Chem Differential Eq HW Solutions Fall 2011 111

# Chem Differential Eq HW Solutions Fall 2011 111 - | f x-t g...

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Section 7.2 The Fourier Transform 111 (c) With the help of Theorem 3 f * g = F - 1 ± - i w 2 e - 3 w 2 4 ² = F - 1 ± i 1 2 4 6 d dw e - 3 w 2 4 ² = 1 2 2 3 F - 1 ± i d dw e - 3 w 2 4 ² = 1 2 2 3 x F - 1 ± e - 3 w 2 4 ² = 2 3 3 xe - 1 3 x 2 . In computing F - 1 ± e - 3 w 2 4 ² , use Exercise 10(a) and (5) to obtain F - 1 ± e - aw 2 ² = 1 2 a e - ( - x ) 2 4 a = 1 2 a e - x 2 4 a . 57. Recall that f is integrable means that ³ -∞ | f ( x ) | dx < . If f and g are integrable, then f * g ( x )= 1 2 π ³ -∞ f ( x - t ) g ( t ) dt. So, using properties of the integral: ³ -∞ | f * g ( x ) | dx = ³ -∞ ´ ´ ´ ´ 1 2 π ³ -∞ f ( x - t ) g ( t ) dt ´ ´ ´ ´ dx 1 2 π ³ -∞ ³ -∞
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Unformatted text preview: | f ( x-t ) g ( t ) | dxdt (Interchange order of integration. ≤ 1 √ 2 π ³ ∞-∞ = ± ∞-∞ | f ( x ) | dx µ ¶· ¸ ³ ∞-∞ | f ( x-t ) | dx | g ( t ) | dt (Change variables in the inner integral X = x-t. ) ≤ 1 √ 2 π ³ ∞-∞ | f ( x ) | dx ³ ∞-∞ | g ( t ) | dt < ∞ ; thus f * g is integrable....
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