Given: $f \in C^{n+1}[a,b]$, $p_n$ interpolates $f$ at distinct nodes $x_0, \ldots, x_n$.

To Prove: For any $x \in [a,b]$, there exists $\xi_x \in (a,b)$ with: $f(x) - p_n(x) = \frac{f^{(n+1)}(\xi_x)}{(n+1)!} \prod_{i=0}^n (x - x_i)$

Proof:

If $x = x_i$ for some $i$, both sides equal zero (by interpolation), so assume $x \neq x_i$ for all $i$.

Define the error function: $\phi(t) = f(t) - p_n(t) - [f(x) - p_n(x)] \frac{\prod_{i=0}^n (t - x_i)}{\prod_{i=0}^n (x - x_i)}$

Observe that $\phi$ has the following properties:

1. $\phi(x_i) = 0$ for $i = 0, \ldots, n$ (since $f(x_i) = p_n(x_i)$ and the numerator vanishes)
2. $\phi(x) = 0$ (by construction)

Thus $\phi$ has at least $n+2$ zeros in $[a,b]$.

By Rolle's theorem, $\phi'$ has at least $n+1$ zeros in $(a,b)$. Applying Rolle's theorem repeatedly:

• $\phi'$ has $\geq n+1$ zeros
• $\phi''$ has $\geq n$ zeros
• $\vdots$
• $\phi^{(n+1)}$ has $\geq 1$ zero, call it $\xi_x \in (a,b)$

Now compute $\phi^{(n+1)}(t)$: $\phi^{(n+1)}(t) = f^{(n+1)}(t) - p_n^{(n+1)}(t) - [f(x) - p_n(x)] \frac{d^{n+1}}{dt^{n+1}}\left[\frac{\prod_{i=0}^n (t - x_i)}{\prod_{i=0}^n (x - x_i)}\right]$

Since $p_n$ has degree $\leq n$, we have $p_n^{(n+1)}(t) = 0$.

The product $\prod_{i=0}^n (t - x_i)$ is a monic polynomial of degree $n+1$, so: $\frac{d^{n+1}}{dt^{n+1}}\left[\prod_{i=0}^n (t - x_i)\right] = (n+1)!$

Therefore: $\phi^{(n+1)}(t) = f^{(n+1)}(t) - [f(x) - p_n(x)] \frac{(n+1)!}{\prod_{i=0}^n (x - x_i)}$

Setting $t = \xi_x$ where $\phi^{(n+1)}(\xi_x) = 0$: $0 = f^{(n+1)}(\xi_x) - [f(x) - p_n(x)] \frac{(n+1)!}{\prod_{i=0}^n (x - x_i)}$

Solving for $f(x) - p_n(x)$: $f(x) - p_n(x) = \frac{f^{(n+1)}(\xi_x)}{(n+1)!} \prod_{i=0}^n (x - x_i)$

This completes the proof.