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Polynomial sequence uniformly convergent on annulus

  1. Mar 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Can we find a sequence, say p_j(z) such that p_j ---> 1/z uniformly for z is an element of an annulus between 1 and 2, that is 1 < abs(z) < 2?

    Then i am asked to do the same thing but for p_j ---> sin(1/z^2).


    2. Relevant equations
    Not too sure about this, maybe Taylor series/Laurent series expansions.


    3. The attempt at a solution
    So while I have no definite path yet set on proving this what I do have are a few thoughts. On this annulus 1/z is analytic because the point z = 0 is not contained. Also, we can write a Taylor/Laurent series expansion for 1/z.

    However, I do not believe that we are able to do the same thing for the sin sequence because we end up with a larger numerator term which blows up and cause the series to diverge. However, against z = 0 is not contained here so maybe that is false? Am I thinking about this correctly or am I a fool?
     
  2. jcsd
  3. Mar 10, 2009 #2
    The Laurent-series will converge uniformly in both cases, because [tex]\{z,1\le|z|\le 2\}[/tex] is compact. It will not be possible to find a taylor series that converges on the entire annulus, because the radius of convergence can only extend to the origin, where there is a singularity.
     
  4. Mar 10, 2009 #3
    so as long as the function you are tying to converge to is defined over the entire annulus any sequence which approaches that function (the Laurent series of that function) will converge uniformly? and this is relevant to my question because i am seeking a polynomial sequence which is just the Laurent series right?

    also: in an unrelated question; what do you mean the radius of converge can extend only to the origin?
     
  5. Mar 10, 2009 #4
    This is certainly true for the Laurent series, but not for all series I think.

    The terms of the Laurent series are polynomials in z, z^-1, so they are not "polynomials" in the usual sense, but rational functions.
    If you are looking for a series of polynomials (only in z), then the only candidate is the Taylor series, which can't converge on the entire annulus, see below.

    From http://en.wikipedia.org/wiki/Radius_of_convergence#Radii_of_convergence_in_complex_analysis":
    Where "the nearest point where f cannot be defined in a way that makes it holomorphic" is basically the nearest singularity. Both 1/z and sin(1/z^2) have one singularity at the origin.
     
    Last edited by a moderator: Apr 24, 2017
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