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# Hw 2 - Math 445 Solutions to selected problems from HW 2...

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Math 445 - Solutions to selected problems from HW 2 Exercise 4, page 512. We must show that 0 sin( w ) w cos( xw ) dx = f ( x ) = π 2 if 0 x < 1 π 4 if x = 1 0 if x > 1 Solution We must show 1 that the integral in the above is the Fourier cosine 2 integral of f ( x ), namely 0 A ( w ) cos( wx ) dw . Thus we just must check that A ( w ) = sin( w ) w . But A ( w ) = 2 π 0 f ( x ) cos( xw ) dx = 2 π 1 0 π 2 cos( xw ) dx = sin( xw ) w 1 0 = sin( w ) w . Note that the value of 0 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 π 4 . Exercise 11, page 517. Find F s ( e - πx ) 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

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F s ( e - πx ) = 2 π 0 e - πx sin( wx ) dx = 2 π e - πx - π sin( wx ) 0 + w π 0 e - πx cos( wx ) dx = w π 2 π e - πx - π cos( wx ) 0 - w π 0 e - πx sin( wx ) dx = w π 2 2 π - w 2 π 2 2 π 0 e - πx sin( wx ) dx = w π 2 2 π - w 2 π 2 F s ( e -
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