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Continuous/analytic functions

by lavster
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lavster
#1
May8-11, 11:56 AM
P: 214
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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gb7nash
#2
May8-11, 12:10 PM
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Quote Quote by lavster View Post
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.

Quote Quote by lavster View Post
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.
lavster
#3
May8-11, 02:07 PM
P: 214
exellent -cheers :)

Studiot
#4
May8-11, 02:44 PM
P: 5,462
Continuous/analytic functions

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.
Gib Z
#5
May8-11, 08:44 PM
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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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