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Theorem

Let $\Omega\subseteq \mathbb{C}$ be open and $f\in H(\Omega)$ ($f$ analytic on $\Omega$). If $ C(z_{0},R)\subseteq\Omega$ (where $C(z_0,R)$ is the circle with origin $z_0$ and radius $R$), then we can represent $f$ on $C(z_{0},R)$ as a power series with convergence radius $\geq R$.

Proof

Let $0<r<R$ and $\gamma:[0,2\pi]\rightarrow\Omega$ be the path $\gamma(t)=z_{0}+re^{it}$.

Because $C(z_{0},R)$ is convex and $\gamma$ is in this circle, we have the following from the Cauchy integral formula: $$ f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi)}{\xi-z}\mathrm{d}\xi\ , (z\in C(z_{0},\ r)) $$

$\color{red}{\text{(1) Why not directly on } C(z_{0},\ R)? }$

Because $C(z_{0},\ r)\subseteq \mathbb{C}\backslash \gamma^{*}$ it follows that $f(z)=\displaystyle \sum_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ for a power series with convergence radius $\geq r$. Because $ r\in (0,\ R)$ is chosen arbitrarily and because the the coefficients $c_{n}$ are determined by the differentials $f^{(n)}(z_{0})$, the power series $\displaystyle \Sigma_{n=0}^{\infty}c_{n}(z-z_{0})^{n}$ has the convergence radius $\geq R$.

$\color{red}{\text{(2) Why are the coefficients calculated via }f^{(n)}(z_{0})? }$

$\color{red}{\text{(3) Why does it follow that the convergence radius is } \geq R? }$

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Chris, per Mex's comment to my answer, do you mean that $C(z_0,R)$ denotes the open ball of radius $R$ centered at $z_0$, instead of, as you stated, the circle (i.e., boundary) of the same center and radius? –  J. Loreaux Jun 7 '12 at 0:16
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1 Answer

up vote 1 down vote accepted

For (1), we choose $r<R$ because the path $\gamma_R(t)=z_0+Re^{it}$ for $t\in[0,2\pi]$ may not lie inside the region $\Omega$. This is because while $C(z_o,R)\subset\Omega$, its boundary, which is $\gamma_R^*$, may not be contained within $\Omega$. Now, to see why the integral formula only holds when $z\in C(z_0,r)$, and not on its boundary (which is $\gamma^*$ in your statement), this is simply due to the statement of the Cauchy Integral Formula. That is, it only holds for points that do not lie on $\gamma$.

For (2), the coefficients $c_n$ are determined by the differentials $f^{(n)}(z_0)$ because of the Taylor formula (which we can obtain from the finite Taylor formula if need be).

For (3), we have shown that the power series has radius of convergence $r$ where $r$ is an arbitrary positive number less than $R$. Thus, the radius of convergence is at least $\sup_{0<r<R} r= R$. So the radius of convergence is greater than or equal to $R$.

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@Loreaux, what do you mean by $\gamma_R(t)$ may not lie inside the region $\Omega$? given that $\Omega$ is an open set and $C(z_0,R)\subseteq \Omega$. So his No.1 question is justified and I am still confused what will be the problem if he apply cauchy formula on $C(z_0,R)$? –  Bunuelian Trick Jun 6 '12 at 4:10
    
@Mex: So, perhaps I misunderstood the notation, but I thought that $C(z_0,R)$ denoted the open ball of radius $R$ centered at $z_0$. I see that the OP said it was the circle of radius $R$ (i.e., the boundary of the set I just mentioned), but that doesn't seem to make sense, for example, because the OP claimed that $C(z_0,R)$ is convex. Anyway, given that it denotes the open ball, it stands to reason if we take $\Omega:=C(z_0,R)$ then certainly $\gamma_R*=\partial C(z_0,R)$ which lies entirely outside $\Omega$. –  J. Loreaux Jun 7 '12 at 0:14
    
It was my mistake, you are right J. Loreaux! –  Chris Jun 7 '12 at 0:19
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