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An example regarding asymptotic expansion

  1. Mar 19, 2007 #1
    Hi all.
    This is not strictly a question on differential equation but I think the topic asymptotic expansion is too closely related to D.E. and I think those experts in D.E. would help with my simple question here.

    Please refer to the attached figure. I would like to know how one could come up with the asymptotic relations (1.89) to (1.91) for the function?
    Is that we use the Taylor series to expand the function in power of epsilon with ingoring the exp(-x/epsilon) term completely?

    And why an exp(-x/epsilon) term suddenly appear in relation (1.91)? This term is not come from any kind of expansion but just that the author wants to tell that exp(-x/epsilon) term is too small and it can be added anywhere without loss of the asymptoticity?

    And when the domain becomes 0<=x<=2, ignoring exp(-x/epsilon) is no longer valid?

    Are these the author want to tell?
     

    Attached Files:

  2. jcsd
  3. Mar 30, 2007 #2
    First of all, asymptotic expansions aren't unique. They are defined in terms of functions, i.e., you can expand a function in terms of monomials [itex]x^\alpha[/itex] or in powers of logarithms [itex]\log^\alpha x[/itex]. The therms of the expansion are defined as limits. For further reference check Holme's book on perturbation theory or the classical book of Kevorkian & Cole.

    Taking this in mind, what happens if you expand [itex]e^{-x/\epsilon}[/itex] to second order and then do a Taylor expansion? of what order is the residue?

    What happens if you expand to third order? order of the residue?

    What happens... n order?

    That should answer your first question. Now, the assumption [itex]x=O(1)[/itex] means that [itex]x[/itex] and [itex]\epsilon[/itex] are not of the same order, so the products [itex]\epsilon^\alpha x^\alpha=O(\epsilon^\alpha)O(1)=O(\epsilon^\alpha)[/itex]. This supposition is valid as long as [itex]x[/itex] is not near zero, because if it is, then [itex]x[/itex] is of order [itex]O(\epsilon)[/itex] (or some function of [itex]\epsilon[/itex] that tends to zero as [itex]\epsilon[/itex] does), and your approximation is no longer valid. Thats why you must take the new variable [itex]X=x/\epsilon[/itex], wich is of order [itex]O(1)[/itex] near zero (or one accordingly).

    What can we conclude? That the approximation is not the same near zero than away form it. Near zero, the approx is [itex]f(x/\epsilon)[/itex], while away from it is [itex]f(x)[/itex].

    Question: How do i paste these two smoothly? For the answer check the references above.
     
    Last edited: Mar 30, 2007
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