Proof

First consider a polynomial of degree less than or equal to

. In this case, the data points can be used to construct the Lagrange Polynomial of degree :

,

where, of course, in the product term. Since we assumed that was a polynomial of degree less or equal to ,

and the error term disappears. So,

.

(Again, in the product term!)

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Consider then the case where is of degree less than or equal to . In this case, one can divide this polynomial by to obtain the expression

,

where both and are polynomials of degree less then .

Then,

,

and therefore,

.

Hence,

,

since is of degree less than or equal to . But, since are the zero's of ,

.

This then proves our Theorem:

.

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