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I'm studying analytic functions. It is known that a real analytic function is an infinitely differentiable function such that the Taylor series at any point $x_0 $ in its domain $$ T\left(x\right) = \sum_{n=0}^{\infty}\frac{f^{\left( n\right)}\left( x_{0}\right)}{n!} \left(x -x_{0}\right)^{n} $$

converges to $ f(x) $ for $ x $ in a neighborhood of $ x_0. $

A complex analytic function is obtained by replacing, in the definition above, "real" with "complex".

We also know that a function is complex analytic if and only if it is holomorphic. This is a first difference between complex analytic functions and real analytic functions (in general, a infinitely differentiable function is not real analytic).

My questions is: Is my definition of analytic function correct? And then, there are other differences between real analytic and complex analytic functions?

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  • $\begingroup$ I don't see the world "real" in the "definition above" $\endgroup$ Commented Aug 3, 2019 at 7:41
  • $\begingroup$ A big difference is that being "$\mathbb R$-entire" (i.e. being analytic at every $x\in\mathbb R$) does not imply that the Taylor series around 0 has infinite radius of convergence. $\endgroup$ Commented Nov 2, 2019 at 8:26
  • $\begingroup$ Isn't analytic just another name for holomorphic? $\endgroup$ Commented Aug 18, 2020 at 6:27

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The definition is almost correct, but there is a small problem. That definition only makes sense if $f$ is infinitely differentiable. The usual definition is: $f$ is analytic if, for each $x_0$ in its domain, there is a power series $\sum_{n=0}^\infty a_n(x-x_0)^n$ about $x_0$ such that, in a neighborhood $N$ of $x_0$,$$(\forall x\in N):f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n.$$It can be proved then that $f$ is indeed infinitely differentiable and that$$(\forall x\in N):f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n.$$In other words,$$(\forall n\in\mathbb Z_+):a_n=\frac{f^{(n)}(x_0)}{n!}.$$

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  • $\begingroup$ In this definition, is there any constraint on the neighborhood? i.e. is existing such a neighborhood enough, not requiring the neighborhood to be large enough? $\endgroup$ Commented Aug 18, 2020 at 6:30
  • $\begingroup$ No, there is no constraint. $\endgroup$ Commented Aug 18, 2020 at 6:32
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I think you are asking about real analyticity in the wrong dimension.

If you have a formal power series $\sum_{n=1} a_n x^n$ with real or complex coefficients, then you can derive the properties of the corresponding analytic function by asking its radius of convergence (which is, you know just a property of the sequence $|a_n|$).

Or, phrased differently, any real analytic function extends to a complex analytic one, and the properties of complex analytic functions are somewhat similar to real analytic functions in one dimension. If you more generally allow functions $f:\mathbb{R}\to \mathbb{C}$ to be called real analytic (with the analogous definition), then there really is no difference. The weird part there is that being complex differentiable implies analyticity.

On the other hand, a function $f:\mathbb{C}\to \mathbb{C}$ is, of course, the same as a function $f:\mathbb{R}^2\to \mathbb{C}$ and such a function could reasonably called analytic if there is a certain regular sequence of polynomials converging locally uniformly to $f$ (just like in the definition you've given).

However, these would be polynomials with complex coefficients in two variables, so the question becomes whether there exist $a_{n,k}\in \mathbb{C}$ such that $$ f(x,y)=\sum_{n,k=0}^{\infty} a_{n,k} (x-x_0)^n(y-y_0)^k $$ In more natural complex notation, you might write this as a power series in $z$ and $\bar{z}$ instead of $x$ and $y$ (since any two linearly independent polynomials of degree $1$ form a basis of the monomials of degree 1, and these generate the whole thing).

So what's the difference there? Well, the map $z\mapsto \bar{z}$ is an orientation-reversing involution of $\mathbb{C},$ whereas the identity clearly preserves it, so the difference between a complex analytic function and an analytic function of two variables is that the former is always the limit of orientation-preserving polynomials.

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    $\begingroup$ Complex differentiability at a point does not imply analyticity. $\endgroup$ Commented Oct 8, 2020 at 13:26
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    $\begingroup$ @FaqirChand Good thing I just wrote "complex differentiable", then. :P $\endgroup$ Commented Oct 9, 2020 at 12:31

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