Bound on $(x-x_0)(x-x_1) \dots (x-x_n)$ for equidistant points

Question

I have to estimate the error bound of a lagrange polynomial and to do that i have to show that $|(x-x_1)(x-x_2) \dots (x-x_n)| \leq \frac{1}{4}(n+1)!h^{n+1}$ for equidistant points where $x_{j+1}-x_j = h$ and where $x \in [x_0,x_n]$. Since all $|x-x_j|$ are decreasing in the interval $[x_0,x_j]$ and increasing in the interval $[x_j,x_n]$, $|x_0-x_j|$ must be the maximum of $|x-x_j|$ from $j = 0$ to $j= \frac{N}{2}$ and $|x_n-x_j|$ must be the maximum of all $|x-x_j|$ from $j=\frac{N+1}{2}$ to $j = n$. If I do that though i dont get the right bound so I'm wondering what i doing wrong. Any help would be appreciated.

Answer 1

Locate your $x$ in some interval $[x_k, x_{k+1}]$ and then use the estimate: $$|(x-x_k)(x-{x_{k+1}})| = (x-x_k)(x_{k+1}-x)\leq\dfrac{(x_{k+1}-x+x-x_k)^2}{4} = \dfrac{h^2}{4}.$$

For the remaining terms, you can use much cruder bounds like $|x-x_{k-j}|\leq x_{k+1}- x_{k-j}= (j+1)h$ and so on.