1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Convergence on [-M,M] for any M implies convergence on R?

  1. Oct 30, 2011 #1
    Good day dear fellows. I am given the following series

    [itex] h(x) = \sum_{n=1}^{\infty} \frac{1}{x^2+n^2}. [/itex]

    It is requested that I show that [itex]h(x)[/itex] is continuous on R. I did the following: use the Weirerstrass M-test to show uniform convergence, and then, using the continuity of the functions that are being summed, conclude that [itex] h(x) [/itex] is continuous. The solutions manual agrees with me. So far so good.

    Next I am asked whether [itex] h(x) [/itex] is differentiable, and, if [itex] h(x) [/itex] is differentiable, to investigate if the derivative [itex] h'(x) [/itex] is continuous. I note that that

    [itex] \frac{d}{dx} \frac{1}{x^2+n^2} = \frac{-2x}{(x^2+n^2)^2}.[/itex]

    At this point I don't see how I am going to get uniform convergence out of a summation over this thing, so I take out the solutions manual again. The manual uses the Weirerstrass M-test to show uniform convergence on [itex] [-M,M] [/itex] for arbitrary [itex]M[/itex], and concludes that there must be uniform convergence on R (because M is aribitrary). To be explicit: on any interval [itex] [-M,M] [/itex],

    [itex] |\frac{2x}{x^2+n^2} | \leq \frac{2M}{n^2} [/itex].

    (the square over the denominator can be removed since the denominator is greater than one).

    And, since

    [itex] \sum_{n=1}^{infty} \frac{2M}{n^2} [/itex]

    converges on [-M,M], uniform convergence of the derivatives follows via the weirerstrass M-test. Since [itex]M[/itex] is arbitrary, we then have uniform convergence (and thus continuity) on R.

    I don't see how this is so, since R is not a closed interval..
     
  2. jcsd
  3. Oct 30, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    On the contrary, R is a closed interval. It is just not a bounded interval.

    In any case, "convergence", in a set X, just means the series converges at every point of X. Let p be any point in R. Then there exist M such that -M< p< M. If the series converges on [-M, M] then it converges at p. Since p could be any point in R, the series converges for all R.

    "Uniform convergence" would be a different matter.
     
  4. Oct 30, 2011 #3
    I think I am confused by the notation for intervals. For any finite numbers [itex] a [/itex] and [itex]b [/itex], [itex] (a,b) [/itex] is an open interval, and every closed interval can be written [itex] [c,d], [/itex] where [itex]c[/itex] and [itex]d[/itex] are its endpoints. But this isn't true for unbounded intervals? In other words, [itex] (-\infty, \infty) [/itex] is a closed interval, since it contains all it's limit points (but not its endpoints)? Endpoints of unbounded intervals are not limit points?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Convergence on [-M,M] for any M implies convergence on R?
  1. Convergence of r^n (Replies: 3)

Loading...