Chem Differential Eq HW Solutions Fall 2011 105

# Chem Differential Eq HW Solutions Fall 2011 105 - Section...

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Section 7.1 The Fourier Integral Representation 105 Solutions to Exercises 7.1 1. We have f ( x )= ± 1i f - a<x<a ,( a> 0) 0 otherwise, This problem is very similar to Example 1. From (3), if ω ± = 0, then A ( ω 1 π ² -∞ f ( t ) cos ωtdt = 1 π ² a - a cos = ³ sin ωt πω ´ a - a = 2 sin . If ω = 0, then A (0) = 1 π ² -∞ f ( t ) dt = 1 π ² a - a dt = 2 a π . Since f ( x ) is even, B ( ω ) = 0. For | x = a the function is continuous and Theorem 1 gives f ( x 2 a π ² 0 sin cos ωx ω dω . For x = ² a , points of discontinuity of f , Theorem 1 yields the value 1 / 2 for the last integral. Thus we have the Fourier integral representation of f 2 a π ² 0 sin cos ω = f | x | <a , 1 / 2i f | x | = a , 0i f | x | >a . ± 5. since f ( x e -| x | is even, B ( w ) = 0 for all w , and A ( w 2 π ² 0 e - t cos wtdt = 2 π e - t 1+ w 2 [ - cos wt + w sin ] µ µ µ µ 0 = 2 π 1 w 2 , where we have used the result of Exercise 17, Sec. 2.6, to evaluate the integral.
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