Let $L_i(x)$ be Lagrange basis polynomials for $n+1$ points $(x_0,y_0),\ldots, (x_n,y_n)$. How do you prove that $\sum_{i=0}^n (x-x_i)^pL_i(x)=0$ for $p\leq n$?
[Math] Sum of Lagrange basis polynomials
numerical methodspolynomials
numerical methodspolynomials
Let $L_i(x)$ be Lagrange basis polynomials for $n+1$ points $(x_0,y_0),\ldots, (x_n,y_n)$. How do you prove that $\sum_{i=0}^n (x-x_i)^pL_i(x)=0$ for $p\leq n$?
Best Answer
Use the interpolation property of the functions $L_j$. For any polynomial $f$ of degree at most $n$ the equality
$$f = \sum_{j=0}^nf(x_j)L_j$$
holds. Apply this to $f(x,y)=(y-x)^p$. For any fixed $y$ this is a polynomial in $x$ of degree $p\leq n$ and interpolation gives
$$f(x,y)=\sum_{j=0}^nf(x_j,y)L_j(x)=\sum_{j=0}^n(y-x_j)^pL_j(x).$$
If $0<p\leq n$ then $f(x,x)=0$ and the result follows. For $p=0$ we have $f(x,x)=1$ and the interpolation above becomes
$$1=\sum_{j=0}^nL_j$$
as was already noted in the comment.