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I'm wondering if there is a way to shorten the following formula: $$ y = \ln ( 1 + ( \ln ( 1 + ( \ln ( 1 + ( \ln ( 1 + ( \ln (1+c) )))))))), $$ where $c$ is a constant positive number.

Update: I edited the title to make it clear that the number of ln is not infinite. It varies on my usage but it shall just be less than 30. Although it looks like it's not possible to have a shortcut formula on this as the comment say, I will just leave this question open until someone could convincingly say that it is impossible to have a shortcut formula on this. I have a hunch that there may be an ingenious and shorter way of computing this because seemingly there is a pattern in the formula.

Maximilian Janisch
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d_air
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    I don't think there is a way to simplify it exactly (logarithms of sums are a tricky business at best, but often there just isn't anything one can do). But approximations can definitely be made. – Arthur May 24 '19 at 08:21
  • Your title has an infinite expression yet the one in the question body is finite - which are you looking for? More importantly, where did this come from and what have you tried? – PrincessEev May 24 '19 at 08:24
  • c is a constant number. Obviously, I can always get the result by repeatedly applying ln. My goal here is just to compute it faster. ln(1+.. can be as many as 30 repetitions. – d_air May 24 '19 at 08:29

2 Answers2

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Yes, it (the infinite version) is equal to zero.

Consider $f(x)=\ln(1+x)$ as an iterator function. It has a fixed point at $x=0$. There are numerous ways to show that it is the only such fixed point.

One way might be to consider the function $g=x-\ln(1+x)$, whose roots correspond to fixed points. Clearly for $x\rightarrow -1$, and $x\rightarrow\infty$, $g\rightarrow \infty$. Note that $g$ is differentiable and so it's extrema occur when $g'=0$. $g$ has only one such extrema, where $x=0$ (and $g$ has a root). This $g$ has only one root, at $x=0$, because another would force another turning point. Thus $f$ has only one fixed point, namely $x=0$ (this isn't actualy used in the sequel once all the details are written down).

If $c$ is is positive, then $|f'(c)|<1$ and indeed $|f'(x)|<1$ for all $x>0$. This means that zero is attracting for $x>0$.

Consider now the iterates $c=x_0,\,x_1=f(x_0),\,x_2=f(x_1),\dots$ with $x_n=f(x_{n-1})$.

The concern would be that if $x_{i+1}<0$ then we have a problem.

Suppose $x_i\geq 0$ and $x_{i+1}<0$. This can't happen though because $\ln(1+x_i)\geq \ln(1)=0$. Neither do we have to worry about $f'(0)=1$ because zero is a fixed point.

Dynamical systems machinery shows that the $x_i\rightarrow 0$. The machinery is the following. Notice that $f$ is continuous on $[0,\infty)$ so in particular on $[0,x_i]$. By the Mean Value Theorem, there exists an $a_i\in [0,x_i]$ such that $$f'(a_i)=\frac{f(x_i)-f(0)}{x_i-0}\Rightarrow f(x_i)<f'(a_i)\cdot x_i.$$

We showed above that $f'(x)<1$ for all $x>0$ and so $x_{i+1}<x_i$, strictly.

The limit cannot be $x_\infty\neq 0$ because $f(x_{\infty})<x_\infty$ if $x_\infty>0$ (because of the MVT on $[0,x_{\infty}]$ and $\displaystyle f'(x_{\infty})=\frac{1}{1+x_\infty}<1$).

JP McCarthy
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  • I just want to know if there is a possible way of computing y without applying so many ln. So for example c=9.6, does your explanation above tells that I cannot avoid repeatedly applying many ln operations to get the value of y? – d_air May 24 '19 at 09:16
  • The more times you apply $\ln(1+x)$ the closer you get to zero... no matter what the value of $c\geq0$ is.. it never equals zero in 'finite time' unless $c=0$ in which case the convergence is immediate. – JP McCarthy May 24 '19 at 09:40
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When $c$ is small, the reciprocal of $c$ increases by almost $1/2$ at each step, so $c$ shrinks slowly to zero.

If $c_{k+1}=\ln(1+c_k)$ then its reciprocal is approximately $$c_{k+1}^{-1}\approx c_k^{-1}+\frac12-\frac{c_k}{12}+\frac{c_k^2}{24}-\cdots$$ For the first few steps, while $N\lt c_0^{-1}$, this accumulates to $$c_N^{-1}\approx c_0^{-1}+\frac N2-\frac{Nc_0}{12}\\ c_N\approx c_0-\frac N2c_0^2$$

Eventually, when $N$ is much larger than $c_0^{-1}$, it accumulates to $$c_N^{-1}\approx c_0^{-1}+\frac N2-\frac1{12}\sum c^k\\ \approx c_0^{-1}+\frac N2-\frac1{12}\sum\frac2k\\ \approx \frac N2 -\frac{\ln N}6$$ Note that $c_0$ is not leading order any more, so this answer is approximately the same whatever your starting point.
Then an estimate for $c_N$ is $$c_N\approx \frac 2N+\frac{2\ln N}{3N^2}$$

Empy2
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