Clamfim Lesson 16 Martedì 4 Dicembre 2012 Beta Function. Fourier Transform

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1 CLAMFIM Clamfm Lesson 16 Martedì 4 Dcembre 212 Beta Functon. Fourer Transform professor Danele Rtell 1/17?

2 Euler Beta Theorem If x, y > : Γ(x) Γ(y) Γ(x + y) = 1 s x 1 (1 s) y 1 ds 2/17?

3 Euler Beta Theorem If x, y > : Γ(x) Γ(y) Γ(x + y) = 1 s x 1 (1 s) y 1 ds Ths dentty can be reformulated ntroducng the Euler Beta functon: n such a way B(x, y) = 1 s x 1 (1 s) y 1 ds Γ(x) Γ(y) B(x, y) = Γ(x + y) 2/17?

4 roof We start from Gamma s defnton Γ(x) = varable puttng t = u 2 so that Γ(x) = 2 u 2x 1 e u2 du t x 1 e t dt then change 3/17?

5 roof We start from Gamma s defnton Γ(x) = varable puttng t = u 2 so that Γ(x) = 2 Smlarly Γ(y) = 2 u 2x 1 e u2 du v 2y 1 e v2 dv t x 1 e t dt then change 3/17?

6 roof We start from Gamma s defnton Γ(x) = varable puttng t = u 2 so that Γ(x) = 2 Smlarly Γ(y) = 2 Now use Fubn s Theorem Γ(x)Γ(y) =4 [,+ ) [,+ ) u 2x 1 e u2 du v 2y 1 e v2 dv t x 1 e t dt then change u 2x 1 v 2y 1 e (u2 +v 2) dudv 3/17?

7 Change to polar coordnate u = ρ cos ϑ v = ρ sn ϑ obtanng 4/17?

8 Change to polar coordnate u = ρ cos ϑ v = ρ sn ϑ obtanng Γ(x)Γ(y) = 4 ( ) ( π/2 ρ 2x+2y 1 e ρ2 dρ cos 2x 1 ϑ sn 2y 1 ϑ dϑ ) 4/17?

9 Change to polar coordnate u = ρ cos ϑ v = ρ sn ϑ obtanng Γ(x)Γ(y) = 4 ( = Γ(x + y) ) ( π/2 ρ 2x+2y 1 e ρ2 dρ π/2 2 cos 2x 1 ϑ sn 2y 1 ϑ dϑ cos 2x 1 ϑ sn 2y 1 ϑ dϑ ) 4/17?

10 To end the proof we have to show that B(x, y) = π/2 2 cos 2x 1 ϑ sn 2y 1 ϑ dϑ 5/17?

11 To end the proof we have to show that B(x, y) = π/2 But comng back to Beta s defnton 2 cos 2x 1 ϑ sn 2y 1 ϑ dϑ B(x, y) = 1 we are done puttng s = cos 2 ϑ s x 1 (1 s) y 1 ds 5/17?

12 Show, usng Gamma and Beta functons, that the Lebesgue measure of A = {(x, y) R 2 x 2/3 + y 2/3 1} s 3 8 π 6/17?

13 Show, usng Gamma and Beta functons, that the Lebesgue measure of A = {(x, y) R 2 x 2/3 + y 2/3 1} s 3 8 π Fgure 1: Asterode 6/17?

14 m(a) =4 1 ( 1 x 2/3 ) 3/2 dx 7/17?

15 m(a) =4 1 ) 3/2 1 (1 x 2/3 dx= 6 u 1/2 (1 u) 3/2 du 7/17?

16 m(a) =4 1 ) 3/2 1 (1 x 2/3 dx= 6 u 1/2 (1 u) 3/2 du= 6B(5/2, 3/2) 7/17?

17 Tonell s Theorem Let A R m measurable and f : A R measurable. Defne S as the R p subset where q-sectons of A have postve measure.e.: S = {x R p l q (A x ) > } Then the equalty holds f(x, y) dxdy = A ( ) f(x, y) dy dx S A x 8/17?

18 Fourer Transform From: The Fourer Transform and ts Applcatons lecture notes of professor Brad Osgood Stanford Unversty Let f a real functon of a real varable. The Fourer Transform (FT) of f s the complex valued functon: Ff(s) := e 2π st f(t)dt 9/17?

19 Fourer Transform From: The Fourer Transform and ts Applcatons lecture notes of professor Brad Osgood Stanford Unversty Let f a real functon of a real varable. The Fourer Transform (FT) of f s the complex valued functon: Ff(s) := There s another notaton n use ˆf(s) = e 2π st f(t)dt e 2π st f(t)dt 9/17?

20 A warnng on defntons Our defnton of the Fourer transform s a standard one, but t s not the only one. The queston s where to put the 2π: n the exponental, as we have done; or perhaps as a factor out front; or perhaps left out completely. There s also a queston of whch s the Fourer transform and whch s the nverse,.e., whch gets the mnus sgn n the exponental. All of the varous conventons are n day-to-day use n the professons. 1/17?

21 Followng the helpful summary provded by T. W. Körner n hs book Fourer Analyss, I wll summarze the many varatons. To be general, let s wrte Ff(s) = 1 A e Bst f(t)dt 11/17?

22 Followng the helpful summary provded by T. W. Körner n hs book Fourer Analyss, I wll summarze the many varatons. To be general, let s wrte Ff(s) = 1 A e Bst f(t)dt The choces that are found n practce are A = 2π B = ±1 A =1 B = ±2π A =1 B = ±1 11/17?

23 Followng the helpful summary provded by T. W. Körner n hs book Fourer Analyss, I wll summarze the many varatons. To be general, let s wrte Ff(s) = 1 A e Bst f(t)dt The choces that are found n practce are A = 2π B = ±1 A =1 B = ±2π A =1 B = ±1 The defnton we ve chosen has A = 1 and B = 2π 11/17?

24 A suffcent condton for the exstence of the ntegral s f L 1 (R). One value s easy to compute, and worth pontng out, namely for s = we have Ff() := f(t)dt 12/17?

25 A suffcent condton for the exstence of the ntegral s f L 1 (R). One value s easy to compute, and worth pontng out, namely for s = we have Ff() := f(t)dt The nverse Fourer transform s defned by: F 1 g(t) := e 2π st g(s)ds 12/17?

26 A suffcent condton for the exstence of the ntegral s f L 1 (R). One value s easy to compute, and worth pontng out, namely for s = we have Ff() := f(t)dt The nverse Fourer transform s defned by: F 1 g(t) := e 2π st g(s)ds The Fourer nverson theorem states that F(F 1 g)=g, F 1 (Ff) =f 12/17?

27 Remann Lebesgue Theorem If f L 1 (R) then lm F(s) = s 13/17?

28 Remann Lebesgue Theorem If f L 1 (R) then lm F(s) = s lancherel Theorem If f L 1 (R) L 2 (R) then ˆf L 2 (R) and f(t) 2 dt = ˆf(s) 2 ds 13/17?

29 Remann Lebesgue Theorem If f L 1 (R) then lm F(s) = s lancherel Theorem If f L 1 (R) L 2 (R) then ˆf L 2 (R) and f(t) 2 dt = Theorem If f, g L 1 (R) then ˆf(s)g(s)ds = ˆf(s) 2 ds f(x)ĝ(x)dx 13/17?

30 Example 1: The trangle functon Consder the trangle functon, defned by Λ(x) = max{1 x, }. Notce that the explct expresson of Λ s then Λ(x) = 1 x x 1 otherwse 14/17?

31 Fgure 2: Graph of trangle functon 15/17?

32 For the Fourer transform we compute (usng the fact that sne functon s odd): FΛ(s) = e 2π st Λ(t)dt 16/17?

33 For the Fourer transform we compute (usng the fact that sne functon s odd): FΛ(s) = e 2π st Λ(t)dt = 1 1 (cos(2πst) sn(2πst)) (1 t )dt 16/17?

34 For the Fourer transform we compute (usng the fact that sne functon s odd): FΛ(s) = e 2π st Λ(t)dt = 1 1 (cos(2πst) sn(2πst)) (1 t )dt FΛ(s) = 1 1 cos(2πst) (1 t ) dt 16/17?

35 For the Fourer transform we compute (usng the fact that sne functon s odd): FΛ(s) = e 2π st Λ(t)dt = 1 1 (cos(2πst) sn(2πst)) (1 t )dt FΛ(s) = 1 1 cos(2πst) (1 t ) dt It s worth notng that ths s a general fact: for each even functon f(t)(= f( t)) the followng relaton holds Ff(s) = cos(2πst)f(t)dt =2 cos(2πst)f(t)dt 16/17?

36 Moreover f f(t)(= f( t)) s and odd functon, we have Ff(s) = sn(2πst)f(t)dt = 2 sn(2πst)f(t)dt 17/17?

37 Moreover f f(t)(= f( t)) s and odd functon, we have Ff(s) = sn(2πst)f(t)dt = 2 Let us go back to the FT of the trangle functon: sn(2πst)f(t)dt snce 1 FΛ(s) =2 cos(2πst) (1 t)dt ( ) sn(2πs) 2πs sn(2πs) + cos(2πs) 1 =2 2πs 4π 2 s 2 t cos Atdt = cos(at) A 2 + t sn(at) A 17/17?

38 Moreover f f(t)(= f( t)) s and odd functon, we have Ff(s) = sn(2πst)f(t)dt = 2 Let us go back to the FT of the trangle functon: snce 1 sn(2πst)f(t)dt FΛ(s) =2 cos(2πst) (1 t)dt ( ) sn(2πs) 2πs sn(2πs) + cos(2πs) 1 =2 2πs 4π 2 s 2 t cos Atdt = cos(at) A 2 + t sn(at) A FΛ(s) = 1 cos(2πs) 2π 2 s 2 Now, smplfyng we get 17/17?

39 Clamfm 212/213: Modell 1 secondo parzale 1 Cognome Nome matrcola { y = cos x 1 y 1. 4 pt Trovare la soluzone y = y(x) del problema d Cauchy 2 y() = { y = y + e x 2. 4 pt Trovare la soluzone y = y(x) del problema d Cauchy y() = y = pt Trovare la soluzone y = y(x) del problema d Cauchy 1+x y + x y() = y 4. 5 pt Trovare la soluzone y = y(x) del problema d Cauchy = y + x 2 y 2 x y(1) = 1 y = y pt Trovare la soluzone y = y(x) del problema d Cauchy x y y(1) = pt Sapendo che y(x) = x è soluzone partcolare dell equazone dfferenzale y (x) = 1 ( ) 1 2x y2 (x) x +1 y(x)+ x +4 2 ( ) determnare la soluzone generale dell equazone ( ) 7. 2 pt Data l equazone dfferenzale per y = y(x) y = 2 x 3 y 3 x 2 ( ) la s trasform n una equazone per η = η(ξ) con l cambo d varabl ξ(x, y) =x, η(x, y) =xy e s rsolva l problema d Cauchy che all equazone ( ) abbna la condzone nzale y(1) = 1/2 8. Data una equazone d Rccat n forma generale y = a(x)+b(x)y + c(x)y 2 (R) n cu s suppone che a, b, c, se B(x) è una prmtva d b(x) e coè B (x) =b(x) s dmostr che l cambo d varabl u(x) =e B(x) y trasforma (R) nell equazone d Rccat rdotta: u = a(x)e B(x) + c(x)e B(x) u 2 Applcare po l cambo d varabl qu ntrodotto per rsolvere l problema d Cauchy { y = xe 2x y + xy 2 y() = L eserczo 8 verrà valutato solo se sano stat rsolt gl altr sette esercz assegnat. 3/11/212

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