Talk:Proof that holomorphic functions are analytic

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[edit] Circular logic

From this article:

Suppose f is differentiable everywhere within some open disk centered at a. Let z be within that open disk. Let C be a positively oriented (i.e., counterclockwise) circle centered at a, lying within that open disk but farther from a than z is. Then, using Cauchy's integral formula, we get...

From Cauchy's integral formula:

The proof of this statement uses the Cauchy integral theorem and, just like that theorem, only needs that f is complex differentiable. One can then deduce from the formula that f must actually be infinitely often continuously differentiable, with
f^{(n)}(a) = {n! \over 2\pi i} \oint_C {f(z) \over (z-a)^{n+1}}\, dz.
Some call this identity Cauchy's differentiation formula. A proof of this last identity is a by-product of the proof that holomorphic functions are analytic.

So:

  • To prove that holomorphic functions are analytic, we need
  • The Cauchy integral formula, but to prove that we need
  • The fact that f^{(n)}(a) = {n! \over 2\pi i} \oint_C {f(z) \over (z-a)^{n+1}}\, dz, which is a byproduct of the proof that
  • Holomorphic functions are analytic

I encountered this trying to prove to myself that f^{(n)}(a) = {n! \over 2\pi i} \oint_C {f(z) \over (z-a)^{n+1}}\, dz; unfortunately, Wikipedia doesn't seem to help because of this circular loop :(.

Or maybe I'm just missing something? Domenic Denicola 01:10, 11 December 2005 (UTC)

You misread the quote from CIF: "One can then deduce from the formula that f must actually be infinitely often continuously differentiable, etc." So, no circle is involved: you first prove the Cauchy integral theorem, then deduce from it the Cauchy integral formula (as sketched there), and then deduce that holomorphic functions are analytic (as decribed here). -- EJ 00:18, 18 December 2005 (UTC)

[edit] Area of convergence

The series converges to the correct value on any open disk around a point if that open disk doesn't contain singularities. Can this area be extended (I'm assuming differentiability at all points considered)? E.g. does it follow that the series converges to the correct value everywhere except at singularities? Or is the maximum open disk not containing singularities the biggest area for which such a general statement holds? 82.103.195.147 12:27, 12 August 2006 (UTC)

The latter. See radius of convergence. -- EJ 15:53, 12 August 2006 (UTC)
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