Chem Differential Eq HW Solutions Fall 2011 129

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

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Unformatted text preview: Section 7.8 129 Generalized Functions 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 − −2(x − 1)f (x) dx −1 0 0 0 = −2(x + 1)f (x) +2 −1 −2f (0) + 2 −2 0 −1 0 = 1 1 f (x) dx + 2(x − 1)f (x) f (x) dx 0 1 f (x) dx + 2f (0) − 2 −1 f (x) dx 0 = 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 −∞ 0 = −2 1 f (x) dx + 2 −1 = f (x) dx 0 −2 f (0) − f (−1) + 2 f (1) − f (0) = 2f (−1) − 4f (0) + 2f (1) = 2δ−1 − 4δ0 + 2δ1 , f (x) . Thus φ (x) = 2δ−1 − 4δ0 + 2δ1 . 29. We use (13) and the linearity of the Fourier transform: 1 F 3δ0 − 2δ−2 = √ 3 − 2e2iw . 2π 33. Using the operational property in Theorem 3(i), Section 7.2, we find F x ( U −1 − U 1) = i d F U −1 − U 1 dw = i d i i −√ eiw + √ e−iw dw 2π w 2π w = −i(i) d eiw − e−iw √ w 2π dw = 1 d 2i sin w √ w 2π dw = 2i w cos w − sin w √ w2 2π = i (Recall eiu − e−iu = 2i sin u) 2 w cos w − sin w . π w2 ...
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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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