0
$\begingroup$

With reference to this irrationality proof of $\zeta(3)$, I need to find the maximum of $$f(x_1,x_2,...,x_n)= \frac{x_1(1-x_1)x_2(1-x_2)...x_n(1-x_n)}{1-(1-x_1x_2...x_{n-1})x_n}$$ where $0<x_i<1$ for all $1\leq i\leq n$.

I tried to find an upper bound of $f$ which is not its maximum value as follows: $$1-(1-x_1x_2...x_{n-1})x_n=1-x_n+x_1x_2...x_{n}\geq 2\sqrt{1-x_n}\sqrt{x_1x_2...x_{n}} $$ where the last inequality follows from AM-GM inequality since $1-x_n>0$ and $x_i>0$ for all $1\leq i\leq n$. So we get $$f(x_1,x_2,...,x_n)\leq \frac{\sqrt{x_1}(1-x_1)\sqrt{x_2}(1-x_2)...\sqrt{x_{n-1}}(1-x_{n-1})\sqrt{x_{n}(1-x_{n})}}{2}$$ Since for all $1\leq i\leq n-1$, $\max_{0<x_i<1}{\sqrt{x_i}(1-x_i)}= \frac{1}{\sqrt{3}}(1-\frac{1}{3})$ and $\max_{0<x_n<1}{\sqrt{x_{n}(1-x_{n})}}=\frac{1}{2}$, so we get $$f(x_1,x_2,...,x_n)\leq \frac{(\frac{2}{3\sqrt{3}})^{n-1}}{4}$$ So for $n=3$, $$\frac{x_1(1-x_1)x_2(1-x_2)x_3(1-x_3)}{1-(1-x_1x_2)x_3}\leq \frac{(\frac{2}{3\sqrt{3}})^{2}}{4}\approx 0.037037037$$
In this irrationality proof of $\zeta(3)$ we have $$\frac{x_1(1-x_1)x_2(1-x_2)x_3(1-x_3)}{1-(1-x_1x_2)x_3}\leq (\sqrt{2}-1)^4\approx 0.02943725$$ So the above method does not give the maximum. Any help will be highly appreciated. Thank you!

Edit What is the maximum value of $f(x_1,x_2,...,x_{11})$?

$\endgroup$
6
  • $\begingroup$ Why not try to apply the brute force and compute partial derivatives? The function is $0$ on boundary and is continuous so it must have a maximum inside the cube. Partial derivatives at this point must be $0$. $\endgroup$
    – Salcio
    Aug 12, 2023 at 16:18
  • $\begingroup$ @Salcio Thanks a lot. Nice idea indeed. But then we have to solve equation in $n$ variables which I am not able to solve $\endgroup$
    – Max
    Aug 12, 2023 at 17:06
  • $\begingroup$ It seems that you can restrict to $x_1,\ldots,x_{n-1}\le {1\over 2}\le x_n$ $\endgroup$ Aug 12, 2023 at 17:14
  • $\begingroup$ @Max I hope you finally find a closed form to this maximum! $\endgroup$
    – Cesareo
    Aug 13, 2023 at 13:14
  • $\begingroup$ @Max Why do you change the question to finding the maximum of $f(x_1, x_2, \cdots, x_{11})$? I think you should keep the original and add something like"Also, what is the maximum for $n = 11$?" Because the answers found the maximum for general $n$. $\endgroup$
    – River Li
    Aug 18, 2023 at 23:00

3 Answers 3

2
$\begingroup$

This is a partial answer reducing the problem to two variables, with no differential calculus. Next one variable can be eliminated by studying the partial derivative.

First of all the function is is positive for $0<x_k<1,$ $1\le k\le n$ and vanishes on the boundary $x_k=0$ or $x_k=1$ for some $1\le k\le n.$ Therefore it attains the maximal value inside the region.

Let $n\ge 3.$ Assume the maximum is attained at $(a_1,a_2,\ldots, a_n).$ The expression is of the form $$A_n{a_1a_2[1+a_1a_2-(a_1+a_2)]\over B_n+C_na_1a_2}$$ where $A_n,B_n,C_n>0$ are independent of $a_1$ and $a_2.$ Given the product $a_1a_2$ the maximal value is attained when $a_1+a_2$ is the smallest possible, i.e. $a_1=a_2.$ This reasoning implies that $a_1=a_2=\ldots=a_{n-1}.$ Therefore we are down to the function in two variables $x:=x_1=\ldots =x_{n-1}$ and $y:=x_n$ of the form $$g(x,y):=(1-x)^{n-1}{x^{n-1}y(1-y)\over 1-(1-x^{n-1})y}$$
The equation $\partial g/\partial y=0$ leads to $$x^{n-1}y^2-2y+1=0$$ Eliminating either $y$ or $x$ we obtain a one variable function of either $x$ or $y.$

$\endgroup$
6
  • $\begingroup$ $(+1)$. Elegant answer indeed. How did we get that the expression can be written as $$A_n{a_1a_2[1+a_1a_2-(a_1+a_2)]\over B_n+C_na_1a_2}$$? $\endgroup$
    – Max
    Aug 12, 2023 at 18:42
  • $\begingroup$ $A_n=a_3(1-a_3)\ldots a_n(1-a_n),$ $B_n=1-a_n,$ $C_n=a_3\ldots a_n$ and $a_1a_2(1-a_1)(1-a_2)=a_1a_2[1+a_1a_2-a_1-a_2]$ $\endgroup$ Aug 12, 2023 at 18:52
  • $\begingroup$ Thanks. Please write how is maximum value of $f$ related with your defined function $g$? $\endgroup$
    – Max
    Aug 12, 2023 at 18:55
  • $\begingroup$ The maximum of $f$ is equal the maximum of $g$, $0<x,y<1$ where $x_n=y,$ $x_1=\ldots =x_{n-1}=x.$ $\endgroup$ Aug 12, 2023 at 18:57
  • $\begingroup$ I have simplified the reasoning getting rid of differential calculus. The final function of one variable can be studied numerically. I do not expect easy formulas for its maximal value. $\endgroup$ Aug 13, 2023 at 6:05
1
$\begingroup$

Assuming that the minimum is located in a interior point to $0\le x_i\le 1, i = 1,\cdots,n$ and calling $p_n(x) = \prod_{i=1}^nx_i$ and $q_n(x) = \prod_{i=1}^n(1-x_i)$ we have

$$ f_n(x) = \frac{p_n(x)q_n(x)}{1-x_n+p_n(x)} $$

so having in mind that $\frac{d}{d x_k}p_n(x) = \frac{p_n(x)}{x_k}$ and $\frac{d}{d x_k}q_n(x) =-\frac{q_n(x)}{1-x_k}$ we have for $k\ne n$

$$ \frac{d}{d x_k}f_n(x) = q_n(x)p_n(x)\frac{x_k p_n(x)+x_n-2x_k(x_n-1)-1}{x_k(x_k-1)(1-x_n+p_n(x))^2}=0 $$

and for $k=n$ analogously

$$ \frac{d}{d x_n}f_n(x) =\frac{q_n(x)p_n(x)(x_np_n(x) -(x_n-1)^2)}{x_n(x_n-1)(1-x_n+p_n(x))^2}=0 $$

and eliminating $p_n(x)$ we arrive at

$$ x_k = \frac{x_n}{x_n+1} $$

The general expression for $f_n(x)$ can be written as

$$ f_n(x) = f_n(x_n) = \frac{(1-x_n) x_n^n \left(1-\frac{x_n}{x_n+1}\right)^{n-1}}{(x_n+1)^{n-1} \left(\frac{x_n^n}{(x_n+1)^{n-1}}-x_n+1\right)} $$

The maximum can be obtained with a procedure to obtain the roots of a polynomial by solving

$$ f_n(x_n)'= 0 $$

NOTE

Calculating for $n=3$ we obtained

$$ f_3 = \frac{1-\frac{1}{\sqrt{2}}}{2 \sqrt{2} \left(\frac{1}{2}+\frac{1}{\sqrt{2}}\right) \left(1+\frac{1}{\sqrt{2}}\right)^2}\approx 0.02943725152285942 $$

for $x_3 = \frac{1}{\sqrt{2}}$

For $n = 4$ we obtained

$$ f_4 \approx 0.008964838545691446, \ \ \ x_4 \approx 0.7759185953243913 $$

Attached a MATHEMATICA script to calculate for $n$ and shown in B

Clear[p, q, Fn]
B = {};
p[n_, x_] := x Product[x/(1 + x), {i, 1, n - 1}]
q[n_, x_] := (1 - x ) Product[1 - x/(1 + x), {i, 1, n - 1}]
Fn[n_, x_] := p[n, x] q[n, x]/(1 - x + p[n, x])
For[k = 2, k <= 10, k++,
  Fkmax = -10^10;
  sols = Solve[D[Fn[k, x], x] == 0, x, Reals];
  nsols = Length[sols];
  For[j = 1, j <= nsols, j++,
   xk = x /. sols[[j]];
   If[0 <= xk <= 1, Fk = Fn[k, x] /. sols[[j]], Fk = -10^10];
   If[Fk >= Fkmax, Fkmax = Fk; xkmax = xk]];
  AppendTo[B, {Fkmax, xkmax}]];

B // N
$\endgroup$
6
  • $\begingroup$ $(+1)$ Excellent answer. But what is the maximum value? $\endgroup$
    – Max
    Aug 12, 2023 at 18:43
  • $\begingroup$ @Max Please. See the obtained results. $\endgroup$
    – Cesareo
    Aug 12, 2023 at 21:39
  • $\begingroup$ Thanks. But we do not have a maximum value of our $f$ by one expression for all $n$ $\endgroup$
    – Max
    Aug 13, 2023 at 6:40
  • 1
    $\begingroup$ Without wanting to be flippant, may I know who told you that there is a closed form for this maximum? $\endgroup$
    – Cesareo
    Aug 13, 2023 at 7:57
  • 1
    $\begingroup$ (+) I am surprised that someone downvoted your answer, which is much more transparent than mine $\endgroup$ Aug 13, 2023 at 13:50
1
$\begingroup$

Let $$u := x_1x_2 \cdots x_{n-1} \in (0, 1).$$

By AM-GM, we have \begin{align*} \frac{x_n(1 - x_n)}{1 - (1 - u)x_n} &= \frac{1 + u}{(1 - u)^2} - \left[\frac{1 - (1 - u)x_n}{(1 - u)^2} + \frac{u}{(1 - u)^2}\cdot \frac{1}{1 - (1 - u)x_n}\right]\\ &\le \frac{1 + u}{(1 - u)^2} - 2\sqrt{\frac{1 - (1 - u)x_n}{(1 - u)^2} \cdot \frac{u}{(1 - u)^2}\cdot \frac{1}{1 - (1 - u)x_n}}\\ &= \frac{1 + u}{(1 - u)^2} - \frac{2\sqrt u}{(1 - u)^2}\\ &= \frac{1}{(1 + \sqrt u)^2} \tag{1} \end{align*} with equality if $x_n = \frac{1}{1 + \sqrt u}$.

By AM-GM, we have \begin{align*} &(1 - x_1)(1 - x_2)\cdots (1 - x_{n-1})\\ \le{}& \frac{1}{(n - 1)^{n - 1}}(1 - x_1 + 1 - x_2 + \cdots + 1 - x_{n-1})^{n-1}\\ ={}& \frac{1}{(n - 1)^{n - 1}}\Big[n - 1 - (x_1 + x_2 + \cdots + x_{n-1})\Big]^{n-1}\\ \le{}& \frac{1}{(n - 1)^{n - 1}} \Big[n - 1 - (n - 1)\sqrt[n-1]{x_1x_2 \cdot x_{n-1}}\Big]^{n-1}\\ ={}& \left(1 - \sqrt[n-1]{u}\right)^{n-1}\tag{2} \end{align*} with equality if $x_1 = x_2 = \cdots = x_{n - 1}$.

From (1) and (2), we have $$f(x_1, x_2, \cdots, x_n) \le g(u) := u \cdot \frac{1}{(1 + \sqrt u)^2} \cdot \left(1 - \sqrt[n-1]{u}\right)^{n-1} \tag{3}$$ with equality if $x_1 = x_2 = \cdots = x_{n-1}$ and $x_n = \frac{1}{1 + \sqrt{x_1x_2 \cdots x_{n-1}}}$.

Thus, equivalently, we turn to find the maximum of $g(u)$ subject to $u \in (0, 1)$.
(Note: Since $g(0)= g(1) = 0$, this optimization problem is well-defined.)

Letting $v = u^{\frac{1}{2(n-1)}} \in (0, 1)$, equivalently, we turn to find the maximum of $$h(v) := \frac{v^{2n - 2}(1 - v^2)^{n-1}}{(1 + v^{n - 1})^2}$$ subject to $v \in (0, 1)$. Since $h(0) = h(1) = 0$, this optimization problem is well-defined. Also, the maximum occurs at some $v \in (0, 1)$ with $h'(v) = 0$.

We have $$h'(v) = - \frac{2(n - 1)v^{2n-2}(1 - v^2)^{n - 2}(v^{n + 1} + 2v^2 - 1)}{v(1 + v^{n-1})^3}.$$ Let $v_0 \in (0, 1)$ be the unique real root of $v^{n + 1} + 2v^2 - 1 = 0$ on $(0, 1)$. Note that $v_0$ is the unique real root of $h'(v) = 0$ on $(0, 1)$. Thus, the maximum of $h$ is $h(v_0)$.

Thus, the maximum of $f(x_1, x_2, \cdots, x_n)$ is equal to $h(v_0)$ when $x_1 = x_2 = \cdots = x_{n-1} = v_0^2$ and $x_n = \frac{1}{1 + v_0^{n-1}}$, where $v_0 \in (0, 1)$ be the unique real root of $v^{n + 1} + 2v^2 - 1 = 0$ on $(0, 1)$.

$\endgroup$
2
  • $\begingroup$ $(+1)$ for your answer. Please edit it to answer my edited question. $\endgroup$
    – Max
    Aug 19, 2023 at 8:12
  • 1
    $\begingroup$ @Max In the answers here, you just let $n = 11$ to get the answer. In my answer, the maximizer $v_0$ is the unique real root of $v^{12} + 2v^2 - 1 = 0$ on $(0, 1)$. The maximum is equal to $h(v_0)$. $\endgroup$
    – River Li
    Aug 19, 2023 at 10:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .