Error bound of Lagrange interpolating For $1 >\varepsilon > 0$ I have an interval $I = [-\varepsilon, \varepsilon]$ and $x_0, \dots, x_n \in I$. L is a function assigning to a function f it's Lagrange interpolation polynomials according the sampling points $x_0, \dots, x_n$.
Now I would like to estimate the error of
$L(T^n_0(f) - f)^{(k)}$ on $I$ for all $k \le n$ (individual estimations are ok) where $T^n_0$ is the n-th order Taylorpolinomial of f at the point $0$ and $f \in C^n(\mathbb{R})$.
My problem is the normal error estimation formula for lagrange does not work in my case because there the function has to be $C^{n+1}$.
 A: Let $p_n∈P_n$ be the interpolation of $f$ with $n+1$ different nodal points $x_0,…,x_n$. Then for every $x∈[a,b]$ there exists a $ξ∈(a,b)$, such that
$$f(x)-p(x) = \frac{f^{(n)}(ξ)-a_nn!}{n!ξ - n!\sum_{j=0}^nx_j}\prod_{j=0}^n(x-x_j)$$
with $a_n$ the $n$-th coefficient of $p_n$.  ($p_n=a_nx^n+…$) 
And therefore we get 
$$|f(x)-p(x)| \leqslant \max_{ξ∈(a,b)}\left|\frac{f^{(n)}(ξ)-a_nn!}{n!ξ - n!\sum_{j=0}^nx_j}\right|\prod_{j=0}^n|x-x_j|$$

How useful is that estimate? 
I would say the same as the standard estimate. You still have a derivative term of $f$. And usually $n$ is not huge. Also the sum of nodal points is nothing really bad. So it depends on your problem, if that estimate gives good values. The only thing that might hurt is $ξ$ in the denominator. 

Proof 
In the proof of the standard estimate you have two cases: $x=x_j$ (nothing to prove) and $x\neq x_j$. In the second case you define the function 
$$F(t)=f(t)-p_n(t)-K(x)\prod_{j=0}^n(t-x_j),$$
which contains at least $(n+2)$ different roots in $[a,b]$, namely $x_0,…,x_n$. The definition of that function is possible, since: 
$$\prod_{j=0}^nx-x_j\neq 0\qquad ⇒ \qquad K(x):=\frac{f(x)-p_n(x)}{\prod_{j=0}^nx-x_j}.$$
Now one would apply $(n+1)$-times Rolle's theorem on $F$, which gives you a root $ξ∈(a,b)$ of the $(n+1)$-th derivative, hence $F^{(n+1)}(ξ)=0$.
Now we can't do that, so we do the best we can: 
Apply Rolle's theorem $n$ times on $F$, which results again in a root $ξ∈(a,b)$ of $F^{(n)}$. We get: 
$$0= F^{(n)}(ξ) = f^{(n)}(ξ) -p^{(n)}_n(ξ) - \frac{d^n}{dt^n}[K(x)\prod_{j=0}^n(t-x_j)]\big|_ξ$$
Now $p_n$ is a polynomial of degree $n$, so 
$$p^{(n)}_n(ξ)=a_n\cdot n!$$
We know that $K(x)$ is independent of $t$, so only the product is of interest. It is:
\begin{align*}&&\prod_{j=0}^n(t-x_j)&=t^{n+1} - \underbrace{[t^nx_0 + t^nx_1 +…+t^nx_n]}_{t^n\sum_{j=0}^nx_j} + \mathcal{O}(t^{n-1})\\
&&&=t^{n+1}-t^n\sum_{j=0}^nx_j + \mathcal{O}(t^{n-1}) \\ \\
⇒&&\frac{d^n}{dt^n}\prod_{j=0}^n(t-x_j)&=n!t - n!\sum_{j=0}^nx_j \\ \\
⇒&&\frac{d^n}{dt^n}[K(x)\prod_{j=0}^n(t-x_j)]\big|_ξ &=K(x)\left(n!ξ - n!\sum_{j=0}^nx_j\right).
\end{align*}
So we now have: 
$$K(x) = \frac{f^{(n)}(ξ)-a_nn!}{\left(n!ξ - n!\sum_{j=0}^nx_j\right)}$$
Which will conclude the stated result: 
$$f(x)-p_n(x) = K(x)\prod_{j=0}^n(x-x_j) = \frac{f^{(n)}(ξ)-a_nn!}{\left(n!ξ - n!\sum_{j=0}^nx_j\right)}\prod_{j=0}^n(x-x_j).$$

Please double-check for mistakes.
