Chem Differential Eq HW Solutions Fall 2011 132

# Chem differential eq hw solutions fall 2011 132

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Unformatted text preview: 132 Chapter 7 The Fourier Transform and its Applications 57. Following the method of Example 9, we have = dψ ∗φ dx = (δ−1 − δ1 ) ∗ ( U −1 − U 1 + U 2 − U 3) = d (φ ∗ ψ ) dx δ−1 ∗ U −1 − δ−1 ∗ U 1 + δ−1 ∗ U 2 − δ−1 ∗ U 3 − δ1 ∗ U −1 +δ1 ∗ U 1 − δ1 ∗ U 2 + δ1 ∗ U 3 = 1 √ ( U −2 − U 0 + U 1 − U 2 − U 0 + U 2 − U 3 + U 4 ) 2π = 1 √ (( U −2 − U 0 ) − ( U 0 − U 1 ) − ( U 3 − U 4 )) . 2π d Integrating dx (φ ∗ ψ) and using the fact that φ ∗ ψ equal 0 for large |x| and that there are no discontinuities on the graph, we ﬁnd 1 if − 2 < x < 0 √2π (x + 2) 1 √ 2π (−x + 2) if 0 < x < 1 √1 φ ∗ ψ ( x) = if 1 < x < 3 2π 1 √ (−x + 4) if 3 < x < 4 2π 0 otherwise = 1 √ (x + 2) U −2 − U 0 + (−x + 2) U 0 − U 1 2π + U 1 − U 3 + (−x + 4) U 3 − U 4 ....
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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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