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Chem Differential Eq HW Solutions Fall 2011 129

# Chem Differential Eq HW Solutions Fall 2011 129 - Section...

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Section 7.8 Generalized Functions 129 function, and integrating by parts, we find φ ( x ) ,f ( x ) = - φ ( x ) ,f ( x ) = - -∞ φ ( x ) f ( x ) dx = - 0 - 1 2( x + 1) f ( x ) dx - 1 0 - 2( x - 1) f ( x ) dx = - 2( x + 1) f ( x ) 0 - 1 + 2 0 - 1 f ( x ) dx + 2( x - 1) f ( x ) 1 0 - 2 1 0 f ( x ) dx = - 2 f (0) + 2 0 - 1 f ( x ) dx + 2 f (0) - 2 1 0 f ( x ) dx = 2 ( U - 1 ( x ) - U 0 ( x ) ) ,f ( x ) - 2 ( U 0 ( x ) - U 1 ( x ) ) ,f ( x ) . = 2 ( U - 1 ( x ) - U 0 ( x ) ) - 2 ( U 0 ( x ) - U 1 ( x ) ) ,f ( x ) . Thus φ ( x ) = 2 ( U - 1 ( x ) - U 0 ( x ) ) - 2 ( U 0 ( x ) - U 1 ( x ) ) . Reasoning similarly, we find φ ( x ) ,f ( x ) = - φ ( x ) ,f ( x ) = - -∞ φ ( x ) f ( x ) dx = - 2 0 - 1 f ( x ) dx + 2 1 0 f ( x ) dx = - 2 ( f (0) - f ( - 1) ) + 2 ( f (1) - f (0) ) = 2 f ( - 1) - 4 f (0) + 2 f (1) = 2 δ - 1 - 4 δ 0 + 2 δ 1 ,f ( x ) . Thus φ ( x ) = 2 δ - 1 - 4 δ 0 + 2 δ 1 . 29. We use (13) and the linearity of the Fourier transform: F ( 3 δ 0 - 2 δ - 2 ) = 1 2 π ( 3 - 2 e 2 iw
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