# What does this definition of a polynomial mean?

In a book I am reading, there is the following definition for a polynomial:

A function $$p: \mathbb{F} \rightarrow \mathbb{F}$$ is called a polynomial with coefficients in $$\mathbb{F}$$ if there exists $$a_0, \ldots, a_m \in \mathbb{F}$$ such that

$$p(z) = a_0 + a_1z + a_2z^2 + \ldots + a_mz^m$$

for all $$z \in \mathbb{F}$$.

However there aspects of this that do not make sense to me. For example, I can not think of any examples where you have a polynomial $$p(z) = a_0 + a_1z + a_2z^2 + \ldots + a_mz^m$$, with coefficients $$a_0, \ldots, a_m \in \mathbb{F}$$ but for only some $$z \in \mathbb{F}$$? Does that even make sense?

What precisely is this definition saying, as it seems to differ slightly (at least in terms of wording) from other definitions, e.g. here.

• Well, is $1+x+sin(\pi\cdot x)$ a polinomial? Take $\mathbb{F}=\mathbb{R}.$ Commented Jun 17, 2020 at 15:38
• @Phicar No, but isn't that just because it isn't written in the form $p(z) = a_0 + a_1z + a_2z^2 + \ldots + a_mz^m$? Commented Jun 17, 2020 at 15:40
• One thing the definition seems to exclude is piecewise functions, that are defined as a polynomial for part of the domain, and some other kind(s) of functions elsewhere. Commented Jun 17, 2020 at 15:50

A function $$f: \mathbb{F} \rightarrow \mathbb{F}$$ is called a polynomial with coefficients in $$\mathbb{F}$$ if there exists a polynomial $$p$$ with coefficients in $$\mathbb{F}$$ such that $$f(x) = p(x)$$ for all $$x \in \mathbb{F}.$$
The function $$p(z) = \sin z$$ satisfies the condition that there are $$a_0, \ldots, a_m \in \mathbb{R}$$ such that $$p(z) = a_0 + a_1 z + a_2 z^2 + \cdots + a_m z^m$$ for only some $$z \in \mathbb{R}.$$ One can for example take $$m=0,\ a_0=0$$ and those "some" $$z$$ to be $$\{n\pi \mid n\in\mathbb{Z}\}.$$ But there are no $$a_0, \ldots, a_m \in \mathbb{R}$$ such that $$p(z) = a_0 + a_1 z + a_2 z^2 + \cdots + a_m z^m$$ for all $$z \in \mathbb{R}.$$ Therefore $$p$$ is not a polynomial function.