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[parent] proof of radius of convergence of a complex function (Proof)

Without loss of generality, it may be assumed that $ z_0 = 0$.

Let $ c_n$ denote the coefficient of the $ n$-th term in the Taylor series of $ f$ about 0. Let $ r$ be a real number such that $ 0 < r < R$. Then $ c_n$ may be expressed as an integral using the Cauchy integral formula.

$\displaystyle c_n = {1 \over 2 \pi i} \oint_{\vert z\vert = r} {f(z) \over z^{n... ...\over 2 \pi r^n} \int_{-\pi}^{+\pi} e^{-n \theta} f(r e^{i \theta}) \, d \theta$

Since $ f$ is analytic, it is also continuous. Since a continuous function on a compact set is bounded, $ \vert f\vert < B$ for some constant $ B > 0$ on the circle $ \vert z\vert = r$. Hence, we have

$\displaystyle \vert c_n\vert = {1 \over 2 \pi r^n} \left\vert \int_{-\pi}^{+\pi... ... d \theta \le {1 \over 2 \pi r^n} \int_{-\pi}^{+\pi} B d \theta = {B \over r^n}$

Consequently, $ \sqrt[n]{c_n} \le \sqrt[n]{B} / r$. Since $ \lim_{n \to \infty} \sqrt[n]{B} = 1$, the radius of convergence must be greater than or equal to $ r$. Since this is true for all $ r < R$, it follows that the radius of convergence is greater than or equal to $ R$.



"proof of radius of convergence of a complex function" is owned by rspuzio. [ full author list (2) ]
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Cross-references: radius of convergence, circle, bounded, compact set, continuous, analytic, Cauchy integral formula, integral, real number, Taylor series, term, coefficient, without loss of generality

This is version 6 of proof of radius of convergence of a complex function, born on 2004-10-03, modified 2006-11-23.
Object id is 6279, canonical name is ProofOfRadiusOfConvergenceOfAComplexFunction.
Accessed 1318 times total.

Classification:
AMS MSC: 30B10 (Functions of a complex variable :: Series expansions :: Power series )
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