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$\underline{Lemma}$: Let $f:[a,b] \mapsto \mathbb{R}$, $n$ times differenciable. and let $y_i = f(x_i)$ for $x_0, x_1, x_2, ..., x_n \in [a,b]$ distincts. Then, there exist $\xi \in [a,b]$ s.t. $\delta^ny[x_0, ..., x_n] = \dfrac{f^{(n)}(\xi)}{n!}$. (where $\delta^ny[x_0, ..., x_n]$ is the divided difference)

$\underline{Proof}$: With no loss of generality: $n>1$ and $a=x_0

Let $p_n(x) = c_0 +c_1(x-x_0) + ... + c_n(x-x_O)...(x-x_{n-1})$ the interoplation polynomial for $(x_i, y_i)$

Let $d(x) = f(x) - p_n(x) \rightarrow d(x_i) = 0, \forall i = 0, ..., n$

$d(x)$ has $n+1$ roots.

Using Rolle's theorem, we come to the conclusion that $d^{(n)}$ has $1$ root. Let us call this zero $\xi$

Now $d^{(n)}(\xi) = 0 \implies f^{(n)} (\xi) = p_n^{(n)}(\xi) = n!C_n$

Therefore $\dfrac{f^{(n)}(\xi)}{n!} = C_n = \delta^ny[x_0, ..., x_n]$

I am having trouble understanding the following: Why is $n!C_n = p_n^{(n)}(\xi)$?

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    Well first agree on 2 things: 1) when you derive P_n(x) n times all of the factors containing c_0,c_1,...,c_n-1 disappear. 2) (x-x0)...(x-x_n-1) is a polynomial of degree n and its leading factor is x^n. when you derive this polynomial n times you get n!. – 2017-01-11
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    @zokomoko Yes, of course, thank you – 2017-01-11

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