Let's define the sequence $\{s_n\}$ recursively as $$s_1=\sqrt2,\ \ \ s_{n+1}=\sqrt2^{\,s_n}.$$ Or, in other words, $$s_n=\underbrace{\sqrt2^{\sqrt2^{\ .^{\ .^{\ .^{\sqrt2}}}}}}_{n\ \text{levels}}.$$ The sequence is monotonically growing, and rapidly converges to a limit $$\lim\limits_{n\to\infty}s_n=2.$$ I'm interested in estimating its convergence speed.

Based on numerical data, I conjectured that $$\ln\left(2-s_n\right)=n\ln\ln2+c_{\sqrt2}+O\big(\left(\ln2\right)^n\big)$$ for some constant $c_{\sqrt2}\approx-0.458709787761420587059021...$

Could you suggest possible approaches to prove (or refute) this conjecture?
I am also interested in a possible closed form of the constant $c_{\sqrt2}$.

Update: We can try to generalize this problem to other bases beyond $\sqrt2$. Let's use a usual notation for tetration $${^n}a=\underbrace{a^{a^{\ .^{\ .^{\ .^a}}}}}_{n\ \text{levels}}.$$ It's known that for all $1/e^e<a<e^{1/e}$ there exists a limit$${^\infty}a=\lim\limits_{n\to\infty}{^n}a=e^{-W\left(-\ln a\right)},$$ where $W(z)$ is the Lambert $W$ function, the inverse of the function $x\mapsto x\,e^x$.

I conjecture that for all $1<a<e^{1/e}$ $$\ln\left({^\infty}a-{^n}a\right)=n \ln\ln\left({^\infty}a\right)+c_a+O\left(e^{n\ln\ln\left({^\infty}a\right)}\right),$$ where $c_a$ is some constant that depends on $a$ but not on $n$ (also note that $\ln\ln\left({^\infty}a\right)<0$, so the last term is exponentially small).

  • $\begingroup$ Did you consider the possibility to replace the function $f(x)=\sqrt 2 ^x$ by $g(x)=2^x-1$ and then $f(x) = 2\cdot (g(x/2-1)+1)$ which is valid for iterations $f^{[n]}(x) = 2\cdot (g^{[n]}(x/2-1)+1) $ as well. Then this can be used for the limit to infinity equivalently; but $g(x)$ has a power series without constant term and a Schröder-function (which resembles the properties at infinity with its taylor-coefficients). Possibly one can use this transformation to understand your limit-expression and the constant as well; (I've not much time now, so I can't be of more help here at the moment) $\endgroup$ – Gottfried Helms May 24 '14 at 7:34
  • $\begingroup$ I have asked a question about the proof of the 'well known () there exist a limit' here : math.stackexchange.com/questions/890319/…. Perhaps you have a nice proof. $\endgroup$ – Free X Aug 8 '14 at 8:19
  • $\begingroup$ Let $s=2 \dot \exp(c_{\sqrt2}) \approx 1.26419732$. I've worked with this number some time ago; it occurs by the method which I mentioned in my earlier comment. Let $\sigma(x)$ be the schröder-function for the exponential to base $\sqrt 2$ centered around the fixpoint $t=2$. Then the same value occurs for $w=\sigma (x/t-1)$ at $x=1$ and (for some reason I don't remember) I looked for the value $-4 \cdot w$ which is equal to the number $s$ to the number of digits which you've provided. So your number $c_{\sqrt 2}$ has in some sense "canonical" relevance (and can be generalized to other bases) $\endgroup$ – Gottfried Helms Sep 18 '14 at 19:54
  • $\begingroup$ Also the number $w = -s/4 \approx −0.316049330525... $ occurs in my answer mathoverflow.net/questions/71429/… in which I explained the application of the schröder-function a bit more. $\endgroup$ – Gottfried Helms Sep 18 '14 at 19:56

This is the fixed point iteration method for solving an equation of the form $x=f(x)=(\sqrt{2})^x$ by iterating $x_{n+1}=f(x_n)$, and it is very commonly discussed in numerical analysis textbooks.

In general, if the solution is $x_\infty = f(x_\infty)$, the error of the $n$-th term behaves as $$ \epsilon_{n+1} = x_{n+1}-x_\infty = f(x_n) - f(x_\infty) \approx f'(x_\infty)(x_n - x_\infty) = f'(x_\infty)\epsilon_n, $$ $$ \epsilon_n = (f'(x_\infty))^n \times O(1), $$ which is where the $n\log\log 2 = n\log f'(2)$ comes from in your data.

The constant term in $\log\epsilon_n - n \log f'(x_\infty)$ is determined not just by the method or the equation, but also by the initial condition $s_1=\sqrt2$, which is why it is very often not analysed. The initial guess at the solution can be anything at all, really, as long as it's close enough to the solution for the method to converge. Thus if the method converges to the same value from most starting values, why bother singling out a specific arbitrarily-chosen one?

In this case the limit $c_{\sqrt2} = \lim y_n$ comes from the recurrence relation $$ y_0 = 1, \qquad \log y_{n+1} = \frac{x_\infty}{(f'(x_\infty))^{n+1}}\Big(f\big((f'(x_\infty))^n\log y_n\big)-1\Big). $$ This limit is easy enough to compute numerically, but there might not be a way to do it analytically.

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Here's proof of a weaker result. Let $r_n = 2 - s_n$. $$ r_{n+1} = 2 - s_{n+1}= 2 - \sqrt{2}^{s_n}= 2\left(1 - \sqrt{2}^{s_n-2}\right)= 2\left(1 - \sqrt{2}^{\,-r_n}\right) $$ Now, let $\alpha = \ln 2$. $$ \sqrt{2}^{\,-r_n}= \left(e^{\ln \sqrt{2}}\right)^{-r_n}= e^{-\frac{1}{2}\alpha r_n}= 1-\frac{1}{2}\alpha r_n + O\left(r_n^2\right) $$ and so $$ r_{n+1}=\alpha\,r_n + \phi\left(r_n\right),\quad \phi\left(s\right) = O\left(s^2\right) $$ Hence, $$ \ln r_{n+1}= \ln\left(\alpha\,r_n + O\left(r^2_n\right)\right)= \ln\alpha + \ln r_n + \ln\left(1+O\left(r_n\right)\right)= $$ $$ =\ln\alpha + \ln r_n + \psi\left(r_n\right),\quad \psi(s)=O\left(s\right) $$ and $$ \ln r_n = \ln r_1 + \left(n-1\right) \ln \alpha + \sum_{k=2}^n\psi\left(r_k\right) $$ This last sum is convergent (proof below), so the rate of convergence is determined by behaviour of $\psi\left(r_n\right)$. Since $\psi\left(s\right)=O(s)$, it's in fact determined by behaviour of $r_n$.

Lemma For any $\varepsilon > 0$, we have $\left|r_n\right|=O\left(\left(\alpha + \varepsilon\right)^n \right)$

Proof Let $\varepsilon > 0$. As $r_n\to 0$ and $\phi\left(s\right)=O\left(s^2\right)$ for sufficiently large $n$ we have $\left|\phi\left(r_n\right)\right|<\varepsilon\, \left|r_n\right|$, and so $$ \left|r_{n+1}\right|= \left|\alpha\,r_n+\phi\left(r_n\right)\right| < \left(\alpha + \varepsilon\right)\left|r_n\right| $$ and it is clear we can bound $\left|r_n\right|$ by geometric sequence with ratio $\alpha + \varepsilon$.

Using this, we can easily bound the tail of the sum by $O\left(\left(\alpha + \varepsilon\right)^n\right)$, and so, to conclude $$ \ln\left(2 - s_n\right) = n\ln\ln 2 + c + O\left(\left(\ln 2 + \varepsilon\right)^n\right) $$ for every $\varepsilon > 0$, where $c=\ln\left(2 - \sqrt{2}\right) - \ln\ln 2 + S$, where $S$ is sum of the series discussed above.

Of course, that's strictly weaker than the conjuecture from question. Still, I believe it's close enough to be at least relevant.

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  • $\begingroup$ You seem to have a very advanced knowledge of real analysis, where did you get it? (i mean the textbooks you used) $\endgroup$ – user153330 Jan 3 '15 at 14:34
  • $\begingroup$ I'm personally very fond of Rudin's Principles of Matematical Analysis, that's where I got my core knowledge of the subject. Note that the above answer does not really use advanced analysis, it's rather just a skillful (if I may say so) usage of few simple concepts. I guess seeing and using similar techniques in related areas (i.e. functional analysis) helped me to develop such ability. That said, I'm a math hobbyist, CS student with little formal pure mathematical training - not much of a textbook authority, and there are tons of far more knowledgeable and skilled people here. $\endgroup$ – Marcin Łoś Jan 3 '15 at 15:36
  • $\begingroup$ Excellent, thanks for that : ) $\endgroup$ – user153330 Jan 3 '15 at 21:00

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