Conjecture: If $f(x)=\prod\limits_{k=1}^n(x-k)$ then $\lim\limits_{n\to\infty}\frac1n [\text{largest root of $f(x)=f'(x)$}]=\frac{e}{e-1}$.

Question

I was thinking about the polynomial $f(x)=\prod\limits_{k=1}^n(x-k)$. I noticed that if we draw the graphs of $y=f(x)$ and $y=f'(x)$ together, the $x$-coordinate of the rightmost intersection seems always to be approximately $1.6n$.

Further numerical investigation suggests that the largest root of $f(x)=f'(x)$ approaches approximately $1.58198n$ as $n\to\infty$.

I conjecture that$\lim\limits_{n\to\infty}\frac1n [\text{largest root of $f(x)=f'(x)$}]=\frac{e}{e-1}=1.5819767...$.

Is my conjecture true?

The coefficients in the expansion of $f(x)$, in decreasing powers of $x$, are found in the sequence of Stirling numbers of the first kind.

The coefficients in the expansion of $f'(x)$, in decreasing power of $x$, are given by A196837.

Other than that, I don't know how to approach approximating the largest root of $f(x)=f'(x)$.

Answer 1

Note that

$$ \frac{f'(x)}{f(x)} = \sum_{k=1}^{n} \frac{1}{x-k}. \tag{1} $$

Using this, it is easy to check that there is a unique root $x = x_n$ of $ f'(x)/f(x) = 1 $ in the interval $(n, \infty)$. On the other hand, recognizing the sum in the RHS of $\text{(1)}$ as a sum of areas of rectangles, we easily obtain

$$ g(x) \leq \frac{f'(x)}{f(x)} \leq g(x-1), $$

where $g(x) = \int_{0}^{n} \frac{\mathrm{d}t}{x - t} = \log\left(\frac{x}{x-n}\right)$. This tells that

$$ \text{[root of $g(x) = 1$]} \leq x_n \leq \text{[root of $g(x-1) = 1$]}, $$

or equivalently,

$$ \frac{e}{e-1} n \leq x_n \leq \frac{e}{e-1} n + 1. $$

Therefore OP's conjecture is true.

Answer 2

$$f(x)=\prod\limits_{k=1}^n(x-k)=\frac{(x-n) }{x}\,(-n+x+1)_n$$ $$f'(x)=\frac{n+x(x-n)\left(H_x-H_{x-n}\right) }{x^2}\,\,(-n+x+1)_n$$ Solving $$\frac{f'(x)}{f(x)}=1$$ reduces to solving $$H_x-H_{x-n}+\frac{n}{x (x-n)}=1$$

Let $x=kn$ and use the asympotics of the harmonic numbers with $k>1$ to obtain $$\log \left(\frac{k}{k-1}\right)+\frac{1}{2 (k-1) k n}+O\left(\frac{1}{n^2}\right)=1$$

So, for an infinite value of $n$ $$\log \left(\frac{k}{k-1}\right)=1\quad\implies\quad k=\frac{e}{e-1}$$

We could even go further and obtain $$\color{blue}{k= \frac{e}{e-1}+\frac{1}{2 n}+24(e+1)\sum_{p=1}^\infty (-1)^p\left(\frac{e-1}{e}\right)^{2 p-1}\frac {a_p}{n^{2p}}}$$ where the very first $a_p$ are $$\left( \begin{array}{cc} p & a_p \\ 1 & 1 \\ 2 & \frac{27-10 e+27 e^2}{240} \\ 3 & \frac{7625-2268 e+7478 e^2-2268 e^3+7625 e^4}{120960 } \\ \end{array} \right) $$

Even if it does not mean too much, for the above truncated series, $$g(x)=H_x-H_{x-n}+\frac{n}{x (x-n)}-1\sim \frac 2 {9 n^8 }$$

Using $n=5$, the series leads to $k=\color{red}{1.67810}78$