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Proving the uniqueness of a polynomial

  1. May 17, 2006 #1
    If a polynomial [tex]p(x)=a_0+a_1x+a_2x^2+ \ldots +a_{n-1}x^{n-1}[/tex] is zero for more than [tex]n-1[/tex] x-values, then [tex]a_0=a_1= \ldots =0[/tex]. Use this result to prove that there is at most one polynomial of degree [tex]n-1[/tex] or less whose graph passes through n points in the plane with distinct x-coordinates.

    Let p(x) be a polynomial of degree n-1 or less such that [tex]p(x_1)=y_1, p(x_2)=y_2, \ldots,p(x_n)=y_n[/tex]. Thus if there are n distinct points, then there will be n equations. If for each of these equations we subtract [tex]y_i[/tex] then we have n root to a polynomial of degree [tex]n-1[/tex] or less. So for any given root we have [tex]a_jx_i^{j}[/tex] where i denotes the coordiante of [tex](x_i,y_i)[/tex] and j denote the subscript for the coefficents of the terms of p(x). Since we can subtract [tex]y_i[/tex] from any of the n terms of the polynomial we have in general [tex]a_jx_i^{j}=y_j[/tex] where [tex]a_j[/tex] is the only variable. Since for any given point this equation has a unique solution for [tex]a_j[/tex], and since the coefficents make up the polynomial p(x), p(x) is unique.

    Does anyone see a problem with the above argument?

    Thanks in advance.
     
    Last edited: May 17, 2006
  2. jcsd
  3. May 17, 2006 #2

    0rthodontist

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    I don't understand what you're saying. You say "we have n roots to a polynomial of degree n-1 or less"--it looks like you have n roots to n possibly different polynomials.

    I think they want you to start (with the assumption that all polynomials are of degree n-1 or less) by letting P be a polynomial that passes through the n points. Now let Q be another such polynomial. Then what are the roots of P - Q, and what does that tell you?
     
    Last edited: May 17, 2006
  4. May 17, 2006 #3
    I mean the polynomial p(x) that satisfies the conditions that it passes through n distinct points. If you take p(x) and put [tex]x_i[/tex] in for x, then you must get [tex]y_i[/tex] since that is how I defined p(x). Thus you have n different equations with each solution being [tex]y_i[/tex]. Since you can subtract the y value from both sides you then have n equations with n roots.

    I started down this path, but I didn't see how to finish using the given information. Then I saw this path, which I think works, but I want to make sure there isn't something I'm overlooking.
     
  5. May 17, 2006 #4

    0rthodontist

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    n equations, but not n roots to any particular single polynomial.

    I reworded my other post in a way that may be clearer.
     
  6. May 17, 2006 #5
    Let me better define my function.

    Let [tex]p:R->R[/tex] be defind such that [tex]p(x_i)=y_i[/tex] and p(x) is a polynomial, and the coefficents of p(x) are not necessarily distinct.

    True, but we are just talking about p(x) (now better defined) and we know that the [tex]a_jx_i^j-y_i=0[/tex] where j=0,1,...,n-1 and i=1,2,...,n. Wouldn't that be enough to conclude then that the coefficents of p(x) are distinct? If they are distinct then p(x) has to be distinct for the given points that it passes through, no? I do realize that this could be proved in another manner, but how can I prove it using the information given in the problem?

    Sorry if I'm being stubborn and all, but I honestly believe that my argument is correct. Of course, I've been wrong many times before so I could well be wrong this time too.
     
  7. May 17, 2006 #6

    0rthodontist

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    It is not necessarily true that a_j x_i^j - y_i = 0 for any i or j.

    I think the problem is really asking for you to do it the way I suggested. You said you had tried something like that--where are you getting stuck?
     
  8. May 18, 2006 #7
    Why not? It's stated in the problem that if you have p(x) of degree n-1 with n roots, then all coefficents are 0. So if you have p(x_i)=y_i then consider p(x_i)-y_i=0. Here there are n polynomials all of degree n-1 or less, and together there are n roots. Therefore the coeffiencts of p are all zero. For example look at the constant term [tex]a_0*x_i^0=0[/tex]. But only when we subtract [tex]y_i[/tex] from the RHS do we know this. So wouldn't it be correct to say that we can write any term of this polynomial as [tex]a_jx_i^j-y_i=0[/tex]?

    For example:
    [tex]
    p(x_i)=a_0+a_1x_i^1+a_2x_i^2+ \ldots +a_{n-1}x_i^{n-1}=y_i
    [/tex]

    Now if we subtract the y's from both sides we get a new polynomial such that for some different or possibly the same x's p(x)=0.

    Now this this new polynomial has n roots and degree n-1. But what difference does it make if I make the new polynomial
    [tex]
    p(x_i)=(a_0-y_i)+a_1x_i^1+a_2x_i^2+ \ldots +a_{n-1}x_i^{n-1}=0
    [/tex]
    or
    [tex]
    p(x_i)=a_0+(a_1x_i^1-y_i)+a_2x_i^2+ \ldots +a_{n-1}x_i^{n-1}=0
    [/tex]

    It doesn't, but in each case we have an equation [tex]a_jx_i^j-y_j=0[/tex].
     
    Last edited: May 18, 2006
  9. May 18, 2006 #8
    I see the problem.....

    You are correct (imagine that, I'm wrong) that [tex]a_jx_1^j-y_i=0[/tex] is not true for all i and j....only for j=0.

    I guess I'm gonna have to go back to the method you suggest. I'm gonna work on that....I'll post what I come up with.

    Thanks
     
    Last edited: May 18, 2006
  10. May 18, 2006 #9

    matt grime

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    Woah, you're making this very hard on yourself: if p and q are two polys of degree n-1 passing through those n points then....

    hint if you have to show U=V ever and it is possible it is often easiest to show U-V=0
     
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