Chem Differential Eq HW Solutions Fall 2011 131

Chem Differential Eq HW Solutions Fall 2011 131 - Section...

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Section 7.8 Generalized Functions 131 Hence F ( f ( x )) = - i w F ( f ± ( x )) = 1 2 πw ± e 2 iw w - 2 e iw w + e - iw w - ie iw +2 ie - 2 iw ² . 45. You may want to draw a graph to help you visualize the derivatives. For f ( x ) = sin x if | x | and 0 otherwise, we have f ”( x ) = cos x if | x | and 0 otherwise. Note that since f is continuous, we do not add delta functions at the endpoints x = ± π when computing f ± .F o r f ±± , the graph is discontinuous at x = ± π and we have f ±± ( x )= - δ - π + δ π - sin x if | x |≤ π and 0 otherwise. Thus f ±± ( x )= - δ - π + δ π - f ( x ) for all x. Taking the Fourier transform, we obtain F ( f ±± ( x )) = F ( - δ - π + δ π - f ( x )) ; - w 2 F ( f ( x )) = - 1 2 π e iπw + 1 2 π e - iπw -F ( f ( x )) ; (1 + w 2 ) F ( f ( x )) = 1 2 π =2 cos( πw ) ³ ´µ ( e iπw + e - iπw ) ⇒F ( f ( x )) = 1 2 π cos( πw ) 1+ w 2 . 49. From Example 9, we have f ± = δ - 1 - δ 1 . So from d
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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