# Real analytic functions VS complex analytic functions

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?

• I don't see the world "real" in the "definition above" Aug 3, 2019 at 7:41
• 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. Nov 2, 2019 at 8:26
• Isn't analytic just another name for holomorphic? Aug 18, 2020 at 6:27

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,$$(\forall n\in\mathbb Z_+):a_n=\frac{f^{(n)}(x_0)}{n!}.$$

• 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? Aug 18, 2020 at 6:30
• No, there is no constraint. Aug 18, 2020 at 6:32

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.

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