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Prove: polynomial is uniquely defined by three of its values

  1. Feb 7, 2017 #1

    TheBlackAdder

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    1. The problem statement, all variables and given/known data
    Algebra - I.M. Gelfand, 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(x1) = Q(x1), P(x2) = Q(x2), P(x3) = Q(x3) for three different numbers x1, x2 and x3, then the polynomials P(x) and Q(x) are equal.

    2. Relevant equations


    3. The attempt at a solution
    The solution is simple (it's in the book) but I again approached it in a convoluted way in which I got stuck.
    P(x) = ax2+bx+c, Q(x) = dx2+ex+f
    P(x1) = Q(x1), P(x2) = Q(x2), P(x3) = Q(x3)

    P(x1) - Q(x1) = 0
    P(x1) - Q(x1) = P(x2) - Q(x2)
    (P(x1) - Q(x1)) - (P(x2) - Q(x2)) = 0
    (P(x1) - Q(x1)) - (P(x2) - Q(x2)) = P(x3) - Q(x3)
    (P(x1) - Q(x1)) - (P(x2) - Q(x2)) - (P(x3) - Q(x3)) = 0
    P(x1) - Q(x1) - P(x2) + Q(x2) - P(x3) + Q(x3) = 0
    P(x1) - P(x2) - P(x3) = Q(x1) - Q(x2) - Q(x3)

    $$(ax_{1}^{2}+bx_{1}+c)-(ax_{2}^{2}+bx_{2}+c)-(ax_{3}^{2}+bx_{3}+c)=(dx_{1}^{2}+ex_{1}+f)-(dx_{2}^{2}+ex_{2}+f)-(dx_{3}^{2}+ex_{3}+f)
    $$$$ax_{1}^{2}+bx_{1}+c-ax_{2}^{2}-bx_{2}-c-ax_{3}^{2}-bx_{3}-c=dx_{1}^{2}+ex_{1}+f-dx_{2}^{2}-ex_{2}-f-dx_{3}^{2}-ex_{3}-f$$$$a(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+b(x_{1}-x_{2}-x_{3})-c=d(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+e(x_{1}-x_{2}-x_{3})-f$$$$
    a(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+b(x_{1}-x_{2}-x_{3})-c-d(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})-e(x_{1}-x_{2}-x_{3})+f=0$$$$(a-d)(x_{1}^{2}-x_{2}^{2}-x_{3}^{2})+(b-e)(x_{1}-x_{2}-x_{3})-c+f=0$$Here I thought I was going to be able to prove you can only get 0 if a=d, b=e and c=f, but
    $$(2-1)(4_{1}^{2}-3_{2}^{2}-2_{3}^{2})+((-1)-(-2))(4_{1}-3_{2}-2_{3})-8+6=0$$$$
    (1)(16-9-4)+(1)(-1)-2=0$$$$
    (1)(3)+(1)(-1)-2=0$$$$
    3-1-2=0$$
    I realise I'm making a mistake by plugging in numbers, because I'm changing the given property that P(x3)=Q(x3)
    $$P(2)=2*2^{2}-1*2+8=14\neq Q(2)=2^{2}-2*2+6=6$$

    Again, is it possible to prove it in a similary fashion as above? I understand the given solution below, but the simplicity didn't dawn on me :(

    The solution:
    Consider a polynomial R(x) = P(x) - Q(x). Its degree does not exceed 2. On the other hand, we know that R(x1)=R(x2)=R(x3)=0;
    in other words, x1, x2 and x3 are roots of the polynomial R(x). But a polynomial of degree not exceeding 2, as we know, cannot have more than 2 roots, unless it is equal to zero. Therefore R(x) is equal to zero and P(x) = Q(x).
     
  2. jcsd
  3. Feb 7, 2017 #2

    PeroK

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    There is a difference between mathematical "tricks" that are non-obvious insights that simplify a problem and good mathematical technique, which is standard and frequently repeatable, that simplifies a problem.

    In this case, looking at the polynomial ##P-Q## is good technique and is something you must try to learn. It's not a one-off insight, it's a very common idea to tackle any problem where two things are equal.

    PS note that you can also simply generalise the problem to polynomials of any order.
     
  4. Feb 8, 2017 #3

    epenguin

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    I think proving 'a polynomial of degree not exceeding 2 is defined uniquely by three of its values' can be done easier.

    There is a key point ppl often don't realise. We think there is linear algebra, and then there is polynomial algebra which is a different thing.

    Whereas in fact a lot of polynomial algebra, most of what you can do, most of what you do do, is really linear algebra!

    One is in the habit of thinking of a, b, c as given knowns and x as unknown; here you just have to think of a, b, c as 'unknowns' and you have plenty of 'knowns', x1, x12, P(x1), x2, ... and looking at it that way, the problem should be something very familiar.
     
    Last edited: Feb 9, 2017
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