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Unformatted text preview: Math 20D First Midterm Solutions April 21, 2004 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. cos x = 1 − x2 /2 + x4 /24 + · · · sin(x2 ) = x2 − · · · f (x) = (1 − x2 /2 + x4 /24 − · · ·)(x2 − · · ·) = x2 − x4 /2 + · · · Thus a1 = a3 = a5 = 0, a2 = 1 and a4 = −1/2. 2. (a) The ratio test shows that the series converges for all x. (R = ∞) (b) The root test gives 2x − 3 < 1, which is equivalent to x − 3 < 1/2, so the radius of convergence is 1/2. (c) The ratio or root test has a limit of x and so R = 1. 3. Often more than one test can be used. (a) The series diverges because the terms do not go to zero: an  = (9/8)n /108n2 → ∞. You could also use the ratio or the root test. (b) It converges. Perhaps the easiest test to use in the integral test with 1/2 1/2 f (x) = x−1/2 e−x since f (x) dx = 2e−x + C . (c) Since sin x is an increasing function when x is small, sin(1/n) is a decreasing function. By the alternating series test, we have convergence. Comparison with 1/n shows that we do not have absolute convergence. (d) The ratio test has a limit of 1/4 and so the series converges. 4. (a) Using the series for f (x): f (−x) = an (−x)n = (−1)n an xn . n Thus bn = (−1) an (b) Since power series are unique and f (−x) = f (x), the two series have the same coeﬃcients; that is, bn = an . From (a), an = (−1)n an , which gives us an = −an for odd n. Thus an = 0 when n is odd. ...
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This note was uploaded on 06/18/2009 for the course MATH 20D taught by Professor Mohanty during the Fall '06 term at UCSD.
 Fall '06
 Mohanty
 Math, Differential Equations, Equations

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