8.5 - π 2(2 n 1 2 which becomes π 2 8 = 1 1 2 1 3 2 1 5 2...

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Math 2940 Solutions, Fall 2011 Section 8 . 5 2 ) Z 1 - 1 1 · x dx = x 2 / 2 | 1 - 1 = 1 / 2 - 1 / 2 = 0. Z 1 - 1 1 · ( x 2 - (1 / 3)) dx = ( x 3 - x ) / 3 | 1 - 1 = 0 - 0 = 0. Z 1 - 1 x · ( x 2 - (1 / 3)) dx = x 4 / 4 - x 2 / 6 | 1 - 1 = (1 / 2 - 1 / 3) - (1 / 2 - 1 / 3) = 0. 4 ) We need Z 1 - 1 1 · ( x 3 - cx ) dx = x 4 / 4 - cx 2 / 2 | 1 - 1 to be 0, but it is for any value of c . We also need We need Z 1 - 1 x · ( x 3 - cx ) dx = x 5 / 5 - cx 3 / 3 | 1 - 1 to be 0. This integral is (2 / 5) - (2 c/ 3) so c = 3 / 5. Finally, Z 1 - 1 ( x 2 - 1 / 3)( x 3 - 2 x/ 5) dx = Z 1 - 1 x 5 - 11 x 3 / 15 + 2 x/ 15 dx = x 6 / 6 - 11 x 4 / 60 + x 2 / 15 | 1 - 1 = (1 / 6 - 11 / 60 + 1 / 15) - (1 / 6 - 11 / 60 + 1 / 15) = 0. 6 ) We know the a i = 0 for all i , b 2 n = 0 and b 2 n +1 = 4 / ( π (2 n + 1)). Thus formula 6 becomes 1 = 1 2 X n =1 16
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Unformatted text preview: π 2 (2 n + 1) 2 which becomes π 2 / 8 = 1 1 2 + 1 3 2 + 1 5 2 + 1 7 2 + ... More stuff on the next page! 7 ) See the picture above. 9(a) a = 1 2 π Z π 1 dx = 1 / 2, a k = 1 π Z π cos kx dx = sin kx k | π =-k = 0 and b k = 1 π Z π sin kx dx =-cos kx k | π =-cos kπ + 1 k = ± k even 2 /k k odd ....
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This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell.

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8.5 - π 2(2 n 1 2 which becomes π 2 8 = 1 1 2 1 3 2 1 5 2...

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