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Continuous/analytic functions

  1. May 8, 2011 #1
    if a function is continuous, does this mean that it is analytic. And if a function is analytic does this mean it is continuous?

  2. jcsd
  3. May 8, 2011 #2


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    No. For example, f(x) = |x| is not differentiable at x = 0. However, it is continuous.

    Yes. Every analytic function has the property of being infinitely differentiable. Since the derivative is defined and continuous, the function is continuous everywhere.
    Last edited: May 8, 2011
  4. May 8, 2011 #3
    exellent -cheers :)
  5. May 8, 2011 #4
    An analytic function is a function that can can be represented as a power series polynomial (either real or complex).

    That is it posesseses a Taylor/Mclaurin expansion.
  6. May 8, 2011 #5

    Gib Z

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    Not exactly true. A function is analytic at a point [itex] z_0 [/itex] if it is smooth (infinitely differentiable) there, and it's Taylor Series centered at [itex] z_0 [/itex] converges to the function on some open set containing [itex] z_0 [/itex].

    Merely being smooth is not enough - For example
    [tex] f(x)=\begin{cases}\exp(-1/x) \mbox{ if } x> 0 \\ 0 \mbox{ if }x\le0,\end{cases} [/tex]

    This function is smooth at 0, with all its derivatives there being 0. Thus, it has a Taylor Series expansion at x=0, [tex] \sum_{n=0}^{\infty} \frac{0}{n!} x^n = 0 [/tex], but that does not coincide with the value of the function for any positive x, so f(x) is smooth (and has a Taylor Expansion), but is not analytic.
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