Math 293 solution set 11

Math 293 solution set 11 - Math 293 Solutions to Problem...

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Math 293 Solutions to Problem Set 11. § 10.3 #1 . Both terms of x 3 +sin 2 x are odd, so the whole function is odd. In other words, ( - x ) 3 + sin( - 2 x ) = - x 3 - sin 2 x = - ( x 3 + sin 2 x ). #2 . sin 2 ( - x ) = ( - sin x ) 2 = (sin x ) 2 = sin 2 x , so this function is even. #3 . Replacing x by - x in (1 - x 2 ) - 1 / 2 does not change the value of this function, so it is even. #10 . The function f ( x ) = | x | is even, so its Fourier series will have only cosine terms (and the constant term a 0 2 ). We have a n = 1 π R π - π f ( x ) cos nx dx = 2 π R π 0 x cos nx dx since the function is even. Using integration by parts, we get 2 π £ x sin nx n fl fl π 0 - R π 0 sin nx dx / . Evaluating the first term at 0 and π gives 0, so we are left with a n = - 2 π R π 0 sin nx dx = 2 π ( cos nx n 2 fl fl π 0 ) = 2 πn 2 ( cos - 1 ) . The factor cos - 1 is 0 if n is even and - 2 if n is odd. Thus the nonzero a n ’s are a 2 n - 1 = - 4 π (2 n - 1) 2 . When n = 0 these formulas do not work since they would involve dividing by 0, so we do a separate calculation: a 0 = 2 π R π 0 x dx = · · · = π . Now we can write the final answer, the Fourier cosine series
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