Chem Differential Eq HW Solutions Fall 2011 124

# Chem Differential Eq HW Solutions Fall 2011 124 - = 1 2 ²...

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124 Chapter 7 The Fourier Transform and its Applications Solutions to Exercises 7.6 1. The even extension of f ( x ) is f e ( x ) = 1 if - 1 <x< 1, 0 otherwise. The Fourier transform of f e ( x ) is computed in Example 1, Sec. 7.2 (with a = 1). We have, for w 0, F c ( f )( w ) = F ( f e )( w ) = 2 π sin w w . To write f as an inverse Fourier cosine transform, we appeal to (6). We have, for x> 0, f ( x ) = 2 π 0 F c ( f )( w ) cos wxdw, or 2 π 0 sin w w cos wxdw = 1 if 0 <x< 1, 0 if x> 1 , 1 2 if x = 1 . Note that at the point x = 1, a point of discontinuity of f , the inverse Fourier transform is equal to ( f ( x +) + f ( x - )) / 2. 5. The even extension of f ( x ) is f e ( x ) = cos x if - 2 π<x< 2 π , 0 otherwise. Let’s compute the Fourier cosine transform using definition (5), Sec. 7.6: F c ( f )( w ) = = 2 π 2 π 0 cos x cos wxdx = 2 π 2 π 0 1 2 [cos( w + 1) x + cos( w - 1) x ] dx = 1 2 2 π sin( w + 1) x w + 1 + sin( w - 1)
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Unformatted text preview: = 1 2 ² 2 π ´ sin 2( w + 1) π w + 1 + sin 2( w-1) π w-1 µ ( w ± = 1) = 1 2 ² 2 π ´ sin 2 πw w + 1 + sin 2 πw w-1 µ ( w ± = 1) = ² 2 π sin2 πw w w 2-1 ( w ± = 1) . Also, by l’Hospital’s rule, we have lim w → ² 2 π sin 2 πw w w 2-1 = √ 2 π, which is the value of the cosine transform at w = 1. To write f as an inverse Fourier cosine transform, we appeal to (6). We have, for x > 0, 2 π ³ ∞ w w 2-1 sin 2 πw cos wxdw = ± cos x if 0 < x < 2 π , i f x > 2 π. For x = 2 π , the integral converges to 1 / 2. So 2 π ³ ∞ w w 2-1 sin2 πw cos 2 πwdw = 1 2 ....
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