Interpolating a function, to estimate the error, knowledge of the function $$\omega(x)=\prod_{i=0}^k (x-x_i)$$ with $x_i$ being the sampling points, is required. In the equidistant case, this would be: $$x_i=a+\frac{b-a}{k}\cdot i$$
Now I want to calculate/estimate $$E_\omega=\max_{x\in[a,b]}|\omega(x)|.$$ It is easy to see, that $E_\omega\leq(b-a)^{k+1}$ but this is in general very imprecise.
Can one calculate $E_\omega$ in general or at least give a better estimate?
By induction, I've proved that $$\frac{(k-1)!}{4 k^{k+1}}(b-a)^{k+1} \le E_\omega\le \frac{k!}{4 k^{k+1}}(b-a)^{k+1}.$$ We may first "normalize" the expression to $E_\omega =(\frac{b-a}{k})^{k+1} |\prod_{i=0}^k (\frac{k}{b-a} x-i)|$, here $x\in(0,b-a)$. When $k=1$, it is obvious that the maximum for function $x(x-1)$ is achieved at $x=1/2$. For $P_k(x)=\prod_{i=0}^k (x-i)$, we claim that the maximum is achieved at $x\in(0,1/2]$. Since if $k$ holds, for $P_{k+1}=P_k\cdot|x-(k+1)|$, when $x>1/2$, let $y_k$ be the point that achieves maximum of $P_k$. Then $P_{k+1}(y_k)|y_k-(k+1)|>P_k(x)|x-(k+1)|$. In addition, we have proved that $\max P_{k+1} \le \max P_k \cdot (k+1)$. With similar method $\max P_{k+1} \ge \max P_k \cdot (k+1/2)$ Thus by seeing that $P_1=1/4$, we see $$\frac{\prod_{i=2}^{k} (i-1/2)}{4 k^{k+1}}(b-a)^{k+1} \le E_\omega\le \frac{k!}{4 k^{k+1}}(b-a)^{k+1}.$$ PS: This is the evaluation starting at $k=1$, if you calculate from $k=2,k=3...$ numerically, you may get a more precise estimate. (because the root for derivative can be found in more strict interval, say for example, $x\in (0,0.2]$.) Also, the innitial value for iteration is changed.