Math450hw5sol - MATH 450 Section 002 Homework 5 Solutions...

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Unformatted text preview: MATH 450 Section 002 Homework 5 Solutions Winter 2009 Section 17.9, p. 919, #2. (b) a ( ) = 1 Z - f ( x ) cos( x )d x = 1 Z L x cos( x )d x = 1 x sin( x ) L x =0- 1 Z L sin( x ) d x = L sin( L ) + 1 cos( x ) 2 L x =0 = L sin( L ) + 1 cos( L ) 2- 1 2 = L sin( L ) + cos( L )- 1 2 . b ( ) = 1 Z - f ( x ) sin( x )d x = 1 Z L x sin( x )d x =- 1 x cos( x ) L x =0 + 1 Z L cos( x ) d x =- L cos( L ) + 1 sin( x ) 2 L x =0 =- L cos( L ) + 1 sin( L ) 2 = sin( L )- L cos( L ) 2 . So the Fourier integral representation of f is f ( x ) = Z L sin( L ) + cos( L )- 1 2 cos( x ) + sin( L )- L cos( L ) 2 sin( x ) d . 1 (g) Integration by parts twice yields, Z e- x cos( x )d x = (- e- x cos( x ) ) - Z e- x sin( x )d x = 1- (- e- x sin( x ) ) - Z e- x 2 cos( x )d x = 1- 2 Z e- x cos( x )d x. Thus a ( ) = 1 Z e- x cos( x )d x = 1 (1 + 2 ) . The calculation above gives us also Z e- x sin( x )d x = 1 1- Z e- x cos( x )d x = 1 - 1 (1 + 2 ) = 2 (1 + 2 ) = 1 + 2 . Thus b ( ) = 1 Z e- x sin( x )d x = (1 + 2 ) . So we have f ( x ) = Z 1 (1 + 2 ) cos( x ) + (1 + 2 ) sin( x ) d . Section 17.10, p. 932, #6 (a). 2 F{ 4 x 2 e- 3 | x | } = 4 F{ x 2 e- 3 | x | } = (17) 4 i 2 d 2 d 2 F{ e- 3 | x | } = (4)- 4 d 2 d 2 6 2 + 9 . (m) F- 1 { 1 2 + i + 2 } = F- 1 { 1 - i 1 + 2 i } = F- 1 { i 1 + i- i 2- i } = (21) F- 1 { i 1 + i }F- 1 {- i 2- i } = (2)+(3) H ( x ) e- x ? H (- x ) e 2 x . Section 17.10, p. 932, #12. (a) Fourier transform gives ( i ) 2 u- 2 u =- f. Thus u ( ) = 1 2 + 2 f ( ) . The convolution theorem implies now that u ( x ) = 1 2 | | e-| || x | ? f ( x ) ....
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Math450hw5sol - MATH 450 Section 002 Homework 5 Solutions...

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