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Analytic function definiton

by JamesGoh
Tags: analytic, definiton, function
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JamesGoh
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Aug24-11, 12:13 AM
P: 140
From my lecture notes I was given, the definiton of an analytic function was as follows:

A function f is analytic at xo if there exists a radius of convergence bigger than 0 such that f has a power series representation in x-xo which converges absolutely for [x-xo]<R

What I undestand is that for all x values, |x-xo| must be less than R (radius of convergence) in order for f to be analytic at xo.

Convergence in a general sense is when the sequence of partial sums in a series approaches a limit

Is my understanding of convergence and analytic functions correct ?
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Fredrik
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Aug24-11, 02:31 AM
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Quote Quote by JamesGoh View Post
What I undestand is that for all x values, |x-xo| must be less than R (radius of convergence) in order for f to be analytic at xo.
What you're saying here would imply that the truth value ("true" or "false") of the statement "f is analytic at x0" depends on the value of some variable x. It certainly doesn't. It depends only on f and x0. (What you said is actually that if |x-x0|≥R, then f is not analytic at x0).

I'm a bit surprised that your definition says "converges absolutely". I don't think the word "absolutely" is supposed to be there. But then, in [itex]\mathbb C[/itex], a series is convergent if and only if it's absolutely convergent. So if you're talking about functions from [itex]\mathbb C[/itex] into [itex]\mathbb C[/itex], then it makes no difference if the word "absolutely" is included or not.

What the definition is saying is that there needs to exist a real number R>0 such that for all x with |x-x0|<R, there's a series [tex]\sum_{n=0}^\infty a_n \left( x-x_0 \right)^n[/tex] that's convergent and =f(x).

I like Wikipedia's definitions by the way. Link.


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