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Algebra: If P(x), degree n, shares n points with x^n, then it is x^n

  1. Nov 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that if a Polynomial P(x) of degree n, shares n points with xn, then P(x)=xn.

    (a more general proof would be almost the same but for two polynomials, but I think I've proved this if this is proved)


    2. Relevant equations
    (FTA) Fundamental Theorem of Algebra - Every P(x) of degree n has at most n roots.

    3. The attempt at a solution
    An idea was to make an analogy between the proof of FTA and this probem. To use for example the factorization: P(x)= ∏(x-ri) Which when proved valid, proves the FTA too. But I don't understand the proofs given online too well.

    But also if we accept FTA, we could also possibly analyze P(x)-xn which shall be called D(x).
    D(x) has degree (n-1) since xn-xn=0.
    Also since P(x) and xn have n equal points, then D(x) has n zeroes.
    But FTA says that the D(x) of degree (n-1) has at most n-1 zeroes. But it has n zeroes. Therefore it must equal 0 itself. P(x)-xn=0 ⇔ P(x)=xn

    Is this correct? Is there any simpler way to prove or address this or the more general problem?
     
    Last edited: Nov 12, 2012
  2. jcsd
  3. Nov 12, 2012 #2

    Dick

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    Looking at D(x) is the right way to do it. BTW D(x) doesn't necessarily have degree (n-1). The largest degree it could have is n-1, but it might have less.
     
  4. Nov 12, 2012 #3

    Ray Vickson

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    You can prove the result without knowing the FTA. If you have
    [tex] x_i^n + a_{n-1} x_i^{n-1} + \cdots + a_0 = x_i^n, i = 1,2,\ldots,n\\
    \text{where } x_1 < x_2 < \cdots < x_n. [/tex]
    you need to show that this gives you
    [tex] a_{n-1} = a_{n-2} = \cdots = a_0 = 0.[/tex]
    Note that you have a set of n equations in the n unknowns aj, and the coefficient matrix is a Vandermonde matrix.

    RGV
     
  5. Nov 16, 2012 #4
    I've looked the Vandermonde matrix up, but I had never seen it before.
    I've set a system which is equivalent to say P(aj)=ajn
    If we subtract xn from both sides. We get p(aj)=0 , p of degree n-1
    If we set all the equations we get a square matrix. Whose determinant can be taken.
    If VC (Vandermonde times coefficients) is a matrix that represents the system of all equations p(aj)=0
    Then we get VC=[0] ⇔ |VC|=0
    V has no null lines and they're not linearly dependent.(actually one a could be zero)
    => |V|≠0, therefore C ought to be 0 since all it does is make a linear combination of the columns of V.
    Or possibly another is to argue that either V is not invertible and |V|=0 which is false, or |V| is invertible and V-1VC=0⇔C=0

    PS:I'm a bit tired today, sorry for the trouble with clearly formalizing my thoughts.
    Thank you for the help so far.
     
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