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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 Mtest. 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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 Spring '10
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 Math, dx, lim Fk

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