Is Analyticity Equivalent to Continuity?

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Discussion Overview

The discussion centers on the relationship between continuity and analyticity of functions, exploring whether continuity implies analyticity and vice versa. It includes theoretical considerations and examples to illustrate the points made.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants question whether continuity of a function implies that it is analytic, with one example provided: the function f(x) = |x|, which is continuous but not differentiable at x = 0.
  • Others argue that every analytic function is continuous, as analytic functions are infinitely differentiable, and thus their derivatives are defined and continuous.
  • One participant defines an analytic function as one that can be represented by a power series polynomial, mentioning the existence of a Taylor/Maclaurin expansion.
  • A later reply challenges the definition of analyticity, stating that a function is analytic at a point if it is smooth and its Taylor series converges to the function in some neighborhood of that point. An example is given of a function that is smooth but not analytic, illustrating that smoothness alone does not guarantee analyticity.

Areas of Agreement / Disagreement

Participants express differing views on the implications of continuity and analyticity, with no consensus reached on whether continuity implies analyticity or the conditions under which a function is considered analytic.

Contextual Notes

Limitations include the dependence on definitions of continuity and analyticity, as well as the specific conditions under which functions are evaluated for differentiability and convergence of Taylor series.

lavster
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if a function is continuous, does this mean that it is analytic. And if a function is analytic does this mean it is continuous?

thanks
 
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lavster said:
if a function is continuous, does this mean that it is analytic.

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

lavster said:
And if a function is analytic does this mean it is continuous?
thanks

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:
exellent -cheers :)
 
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.
 
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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