Let $a=x_0,x_1,...,x_n=b$ are $n+1$ points which are equally spaced in $[a,b]$. The distance between consecutive terms is $h= \frac{b-a}{n}$ and $x \in [a,b]$
Show that $ \biggr|\prod_{i=}^n (x-x_i) \biggr| \leq \frac{h^{n+1}.n!}{4} $
I have written
$ \biggr|\prod_{i=}^n (x-x_i) \biggr| = \biggr| (x-x_0)(x-x_0-h)(x-x_0-2h)...(x-x_0-nh) \biggr|$
I know it is so easy but I couldn’t see in no way. I think I have written something unnecessary. Thanks for any help..
Argue that the maximal error occurs in the boundary intervals. Set $x=sh$. Then $$\prod_{k=0}^n|x-x_k|=h^{n+1}|s(1-s)|\cdot|2-s|\cdots|n-s|$$ and the bound should be obvious.
More generally, for $x=mh+s$, $m=0,...,n$, you get
$$h^{n+1}|m+1-(1-s)|⋯|2-(1-s)|⋅|s(1-s)|⋅|2-s|⋯|n-m-s|\le \frac14h^{n+1}(m+1)!(n-m)!$$
Take the next term into the maximum calculation, thus maximizing $s(1-s)(2-s)$ and find the maximum for $s\in [0,\frac32]$ at the interval ends or the root of $$ 0=3s^2-6s+2\iff s=1-\frac1{\sqrt3} $$ so that the bound can be decreased to $$ h^{n+1}\frac{n!}{3\sqrt3}. $$