20d_04s_1s

20d_04s_1s - Math 20D First Midterm Solutions 1 While one could differentiate five times it is easier to multiply series In the following

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

This note was uploaded on 06/18/2009 for the course MATH 20D taught by Professor Mohanty during the Fall '06 term at UCSD.

Ask a homework question - tutors are online