Complex-analytic implies holomorphic - Revision history

Vipul at 16:08, 12 September 2008

2008-09-12T16:08:13Z

← Older revision		Revision as of 16:08, 12 September 2008
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	Then, <math>f</math> is a [[holomorphic function]]: it is complex-differentiable, and the complex differential is a continuous function. In fact, <math>f</math> is differentiable infinitely often.		Then, <math>f</math> is a [[holomorphic function]]: it is complex-differentiable, and the complex differential is a continuous function. In fact, <math>f</math> is differentiable infinitely often.

	==~~Proof~~==		==Facts used==

	~~We will show the following somewhat stronger fact~~: ~~if <math>f</math>~~ is ~~a complex-analytic function at <math>z_0</math> with a power series:~~		# [[uses::Power series is infinitely differentiable in disk of convergence]]

	~~<math>f(z) :~~= ~~\sum_{n~~=~~0}^\infty a_n(z - z_0)^n</math>~~		==Proof==

	By basic facts of power series, <math>f</math> converges absolutely, and uniformly on every disc of radius strictly smaller than its radius of convergence. We'll show that <math>f'</math> exists on the open disc of the radius of convergence, and is given by the following absolutely convergent power series:

	~~<math>f'(z) :~~= ~~\sum_{n~~=~~0}^\infty na_n(z - z_0)^{n-1}</math>~~

	Note that the power series on the right clearly has the same radius of convergence as the power series for <math>f</math>, so we can concentrate on showing that it equals <math>f</math>. Observe that if <math>s_n</math> is the partial sum till the <math>n^{th}</math> term of <math>f</math>, then <math>s_n'(z)~~</math> indeed equals the sum, till the <math>n^{th}</math> term, for <math>f'(z)</math>. Thus, what we really need to show is that the error term goes to zero~~.		The proof follows directly from fact (1).

Vipul: 4 revisions

2008-05-18T19:10:52Z

4 revisions

← Older revision	Revision as of 19:10, 18 May 2008
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Vipul at 18:57, 26 April 2008

2008-04-26T18:57:05Z

← Older revision		Revision as of 18:57, 26 April 2008
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			{{basic fact}}
	==Statement==		==Statement==

Vipul at 22:56, 14 April 2008

2008-04-14T22:56:20Z

← Older revision		Revision as of 22:56, 14 April 2008
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	==Proof==		==Proof==

	We ~~guess~~ that the power series ~~actually represents~~ the ''~~Taylor expansion''~~		We will show the following somewhat stronger fact: if <math>f</math> is a complex-analytic function at <math>z_0</math> with a power series:

			<math>f(z) := \sum_{n=0}^\infty a_n(z - z_0)^n</math>

			By basic facts of power series, <math>f</math> converges absolutely, and uniformly on every disc of radius strictly smaller than its radius of convergence. We'll show that <math>f'</math> exists on the open disc of the radius of convergence, and is given by the following absolutely convergent power series:

			<math>f'(z) := \sum_{n=0}^\infty na_n(z - z_0)^{n-1}</math>

			Note that the power series on the right clearly has the same radius of convergence as the power series for <math>f</math>, so we can concentrate on showing that it equals <math>f</math>. Observe that if <math>s_n</math> is the partial sum till the <math>n^{th}</math> term of <math>f</math>, then <math>s_n'(z)</math> indeed equals the sum, till the <math>n^{th}</math> term, for <math>f'(z)</math>. Thus, what we really need to show is that the error term goes to zero.

Vipul at 22:29, 14 April 2008

2008-04-14T22:29:52Z

← Older revision		Revision as of 22:29, 14 April 2008
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	==Proof==		==Proof==

	~~To execute this proof, we need to argue~~ that		We guess that the power series actually represents the ''Taylor expansion''

Vipul: New page: ==Statement== Suppose $U \subset \mathbb{C}$ is a domain and $f:U \to \mathbb{C}$ is a complex-analytic function: for every point $z_0 \in U$ , there e...

2008-04-14T20:24:24Z

New page: ==Statement== Suppose <math>U \subset \mathbb{C}</math> is a domain and <math>f:U \to \mathbb{C}</math> is a complex-analytic function: for every point <math>z_0 \in U</math>, there e...

New page

==Statement==

Suppose <math>U \subset \mathbb{C}</math> is a [[domain]] and <math>f:U \to \mathbb{C}</math> is a complex-analytic function: for every point <math>z_0 \in U</math>, there exists a real number <math>r > 0 </math> and a power series <math>\sum_{n=0}^\infty a_n (z - z_0)^n</math> such that the power series converges and agrees with <math>f</math> in the ball of radius <math>r</math>.

Then, <math>f</math> is a [[holomorphic function]]: it is complex-differentiable, and the complex differential is a continuous function. In fact, <math>f</math> is differentiable infinitely often.

==Proof==

To execute this proof, we need to argue that