Is Solving an Integral Equation as Easy as a Fourier Transform?

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In summary, the conversation discusses solving an integral equation by converting it to a Fourier transform. The equation is g(x)=Int(-1,1)exp(-xy)f(y)dy and can be written as g(x)=Int(-infinite,infinite)exp(-xy)f(y)W(y)dy, where W(y) is defined as 1 if -1<y<1 and 0 elsewhere. The person in the conversation expresses surprise at how easy it seems to solve the integral equation. However, there is confusion about whether the Fourier transform requires an additional term of exp(-ixy).
  • #1
eljose79
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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...
 
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  • #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?
 
  • #3
think he forgot the i in the exp
but i don't know if you can add the W(y)
 

Related to Is Solving an Integral Equation as Easy as a Fourier Transform?

1. What does "This can not be so easy" mean?

"This can not be so easy" is a phrase often used to express disbelief or skepticism about something that appears to be simple or effortless.

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5. How can I determine if something is truly easy or not?

It is important to approach tasks and concepts with an open mind and to not make assumptions about their level of difficulty. Instead, gather information, ask questions, and actively engage with the subject to determine its level of ease or difficulty for yourself.

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