Lagrange interpolation $f(x)-p_n(x)=f[x_0,x_1,...,x_n,x]\prod_{i=0}^n(x-x_i)$

Question

Let $f[x_0,x_1,...,x_k]$ be difference quotient with nodes $x_0,...,x_k$ (as in Newton's interpolation).

Let $p_n$ be Lagrange's interpolation polynomial of function $f$.

Prove for all $x$: $$ f(x)-p_n(x)=f[x_0,x_1,...,x_n,x]\prod_{i=0}^n(x-x_i) $$ (I don't need to prove it when $x=x_i$)

I have tried to prove it using this theorem: $$ f(x)-p(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n(x-x_i) $$ but my teacher said, that I don't need that, and that there is simpler proof. Do you have any ideas?

Answer 1

You can just add an interpolation node... Let $p_n$ be the interpolating polynomial at $x_0, \cdots, x_n$ and $p_{n+1}$ the interpolating polynomial at $x_0, \cdots, x_n, x$, then we have

$$ p_{n+1}(t) = p_n(t) + f[x_0,\cdots, x_n, x](t-x_0)\cdots(t-x_n) $$

Evaluating at $t=x$, you get simply

$$ \underbrace{p_{n+1}(x)}_{=f(x)} = p_n(x) + f[x_0,\cdots, x_n, x](x-x0)\cdots(x-x_n) $$

or

$$ f(x)-p_n(x)=f[x_0,\cdots, x_n, x](x-x0)\cdots(x-x_n). $$

Note: The divided differences are precisely obtained in such a way that $$ p_{k+1}(x) = p_k(x) + c (x-x_0)\cdots (x-x_k), $$ where $p_{k+1}$ is the interp.pol. at $x_0,\cdots x_k$, $p_{k}$ is the interp. pol. at $x_0,\cdots, x_{k-1}$ and the constant $c$ is defined as $f[x_0, \cdots, x_k]$.