soln3 - R. Balan Homework #3 Solutions MATH 464 1. (1 - 2x)...

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R. Balan Homework #3 Solutions MATH 464 1. π 2 (1 - 2 x ) = X k =1 sin(2 πkx ) k 2. π 2 6 (1 - 6 x + 6 x 2 ) = X k =1 cos(2 πkx ) k 2 3. π 3 3 ( x - 3 x 2 + 2 x 3 ) = X k =1 sin(2 πkx ) k 3 4. π 4 90 (1 - 30 x 2 + 60 x 3 - 30 x 4 ) = X k =1 cos(2 πkx ) k 4 5. f ( x ) = 1 8 + X k 6 =0 ± i ( - 1) k 4 πk + ( - 1) k - 1 4 π 2 k 2 ² e 2 πikx = 1 8 - 1 2 π X k> 0 ( - 1) k k sin (2 πkx ) - 1 π 2 X p> 0 1 (2 p + 1) 2 cos (2 π (2 p + 1) x ) To have pointwise convergence, a = f (0 . 5 - )+ f (0 . 5) 2 = 1 4 . 6. f ( x ) = 1 2 - 4 π 2 X n 0 1 (2 n + 1) 2 cos(4 π (2 n + 1) x ) The Fourier series converges pointwise at every x . 7. f ( x ) = 1 2 + 2 π X n 0 sin(4 π (2 n + 1) x ) 2 n + 1 The Fourier series converges to f(x) at all points in (0 , 1) except at x = 1 4 , 1 2 , 3 4 . 8. sin(2 πsx ) = 2 sin( πs ) π ± 1 1 2 - s 2 sin(2
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