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What are asymptotic expressions for the coordinates of the turning point in $x\in(0,1)$ on $y=|x(x-1)(x-2)\dots(x-n)|$ as $n\to\infty$ ?

Here is the graph for $n=10$.

enter image description here

The turning point in $x\in(0,1)$ on $y=\color{red}{\frac{\log{n}}{n!}}|x(x-1)(x-2)\dots(x-n)|$ has the following approximate coordinates:

$n=10: (0.2854, 0.2644)$
$n=20: (0.2450, 0.2877)$
$n=40: (0.2131, 0.3031)$
$n=80: (0.1879, 0.3137)$
$n=160: (0.1675, 0.3215)$

I'm guessing the asymptotic expressions will be something like $\left(\frac{1}{\log{n}},\frac{n!}{k\log{n}}\right)$ for some constant $k$.

This is a follow-up question to a question about the total area of the regions enclosed by $y=|x(x-1)(x-2)\dots(x-n)|$ and the $x$-axis.

Dan
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  • A tiny remark : there is no need to use an absolute value. – Jean Marie Mar 15 '23 at 22:54
  • @JeanMarie Without absolute value, the turning point could be above or below the $x$-axis, depending on the parity of $n$. – Dan Mar 15 '23 at 22:58
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    Thanks. I didn't see this side (intended pun) of the question ... – Jean Marie Mar 15 '23 at 23:11
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    $$f(x)=|x(x-1)\dots(x-n)|=\frac{x\Gamma(1-x)(1-x)\dots(n-x)}{\Gamma(1-x)}\frac{\Gamma(x)}{\Gamma(x)}=\frac{\sin(\pi x)}\pi\Gamma(1+x)\Gamma(n+1-x)$$ $$f'(x_0)=0,\Rightarrow,\pi\cot\pi x_0+\psi(1+x_0)=\psi(n+1-x_0)$$ For $n\to\infty,,x_0\to0$. Given that $\psi(n)\sim\ln n,(,\gg1)$ and $\psi(1)=-\gamma$ $$\frac1{x_0}-\gamma\approx\ln n,\Rightarrow,x_0\approx\frac1{\gamma+\ln n}$$ For $n=160,,x_0\approx0.1769...$ – Svyatoslav Mar 16 '23 at 03:40
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    An asymptotic expansion is as follows: $$ x_0 \sim \frac{1}{{\log (n) + \gamma }} - \frac{{\pi ^2 }}{6}\frac{1}{{(\log (n) + \gamma )^3 }} + \frac{{2\pi ^4 }}{{45}}\frac{1}{{(\log (n) + \gamma )^5 }} - \ldots $$ as $n\to +\infty$. – Gary Mar 16 '23 at 07:37

1 Answers1

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As @Svyatoslav already wrote in comments, you need to find the zero of function $$g(x)=\pi \cos (\pi x)-\sin (\pi x) \big(\psi ^{(0)}(n-x+1)-\psi ^{(0)}(x+1)\big)$$ for large values of $n$ that is to say for small $x$ as you noticed.

Expanded as a series around $x=0$ $$g(x)=\pi -\pi H_n\,x+O\left(x^2\right)$$ gives, as a first estimate $$\color{blue}{x_0=\frac{1}{H_n}}$$

It is easy to show that $g(x_0)<0$ and $g''(x_0) >0$. So, using Newton method, by Darboux theorem, $x_1$ is an overestimate of the solution but not too bad $$\left( \begin{array}{ccc} n & x_1 & \text{solution} \\ 10 & 0.284409 & 0.285421 \\ 20 & 0.244703 & 0.244995 \\ 30 & 0.225306 & 0.225460 \\ 40 & 0.213043 & 0.213144 \\ 50 & 0.204299 & 0.204372 \\ 60 & 0.197609 & 0.197666 \\ 70 & 0.192250 & 0.192296 \\ 80 & 0.187813 & 0.187853 \\ 90 & 0.184051 & 0.184085 \\ 100 & 0.180800 & 0.180829 \\ \end{array} \right)$$

Now, the second iterate $$\left( \begin{array}{ccc} n & x_2 & \text{solution} \\ 5 & 0.3365508733 & 0.3365534733 \\ 10 & 0.2854211662 & 0.2854213370 \\ 15 & 0.2606074001 & 0.2606074400 \\ 20 & 0.2449953893 & 0.2449954044 \\ 25 & 0.2339092745 & 0.2339092819 \\ 30 & 0.2254596051 & 0.2254596093 \\ 35 & 0.2187123217 & 0.2187123217 \\ \end{array} \right)$$

Since you need the first derivative $$g'(x)=\pi \cos (\pi x) (\psi ^{(0)}(x+1)-\psi ^{(0)}(n-x+1))+$$ $$\sin (\pi x) \left(\psi ^{(1)}(n-x+1)+\psi ^{(1)}(x+1)-\pi ^2\right)$$

Claude Leibovici
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