Let $f\in C^{n+1}([a,b])$, $x_0,\dotso, x_n\in[a,b]$ pairwise different and $p\in \mathbb{P}_n$ the interpolation polynomial (of degree n) with $f(x_j)=p(x_j)$ for all $0\leq j\leq n$
Show that for every $j$ there exists $\xi_j\in [a,b]$ with $f'(x_j)-p'(x_j)=\prod_{i\neq j} (x_i-x_j)\cdot\frac{f^{(n+1)}(\xi_j)}{(n+1)!}$
I struggle to find the idea for a proof. The only thing I got so far is, that I want to define a 'help-function', which might look like this:
$h(x)=f(x)-p(x)-\frac{f'(x_j)-p'(x_j)}{\prod_{i\neq j}(x_j-x_i)}\cdot\prod_{i=0}^n (x-x_i)$
But this does not do it yet. And I doubt, that it is a correct approach anyways. I wanted to give a function $h$ with at least $n+2$ roots. The function above has at least $n+1$ roots $x_0,\dotso, x_n$.
I also thought about using some mean-value-theorem. Of the differential calculus or even for integrals, but never thought it through.
Can you give me a hint for this problem? I would like to solve it alone, but I am stuck here. Is the idea with a 'help-function' correct?
Thanks in advance.
You can begin by the corresponding formula for the interpolation error, which is $$f(x)-p(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\cdot \prod_{i=0}^n(x-x_i),$$ that implies $$f'(x)-p'(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\cdot \frac d{dx}\left(\prod_{i=0}^n(x-x_i)\right).$$
The derivative of the product has $n+1$ terms, but only one of them does not contain the factor $(x-x_j)$ (for any $j$ you choose); that term is in fact $$\prod_{i\neq j}(x-x_i),$$ so when you evaluate in $x=x_j$ this is the only term that is non-zero and you get the desired formula. (*)
(*) Actually, you get the formula with $(x_j-x_i)$ instead of $(x_i-x_j)$. I believe this is a typo, because for odd values of $n$ this changes the sign of the expression.