# The definition of "entire function"

I am reading about the definition of "entire functions" : "If a complex function is analytic at all finite points of the complex plane $\mathbb{C}$, then it is said to be entire ..."

In fact, I'd like to understand this definition. Thus I wish a help to respond my questions.

1. Are all analytic functions on $\mathbb{C}$ entire?

2. Why do we need to use this definition?

• The question in point 1 doesn't make sense. Since this is a question about understanding, you should probably clarify what you mean rather than leave it to people to guess. (many won't even realize they're guessing!) Unless, I suppose, the question you need answered is what's wrong with your grammar (but if so you should clarify!).
– user14972
Commented Aug 21, 2017 at 13:07
• Entire functions are very nice - they can all be written as a power series that converges everywhere on $\mathbb C$. They extend the ring of polynomials in useful ways. Commented Aug 21, 2017 at 13:12
• Be careful with prepositions. We say a function is analytic at a point, or we say a function is analytic on a set (if it is analytic at every point in the set.) So an entire function is analytic on $\mathbb C$, not at $\mathbb C$. Commented Aug 21, 2017 at 13:13
• Thank you @ThomasAndrews ! I repaired that point! Commented Aug 22, 2017 at 0:25
• @Hurkyl : Thank you for the advice. I will write carefully next time! Commented Aug 22, 2017 at 0:27

There are 3 definitions of entire functions, all equivalent :

1. $f(z) = \sum_{n=0}^\infty a_n z^n$ is entire iff it converges for every $z \in \mathbb{C}$ (see the radius of convergence of power series)

2. $f$ is entire iff it is everywhere analytic, that is for every $z_0 \in \mathbb{C}$ there is $r > 0$ such that $f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$ for $|z-z_0| < r$.

3. $f$ is entire iff it is everywhere holomorphic.

The Cauchy integral formula for analytic functions (not difficult) lets us show $2. \implies 1.$ And the Cauchy integral formula for holomorphic functions (harder) lets us show $3.\implies 2.$

Note how this doesn't work when $\mathbb{C}$ is replaced by $\mathbb{R}$ : $\ \ f(x)=\frac{1}{1+x^2}$ is analytic on $\mathbb{R}$ but its Taylor series has a finite radius of convergence because of the singularity at $\pm i$.

• Oh, thank you for your clear explanation to me! @reuns Commented Aug 22, 2017 at 0:32

Let $f: \mathbb{C} \to \mathbb{C}$ be a complex function.

Then $f$ is an entire function $\iff f$ can be given by an everywhere convergent power series: $$\displaystyle f: \mathbb{C} \to \mathbb{C}: f \left({z}\right) = \sum_{n \mathop = 0}^\infty a_n z^n; \quad \lim_{n \mathop \to \infty} \sqrt [n] {\left|{a_n}\right|} = 0$$

So if $f$ is entire then this means $f$ is holomorphic on $\mathbb{C}$. It must be analytic at every point of $\mathbb C$. In order for that to be true, the function must be defined at every point of $\mathbb C$.

If $G$ is an open set in $\mathbb C$, then we write $H(G)$ for the set of all analytic functions $g:G \to \mathbb C$.

A function $f \in H( \mathbb C)$ is called entire.