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Asymptotic analysis question

  1. May 3, 2008 #1
    let be [tex] f(x) \sim g(x) [/tex] , in the sense that for big x f(x) is asymptotic to g(x) , my question if what happens to their Laplace transform ??

    i belive that [tex] \int _{0}^{\infty}dt f(t)exp(-st) \approx \int _{0}^{\infty}dt g(t)exp(-st) [/tex]

    in first approximation the Laplace transform of f(x) and the Laplace transform of g(x) must be equal.

    another question if we had a Linear operator L so we can define its inverse L^{-1} is it true that [tex] f(x) \sim L(g(x)) \rightarrow L^{-1} f(x)= g(x) [/tex]
  2. jcsd
  3. May 3, 2008 #2


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    The answer to your first question is "what do you mean by [itex]\approx[/itex]?". The difference between f and g may be extremely large for small x: and that difference will show up in the integral.

    As for your second question- isn't that the definition of "inverse"?
  4. May 4, 2008 #3
    Yes Hallsoftivy L^{-1} means the inverse operator, but is it valid even for asymptotic relations ?? , and as for the Laplace integral if f(x) is asymptotic to g(x) and the Laplace transform of f(x) and g(x) are denoted by F(s) G(s) then would be true that [tex] F(s) \sim G(s) [/tex] ?, but i believe that having a kernel inside the integral transform exp(-st) for big values of 's' the remainder of the asymptotic relation could be make smoother
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