Show that the error of quadratic interpolation $\xi$ in equidistant points, the values of which are obtained by rounding and the error is lesser or equal to $\epsilon$, is given by:
$$\xi\le|E|+\frac{5}{4}\epsilon$$
I got as far as $$\prod^n_{i=0} (x-x_i)\frac{f^5}{5!} \le |E|+\frac{5}{4}\epsilon$$
I don't know what |E| is. Help?
You know that $|\hat y_i-f(x_i)|\le ϵ$, where $\hat y_i$ are the rounded function values.
Now \begin{alignat}{1} \left|f(x)-\sum_{i=0}^n\hat y_iL_i(x)\right| &\le \left|f(x)-\sum_{i=0}^n f(x_i)L_i(x)\right|&+\sum_{i=0}^n |\hat y_i-f(x_i)|·|L_i(x)|\\ &\le |E(x)|&+ϵ\sum_{i=0}^n|L_i(x)| \end{alignat} where $L_i(x)$ is the Lagrange interpolation kernel for position $x_i$. The graph of the sum in the last term for $x_i=i$, $i=0,1,2=n$ is