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Analytic Function |
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| Dec9-07, 11:20 PM | #1 |
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Analytic Function
A theorem states:
A continuously differentiable function f(z) is analytic iff the differential f(z)dz is closed. Isn't continuously differentiable the DEFINITION of analytic? |
| Dec9-07, 11:44 PM | #2 |
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Recognitions:
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An alternative definition appears to be "an analytic function is a function that is locally given by a convergent power series."
See http://en.wikipedia.org/wiki/Analytic_function |
| Dec9-07, 11:51 PM | #3 |
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But nevertheless it must be equivalent to "continuously differentiable". This theorem seems to say that "an analytic function is analytic iff the differential fdz is closed".
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| Dec9-07, 11:53 PM | #4 |
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Recognitions:
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Analytic Function
Different texts my have different definitions of the same object or property. A definition in one text can be a result in another.
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| Dec9-07, 11:58 PM | #5 |
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The definition given in my text is "a function f(z) is analytic on the open set U if f(z) is complex differentiable at each point of U and the complex derivative f'(z) is continuous on U."
A while later came the theorem "a continuously differentiable function f(z) on a domain D is analytic iff the differential f(z)dz is closed." I see circularity here. (We're talking about complex functions here) |
| Dec10-07, 03:27 AM | #6 |
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What does the proof say?
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| Dec11-07, 12:25 PM | #7 |
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"Left as exercise."
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| Dec13-07, 04:22 PM | #8 |
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"Continuously differentiable" is not the same as "complex differentiable," as far as I know. Make sure you've checked how the book defines those terms, and what sort of function you're looking at.
If a function is complex analytic, then it is smooth (ie ANY derivative is continuous, not just the first one). Generally "continuously differentiable" only means the first derivative is continuous (but maybe your book is different). If a function is real-analytic, the usual definition is that the Taylor series at any point converges to the function in a neighborhood of that point, or something like that. |
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