a question on fourier transforms by Akhila Raman

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akhila raman

Posted: Sat Mar 17, 2007 11:29 pm

Guest

i noticed F.T. has its share of curiosity:
http://mathworld.wolfram.com/FourierTransform1.html

let us do this without using duality property.
we want the Fourier transform of f(x)=1:

F(k)= [Lt X->inf] [int -X,X] exp(-i*2*pi*k*x) dx

= [Lt X->inf] sin(2*pi*k*X)/(pi*k)

if we wish to argue F(k)= delta(k), then by definition
of delta function, for k # 0, F(k)=0.
[ http://mathworld.wolfram.com/DeltaFunction.html ]

that means for k # 0,
[Lt X->inf] sin(2*pi*k*X) = 0 !

which says for X->inf, a sinusoid goes to zero, which
is not quite right by the definition of sinusoidal
function. it should remain oscillating between [-1,1] as
X-> infinity by definition...

-Akhila Raman

Gordon Sande

Posted: Sun Mar 18, 2007 12:27 pm

Guest

On 2007-03-18 01:29:42 -0300, "akhila raman" <akhila.raman@gmail.com> said:

Quote:

i noticed F.T. has its share of curiosity:
http://mathworld.wolfram.com/FourierTransform1.html

let us do this without using duality property.
we want the Fourier transform of f(x)=1:

The usual integral FT is L_2 to L_2. Your function is not
in L_2 so the results to not obtain.

You need Generalized Harmoninc Analysis if you want to
use your function. If you want to play the delta function
game then you will need to bone up on distribution theory.

Assumptions matter. Particularly when fussing about with
extreme cases.

Quote:

F(k)= [Lt X->inf] [int -X,X] exp(-i*2*pi*k*x) dx

= [Lt X->inf] sin(2*pi*k*X)/(pi*k)

if we wish to argue F(k)= delta(k), then by definition
of delta function, for k # 0, F(k)=0.
[ http://mathworld.wolfram.com/DeltaFunction.html ]

that means for k # 0,
[Lt X->inf] sin(2*pi*k*X) = 0 !

which says for X->inf, a sinusoid goes to zero, which
is not quite right by the definition of sinusoidal
function. it should remain oscillating between [-1,1] as
X-> infinity by definition...

-Akhila Raman

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