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Validating a Delta Sequence

  1. Oct 6, 2012 #1
    The problem statement, all variables and given/known data
    Show that [tex] \delta_n(x) = ne^{-nx} \quad \mathrm{for}\quad x>0[/tex][tex] \qquad = 0 \quad \mathrm{for}\quad x<0[/tex]

    satisfies [tex]\lim_{n\longrightarrow\infty}\int_{-\infty}^\infty \delta_n(x)f(x)\mathrm{d}x = f(0)[/tex]

    The attempt at a solution
    The hint says to replace the upper limit ([itex]\infty[/itex]) with [itex]c/n[/itex], where [itex]c[/itex] is "large but finite", and then use the mean value theorem of integral calculus.

    I do not understand how this replacement in the hint is allowable. Since [itex]n\longrightarrow\infty[/itex], [itex]c/n\longrightarrow0[/itex], not [itex]\infty[/itex]. Even if this is okay, how does it aid using the mean value theorem of integral calculus?
     
  2. jcsd
  3. Oct 6, 2012 #2
    Are there any restrictions on f(x)?
     
  4. Oct 7, 2012 #3
    It is only stated that [itex]f(x)[/itex] is a well-behaved function. It seems that this would be easier to prove if [itex]\lim_{x\longrightarrow\pm\infty}f(x)[/itex] was at least finite, but no such assumption is made.
     
  5. Oct 7, 2012 #4
    Then assume the integral of f(x) over the entire real line is finite. Given that, you should be estimate the "left-over" as you restrict integration to c/n.
     
  6. Oct 7, 2012 #5
    Why can this be assumed? Not even a function as well-behaved as [itex]f(x)=x^2[/itex] satisfies this.
     
  7. Oct 7, 2012 #6
    ## e^{x^{100}} ## is also well behaved, but will it be acceptable in this case?
     
  8. Oct 7, 2012 #7
    Sure, if [itex]f(x) = e^{x^{100}}[/itex], then I do not know how I would answer the problem. Any suggestions?
     
  9. Oct 7, 2012 #8
    As I said, you should select a class of functions that are admissible. Those whose integral over the entire real line is finite obviously fit the bill. Those the product of which with exp(-x) has the same property would work, too.
     
  10. Oct 8, 2012 #9
    Evidently, however, the proof can be made for any well-behaved function [itex]f(x)[/itex]. I understand how to prove this for special cases, but I am confused on how to prove this in general.
     
  11. Oct 8, 2012 #10
    ## \delta_1 = e^{-x} ## so you must have at least ## \int_0^{\infty} f(x)e^{-x} dx < \infty## for any "well-behaved" function.
     
  12. Oct 20, 2012 #11
    If there exists a Taylor series expansion of f(x) of the form [tex]f(x)=\sum_{k=0}^\infty f^{(k)}(0)\frac{x^k}{k!}[/tex]

    does that imply that ## \int_0^{\infty} f(x)e^{-x} dx < \infty##?

    Thanks for your help.
     
  13. Oct 20, 2012 #12

    HallsofIvy

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    You are asked to show this for n a fixed constant. There is nothing said about "n going to infinity" and no reason to worry about that.
     
  14. Oct 20, 2012 #13
    Hi HallsofIvy. I confess I do not understand what you mean. Certainly the following says that n goes to infinity, no?

    [tex]\lim_{n\longrightarrow\infty}\int_{-\infty}^\infty \delta_n(x)f(x)\mathrm{d}x[/tex]
     
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