Chem Differential Eq HW Solutions Fall 2011 130

# Chem Differential Eq HW Solutions Fall 2011 130 - i e-iw w...

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130 Chapter 7 The Fourier Transform and its Applications The formula is good at w = 0 if we take the limit as w 0. You will get F ( x ( U - 1 -U 1 ) ) = lim w 0 i ± 2 π ² w cos w - sin w w 2 ³ = i ± 2 π lim w 0 - w sin w 2 w =0 . (Use l’Hospital’s rue.) Unlike the Fourier transform in Exercise 31, the transform here is a nice continuous function. There is a major diﬀerence between the trans- forms of the two exercises. In Exercise 31, the function is not integrable and its Fourier transform exists only as a generalized function. In Exercise 33, the function is integrable and its Fourier transform exists in the usual sense of Section 7.2. In fact, look at the transform in Exercise 31, it is not even de±ned at w =0 . An alternative way to do this problem is to realize that φ ± ( x )= - δ - 1 - δ 1 + U - 1 -U 1 . So F ( φ ± ( x ) ) = F ( - δ - 1 - δ 1 + U - 1 -U 1 ) = 1 2 π ´ - e iw - e - iw - i e iw w
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Unformatted text preview: i e-iw w µ . But F ( φ ± ( x ) ) = iw F ( φ ( x ) ) . So F ( φ ( x ) ) = i w F ( e iw + e-iw + i e iw w-i e-iw w ) = i √ 2 π w ² 2 cos w + i w (2 i sin w ) ³ = i ± 2 π ² w cos w-sin w w 2 ³ . 37. Write τ 2 ( x ) = x 2 U ( x ), then use the operational properties F ( τ 2 ( x ) ) = F ( x 2 U ( x ) ) =-d 2 dw 2 F ( U ( x )) =--i √ 2 π d 2 dw 2 ² 1 w ³ = 2 i √ 2 π 1 w 3 = i ± 2 π 1 w 3 . 41. We have f ± ( x ) =-U-2 ( x ) + 2 U-1 ( x )- U 1 ( x ) + δ-1 ( x )-2 δ 2 ( x ) . So F ( f ± ( x )) = F (-U-2 ( x ) + 2 U-1 ( x )- U 1 ( x ) + δ-1 ( x )-2 δ 2 ( x )) =--i √ 2 π e 2 iw w + 2-i √ 2 π e iw w--i √ 2 π e-iw w + 1 √ 2 π e iw-2 √ 2 π e-2 iw ....
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