Find polynomials which satisfy given conditions

by RoRi

July 15, 2015

Let $p$ be a polynomial of degree at most 2. Find all $p$ that satisfying the given conditions.

$p(x) = p(1-x)$ .
$p(x) = p(1+x)$ .
$p(2x) = 2p(x)$ .
$p(3x) = p(x+3)$ .

Since $p$ is a polynomial of degree at most 2 we may write,

$p(x) = ax^2 + bx + c,$

for $a,b,c \in \mathbb{R}$ constants.

We substitute,

$\begin{align*} p(x) = p(1-x) &&\implies&& ax^2 +bx + c &= a(1-x)^2 + b(1-x) + c \\ && \implies && ax^2 + bx + c &= ax^2 + (-2a-b)x + (a+b+c). \end{align*}$

Equating like powers of $x$ , we have,

$a = a, \qquad b = -2a -b \ \implies \ a = -b, \qquad c = a + b + c \ \implies \ c \text{ is arbtirary}.$

Thus,

$p(x) = -bx^2 + bx + c = bx(1-x) + c.$

Or, since these are arbitrary constants we can relabel them,

$p(x) = ax(1-x) + b.$
Again, we substitute,

$\begin{align*} p(x) = p(1+x) && \implies && ax^2 + bx + c &= a(1+x)^2 + b(1+x) + c \\ &&\implies && &= ax^2 + (2a+b)x + (a+b+c). \end{align*}$

Again, equating like powers of $x$ ,

$a = a, \qquad b = 2a + b \ \implies \ a = 0, \qquad c = a+b+c \ \implies \ b = 0.$

Hence,

$p(x) = c, \qquad c \text{ an arbitrary constant}.$
Substituting,

$p(2x) = 2p(x) \qquad \implies \qquad 4ax^2 + 2bx + c = 2ax^2 + 2bx + 2c.$

Equating like powers of $x$ ,

$4a = 2a \ \implies \ a = 0, \qquad 2b = 2b \ \implies \ b \text{ arbitrary}, \qquad c = 2c \ \implies \ c = 0.$

Thus,

$p(x) = bx, \qquad b \text{ arbitrary}.$
Substituting,

$p(3x) = p(x+3) \qquad \implies \qquad 9ax^2 + 3bx + c = ax^2 + (6a+b)x + (9a+3b+c).$

Equating like powers of $x$ ,

$9a = a \ \implies \ a = 0, \qquad 3b = 6a +b \ \implies \ b = 0, \qquad c = 9a + 3b + c = c \ \implies \ c \text{ arbitrary}.$

Hence,

$p(x) = c \qquad \text{ for } c \text{ an arbitrary constant}.$

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