HW2 - Page 512, problem 2: πx / 2 0 < x < 1...

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Unformatted text preview: Page 512, problem 2: πx / 2 0 < x < 1 ∞ 1 2 2 πv sin(ω ) − ω ⋅ cos(ω ) f ( x) = π / 4 x = 1 ⇒ B (ω ) = ∫ f (v ) sin(ωv)dv = ∫ sin(ωv)dv = π0 π02 ω2 0 x >1 ∞ ∞ sin(ω ) − ω ⋅ cos(ω ) sin(ωx)dω ω2 0 ⇒ f ( x ) = ∫ B (ω ) sin(ωx)dω = ∫ 0 Page 512, problem 9: ∞ 1 x 0 < x < 1 2 2 2 cos(ω ) + ω ⋅ sin(ω ) − 1 f ( x) = ⇒ A(ω ) = ∫ f (v ) cos(vω )dv = ∫ v ⋅ cos(vω )dv = π0 π0 π ω2 0 x > 1 ∞ 2 cos(ω ) + ω ⋅ sin(ω ) − 1 cos(ωx)dω π∫ ω2 0 Page 512, problem 15: ⇒ f ( x) = ∞ π sin( x) 0 < x < π 2 2 2 f ( x) = ⇒ B (ω ) = ∫ f (v) sin(vω )dv = ∫ sin(v ) sin(vω ) dv = sin(πω) x >π π0 π0 1− ω2 0 ∞ 2 sin(πω) sin( xω )dω 2 0 1−ω Page 517, problem 5: ⇒ f ( x) = ∫ ∞ ˆ 1 = 2 cos(ωx)dω. f c−1 2 π ∫ 1+ ω2 1 + ω 0 From problem 3 on page 512: ∞ cos(ωx) π −x 1 2 π −x π −x −1 ∫ 1 + ω 2 dω = 2 e ⇒ fˆc 1 + ω 2 = π 2 e = 2 e 0 Page 517, problem 11: ( ) ˆ f s e −πx = ∞ ∞ 2 −πx 2 −ω π 2 ω −πx −πx ∫ e sin(ωx)dx = π π 2 + ω 2 e cos(ωx) − π 2 + ω 2 e sin(ωx) 0 = π π 2 + ω 2 π0 Page 528, problem 8: x ⋅ e − x − 1 < x < 0 ˆ f ( x) = ⇒ f (ω ) = 0 otherwise = = 1 ( − 1 − iω ) xe ( −1−iω ) x − e ( −1−iω ) x ( − 1 − iω ) 2 2π 1 2π − 1 − iωe1+iω ( 1 + iω ) 2 0 −1 1 2π ∞ ∫ −∞ f ( x)e −iωx dx = 1 2π 0 ∫ x⋅e −1 −x e −iωx dx ...
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HW2 - Page 512, problem 2: πx / 2 0 &amp;lt; x &amp;lt; 1...

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