|Apr4-03, 09:24 AM||#1|
this can not be so easy
suppose we want to solve the integral equation
g(x)=Int(-1,1)exp(-xy)f(y)dy (2) then this is equal to
g(x)=Int(-infinite,infinite)exp(-xy)f(y)W(y)dy where the W(y) function is defined like that:
W(y) is 1 iif -1<y<1 and 0 elsewhere.
then we would have that the equation (2) is a fourier transform and inverting we can get f(x)W(x).
Is all that?..i can not believe that solving an integral equation could be so easy...
|Apr4-03, 10:23 AM||#2|
I'm afraid I don't quite understand your question.
You refer to g(x)=Int(-1,1)exp(-xy)f(y)dy as an "integral equation" so I presume you mean that g is given and you want to find f.
How does converting to the integral from -inf to inf make this a Fourier transform? Doesn't the Fourier transform have an exp(-ixy) in it?
|Apr5-03, 06:18 AM||#3|
think he forgot the i in the exp
but i don't know if you can add the W(y)
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