ps4sol

# ps4sol - ACM 95/100c Problem Set 4 Solutions Lei Zhang Each...

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Unformatted text preview: ACM 95/100c Problem Set 4 Solutions Lei Zhang April 22, 2006 Each problem is worth 10 points - each part of a multi-part problem is weighted equally Problem 1 (Collaboration allowed) The following problem examines the Fourier transform for the δ-function but also interprets the δ-function as an extension of the Kronecker δ that appears when we examine the orthogonality of Sturm-Liouville eigenfunctions on finite domains (like sines and cosines). (a) For what α does the function α exp(- β ( x- x ) 2 ) have unit area for-∞ < x < ∞ ? (b) Show that in the limit as β → ∞ ,the resulting function in part (a) satisfies the properties of the Dirac delta function δ ( x- x ). (c) Obtain the Fourier transform of δ ( x- x ) in two ways: i. Take the transform of part (a) and take the limit as β → ∞ . ii. Use the integration properties of the Dirac delta function. (d) Show that the transform of δ ( x- x ) is consistent with the following idea: Transforms of sharply peaked functions are spread out (that is they contain lots of frequencies). (e) Interpret the following equation δ ( ω- ω ) = 1 2 π Z ∞-∞ exp(- ix ( ω- ω )) dx. as a type of orthogonality relationship for the “eigenfunctions” exp(- iωx ). (f) Use this orthogonality relationship to compute F ( ω ) if f ( x ) = Z ∞-∞ F ( ω )exp(- iωx ) dω. This is a way of computing the “Fourier coefficients of f ( x )” otherwise known as the Fourier transform of f ( x ) using the appropriate orthogonality condition and the appropriate eigenfunctions. Solution 1 (a) For the function to have unit area for-∞ < x < ∞ means that 1 Z ∞-∞ α exp £- β ( x- x ) 2 / dx = Z ∞-∞ α exp (- βy 2 ) dy = Z ∞-∞ α √ β exp (- z 2 ) dz = α √ β √ π = 1 . (1) So, α = r β π . (2) (b) Let f ( x ) ≡ r β π exp £- β ( x- x ) 2 / . (3) For x 6 = x , lim β →∞ f ( x ) = 0 , (4) since exp £- β ( x- x ) 2 / decays in β faster than √ β grows. Also, f ( x ) = q β π becomes unbounded as β → ∞ . Finally, by part a), Z ∞-∞ f ( x ) dx = 1 (5) for all β . Thus, as β → ∞ , f ( x ) satisfies the properties of δ ( x- x ). 2 (c) i. 1 2 π Z ∞-∞ r β π exp £- β ( x- x ) 2 / exp( iωx ) dx = r β π exp (- βx 2 ) 1 2 π Z ∞-∞ exp •- β x 2- 2 xx- i ω β x ¶‚ dx = r β π exp (- βx 2 ) 1 2 π Z ∞-∞ exp "- β x- x- i ω 2 β ¶ 2 + β x + i ω 2 β ¶ 2 # dx = r β π exp- βx 2 + βx 2 + iωx- ω 2 4 β ¶ 1 2 π Z ∞-∞ exp "- β x- x- i ω 2 β ¶ 2 # dx = 1 2 π exp iωx- ω 2 4 β ¶ r β π Z ∞-∞ exp (- βy 2 ) dy = 1 2 π exp iωx- ω 2 4 β ¶ → 1 2 π exp( iωx ) as β → ∞ . (6) ii. By the properties of the Dirac δ function, 1 2 π Z ∞-∞ δ ( x- x )exp( iωx ) dx = 1 2 π exp( iωx ) . (7) Note (not necessary for students to state): It requires complex analysis to justify Z ∞-∞ exp "- β x- x + i ω 2 β ¶ 2 # dx = Z ∞-∞ exp (- βy 2 ) dy. (8) This justification is not required here, though. See Haberman’s “Appendix to 10.3” for details.This justification is not required here, though....
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ps4sol - ACM 95/100c Problem Set 4 Solutions Lei Zhang Each...

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