5
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.