This can not be so easy

  • Thread starter eljose79
  • Start date
  • #1
eljose79
214
1
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...
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
43,021
970
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
ottjes
24
0
think he forgot the i in the exp
but i don't know if you can add the W(y)
 

Suggested for: This can not be so easy

  • Last Post
Replies
7
Views
374
  • Last Post
Replies
5
Views
540
Replies
3
Views
291
Replies
3
Views
539
Replies
4
Views
394
Replies
15
Views
157
Replies
2
Views
492
Top