Ritelli, Daniele

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Almae Matris Studiorum Campus

1/17 P!i?"##$$$!"#$#CLAMFIM 2012-2013Modelli @ ClamfimLesson 16Marted`ı 4 Dicembre 2012Beta Function. Fourier Transformprofessor Daniele Ritelliwww.unibo.it/faculty/daniele.ritelli2/17 P!i?"##$$$!"#$#Euler Beta Theorem If x, y > 0:Γ(x)Γ(y)Γ(x+ y)=∫ 10s x−1 (1− s) y−1ds2/17 P!i?"##$$$!"#$#Euler Beta Theorem If x, y > 0:Γ(x)Γ(y)Γ(x+ y)=∫ 10s x−1 (1− s) y−1dsThis identity can be reformulated introducing the Euler Beta function:B(x, y) =∫ 10s x−1(1− s) y−1dsin such a wayB(x, y) =Γ(x)Γ(y)Γ(x+ y)3/17 P!i?"##$$$!"#$#ProofWe start from Gamma’s definition Γ(x) =∫ +∞0tx−1e−tdt then changevariable putting t = u2 so thatΓ(x) = 2∫ +∞0u2x−1e−u2du3/17 P!i?"##$$$!"#$#ProofWe start from Gamma’s definition Γ(x) =∫ +∞0tx−1e−tdt then changevariable putting t = u2 so thatΓ(x) = 2∫ +∞0u2x−1e−u2duSimilarlyΓ(y) = 2∫ +∞0v2y−1e−v2dv3/17 P!i?"##$$$!"#$#ProofWe start from Gamma’s definition Γ(x) =∫ +∞0tx−1e−tdt then changevariable putting t = u2 so thatΓ(x) = 2∫ +∞0u2x−1e−u2duSimilarlyΓ(y) = 2∫ +∞0v2y−1e−v2dvNow use Fubini’s TheoremΓ(x)Γ(y) = 4∫∫[0,+∞)×[0,+∞)u2x−1v2y−1e−(u2+v2)dudv4/17 P!i?"##$$$!"#$#Change to polar coordinateu = ρ cos ϑv = ρ sin ϑobtaining4/17 P!i?"##$$$!"#$#Change to polar coordinateu = ρ cos ϑv = ρ sin ϑobtainingΓ(x)Γ(y) = 4(∫ +∞0ρ2x+2y−1e−ρ2dρ)(∫ pi/20cos2x−1 ϑ sin2y−1 ϑ dϑ)4/17 P!i?"##$$$!"#$#Change to polar coordinateu = ρ cos ϑv = ρ sin ϑobtainingΓ(x)Γ(y) = 4(∫ +∞0ρ2x+2y−1e−ρ2dρ)(∫ pi/20cos2x−1 ϑ sin2y−1 ϑ dϑ)= Γ(x+ y)∫ pi/202 cos2x−1 ϑ sin2y−1 ϑ dϑ5/17 P!i?"##$$$!"#$#To end the proof we have to show thatB(x, y) =∫ pi/202 cos2x−1 ϑ sin2y−1 ϑ dϑ5/17 P!i?"##$$$!"#$#To end the proof we have to show thatB(x, y) =∫ pi/202 cos2x−1 ϑ sin2y−1 ϑ dϑBut coming back to Beta’s definitionB(x, y) =∫ 10sx−1 (1− s)y−1dswe are done putting s = cos2 ϑ6/17 P!i?"##$$$!"#$#Show, using Gamma and Beta functions, that the Lebesgue measureof A = {(x, y) ∈ R2 | x2/3 + y2/3 ≤ 1} is 38√pi6/17 P!i?"##$$$!"#$#Show, using Gamma and Beta functions, that the Lebesgue measureof A = {(x, y) ∈ R2 | x2/3 + y2/3 ≤ 1} is 38√pi!1.0 !0.5 0.5!1.0!0.50.5Figure 1: Asteroide7/17 P!i?"##$$$!"#$#m(A) = 4∫ 10(1− x2/3)3/2dx7/17 P!i?"##$$$!"#$#m(A) = 4∫ 10(1− x2/3)3/2dx= 6∫ 10u1/2(1− u)3/2du7/17 P!i?"##$$$!"#$#m(A) = 4∫ 10(1− x2/3)3/2dx= 6∫ 10u1/2(1− u)3/2du= 6B(5/2, 3/2)8/17 P!i?"##$$$!"#$#Tonelli’s Theorem Let A ⊂ Rm measurable and f : A→ R measur-able. Define S as the Rp subset where q-sections of A have positivemeasure i.e.:S = {x ∈ Rp | "q(Ax) > 0}Then the equality holds∫A|f(x, y)| dxdy =∫S(∫Ax|f(x, y)| dy)dx9/17 P!i?"##$$$!"#$#Fourier TransformFrom: The Fourier Transform and its Applications lecture notesof professor Brad Osgood Stanford UniversityLet f a real function of a real variable. The Fourier Transform (FT)of f is the complex valued function:Ff(s) :=∫ +∞−∞e−2pi istf(t)dt9/17 P!i?"##$$$!"#$#Fourier TransformFrom: The Fourier Transform and its Applications lecture notesof professor Brad Osgood Stanford UniversityLet f a real function of a real variable. The Fourier Transform (FT)of f is the complex valued function:Ff(s) :=∫ +∞−∞e−2pi istf(t)dtThere is another notation in usefˆ(s) =∫ +∞−∞e−2pi istf(t)dt10/17 P!i?"##$$$!"#$#A warning on definitions Our definition of the Fourier transformis a standard one, but it’s not the only one. The question is where toput the 2pi: in the exponential, as we have done; or perhaps as a factorout front; or perhaps left out completely. There’s also a question ofwhich is the Fourier transform and which is the inverse, i.e., whichgets the minus sign in the exponential. All of the various conventionsare in day-to-day use in the professions.11/17 P!i?"##$$$!"#$#Following the helpful summary provided by T. W. Ko¨rner in his bookFourier Analysis, I will summarize the many variations. To be general,let’s writeFf(s) = 1A∫ +∞−∞eiBstf(t)dt11/17 P!i?"##$$$!"#$#Following the helpful summary provided by T. W. Ko¨rner in his bookFourier Analysis, I will summarize the many variations. To be general,let’s writeFf(s) = 1A∫ +∞−∞eiBstf(t)dtThe choices that are found in practice areA =√2pi B = ±1A = 1 B = ±2piA = 1 B = ±111/17 P!i?"##$$$!"#$#Following the helpful summary provided by T. W. Ko¨rner in his bookFourier Analysis, I will summarize the many variations. To be general,let’s writeFf(s) = 1A∫ +∞−∞eiBstf(t)dtThe choices that are found in practice areA =√2pi B = ±1A = 1 B = ±2piA = 1 B = ±1The definition we’ve chosen has A = 1 and B = −2pi12/17 P!i?"##$$$!"#$#A sufficient condition for the existence of the integral is f ∈ L1(R).One value is easy to compute, and worth pointing out, namely fors = 0 we haveFf(0) :=∫ +∞−∞f(t)dt12/17 P!i?"##$$$!"#$#A sufficient condition for the existence of the integral is f ∈ L1(R).One value is easy to compute, and worth pointing out, namely fors = 0 we haveFf(0) :=∫ +∞−∞f(t)dtThe inverse Fourier transform is defined by:F−1g(t) :=∫ +∞−∞e2pi istg(s)ds12/17 P!i?"##$$$!"#$#A sufficient condition for the existence of the integral is f ∈ L1(R).One value is easy to compute, and worth pointing out, namely fors = 0 we haveFf(0) :=∫ +∞−∞f(t)dtThe inverse Fourier transform is defined by:F−1g(t) :=∫ +∞−∞e2pi istg(s)dsThe Fourier inversion theorem states thatF(F−1g) = g, F−1(Ff) = f13/17 P!i?"##$$$!"#$#Riemann Lebesgue Theorem If f ∈ L1(R) thenlim|s|→∞F(s) = 013/17 P!i?"##$$$!"#$#Riemann Lebesgue Theorem If f ∈ L1(R) thenlim|s|→∞F(s) = 0Plancherel Theorem If f ∈ L1(R) ∩ L2(R) then fˆ ∈ L2(R) and∫ ∞−∞|f(t)|2dt =∫ ∞−∞|fˆ(s)|2ds13/17 P!i?"##$$$!"#$#Riemann Lebesgue Theorem If f ∈ L1(R) thenlim|s|→∞F(s) = 0Plancherel Theorem If f ∈ L1(R) ∩ L2(R) then fˆ ∈ L2(R) and∫ ∞−∞|f(t)|2dt =∫ ∞−∞|fˆ(s)|2dsTheorem If f, g ∈ L1(R) then∫ ∞−∞fˆ(s)g(s)ds =∫ ∞−∞f(x)gˆ(x)dx14/17 P!i?"##$$$!"#$#Example 1: The triangle function Consider the “triangle func-tion”, defined by Λ(x) = max{1 − |x|, 0}. Notice that the explicitexpression of Λ is thenΛ(x) =1− |x| |x| ≤ 10 otherwise15/17 P!i?"##$$$!"#$#!1.5 !1.0 !0.5 0.5 1.00.20.40.60.81.0Figure 2: Graph of triangle function16/17 P!i?"##$$$!"#$#For the Fourier transform we compute (using the fact that sine func-tion is odd):FΛ(s) =∫ +∞−∞e−2pi istΛ(t)dt16/17 P!i?"##$$$!"#$#For the Fourier transform we compute (using the fact that sine func-tion is odd):FΛ(s) =∫ +∞−∞e−2pi istΛ(t)dt =∫ 1−1(cos(2pist)− i sin(2pist)) (1− |t|) dt16/17 P!i?"##$$$!"#$#For the Fourier transform we compute (using the fact that sine func-tion is odd):FΛ(s) =∫ +∞−∞e−2pi istΛ(t)dt =∫ 1−1(cos(2pist)− i sin(2pist)) (1− |t|) dtFΛ(s) =∫ 1−1cos(2pist) (1− |t|) dt16/17 P!i?"##$$$!"#$#For the Fourier transform we compute (using the fact that sine func-tion is odd):FΛ(s) =∫ +∞−∞e−2pi istΛ(t)dt =∫ 1−1(cos(2pist)− i sin(2pist)) (1− |t|) dtFΛ(s) =∫ 1−1cos(2pist) (1− |t|) dtIt is worth noting that this is a general fact: for each even functionf(t)(= f(−t)) the following relation holdsFf(s) =∫ +∞−∞cos(2pist)f(t)dt = 2∫ +∞0cos(2pist)f(t)dt17/17 P!i?"##$$$!"#$#Moreover if f(t)(= −f(−t)) is and odd function, we haveFf(s) = −i∫ +∞−∞sin(2pist)f(t)dt = −2i∫ +∞0sin(2pist)f(t)dt17/17 P!i?"##$$$!"#$#Moreover if f(t)(= −f(−t)) is and odd function, we haveFf(s) = −i∫ +∞−∞sin(2pist)f(t)dt = −2i∫ +∞0sin(2pist)f(t)dtLet us go back to the FT of the triangle function:FΛ(s) = 2∫ 10cos(2pist) (1− t) dt= 2(sin(2pis)2pis− 2pis sin(2pis) + cos(2pis)− 14pi2s2)since∫t cosAtdt =cos(At)A2+t sin(At)A17/17 P!i?"##$$$!"#$#Moreover if f(t)(= −f(−t)) is and odd function, we haveFf(s) = −i∫ +∞−∞sin(2pist)f(t)dt = −2i∫ +∞0sin(2pist)f(t)dtLet us go back to the FT of the triangle function:FΛ(s) = 2∫ 10cos(2pist) (1− t) dt= 2(sin(2pis)2pis− 2pis sin(2pis) + cos(2pis)− 14pi2s2)since∫t cosAtdt =cos(At)A2+t sin(At)ANow, simplifying we getFΛ(s) = 1− cos(2pis)2pi2s2Clamfim 2012/2013: Modelli 1 secondo parziale 1CognomeNomematricola1. 4 pt Trovare la soluzione y = y(x) del problema di Cauchy{y′ = − cosx√1− y2y(0) = 02. 4 pt Trovare la soluzione y = y(x) del problema di Cauchy{y′ = y + exy(0) = 03. 4 pt Trovare la soluzione y = y(x) del problema di Cauchyy′ =11 + xy + xy(0) = 04. 5 pt Trovare la soluzione y = y(x) del problema di Cauchyy′ = y +√x2 − y2xy(1) = 15. 5 pt Trovare la soluzione y = y(x) del problema di Cauchyy′ = −y2 +1xyy(1) = 26. 6 pt Sapendo che y(x) = x e` soluzione particolare dell’equazione differenzialey′(x) =12xy2(x)−(1x+ 1)y(x) +x+ 42(!)determinare la soluzione generale dell’equazione (!)7. 2 pt Data l’equazione differenziale per y = y(x)y′ =2x3y− 3x2(•)la si trasformi in una equazione per η = η(ξ) con il cambio di variabili ξ(x, y) = x, η(x, y) = xye si risolva il problema di Cauchy che all’equazione (•) abbina la condizione iniziale y(1) = 1/28. Data una equazione di Riccati in forma generaley′ = a(x) + b(x)y + c(x)y2 (R)in cui si suppone che a, b, c "= 0, se B(x) e` una primitiva di b(x) e cioe` B′(x) = b(x) si dimostriche il cambio di variabili u(x) = e−B(x)y trasforma (R) nell’equazione di Riccati ridotta:u′ = a(x)e−B(x) + c(x)eB(x)u2Applicare poi il cambio di variabili qui introdotto per risolvere il problema di Cauchy{y′ = xe−2x − y + xy2y(0) = 0L’esercizio 8 verra` valutato solo se siano stati risolti gli altri sette esercizi assegnati.30/11/20120000662032 Altini Luca 25,50000640225 Baravelli Emanuela 26,50000673676 Brighi Alberto 280000676637 Caligiuri Giorgia 280000661808 Carloni Carolina 280000675448 Ceccarelli Giacomo 29,50000672247 Ciabatti Francesco 300000599130 Colonnese Giusy 260000602324 Di Domenico Danila 250000661583 Di Girolamo Jonathan 290000633861 Donadel Riccardo 28Faggioli Alessia 290000645870 Finzi Marina 280000607428 Fontana Stefano 270000666534 Furlotti Luca 290000661663 Galato Gianluca 27,50000661924 Gentile Francesco 29,5Graziosi Marco 280000638471 Lalli Gianluca 260000676010 Lamuraglia Nicola 260000632288 Lazar Valeriu 260000665721 Lenzi Francesca 29,5Luisi Mattia 26,50000654365 Luna Costanza 160000662683 Manserra Valentina 28,50000676306 Marconi Alberto 29,50000669193 Mariani Marco 29,50000644195 Marinelli Mariangela 270000663017 Marzotto Mario 290000673783 Messina Dario 28,5Metti Cristian 270000661814 Minghetti Riccardo 27,50000677441 Nanni Matteo 24,50000668649 Paolucci Antonio 27,5Pentassuglia Gianni 25,50000667189 Quici Laura 30 +0000639500 Radovani Stivi 26Rettino Soledania 240000673238 Rinaldi Francesco 25,50000666344 Rinaldi Elena 30 ++0000664450 Salaroli Claudio 290000666734 Valenti Valentina 270900047959 Votekova Anastasia 28

Lezione 16, 4 dicembre 2012

http://campus.unibo.it/97723/1/L16%2820121204%29.pdf

Lezione 16, 4 dicembre 2012

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