What is Polynomial, degree? — Mathematics

Definition 8.1 University Mathematics — Year 1 · Chapter 8 — Polynomials

A polynomial with coefficients in $K$ is a formal sum

P = a_0 + a_1 X + a_2 X^2 + \dots + a_n X^n = \sum_{k} a_k X^k,

with $a_k \in K$ all zero from some index on. With the natural addition and the product

\Bigl(\sum_i a_i X^i\Bigr)\Bigl(\sum_j b_j X^j\Bigr) = \sum_k \Bigl(\sum_{i+j=k} a_i b_j\Bigr) X^k,

the set $K[X]$ is a commutative ring. The degree $\deg P$ of $P \neq 0$ is the largest $n$ with $a_n \neq 0$ ; $a_n$ is the leading coefficient ( $P$ is monic when $a_n = 1$ ), and by convention $\deg 0 = -\infty$ . Every polynomial defines a function $x \mapsto P(x)$ on $K$ by substitution.

Examples

Example 8.4

Divide $A = X^4 + X^3 - 2X + 1$ by $B = X^2 + 1$ :

X^4 + X^3 - 2X + 1 = (X^2 + 1)(X^2 + X - 1) + (-3X + 2).

(Compute: subtract $X^2 B$ , then $X B$ , then $-B$ ; the remainder $-3X + 2$ has degree $1 < 2$ .)

Example 8.9 (The auxiliary-polynomial trick)

Let $P$ be the polynomial of degree $\leq n$ with

P(k) = \frac{k}{k+1} \qquad (k = 0, 1, \dots, n) ;

it exists and is unique by Lagrange interpolation below. What is $P(n+1)$ ? Clear denominators: the polynomial $Q = (X+1)P - X$ has degree $\leq n + 1$ and vanishes at the $n + 1$ points $0, 1, \dots, n$ , so by Theorem 8.7

Q = c\,X(X-1)(X-2)\cdots(X-n)

for some constant $c$ . Evaluate where $Q$ is known independently: at $X = -1$ , $Q(-1) = 0 \cdot P(-1) + 1 = 1$ , while the product equals $(-1)(-2)\cdots(-1-n) = (-1)^{n+1}(n+1)!$ ; hence $c = \frac{(-1)^{n+1}}{(n+1)!}$ . Now evaluate at $X = n + 1$ :

(n+2)\,P(n+1) - (n+1) = Q(n+1) = c\,(n+1)! = (-1)^{n+1} ,

so $P(n+1) = \dfrac{(n+1) + (-1)^{n+1}}{n+2}$ : equal to $1$ for odd $n$ , and to $\frac{n}{n+2}$ for even $n$ — the interpolating polynomial does not continue the pattern $\frac{n+1}{n+2}$ . The trick to remember: encode the data as roots of an auxiliary polynomial, identify the unknown constant at a point outside the data, and harvest.

Example 8.13 (Detecting multiple roots with a gcd)

When no root is known, Proposition 8.11 still gives a global multiple-root detector: $a$ is a multiple root of $P$ iff it is a common root of $P$ and $P'$ , so $P$ has a multiple root (in $\C$ ) iff $\gcd(P, P') \neq 1$ — computable by the Euclidean algorithm without solving anything. Sample: $P = X^3 - 3X + 2$ , $P' = 3X^2 - 3 = 3(X - 1)(X + 1)$ . Testing the roots $\pm1$ of $P'$ inside $P$ : $P(1) = 0$ but $P(-1) = 4$ , so

\gcd(P, P') = X - 1 :

the root $1$ is multiple; dividing twice, $P = (X - 1)^2(X + 2)$ . The gcd even reports the full set of multiple roots, each with multiplicity lowered by one — the fact every computer-algebra system exploits to “square-free factorize” before any root-hunting, and the polynomial twin of Exercise 8.9’s no-multiple-root arguments.

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Examples

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