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Proving theorem for polynomials

  1. May 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove the following statement:

    Let f be a polynomial, which can be written in the form
    fix) = a(n)X^(n) + a(n-1)X^(n-1) + • • • + a0
    and also in the form
    fix) = b(n)X^(n) + b(n-1)X^(n-1) + • • • + b0

    Prove that a(i)=b(i) for all i=0,1,2,...,n-1,n

    2. Relevant equations
    3. The attempt at a solution

    0 = f(x) - f(x) = (a(n)-b(n))X^(n) + (a(n)-b(n))X^(n) + ... + a0 - b0
    Need to prove that Di = (a(i)-b(i)) = 0 for i = 0,1,2,3,4,...n

    But I do not know how.

    This is the answer from book, using the method of contradiction:
    Suppose that there exists some index i such that Di does not equal 0.
    Let m be the largest of these indices, so that we can write
    0 = D(m)X^(m) + ... + D0 for all x and D(m) does not equal 0. This contradicts Theorem 2. Therefore we conclude that Di = 0 for all i = 1, . . . , n, thus proving the theorem.

    Theorem 2 is as stated:
    Let f be a polynomial. Let a0,a1,...,a(n-1),a(n) be numbers such that a(n) does not equal 0, and such that we have: f(x) = a0 + a1X + ... + a(n)X^(N) for all x. Then f has at most n roots.

    I don't understand the contradiction between theorem 2 and the statement regarding D(m). Could someone explain?
  2. jcsd
  3. May 24, 2016 #2


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    The polynomial with the D coefficients is zero for all x and therefore every real number is a root of it. Therefore it has more than m roots. But theorem 2 says it cannot have more than m roots.
  4. May 24, 2016 #3
    Thank you, answers it perfectly
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