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A function that is analytic at a point is one that can be represented by a Taylor or Maclaurin series? We also say that the radius of convergence should be positive. What if it was negative? What that would change and why? I don't quite understand what it means exactly, can you explain in layman's terms?

Gladstone Asder
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A function $f$ is analytic in some closed set $D$ if there exists an open set $U$ such that $D \subset U$ and $f$ is analytic in $U$. Hence for a complex valued function, it means that in $U$ it satisfies the Cauchy-Riemann equation and the first real partial derivatives exist in $U$,or, equivalently, that it can be represented by a power series in some neighborhood contained in $U$.

user40276
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A MacLaurin series is nothing more than a Taylor series around $z=0$ instead of around $z=a$. So you can stop thinking about the term "MacLaurin".

A negative radius of convergence doesn't make sense : the radius of convergence is defined as a limit of positive real numbers, thus must be non-negative or infinite (thanks to Andres for making that clear by what I meant). Also, the radius of convergence is the biggest radius for which the Taylor expansion of your function converges, so geometrically it doesn't make sense to say that this radius is negative.

Hope that helps,

Patrick Da Silva
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  • "thus must be positive" or zero, or $\infty$. Also, it is a limit of non-negative real numbers (they may fail to be strictly positive). – Andrés E. Caicedo Jun 04 '13 at 22:31
  • @Andres : Actually when we work out $\lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|}$ we do need them to be strictly positive and if there are some zero terms (like in the expansion of $\sin$ or $\cos), we usually restrict our attention to the positive coefficients. I "handwavely" included zero/infinity in my description of a positive radius, but thanks for pointing that out. I was mostly focused on pointing out that it couldn't be negative... – Patrick Da Silva Jun 04 '13 at 22:36
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    No, the right expression to compute is $\limsup_{n\to\infty}|a_n|^{1/n}$, where $a_n$ is the coefficient of $(x-a)^n$ if we are expanding about $a$; its inverse in the extended real line is the radius of convergence. The limit you wrote may fail to exist, for several reasons. – Andrés E. Caicedo Jun 04 '13 at 22:40