I would like to find out what the sum estimates and prove that it estimates that function. $$\sum_{j=0}^ml_j(x)*x_j^k=?$$ From the Lagrange interpolation polynomials we know that $$l_k(x) = \prod_{i=0}^m\frac{x-x_j}{x_m-x_j}$$
The sum estimates the $x^k$ as per my assumption, but how can I prove that this Lagrange interpolation polynomials estimates the $f(x)=x^k$ function?
Suppose you meant to write $$ l_{\,j} (x) = \prod\limits_{0\, \le \,i\, \ne \,j\, \le \,m} {{{x - x_{\,i} } \over {x_{\,j} - x_{\,i} }}} \quad \left| {\;x_{\,0} \ne x_{\,1} \ne \cdots \ne x_{\,m} } \right. $$ The $l_{\,j} (x)$ are polynomials in $x$, of degree $m$, which evaluate to $1$ in ${x_{\,j} }$ and are null in the other points.
If you combine them with the values that a polynomial of degree $k <=m$ takes on $x_{\,0} ,x_{\,1} , \cdots ,x_{\,m} $ you will get that polynomial.
So $$ \quad x^{\,k} = \sum\limits_{0\, \le \,j\, \le \,m} {l_{\,j} (x)\;x_{\,j} ^{\,k} } \quad $$ if $k<=m$