12
$\begingroup$

I was messing around with the infinitely nested logarithm $f(x)=\ln\left(x+\ln\left(x^2+\ln\left(x^3+\ln\left(\dots\right)\right)\right)\right)$ on Desmos when I decided to take the integral from $x = 1$ to $x = 2$. Here's what I got where $n$ represents the number of times the logarithm is nested. For example, for $n=1$, it is $f_1(x)=\ln\left(x\right)$ and for $n=2$, it is $f_2(x)=\ln\left(x+\ln\left(x^2\right)\right)$.

$$\begin{array}{c|c} \text{n} & \int_{1}^{2}{f_n(x)}dx \\ \hline 1 & 0.38629436112 \\ \hline 2 & 0.769701324476 \\ \hline 3 & 0.923253197047 \\ \hline 4 & 0.968765188969 \\ \hline 5 & 0.982950647903 \\ \hline 10 & 0.992710069039 \\ \end{array}$$

$\therefore\text{I conjecture that it converges to $1$}$

Can someone help me analytically prove or refute my conjecture?

I lack the skills to do it.

P.S. I could not figure out a closed form for the infinitely nested logarithm but I assume finding one would make this a lot easier to prove.

$\endgroup$
3
  • 1
    $\begingroup$ This is fascinating. The value of 𝑓𝑛 at 𝑥=2 agrees to at least 14 decimal places when 𝑛=15 vs when 𝑛=12. When 𝑛=15 the integral numerically evaluates (using scipy) to .993. I'm wondering if this is a numerical coincidence and the true value is not 1. $\endgroup$
    – hunter
    Dec 20, 2023 at 3:03
  • $\begingroup$ @hunter Seems likely. On Desmos, for $n=30$, it numerically evaluated to $0.993409086452$. $\endgroup$ Dec 20, 2023 at 3:23
  • 1
    $\begingroup$ @hunter It seems like it is ridiculously close to $\cos\left(\tanh\left(\frac{3}{26}\right)\right)\approx 0.9934090894200$ which seems extremely unlikely but still interesting $\endgroup$ Dec 20, 2023 at 3:28

2 Answers 2

5
$\begingroup$

My guess is, it is not having exactly 1 as limit, but has a limit smaller than that. But, I think showing the integral is convergent is also fun.

  1. $\ln(z)\le z-1$ for $z\ge 1$.

  2. $x^n + a \le (x + \frac{a}{nx^{n-1}})^n$, if $x\ge 1$ and $a \ge 0$.

  3. For any $n$,

$$\ln(x^{n-1} + n \ln x)\le \ln(x^{n-1} + n (x-1))\le \ln\left(\left(x + \frac{n(x-1)}{(n-1)x^{n-2}}\right)^{n-1}\right)\le (n-1)(x - 1)\left(1 + \frac{n}{(n-1)x^{n-2}}\right)$$

and the next term

$$\ln\left(x^{n-2} + \ln(x^{n-1} + n \ln x) \right)\le (n-2) (x - 1)\left(1 + \frac{n-1}{(n-2)x^{n-3}} \left(1+ \frac{n}{(n-1)x^{n-2}}\right)\right)$$

thus

$$f_n(x)\le (x-1)\left(1 + \frac{2}{1}\cdot\left(1 + \frac{3}{2x}\cdot\left(1 + \frac{4}{3x^2}\cdots\right)\right)\right)\le (x-1)\left(1 + \frac{2}{x^{0}} + \frac{3}{x^{1}}+\cdots +\frac{n-1}{x^{(n-3)(n-2)/2}}+\frac{n}{x^{(n-2)(n-1)/2}}\right)$$

If you integrate a term $(x-1)\frac{m}{x^{(m-1)(m-2)/2}}$ on [1,2], that is

$$\le m \frac{1 - \frac{(m-1)(m-2)}{2}2^{1-(m-1)(m-2)/2}}{\left(\frac{(m-1)(m-2)}{2} - 2\right)^2} = O(m^{-3}) + O(m^{-1} 2^{-(m-1)(m-2)/2})$$

This sum over $m$ is finite. If you calculate carefully, you should be able to get a far-from-optimal bound easily.

I believe if you stop the summation earlier in the tail, and compute accurately for the low degree terms, it might be able to give a bound below one.

$\endgroup$
3
$\begingroup$

Comment.
Of course, $f_{n+1}(x) > f_n(x)$ for all $x$. It seems $f_n(x)$ converges pointwise, say to $f(x)$.

A

By monotone convergence, we want to compute $\int_1^2 f(x)\,dx$.

$\endgroup$
2
  • 2
    $\begingroup$ I had attempted but failed to find a closed form for the logarithm. $\endgroup$ Dec 20, 2023 at 3:11
  • $\begingroup$ is the convergence uniform ? $\endgroup$
    – pie
    Dec 20, 2023 at 5:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .