Let $n \in \Bbb{N}$ and $x \in [0,1]$ prove $$ \left|\prod_{i=0}^n \left(x - \frac{i}{n}\right)\right| \le \frac{n!}{4n^{n+1}}$$

I manage to show that $\left| (x-\frac{n-1}{n})(x-\frac{n}{n})\right| \le \frac{1}{4n^2}$ by taking derivative and finding the maximum, but I didn't manage to show that the rest $\le \frac{n!}{n^{n-1}}$

  • $\begingroup$ have you tried induction? $\endgroup$ – Dr. Sonnhard Graubner Dec 31 '14 at 22:11
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    $\begingroup$ The right side makes it look like Stirling's approximation would help a lot ... $\endgroup$ – Zubin Mukerjee Dec 31 '14 at 22:13
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    $\begingroup$ Two perhaps trivial comments. 1) The given inequality is the same as $\left\lvert \prod_{i=0}^n (nx-i)\right\rvert \le \frac{n!}{4}$. 2) From some experimentation, it appears that the maximum value of this polynomial as a fraction of $\frac{n!}{4}$ declines, perhaps to zero, as $n$ increases, so this is really not a sharp bound at all. For example, when $n=25$, the ratio is about $0.36$; at $n=100$, it is about $0.27$. This makes sense, since the larger $n$ is, the more opportunities there are for $nx$ to be close to an integer. $\endgroup$ – rogerl Dec 31 '14 at 23:44

It seems the following.

For $x\in [0;1]$ put $$f(x)=\left|\prod_{i=0}^n \left(x - \frac{i}{n}\right)\right|.$$

Since $f$ is a continuous function and a set $[0,1]$ is compact, there exists a point $x\in [0,1]$ such that $f(x)=\sup\{f(y):y\in [0,1]\}$. Since $f(y)=f(1-y)$ for each $y\in [0,1]$, without loss of generality we may suppose that $x\le 1/2$. If $n=1$ then $$f(x)=x(1-x)\le\left(\frac{x+1-x}2\right)=\frac 14=\frac{1!}{4\cdot 1^2}.$$ So we suppose that $n\ge 2$. If $x>\frac 1n$ then $$\frac {f\left(x-\frac 1n\right)}{f(x)}=\frac{1-\frac 1n -x}{x}>1.$$

So $x\in\left(0,\frac 1n\right)$. Then

$$f(x)=x\left(\frac{1}{n}-x\right) \left(\frac{2}{n}-x\right)\dots \left(\frac{n}{n}-x\right)\le \frac 1{4n^2}\frac 2n\cdots\frac nn=\frac{n!}{4n^{n+1}}.$$

The graph of the function $f(x)$ for $n=8$:

enter image description here


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