4
$\begingroup$

A triangle is drawn in a region bounded by the $x$-axis and a polynomial curve with all real roots, with one side of the triangle on the $x$-axis, as shown in this example:

enter image description here

What is the supremum of the ratio $R$ of the triangle's area to the region's area?

(We require that the polynomial has all real roots, because otherwise the answer would be trivial: the supremum would be $1$, for example with $y=x-x^n$ and letting $n\to\infty$.)

I considered the largest area triangle drawn in the region bounded by the $x$-axis and the polynomial $y=(-1)^{n-1}(x^2-1)\left(\left(\frac{x}{a}\right)^2-1\right)^n, a>1$, from $x=-1$ to $x=1$.

enter image description here

Numerical investigation suggests that for any given $a$ value, the ratio $R$ is maximized when $n\approx 1.880001a^2$.

$\color{blue}{a=10}$:

$\space R\approx0.8953810$ with $y=(x^2-1)\left(\left(\frac{x}{10}\right)^2-1\right)^{\color{red}{187}}$

$\space R\approx0.8953823$ with $y=-(x^2-1)\left(\left(\frac{x}{10}\right)^2-1\right)^{\color{red}{188}}$

$\space R\approx0.8953822$ with $y=(x^2-1)\left(\left(\frac{x}{10}\right)^2-1\right)^{\color{red}{189}}$

$\color{blue}{a=100}$:

$\space R\approx0.8955295$ with $y=-(x^2-1)\left(\left(\frac{x}{100}\right)^2-1\right)^{\color{red}{18799}}$

$\space R\approx0.8955296$ with $y=(x^2-1)\left(\left(\frac{x}{100}\right)^2-1\right)^{\color{red}{18800}}$

$\space R\approx0.8955295$ with $y=-(x^2-1)\left(\left(\frac{x}{100}\right)^2-1\right)^{\color{red}{18801}}$

$\color{blue}{a=1000}$:

$\space R\approx0.89553095$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880000}}$

$\space R\approx0.89553105$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880001}}$

$\space R\approx0.89553101$ with $y=(x^2-1)\left(\left(\frac{x}{1000}\right)^2-1\right)^{\color{red}{1880002}}$

I conjecture that $\text{sup}(R)$ is approached when $a\to\infty$ (for each value of $a$, we choose the value of $n$ that maximizes $R$), and that $\text{sup}(R)\approx 0.895531$.

Improve this question
$\endgroup$
13
  • 1
    $\begingroup$ @JosephO'Rourke I now realize that my last sentence ("I posted this question on...) was misleading. I posted two questions on MSE: the first was the one you linked to, the second was the same as this question. I deleted the second question. (It indeed received no answer and few views. Well, to be precise, someone answered it but then deleted their answer because they misread the question.) I'm sorry, didn't mean to be misleading. $\endgroup$
    – Dan
    Commented Mar 2, 2023 at 14:39
  • 1
    $\begingroup$ Interesting question. It is useful to cross-reference when posting related questions. $\endgroup$
    – Joseph O'Rourke
    Commented Mar 2, 2023 at 14:41
  • 1
    $\begingroup$ @GerryMyerson Point taken. I'll try not to abuse the edit function. $\endgroup$
    – Dan
    Commented Mar 3, 2023 at 12:20
  • 1
    $\begingroup$ @PavelGubkin The optimal triangle does not always have vertices at $(-1,0)$ and $(1,0)$ (but it always has a vertex at $(0,1)$). The length of the (horizontal) base of the optimal triangle is at most $2$; it depends on $a$ and $n$. $\endgroup$
    – Dan
    Commented Mar 3, 2023 at 13:15
  • 2
    $\begingroup$ But actually that last condition is not even necessary or useful. So, to recap, I think a problem which is very similar to the original, but for which it may be relativily easy to work out and explicity solution, would be that of minimizing $\int_0^1 \big(P(x)e^{-ax^2}-1+x\big)^2dx$ over real $a\ge 0$ and $P$ of the form $P(x)=\prod_{1\le i \le n} (1-a_i x^2)$ for $0\le a_i \le1$. $\endgroup$
    – Yaakov Baruch
    Commented Mar 5, 2023 at 9:12

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .