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Function can be represented by a Taylor series

  1. Oct 30, 2009 #1
    If a function can be represented by a Taylor series at x0, but only at this point, (radius of convergence = 0), is it considered analytic there?
     
  2. jcsd
  3. Oct 30, 2009 #2

    mathman

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    Re: Analyticity

    No. All you have is f(x0)=f(x0)
     
  4. Oct 30, 2009 #3
    Re: Analyticity

    Could you please elaborate on your response? Not sure I follow...
     
  5. Oct 31, 2009 #4

    Office_Shredder

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    Re: Analyticity

    Given any function (any function at all, seriously), the "Taylor Series" around x0 with 0 radius of convergence is

    f(x)=f(x0). This is pretty pointless
     
  6. Oct 31, 2009 #5

    HallsofIvy

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    Re: Analyticity

    In order to be "analytic" at a point, the Taylor's series for the function, around that point, must converge to the function in some neighborhood of the function.

    And it depends upon what you mean by "represented by the Taylor's series".

    The function
    [tex]f(x)= e^{-\frac{1}{x^2}[/tex]
    if [itex]x\ne 0[/itex], f(0)= 0, has all derivatives at 0 equal to 0 and so its Taylor's series, about x= 0, exists, has infinite radius of convergence, but is equal to f only at x= 0. That function is NOT "analytic".
     
  7. Oct 31, 2009 #6
    Re: Analyticity

    I have a question about analyticity: Suppose I want to show that a function [itex]f(z)[/itex] is analytic in some open subset [itex]\Omega[/itex] of the complex plane. Is it enough to show that [itex]f[/itex] has a power series representation that converges for every [itex]z[/itex] in [itex]\Omega[/itex]?
     
  8. Oct 31, 2009 #7

    mathman

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    Re: Analyticity

    Only when the convergence is to f(z) itself. As previously noted the power series for e-1/x2 is all 0, not the function itself.
     
  9. Nov 2, 2009 #8
    Re: Analyticity

    Nope. Analyticity is a neighborhood property.
     
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