7 let f t et and t for t 1 0 for t 1 g t compute

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Unformatted text preview: ∞ as f is real. Therefore ∞ ∞ 2 f (x)dx = f (x) −∞ 1 = 2π ∞ −∞ ∞ ∞ iωx f (ω ) f (x)e −∞ −∞ 1 f (ω)eiωx dω dx −∞ 2π ∞ 1 dx dω = 2π 1 f (f ) dω = 2π −∞ ∗ ∞ |f |2dω. −∞ 4. (Problem 1 of 3.7) Let f (t) = e−t and t for |t| ≤ 1 0 for |t| > 1. g (t) = Compute f ∗ g . Answer 1 f ∗ g (t) = xe−(t−x) dx = 2e−(t+1) . −1 5. (difficult!) ∞ Suppose that f is continuously differentiable in (−∞, ∞) and −∞ |f |dx < ∞, use Riemann-Le...
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This document was uploaded on 03/02/2014 for the course MATH Math4052 at HKUST.

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