soln4 - R. Balan Homework #4 Solutions MATH 464 1. f1 =...

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R. Balan Homework #4 Solutions MATH 464 1. f 1 = 1 [1 , 2] ,F 1 ( ω ) = e - 3 πiω sin ( πω ) πω 2. f 2 ( x ) = x + 3 2 1 [ - 3 , - 1) + 1 [ - 1 , 1) + 3 - x 2 1 [1 , 3) , F 2 ( ω ) = sin (2 πω ) sin (4 πω ) 2 π 2 ω 2 3. f 3 ( x ) = e - a | x | , F 3 ( ω ) = 2 a 4 π 2 ω 2 + a 2 4. f 4 ( x ) = 1 α 2 + x 2 = π α F 3 ( x ) | a =2 πα , F 4 ( ω ) = π α e - 2 πα | ω | Then replace α by a : f ( x ) = 1 x 2 + a 2 F ( ω ) = π a e - 2 πa | ω | 5. f 5 ( x ) = sin (2 πx ) f 1 ( x ) = 1 2 i ( e 2 πix f 1 ( x ) - e - 2 πix f 1 ( x )) F 5 ( ω ) = 1 2 i [ F 1 ( ω - 1) - F 1 ( ω + 1)] = e - 3 πiω sin ( πω ) ( ω 2 - 1) 6. f 6 ( x ) = e -| x | cos (2 παx ) = 1 2 ( e 2 πiαx + e - 2 πiαx ) f 3 ( x ) | a =1 F 6 ( ω ) = 1 2 ( F 3 ( ω - α ) + F 3 ( ω + α )) = 1 1 + 4 π 2 ( ω - α ) 2 + 1 1 + 4 π 2 ( ω + α ) 2 Then replace α by a : f ( x ) = e -| x | cos (2 πax ) F ( ω ) = 1 1 + 4 π 2 ( ω - a ) 2 + 1 1 + 4 π 2 ( ω + a ) 2 7. f 7 ( x ) = cos (2 πx ) x 2 + a 2 = 1 2 ( e 2 πix + e - 2 πix ) f 4 ( x ) F 7 ( ω ) = 1 2 ( F 4 ( ω - 1) + F 4 ( ω + 1)) = π 2 a h e - 2 πa | ω - 1 | + e - 2 πa | ω +1 | i 8. f 8 ( x ) = 1 ax 2 + bx + c = 1 a 1 ( x + q ) 2 + r 2 , F 8 ( ω ) = 1 a e 2 πiqω F 4 ( ω ) | a in 4 - >r where q = b 2 a , r = q c a - b 2 4 a 2 . Thus F 8 ( ω ) = π ar e 2 πiqω - 2 πr | ω |
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R. Balan Homework #4 Solutions MATH 464 9. Use F 4 and Poisson formula to get: X n = -∞ 1 ( x - n ) 2 + a 2 = X n = -∞ f 4 ( x - n ) = X k = -∞ e 2 πikx F 4 ( k ) = π a X k = -∞ e - 2 πa | k | +2 πikx Then simplify the sum as follows: X k =0 e - k 2 π ( a - ix ) = 1 1 - e - 2 π ( a - ix ) k = - 1 X k = -∞ e k 2 π ( a +
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This note was uploaded on 05/05/2010 for the course MATH 464 taught by Professor Staff during the Spring '08 term at Maryland.

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soln4 - R. Balan Homework #4 Solutions MATH 464 1. f1 =...

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