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soln5 - R Balan Homework#5 Solutions MATH 464 I 1 F(0 = 1 2...

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Unformatted text preview: R. Balan Homework #5 Solutions MATH 464 I . 1. F (0) = 1 2. F (0) =- 2 πi 3. F (1) + F (- 1) = 0 4. Z ∞-∞ sF ( s ) ds = 0 5. F ( s ) = F ( s ) , that means F ( s ) is real valued II . 6. We look for a function f : R → C that satisfies the given equation. Since we have the freedom to impose any condition we want, we make a choice f (- x ) = f ( x ) (that is, we look for an even function). Then its Fourier transform is: F ( s ) = Z ∞-∞ e- 2 πisx f ( x ) dx = Z ∞ ( e- 2 πisx f ( x ) + e 2 πisx f (- x )) dx = Z ∞ ( e- 2 πisx + e 2 πisx f ( x )) dx = 2 Z ∞ cos (2 πxs ) f ( x ) dx Note also: F (- s ) = F ( s ) We want to have F ( s ) = 2 e- s for s > 0. Then, because of the symmetry F (- s ) = F ( s ), we obtain F ( s ) = 2 e-| s | . The function f is now obtained through the synthesis equation (the inverse Fourier transform): f ( x ) = Z ∞-∞ e 2 πisx F ( s ) ds = 2 Z ∞-∞ e 2 πisx e-| s | ds = 4 1 + 4 π 2 x 2 Note that indeed f (- x ) = f ( x ) as we assumed.) as we assumed....
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soln5 - R Balan Homework#5 Solutions MATH 464 I 1 F(0 = 1 2...

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