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continuous/analytic functions |
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| May8-11, 11:56 AM | #1 |
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continuous/analytic functions
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 |
| May8-11, 12:10 PM | #2 |
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| May8-11, 02:07 PM | #3 |
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exellent -cheers :)
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| May8-11, 02:44 PM | #4 |
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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. |
| May8-11, 08:44 PM | #5 |
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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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