Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Delta sequences

  1. Dec 7, 2005 #1
    As a problem I was asked to show that phi, as defined by:
    [tex]\phi_n(t) = \frac{n}{\pi(1+n^2t^2)}[/tex]
    Satisfies the property that for any f with the property to continuious at 0, then:
    [tex]\lim_{n\rightarrow\infty} \int_{-\infty}^{\infty} \phi_n(t)f(t)dt = f(0)[/tex]
    But if we let f be 1/phi, we see that it is continuous, but f(0) = 0 and the above integral is infinity.
    Is this a valid counterexample?
     
  2. jcsd
  3. Dec 7, 2005 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    What is phi? I know what phi_n is, but not phi. (so, no it isn't a counter example, and for 1/phi(0) to be 0 phi would have to be infinity at 0)
     
    Last edited: Dec 7, 2005
  4. Dec 7, 2005 #3
    Well, can I let f = 1/phi_n? Then clearly:
    [tex] f(0) = \frac{\pi(1+n^2 (0)^2)}{n} = 0 [/tex]
    What is wrong with that?
     
  5. Dec 7, 2005 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If you can, it certainly cannot be the same n used by the limit.
     
  6. Dec 7, 2005 #5
    I do not see why that is the case.
     
  7. Dec 7, 2005 #6

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [tex]\lim_{n\rightarrow\infty} \int_{-\infty}^{\infty} \phi_n(t)f(t)dt = f(0)[/tex]
    The n attached to the limit simple only exists within the scope of the limit. It has absolutely no relation to any other n's that might appear elsewhere.


    In fact, many dialects of the language of mathematics expressly forbid making a substitution where the substituted term uses a symbol that is introduced by the context.

    In short, the symbol [itex]\lim_{n \rightarrow \infty}[/itex] introduces the symbol n, therefore such dialects expressly forbid you to make any substitutions inside the limit that contain the symbol n. (Such as your attempt at substituting [itex]f = 1 / \phi_n[/itex])


    For a different, intuitive reason, in the above limit, f is a function constant. It refers to precisely one (unspecified) function of one variable. Not many functions of one variable, and not one function of two variables.

    When you make the naive substitution [itex]f = 1 / \phi_n[/itex], you've replaced f with something that is not a function constant -- the function you're using changes as n changes, which conflicts with the original syntax that specifies that you're supposed to be using the same function f for all n.


    In fact, I'm pretty sure that [itex]f := 1 / \phi_n[/itex] is an ill-formed definition -- the symbol n has no meaning in this context, so it doesn't make sense to define anything in terms of n.
     
    Last edited: Dec 7, 2005
  8. Dec 7, 2005 #7
    Thank you Hurkyl.
     
  9. Dec 8, 2005 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    My view would simply be that in the statement of the result, we pick f and fix it, then we write
    [tex] \int phi_n f dx[/tex]

    as a sequence, this sequence tends to f(0)

    now you want to pick a different f for each term in the sequence. you simply can't do that, it is contradicting the hypotheses of the statement, as well as the other deeper philosophical implications of hurkyl's post.
     
  10. Dec 8, 2005 #9
    Thank you as well Matt.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Delta sequences
  1. Delta function (Replies: 1)

  2. Delta function (Replies: 3)

Loading...