TITLE: Is there a deeper reason why these special points of a function form a geometric sequence?
QUESTION [5 upvotes]: If we consider the function $f(x) = \ln^2(x) - \ln(x)$ over $(0, +\infty)$, we can note the following:

$f(x) = 0$ at two points A and C of abscissas $x_A = 1$ and $x_C = e$ (the roots of the function).
$f'(x) = 0$ at the point B of abscissa $x_B = e^{1/2}$ (the global minimum of the function)
$f''(x) = 0$ (and changes signs) at D of abscissa $x_D = e^{3/2}$ (the inflection point of the function).

These four notable points A, B, C and D on the curve form a geometric sequence of common ratio $r = e^{1/2}$ and first term $u_0 = 1$. Do the next terms carry any significance in the function? Is there a specific, 'deeper' reason behind this? Is this just a fluke, or is this behavior exhibited in a general class of functions? I also noted that the similar function $g(x) = \ln^2(x) + \ln(x)$ has a similar pattern (the intersection points, minimum and inflection point form a geometric sequence with ratio $r$ and first term $v_0 = e^{-1/2}$), which is what led me to suspect that this could be a common theme among a class of functions.

REPLY [2 votes]: First try and find a formula for the nth derivative of $f(x)$:
$$f'(x)=\frac{1}{x}(2\ln x-1)$$
$$f''(x)=-\frac{1}{x^2}(2\ln x-3)$$
$$f'''(x)=\frac{2}{x^3}(2\ln x-4)$$
$$f''''(x)=-\frac{6}{x^4}\big(2\ln x-\frac{14}{3}\big)$$
The derivative at each step will be given by
$$f^{(n)}(x)=(-1)^{n-1}\frac{(n-1)!}{x^n}\big(2\ln x-a_n\big)$$
And we can attempt to define $a_n$ recursively. Suppose we have
$$f^{(n)}(x)=(-1)^{n-1}\frac{(n-1)!}{x^n}\big(2\ln x-a_n\big)$$
and we take the derivative again. We get
$$f^{(n+1)}(x)=(-1)^{n}\frac{n!}{x^{n+1}}\big(2\ln x-a_n-\frac{2}{n}\big)$$
So we have
$$a_1=1$$
$$a_{n+1}=a_n+\frac{2}{n}$$
And if we let $H_k$ represent the kth harmonic number, then
$$a_n=2H_{n-1}+1$$
and so the zero of $f^{(n)}(x)$ occurs at
$$2\ln x-2H_{n-1}-1=0$$
$$2\ln x=2H_{n-1}+1$$
$$\ln x=H_{n-1}+\frac{1}{2}$$
$$x=e^{H_{n-1}+\frac{1}{2}}$$
That's the pattern you were looking for.