Chem Differential Eq HW Solutions Fall 2011 125

Chem Differential Eq HW Solutions Fall 2011 125 - Section...

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Unformatted text preview: Section 7.6 The Fourier Cosine and Sine Transforms 125 9. Applying the definition of the transform and using Exercise 17, Sec. 2.6 to evaluate the integral, 2 π Fs (e−2x )(w) = ∞ e−2x sin wx dx 0 2 e−2x [−w cos wx − 2 sin wx] π 4 + w2 = ∞ x=0 2w . π 4 + w2 = The inverse sine transform becomes f ( x) = 13. We have fe (x) = Fc 1 . 1+x2 1 1 + x2 2 π ∞ 0 w sin wx dw. 4 + w2 So =F 1 1 + x2 = π −w e 2 (w > 0), by Exercise 11, Sec. 7.2. 17. We have fe (x) = Fc cos x 1 + x2 cos x 1+x2 . =F So cos x 1 + x2 = π −|w−1| + e−(w+1) e 2 (w > 0), by Exercises 11 and 20(b), Sec. 7.2. − 21. From the definition of the inverse transform, we have Fc f = Fc 1f. So Fc Fc f = −1 −1 Fc Fc f = f . Similarly, FsFs f = Fs Fs f = f . ...
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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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