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Unformatted text preview: Math 445 - Solutions to selected problems from HW 2 Exercise 4, page 512. We must show that Z sin( w ) w cos( xw ) dx = f ( x ) = 2 if x < 1 4 if x = 1 if x > 1 Solution We must show 1 that the integral in the above is the Fourier cosine 2 integral of f ( x ), namely R A ( w ) cos( wx ) dw . Thus we just must check that A ( w ) = sin( w ) w . But A ( w ) = 2 R f ( x ) cos( xw ) dx = 2 R 1 2 cos( xw ) dx = sin( xw ) w 1 = sin( w ) w . Note that the value of R sin( w ) w cos( xw ) dx at x = 1 is given as the average of the left and right limits of f ( x ) (since we have a jump-discontinuity at x = 1), which is 4 . Exercise 11, page 517. Find F s ( e- x ) by integration. Solution By definition, 1 exercises 3 and 5 on page 512 are solved in exactly the same way as this one 2 make sure you know why cosine and not sine integral 1 F s ( e- x ) = r 2 Z e- x sin( wx ) dx = r 2 e- x- sin( wx ) + w...
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- Fall '07