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Math Help - Polynomial uniqueness proof

  1. #1
    Junior Member PaulAdrienMauriceDirac's Avatar
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    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 P(x) and Q(x) are polynomials of degree not exceeding 2 and P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3) for three different numbers x_1,x_2,\text{ and } x_3, then the polynomials P(x) and Q(x) are equal.

    I'm not very good at proofs, so I have questions. If I show that if P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3) is true and P(x) and Q(x) are polynomials of degree not exceeding 2, then P(x) and 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 V(x)=P(x)-Q(x), After that V(x_1)=V(x_2)=V(x_3)=0, because P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3), so x_1,x_2,x_3 are roots of V(x) which can't have more than two roots. Here I'm confused, I shouldn't have assumed that P(x_1)=Q(x_1),P(x_2)=Q(x_2),P(x_3)=Q(x_3) was true, right?
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    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?
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    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.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial uniqueness proof

    Quote Originally Posted by Deveno View Post
    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 View Post
    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.
    Last edited by PaulAdrienMauriceDirac; November 6th 2012 at 10:22 AM.
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    Re: Polynomial uniqueness proof

    Quote Originally Posted by emakarov View Post
    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.

    *********

    to answer PaulAdreinMauriceDirac's question:

    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).
    Last edited by Deveno; November 6th 2012 at 10:48 AM.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial uniqueness proof

    Quote Originally Posted by Deveno View Post

    to answer PaulAdreinMauriceDirac's question:

    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!
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    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 f(x)= a_1x^2+ b_1x+ c_1. So suppose f(x_1)= y_1, f(x_2)= y_2, and f(x_3)= y_3 and that g(x)= a_2x^2+ b_2x+ c takes on those same values. That gives three equations you can solve to show that a_1= a_2, b_1= b_2, and c_1= c_2.
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    Re: Polynomial uniqueness proof

    Quote Originally Posted by HallsofIvy View Post
    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 f(x)= a_1x^2+ b_1x+ c_1. So suppose f(x_1)= y_1, f(x_2)= y_2, and f(x_3)= y_3 and that g(x)= a_2x^2+ b_2x+ c takes on those same values. That gives three equations you can solve to show that a_1= a_2, b_1= b_2, and 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)
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial uniqueness proof

    Thank you HallsofIvy and Deveno!
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