AMath350.F12.A9.Sol

# form yofxthee with f g the x thus the fourier

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Unformatted text preview: 2 x dx = u f).! ) form yofx,the}e= with f= )g (the .x Thus the Fourier Transform ﬁnd (ˆ((!exists.. 5.31 ( however the respectyto Fourier= g any value(of)y , x exists. g )y,)ySince of xe and in class). ˆ( lim exists, 1 i! second integral diverges (this wasZshown of f x Z Z s i! Z Z 01 Assuming that g is diﬀerentiable, show that ZF {uy (x, y )} = uyx ! , y ). (1 i )x 0 ˆ( x 6=(x) +i g (y ))e i x dx x= iZ x dx(x thei left diverges dx = i value (b) g (y )(f (x)ethisx means theeintegralfon )e lim dx0+(1 (iy )x for anylim . eof y and = f 0, e dx Z (c) dx x = e = i x eg ) s s s 1 i( 2. (a) The Fourier y )e i on the right y ) does )e since the Fourier Transform of f!x)s f (x g ( Tranform of u(x, f (x not dx. the ﬁrst)integral dx = g (y ) converges exist. s Zi s 1 ee 1 Since the Fourier Transform of f (xthe i Fourier was shown in class). Since the a lim = a integral =)eexistsusing integral on the right conThus 3. (a) exists, however the second)e i. x dx diverges, (this Transform of f (xproperty. F{ (x a)} = (x the substitution ) exists. s 1 the integralintegral on the left divergesany value of y , and we imeans the on the left converges for for any value of y and ! 1 i! verges, thus g (y ) 6= 0, this Z Z ˆˆ ˆ (b) Let F{f (x)Tranform(of).Using theˆy (! , y )xof Zpart(a) ). Further, have u(! , y )}== (f (! ). u Clearly u result = g (y )f (! and the Convolution ˆ g y )f ! x, i 2. (a) the Fourier y )e Z x dx = g ((y ) y )i does )not exist. f ( x) g ( f (x e i dx. x Z ˆ Theorem ywe = F{uy (x, )} have f (x)g (y )i x dx = i a y ) e g( f (x)e i x dx = g (y )f (! ). Sincexthe)} = Fourier TransformZ ofdx(=)eexistsusing integral on the right confx the Z the Z 3. (a) F{ ( a ( x a) e , substitution property. Z 1 ˆ x {F{f i of y F 1 { f (x) +i the)) verges, thus t y integral on (the f (x (e i tx }} = g ( any ez ) xz t) and f ( (b)Let u(f (! )e= g (}x=eFig (ydxwithxg (yleft )converges fory )f (xvalue(show, dz = wex t). = )}F{ x ) dx + dx. ˆ( ˆ b) Let F{f (x)}f = )f+! ). ) Using the 6= 0. Use...
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