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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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