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.

  • $\begingroup$ Does $\ln^2 x$ represent $(\ln x)^2$ or $\ln(\ln x)$? $\endgroup$ Jun 5 '17 at 17:13
  • 1
    $\begingroup$ @Frpzzd $\ln^2 (x)$ in my question means $(\ln x)^2$ (the former), sorry for the ambiguity. $\endgroup$
    – Daccache
    Jun 5 '17 at 17:15
  • $\begingroup$ It is doubtful that 4 terms on their own are more than a coincidence. I'd start by looking to see if I could find the next terms in the sequence somewhere in the function. $\endgroup$
    – Paul
    Jun 5 '17 at 17:18
  • $\begingroup$ By a quick calculation, it appears that the term you get from the third derivative is $e^2$ (which is promising), but the term generated from the fourth derivative is $e^{7/3}$, which seems to break the pattern you see. If you have the capabilities it seems possible to find a recurrence relation for these solutions, and perhaps a general solution. $\endgroup$ Jun 5 '17 at 17:26
  • $\begingroup$ Does my answer satisfy you? If so, please consider accepting it. If not, please let me know how I can improve it. :) $\endgroup$ Jun 5 '17 at 21:38

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.


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