Dgdrxrt

Suppose that $x in [0,1]$ and the points $x_1, x_2,ldots x_n$ are evenly spaced in the interval $[0,1]$. I am trying to find a tight bound for the maximum of:

$$(x - x_1)(x-x_2)cdots(x-x_n) $$

I realize that the above expression will be smaller in absolute value than the absolute value of the smallest term $(x-x_i)$ for $ 1le i le n$ since each term will is less than $1$. Thus, if there are $n$ equally spaced points, each distanced $frac1n-1$ apart, then smallest term will be bounded above by $frac12(n-1)$.

I am wondering if there is a way to get a tighter bound than the one I have found above.

One finds that the factors, excluding those for the points directly around $x$, are largest in the first and last interval. As the maxima of third degree polynomials are still accessible, for an $xin[x_1,x_2]$ we get the first bound
$$
|(x-x_1)cdots(x-x_n)|le|(x-x_1)(x-x_2)(x-x_3)|cdot |(x_4-x_1)cdots(x_n-x_1)|
$$
Now set $s=x-x_2$ and $x_2-x_1=h=x_3-x_2$, more generally $x_k-x_1=(k-1)h$, so that $sin[-h,0]$. The cubic factor reads as $(s+h)s(s-h)=s^3-sh^2$ which has its extrema at $s=pmfrachsqrt3$ with value $mpfrac2sqrt3h^39$.

The upper bound of this method is thus
$$
|(x-x_1)cdots(x-x_n)|lefrac2sqrt3h^39cdot 3hcdot 4hcdots(n-1)hlefracsqrt39frac(n-1)!(n-1)^nsim fracsqrt6pi9sqrtn-1e^1-n.
$$

Let $ngeq3$ and let $x_i:=fracin-1$ so that we want to determine the maximum of the polynomial
$$f_n(x):=prod_i=1^n(x-x_i)=prod_i=1^nleft(x-frac1-in-1right),$$
on the interval $[0,1]$. If $n$ is odd then for all $xin(x_1,x_2)$ and $i>2$ we have
$$frac1-in-1<x-x_i<frac2-in-1<0,$$
from which it follows that
$$(x-x_1)(x-x_2)prod_i=3^nfrac2-in-1
<f_n(x)
<(x-x_1)(x-x_2)prod_i=3^nfrac1-in-1.$$
The maximum of $(x-x_1)(x-x_2)$ on the interval $(x_1,x_2)$ is at the midpoint $fracx_1+x_22=frac12(n-1)$, yielding the following bounds for the maximum $M_n$ of $f_n$ on the interval $(x_1,x_2)$:
$$M_n
>left(tfrac12(n-1)-x_1right)left(tfrac12(n-1)-x_2right)prod_i=3^nfrac2-in-1
=frac14frac(n-2)!(n-1)^n.
$$
$$M_n
<left(tfrac12(n-1)-x_1right)left(tfrac12(n-1)-x_2right)prod_i=3^nfrac1-in-1
=frac14frac(n-1)!(n-1)^n,$$
Similarly, if $n$ is even then for all $xin(x_2,x_3)$ and $i>3$ we have
$$(x-x_1)(x-x_2)(x-x_3)prod_i=4^nfrac3-in-1
<f_n(x)
<(x-x_1)(x-x_2)(x-x_3)prod_i=4^nfrac2-in-1,$$
and some basic algebra shows that the maximum of $(x-x_1)(x-x_2)(x-x_3)$ is at $x=frac1+sqrt33(n-1)$, so
$$M_n>left(tfrac1+sqrt33(n-1)-x_1right)
left(tfrac1+sqrt33(n-1)-x_2right)
left(tfrac1+sqrt33(n-1)-x_3right)prod_i=4^nfrac3-in-1
=2fracsqrt39frac(n-3)!(n-1)^n,$$
$$M_n<left(tfrac1+sqrt33(n-1)-x_1right)
left(tfrac1+sqrt33(n-1)-x_2right)
left(tfrac1+sqrt33(n-1)-x_3right)prod_i=4^nfrac2-in-1
=2fracsqrt39frac(n-2)!(n-1)^n.$$

Proof that the maximum is in $(x_1,x_2)$ if $n$ is odd, and in $(x_2,x_3)$ if $n$ is even:

(A bit messy, might clean up later)

It it not hard to see that for $xin(x_k,x_k+1)$
we have $operatornamesgnf(x)=(-1)^n-k$. So the maximum is in some interval $(x_k,x_k+1)$ with $kequiv npmod2$.

The symmetry in the product shows that for all $xin(0,1]$ we have
$$fleft(x-frac1n-1right)
=fracx-x_nx-x_1fleft(xright)
=(1-x^-1)f(x),$$
and of course for $xin[tfrac12,1]$ we have $|1-x^-1|leq1$, so the maximum value of $f$ on the interval $[tfrac12,1]$ is assumed on the interval $(x_n-2,x_n-1)$. It is clear that
$$f(1-x)=(-1)^nf(x),$$
for all $xin[0,1]$, from which it follows that the maximum value of $f$ on the interval $[0,1]$ is assumed on $(x_1,x_2)$ if $n$ is odd, and on $(x_2,x_3)$ if $n$ is even.

The polynomial $f$ of degree $n$ has precisely $n$ distinct roots in the interval $[0,1]$. It follows that the polynomial $f'$ has precisely one root in each interval $(x_k,x_k+1)$ for $kin1,ldots,n-1$. The maximum of $f$ is then assumed at the unique root of $f'$ in the interval $(x_k,x_k+1)$, where $k=1$ if $n$ is odd and $k=2$ if $n$ is even.