KGS math club/solution 5 1

Proof:

Let n be the degree of the polynomial P. Since P has nonzero degree, n>0. The Taylor polynomial of P about the point a is

$F(x)=P(a)+P'(a)(x-a)+\dotsb +\dfrac{P^{(n)}(a)(x-a)^n}{n!}=P(x)$ .

The Taylor polynomial of P about the point b is

$G(x)=P(b)+P'(b)(x-b)+\dotsb +\dfrac{P^{(n)}(b)(x-b)^n}{n!}=P(x))$ .

Note that $F(x)=G(x)$ for all real numbers x, because $F=P=G$ . Also, since $P^{(i)}(a)=P^{(i)}(b)$ for all i, we can rewrite G as $G(x)=P(a)+P'(a)(x-b)+\dotsb +\dfrac{P^{(n)}(a)(x-b)^n}{n!}=P(x))$ .

Now suppose that $a\neq b$ . Let $c=|a-b|$ . Then c is greater than zero. Let $y,z\in\Re$ such that $|y-z|=c$ . Then $P(y)=P(z)$ ; since P is a nonconstant polynomial this implies that there is a turning point in the interval (z,y). Hence, between any two real numbers x,y with $|x-y|=c$ , there exists a turning point. So P has infinitely many turning points. This is a contradiction, since a polynomial only has finitely many turning points. Therefore, $a=b$ .

Alternative Proof:

Derivation decreases the degree of a polynomial by 1. Hence, the $(n-1)$ th derivative of $P(x)$ is a linear polynomial of the form $P^{(n-1)}(x) = rx + s$ for some $r, s$ . By assumption $r\neq0$ , hence the graph of this linear polynomial is not horizontal. The requirement $P^{(n-1)}(a) = P^{(n-1)}(b)$ then implies $a = b$ .

Page

Toolbox

Search

KGS math club/solution 5 1