Math450hw3sol - MATH 450 Section 002 Homework 3 Solutions Winter 2009 Section 17.5 p 880#2(b Since | sin(2 nx n 2 2 n 2 | = | sin(2 nx | | n 2 2 n

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Unformatted text preview: MATH 450 Section 002 Homework 3 Solutions Winter 2009 Section 17.5, p. 880 #2(b). Since | sin(2 nx ) n 2- 2 n + 2 | = | sin(2 nx ) | | n 2- 2 n + 2 | ≤ 1 ( n- 1) 2 + 1 =: M n , we can check whether ∑ M n < ∞ and try to use the M-test. We can try to use the ratio test to check if the series converges, but it does not give an answer. So we will try the integral test. Let f ( x ) = 1 ( x- 1) 2 +1 . Then f is a decreasing function and Z ∞ 1 dx ( x- 1) 2 + 1 = Z ∞ dx 1 + x 2 = π 2 . So the series ∑ M n converges by the integral test and the original series converges uniformly. Section 17.5, p.880 #4(e). Let f ( x ) = ∞ X n =1 e- n sin nx for-∞ < x < ∞ . Since | d dx ( e- n sin nx ) | = | ne- n cos nx | ≤ ne- n = M n , and ∑ M n converges by the root test (and by the ratio test), we have uniformly convergent series and f ( x ) = ∞ X n =1 d dx ( e- n sin nx ) = ∞ X n =1 ne- n cos nx. 1 Section 17.6, p.885 #2(c) (for k = 1 , 2 , 3 ). First we need to compute the Fourier series of cos 2 x . Recall that cos 2 x = cos( x ) cos( x ) = 1 2 (cos( x + x ) + cos( x- x )) = 1 2 (cos(2 x ) + 1) ....
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This note was uploaded on 04/24/2010 for the course MATH 425 taught by Professor K during the Spring '10 term at University of Michigan-Dearborn.

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Math450hw3sol - MATH 450 Section 002 Homework 3 Solutions Winter 2009 Section 17.5 p 880#2(b Since | sin(2 nx n 2 2 n 2 | = | sin(2 nx | | n 2 2 n

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