Chem Differential Eq HW Solutions Fall 2011 110

# Chem Differential Eq HW Solutions Fall 2011 110 - = 1 √ 2...

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110 Chapter 7 The Fourier Transform and its Applications By Theorem 3(i) F ( xe - x 2 )= i d dw ± 1 2 e - w 2 / 4 ² = i 2 2 e - w 2 / 4 . 41. We have F ( 1 1+ x 2 )= ³ π 2 e -| w | . So if w> 0 F ( x 1+ x 2 )( w )= i ´ π 2 d dw e - w = - i ´ π 2 e - w . If w< 0 F ( x 1+ x 2 )( w )= i ´ π 2 d dw e w = i ´ π 2 e w . If w =0 , F ( x 1+ x 2 )(0) = 1 2 π µ -∞ x 1+ x 2 dx =0 (odd integrand). We can combine these answers into one formula F ( x 1+ x 2 )( w )= - i ´ π 2 sgn ( w ) e -| w | . 45. Theorem 3 (i) and Exercise 19: F ( xe - 1 2 ( x - 1) 2 )= i d dw ± F ( e - 1 2 ( x - 1) 2 ) ² = i d dw ± e - iw F ( e - 1 2 x 2 ) ² = i d dw ± e - iw e - 1 2 w 2 ) ² = i d dw ± e - 1 2 w 2 - iw ) ² = i ( - w - i ) e - 1 2 w 2 - iw =( 1 - iw ) e - 1 2 w 2 - iw . 49. h ( ω )= e - ω 2 · 1 1+ ω 2 = F ( 1 2 e - x 2 / 4 ) ·F ( ´ π 2 e -| x | ) . Hence h ( x )= f * g ( x
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Unformatted text preview: ) = 1 √ 2 π µ ∞-∞ 1 √ 2 e-( x-t ) 2 4 ´ π 2 e-| t | dt = 1 2 √ 2 µ ∞-∞ e-( x-t ) 2 4 e-| t | dt. 53. Let f ( x ) = xe-x 2 / 2 and g ( x ) = e-x 2 . (a) F ( f )( w ) =-iwe-w 2 2 , and F ( g )( w ) = 1 √ 2 e-w 2 4 . (b) { * } = { · } =-i w √ 2 e-w 2 4 e-w 2 2 =-i w √ 2 e-3 w 2 4 ....
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