Polynomial: Difference between revisions
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Created page with "== Mathematical definition == Let $P(x)$ be a mathematical function $P:\mathbb{F} \to \mathbb{F}$ If $P(x)$, through algebraic manipulation, can be written in the form of {{#content:Q2150}} Then $P(x)$ is a '''polynomial''' over $\mathbb{F}$ <blockquote> Technically, $0$ is a polynomial. </blockquote> <blockquote> $x+x^{\frac{1}{2}}$ is '''not''' a polynomial as $\frac{1}{2}\notin \mathbb{N}$ </blockquote> == Degree == The degree of a nonzero polynomial $P(x) \,..." |
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== Degree == | == Degree, leading coefficient/term == | ||
The degree of a nonzero polynomial $P(x) \, \deg P=\max\{k: a_k\ne 0\}$. | The degree of a nonzero polynomial $P(x) \, \deg(P)=\max\{k: a_k\ne 0\}$. | ||
Revision as of 13:22, 16 October 2026
Mathematical definition
Let $P(x)$ be a mathematical function $P:\mathbb{F} \to \mathbb{F}$
If $P(x)$, through algebraic manipulation, can be written in the form of
$$P(x)=\sum_{k=0}^{n} a_k x^k, n\in\mathbb{N},a_0, a_1, \dots, a_n\in \mathbb{F}$$
where $x\in \mathbb{F}$ is an independent variable
Then $P(x)$ is a polynomial over $\mathbb{F}$
Technically, $0$ is a polynomial.
$x+x^{\frac{1}{2}}$ is not a polynomial as $\frac{1}{2}\notin \mathbb{N}$
Degree, leading coefficient/term
The degree of a nonzero polynomial $P(x) \, \deg(P)=\max\{k: a_k\ne 0\}$.