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Proving in math

  1. Aug 27, 2004 #1

    Alkatran

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    Let's say I wanted to prove that, given n points, it takes a maximum of a (n-1)th degree polynomial to represent them all. How would I do it? My instinct is to just say because you need a max of (n-1) max/mins ...
     
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  3. Aug 27, 2004 #2

    matt grime

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    what does represent mean in this context? and are you sure you mean "maximum"

    what degree one polynimial "represents" the two points 0 and 1. and point in what space? R, R^2, R^3...?
     
  4. Aug 27, 2004 #3

    Alkatran

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    I mean that, given 1 point that your equation must touch, you need a 0th degree equation. Given 2 you need a 1st, etc...

    For example, if you are given a set of points with 2 elements:
    (a,b), (c,d)
    You need a 1st degree equation, or line.
    y = mx + e
    The correct value of m and e will hit both points.

    Similarly, if you have 3 points, you need a quadratic.
     
  5. Aug 27, 2004 #4

    matt grime

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    In that case, given n points in the plane with distinct x coords, there exists a degree n-1 polynomial passing through them, since a degree n-1 poly has n coefficients and therefore you have a system of n linearly independent equations in n unknowns to solve.

    you don't mean maximum at all since given n points then there is a polynomial of degree r=>n-1 passing through those points (again with distinct x values) which is unique when r=n-1.
     
  6. Aug 27, 2004 #5
    Alkatran,
    If u go through Lagrange Interpolation method, u would see how lagrange came up with an extremely simple way to do it!
     
  7. Aug 27, 2004 #6

    Alkatran

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    I'm aware of how to solve the problem. My question was how do I prove that I will never need a 5th degree equation for 5 points?
     
  8. Aug 27, 2004 #7
    Lagrange Interpolation Method works for any given n points.
    Hence Proved!
     
  9. Aug 27, 2004 #8

    matt grime

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    The proof that the equations formed by substituting in the n points are linearly independent is called the vandermonde determinant.
     
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