higher than five since we only need to keep terms through the fifth degree. cosx =1—mQ/2+m4/24+--- sin(m2] = m —--- fir] — (1 m2/2 : m4/24 ---)(m2 ---j — m2 x4/2 : Thus a1 = a3 = a5 = 0, a2 = 1 and a4 = 71/2. 2. (15 points) Find the radii of convergence of the following power series. Yon must give correct reasons for your answers to receive credit. 0° m” 0° ,2 0° ngxfl‘l'l 2. (a) The ra io test shows that the series converges for all m. (R = 00) (b) The root test gives |2x —3| < 1, which is equivalent to |m — 3| < 1/2, so the radius of convergence is 1/2. (c) The raJio or root test has a limit of lm| and so R = 1. 3. (20 points) Determine if each of the following series is convergent or divergent. If the series is convergent and the terms alternate in sign, determine the series is absolutely or conditionally convergent. Yon mast give correct reasons for yonr answers to receive credit. 8 71.3272 00 [—1) 0° e‘fi (a) g —n323” (bl ; fl 3. Often more than one test can be used. (T1532 )5 (C) Zf—UnsiHU/fl) Cd) m1 n (2n :1 (a) The series diverges because the terms do not go to zero: |an| = {9/8]”/108n2 —> 00. You could also use the ratio or the root test. (b) It converges. Perhaps the easiest test to use in the integral test with f(m] : m‘ifleflvfl2 since f mej dm : 2e‘w1x2 + C. (c) Since sinm is an increasing function when cc is small, sinl:1/n) is a decreasing function. By the alternating series test, we have convergence. Comparison with 21/10. shows that we do not have absolute convergence. (d) The ratio test has a limit of 1/4 and so the series converges. 4. (8 points] Let Eamm” be the Maclaurin series for f(mj and let Zbfim” be the Maclau— rin series for fl—m). [a] Express on in terms of can. [b] Suppose f(£t) is an even function; that is, f(—m) = flim) Show that, whenever n is odd, an = U. 4. (a) Using the series for fix): f(—m] = Zan[—x)n = E[—1)namm”. Thus a. = (71)”afl (b) Since power series are unique and f(—m) = fix), the two series have the same coefficients; that is, b... = am. From (a), on = (—1jnam, which gives us an = —a.2 for odd n. Thus a.,2 = 0 when n is odd. ...
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