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Difference between revisions of "KGS math club/solution 5 1"

(New page: 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 <math>F(x)=P(a)+P'(a)(x-a)+\dotsb +\dfrac{P^{(n)}(a)(x...)
 
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Now suppose that <math>a\neq b</math>. Let <math>c=|a-b|</math>.  Then c is greater than zero.  Let <math>y,z\in\Re</math> such that <math>|y-z|=c</math>.  Then <math>P(y)=P(z)</math>; 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 <math>|x-y|=c</math>, 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, <math>a=b</math>.
 
Now suppose that <math>a\neq b</math>. Let <math>c=|a-b|</math>.  Then c is greater than zero.  Let <math>y,z\in\Re</math> such that <math>|y-z|=c</math>.  Then <math>P(y)=P(z)</math>; 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 <math>|x-y|=c</math>, 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, <math>a=b</math>.
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Alternative Proof:
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Derivation decreases the degree of a polynomial by 1. Hence, the  <math>(n-1)</math>th derivative of <math>P(x)</math> is a linear polynomial of the form <math>P^{(n-1)}(x) = rx + s</math> for some <math>r, s</math>. By assumption <math>r\neq0</math>, hence the graph of this linear polynomial is not horizontal. The requirement <math>P^{(n-1)}(a) = P^{(n-1)}(b)</math> then implies <math>a = b</math>.

Latest revision as of 19:24, 21 December 2015

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 $a = b$.

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