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I Convergence of Taylor series in a point implies analyticity

  1. Dec 10, 2016 #1
    Suppose that the Taylor series of a function ##f: (a,b) \subset \mathbb{R} \to \mathbb{R}## (with ##f \in C^{\infty}##), centered in a point ##x_0 \in (a,b)## converges to ##f(x)## ##\forall x \in (x_0-r, x_0+r)## with ##r >0##. That is
    $$f(x)=\sum_{n \geq 0} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n \,\,\,\,\,\,\,\,\,\,\,\,\,\forall x \in (x_0-r,x_0+r) $$
    (In particular, I do not know in advance anything about the analyticity of ##f(x)## in ##(a,b)##, but only the convergence of the Taylor series in ##x_0##.)

    Does this imply that, chosen a ##x_1 \in (x_0-r,x_0+r)##, the Taylor series of ##f(x)## centered in ##x_1## converges at least in ##(x_1-r',x_1+r')## with ##r'=r-|x_0-x_1|##? That is
    $$f(x)=\sum_{n \geq 0} \frac{f^{(n)}(x_1)}{n!} (x-x_1)^n \,\,\,\,\,\,\,\,\,\,\,\,\,\forall x \in (x_1-r',x_1+r') $$
    If this is true, how can I prove it?
     
  2. jcsd
  3. Dec 11, 2016 #2

    stevendaryl

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    Fooling around with the power series, I see that the conclusion is easily provable if you additionally know the following:
    1. The series for [itex]f(x)[/itex] is absolutely convergent (so that you can reorder terms and get the same answer).
    2. The series for [itex]\frac{df}{dx}, \frac{d^2 f}{dx^2}, ...[/itex] are also absolutely convergent.
    As to point two--can you prove that if [itex]f(x)[/itex] is equal to its Taylor series at a point, then [itex]f^{(n)}(x)[/itex] is equal to its Taylor series at that point?

    As to point one--I'm not sure whether you can get away without assuming that, or not.
     
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