{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online