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Proving in math

by Alkatran
Tags: math, proving
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Alkatran
#1
Aug27-04, 09:41 AM
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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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matt grime
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Aug27-04, 10:18 AM
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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...?
Alkatran
#3
Aug27-04, 10:26 AM
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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.

matt grime
#4
Aug27-04, 10:33 AM
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Proving in math

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.
TenaliRaman
#5
Aug27-04, 10:36 AM
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Alkatran,
If u go through Lagrange Interpolation method, u would see how lagrange came up with an extremely simple way to do it!
Alkatran
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Aug27-04, 10:41 AM
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Quote Quote by TenaliRaman
Alkatran,
If u go through Lagrange Interpolation method, u would see how lagrange came up with an extremely simple way to do it!
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?
TenaliRaman
#7
Aug27-04, 10:46 AM
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Lagrange Interpolation Method works for any given n points.
Hence Proved!
matt grime
#8
Aug27-04, 10:51 AM
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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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