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Power series and Laplace transform

  1. Jul 12, 2014 #1
    As it can be read here, http://en.wikipedia.org/wiki/Laplace_transform#Relation_to_power_series
    the Laplace transform is a continuous analog of a power series in which the discrete parameter n is replaced by the continuous parameter t, and x is replaced by exp(-s).

    Therefore, computing a discrete power series or a continuous laplace transform should converge to the same function, is it right?

    Let's apply it for the simplest case: a(x)=1

    • For the discrete power series it converges to 1/1-x (provided that -1<x<1)

    • For the continuous power series it converges to 1/s (provided that s>0)
    Now, this two should be equivalent right? If you substitute s=-ln(x) you get
    -1/ln(x), which is not the same as 1/1-x.

    What I am doing wrong?
     
  2. jcsd
  3. Jul 12, 2014 #2

    pasmith

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    Homework Helper

    "Replace the discrete parameter n by the continuous paramater t" is not that straightforward.

    Doing that literally would require that
    [tex]
    \int_0^\infty f(t)e^{-st}\,dt = \sum_{n=0}^\infty f(n)e^{-ns}
    [/tex] which, as your example shows, is in general false.
    What is true is that if you define [itex]f(t) = \sum_{n=0}^\infty a_n\delta(t - n)[/itex] where [itex]\delta[/itex] is the Dirac delta distribution then
    [tex]
    \int_{0^{-}}^\infty f(t)e^{-st}\,dt = \sum_{n=0}^\infty a_n e^{-ns}.
    [/tex] Observe that here [itex]f(t) = 0[/itex] for all non-integer [itex]t[/itex] and that [itex]f(t)[/itex] is not technically defined for integer [itex]t[/itex]; it is not the case that [itex]f(n) = a_n[/itex].
     
  4. Jul 12, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    In general, your statement "computing a discrete power series or a continuous laplace transform should converge to the same function" is incorrect. Interchanging two limit process does NOT always give the same result.
     
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