Proving (n-1)th Degree Polynomial Representation of n Points

[Alkatran]

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 ...

 

[matt grime]

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]

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]

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]

Alkatran,
If u go through Lagrange Interpolation method, u would see how lagrange came up with an extremely simple way to do it!

 

[Alkatran]

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]

Lagrange Interpolation Method works for any given n points.
Hence Proved!

 

[matt grime]

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