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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.

Jony Pollon
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  • Well, is $1+x+sin(\pi\cdot x)$ a polinomial? Take $\mathbb{F}=\mathbb{R}.$ – Phicar Jun 17 '20 at 15:38
  • @Phicar No, but isn't that just because it isn't written in the form $p(z) = a_0 + a_1z + a_2z^2 + \ldots + a_mz^m$? – Jony Pollon Jun 17 '20 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. – DreiCleaner Jun 17 '20 at 15:50

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The definition does not define what a polynomial is; it defines what it means for a function to be polynomial. The definition above could be written as follows:

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.

md2perpe
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