Polynomial uniqueness proof

• Nov 6th 2012, 09:30 AM
Polynomial uniqueness proof
This problem is from my Gelfand's Algebra book.

Problem 164. Prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values.

This means that if $\displaystyle P(x)$ and $\displaystyle Q(x)$ are polynomials of degree not exceeding 2 and $\displaystyle P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ for three different numbers $\displaystyle x_1,x_2,\text{ and } x_3,$ then the polynomials $\displaystyle P(x)$ and $\displaystyle Q(x)$ are equal.

I'm not very good at proofs, so I have questions. If I show that if $\displaystyle P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ is true and $\displaystyle P(x)$ and $\displaystyle Q(x)$ are polynomials of degree not exceeding 2, then $\displaystyle P(x)$ and $\displaystyle Q(x)$ are equal, will it prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values?

Is this the way to go?

Let $\displaystyle V(x)=P(x)-Q(x)$, After that $\displaystyle V(x_1)=V(x_2)=V(x_3)=0$, because $\displaystyle P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$, so $\displaystyle x_1,x_2,x_3$ are roots of $\displaystyle V(x)$ which can't have more than two roots. Here I'm confused, I shouldn't have assumed that $\displaystyle P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3)$ was true, right?
• Nov 6th 2012, 10:11 AM
Deveno
Re: Polynomial uniqueness proof
actually, your proof is fine. a polynomial of degree n, cannot have more than n roots.

so you have proved that V(x) is NOT of degree 0,1, or 2. what's left?
• Nov 6th 2012, 10:17 AM
emakarov
Re: Polynomial uniqueness proof
I don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.
• Nov 6th 2012, 10:18 AM
Re: Polynomial uniqueness proof
Quote:

Originally Posted by Deveno
actually, your proof is fine. a polynomial of degree n, cannot have more than n roots.

so you have proved that V(x) is NOT of degree 0,1, or 2. what's left?

Yes, but does that prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values?

Quote:

Originally Posted by emakarov
I don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.

Yes, I think that's right, I'm assuming too much and I don't really understand how I can prove that polynomial of degree not exceeding 2 is defined uniquely by three of its values.
• Nov 6th 2012, 10:40 AM
Deveno
Re: Polynomial uniqueness proof
Quote:

Originally Posted by emakarov
I don't know what comes before this exercise in the book, but relying on the fact that a polynomial of degree n cannot have more than n roots seems to easy. It is possible that the real problem is to prove that fact.

well this is only true in a field, of course, but presumably we are talking about polynomials in R[x].

and one can argue by degree:

proof(by induction): if p(x) in R[x] is of degree n > 0 then p has at most n real roots.

(the reason for requiring n > 0 will be explained later).

base case: n = 1.

in this case p(x) = ax + b, which has the sole root x = -b/a (we may assume a ≠ 0, or else p is not of degree 1). since exactly one root is a stronger condition that at most one root, the theorem holds in this case.

(strong) induction hypothesis: suppose whenever 0 < k < n, deg(p) = k implies p has at most k roots.

let deg(p) = n.

it could be that p has no roots at all. since 0 < n, the theorem holds in this case.

otherwise, let a be a root of p.

writing p(x) = q(x)(x - a) + r(x), where either r(x) is identically 0, or deg(r) < deg(x-a) = 1, we see that r is a constant polynomial, or the 0-polynomial.

so p(x) = q(x)(x - a) + r, for some real number r. thus:

0 = p(a) = q(a)(a - a) + r = q(a)0 + r = r.

hence p(x) = q(x)(x - a).

since deg(p) = deg(q) + deg(x-a), we have:

n = deg(q) + 1, that is:

deg(q) = n-1 < n, so q has at most n-1 roots.

lemma: if p(x) = f(x)g(x) for real polynomials p,f, and g, then if a is a root of p, either a is a root of f, or a is a root of g.

proof: 0 = p(a) = f(a)g(a), so either f(a) = 0, or g(a) = 0.

main proof continued:

let b be any root of p(x) = q(x)(x - a).

then either b is a root of q(x) (of which we have at most n - 1), or b is a root of x - a (which has the single root a).

thus we have at most n - 1 + 1 = n roots of p, QED.

*********

you have proved that if P = Q at 3 points (and deg(P), deg(Q) < 3), P = Q at every point.

the only thing remaining to show is that given any 3 points, there is a polynomial of degree < 3 that goes through those 3 points. can you do this?

**********

OOPS! i almost forgot my explanation for why we take n > 0:

a polynomial of degree 0 is a constant polynomial:

p(x) = c.

unless c = 0, p has no roots at all (which is ok according to our theorem).

but if p(x) = 0, for all x, we have infinitely many roots (very bad).

on the other hand:

p(x) = 0 = 0 + 0x = 0 + 0x + 0x2, etc.

is sort of a SPECIAL polynomial, it really doesn't "have" a degree. it is this special case we seek to avoid by stating deg(p) > 0. depending on which book you read you get:

deg(0) = undefined, or sometimes:

deg(0) = -∞ (this is to make the formula deg(fg) = deg(f) + deg(g) always work out).
• Nov 6th 2012, 11:10 AM
Re: Polynomial uniqueness proof
Quote:

Originally Posted by Deveno

you have proved that if P = Q at 3 points (and deg(P), deg(Q) < 3), P = Q at every point.

the only thing remaining to show is that given any 3 points, there is a polynomial of degree < 3 that goes through those 3 points. can you do this?

I will try to prove it now, thank you for help!
• Nov 6th 2012, 11:25 AM
HallsofIvy
Re: Polynomial uniqueness proof
Because you are specifically asked about "polynomials of degree not greater than 2", I don't think you need to be very "sophisticated"! You can use the fact that any such polynomial can be written in the form $\displaystyle f(x)= a_1x^2+ b_1x+ c_1$. So suppose $\displaystyle f(x_1)= y_1$, $\displaystyle f(x_2)= y_2$, and $\displaystyle f(x_3)= y_3$ and that $\displaystyle g(x)= a_2x^2+ b_2x+ c$ takes on those same values. That gives three equations you can solve to show that $\displaystyle a_1= a_2$, $\displaystyle b_1= b_2$, and $\displaystyle c_1= c_2$.
• Nov 6th 2012, 12:22 PM
Deveno
Re: Polynomial uniqueness proof
Quote:

Originally Posted by HallsofIvy
Because you are specifically asked about "polynomials of degree not greater than 2", I don't think you need to be very "sophisticated"! You can use the fact that any such polynomial can be written in the form $\displaystyle f(x)= a_1x^2+ b_1x+ c_1$. So suppose $\displaystyle f(x_1)= y_1$, $\displaystyle f(x_2)= y_2$, and $\displaystyle f(x_3)= y_3$ and that $\displaystyle g(x)= a_2x^2+ b_2x+ c$ takes on those same values. That gives three equations you can solve to show that $\displaystyle a_1= a_2$, $\displaystyle b_1= b_2$, and $\displaystyle c_1= c_2$.

the nice thing about this suggestion is that it also suggests a way to prove there is a polynomial which goes through our 3 given points.

(think: system of linear equations)
• Nov 8th 2012, 07:21 AM