Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let $$ f(x)=\prod_{i=0}^n\left(x-\frac{i}{n}\right) $$

where $x\in[0,1]$.

How to deduce a bound for it? Particularly, I want to prove $$ |f(x)|\leq\frac{(n+1)!}{n^{n+1}} $$

share|cite|improve this question
up vote 1 down vote accepted

We can do a bit better. Suppose we have $n+1$ points $x_0, \cdots, x_n$ distributed evenly along $[a,b] $ with $x_0=a,x_n=b.$ Then for $x\in [a,b]$, $$ \left|\prod_{i=0}^n (x-x_i)\right|\leq\frac{n!}{4}\left(\frac{b-a}{n}\right)^{n+1} .$$

For your problem this gives $|f(x)|\leq \dfrac{n!}{4n^{n+1}}.$ The bound in the OP is $(n+1)/4$ times larger than this bound.

Hints: By scaling and shifting, it is equivalent to show that for $x\in [0,n]$ we have $|P(x)|\leq n!/4$ where $P(x)=x(x-1)\cdots (x-n).$ To show this, use induction. Note, in the induction step the inequality of the induction hypothesis does not apply over the entire interval for the next case, but we get around this by exploiting some symmetry of $P(x).$

share|cite|improve this answer
I got it, thank you! – hxhxhx88 Dec 18 '12 at 8:02

Hint: start by writing each term as $\frac{nx-i}n$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.