# A limit related to super-root (tetration inverse).

Recall that tetration ${^n}x$ for $n\in\mathbb N$ is defined recursively: ${^1}x=x,\,{^{n+1}}x=x^{({^n}x)}$.

Its inverse function with respect to $x$ is called super-root and denoted $\sqrt[n]y_s$ (the index $_s$ is not a variable, but is part of the notation — it stands for "super"). For $y>1, \sqrt[n]y_s=x$, where $x$ is the unique solution of ${^n}x=y$ satisfying $x>1$. It is known that $\lim\limits_{n\to\infty}\sqrt[n]2_s=\sqrt{2}$. We are interested in the convergence speed. It appears that the following limit exists and is positive: $$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$ Numerically, $$\mathcal L\approx0.06857565981132910397655331141550655423...\tag2$$

Can we prove that the limit $(1)$ exists and is positive? Can we prove that the digits given in $(2)$ are correct? Can we find a closed form for $\mathcal L$ or at least a series or integral representation for it?

• Nov 14, 2015 at 10:55
• Just a couple of days I played with a family of power series for the n'th superroot; it has a surprising form in the limit... Perhaps this small note on my webspace is interesting for this question: go.helms-net.de/math/tetdocs/Wexzal_Superroot.pdf Nov 14, 2015 at 15:52
• "Super root" sounds amazing :) Nov 14, 2015 at 18:58
• Super-roots are given a degree/order. Your first definition defines the Super-root of order $n>1$ ($x>1$, such that ${^n}x=y$). The notation $\lim\limits_{n\to\infty}\sqrt[n]2_s$, however, suggests that you want to measure the speed of convergence of Super-roots of order $\infty$. Super-roots of finite order and of infinite order are different things and the processes that extract them are different. Which ones do you really want to measure the speed of convergence of? Super-roots of finite or infinite order?
– user127032
Nov 15, 2015 at 3:29
• @YiannisGalidakis Thanks for your comment, but I do not see any ambiguity in my question. The super-root of infinite order $\sqrt[\infty]x_s$ is just the limit of the sequence of super-roots of finite order $n$ as $n\to\infty$. As mentioned on the Wikipedia page that I linked, it can be expressed in elementary functions as $\sqrt[\infty]x_s=x^{1/x}$. In my particular case $x=2$ and $\sqrt[\infty]2_s=\sqrt2$. As $n$ grows, the super-roots $\sqrt[n]2_s$ steadily approach this limiting value. I investigate the speed of this convergence. Nov 15, 2015 at 22:30

I have some observations, from where someone more experienced might be able to derive the proof - maybe this is helpful)

Let the iterated functional root ("superroot of order") $B(z,n)$ (which finds the " B "ase of the powertower) be defined as $$\;^n b = z \qquad \to \qquad B(z,n) = b$$ For the following let us always denote $u$ for the $\log(z)$ for notational convenience.
Let us then define the generalization of the Lambert-W-function to higher iterates as a simple conjugate of the base-finding $B()$-function: $$W(u,n) = W(\log(z),n) = \log( B(\exp(u),n)) \tag 1$$ where we can find a formal power series for $W(u,n)$ from the Lagrange series reversion of $u \cdot \exp(u)$, such that $$\begin{array}{} \mathcal {\text{ Taylor:} } & W^{-1}(u,1) = u \\ \mathcal {\text{ Taylor:} }& W^{-1}(u,2) = u \cdot \exp(u) \\ \mathcal {\text{ Taylor:} }& W^{-1}(u,3) = u \cdot \exp(u \cdot \exp(u) ) \\ \end{array} \tag 2$$ Then your limit for a general $z$ and $u = \log(z)$ reads $$\lim_{n \to \infty} {B(z,n)-z^{1/z} \over \log(z)^n} \qquad \text{or} \qquad \lim_{n \to \infty} {\exp(W(u,n))-\exp(u \cdot \exp(-u)) \over u^n} \tag 3$$ Actually we use $z=2$ (and thus $u=\log(2)$) and this reads a bit more friendly: $$\lim_{n \to \infty} {\exp(W(u,n))- \sqrt 2 \over u^n} \tag 4$$

Now, if we look at the formal power series for $W(u,n)$ we find a striking pattern, in that with increasing n the coefficients of the equivalently longer leading part become constant, which can be written like the following $$\begin{array} {} W(u,n) &=& u \Large \left(1-u+{ u^2 \over 2!}-{ u^3 \over 3!} + ... \right. \\ && \qquad + u^{n-1} \Large \left( {a_{n,1} u \over 1!} -{a_{n,2} u^2 \over 2!} + ... \right) \\ && \left. \quad \Large \right)\end{array} \tag 5$$ Let us denote the part in the inner parenthese as $R(u,n)$ $$R(u,n) = {a_{n,1} u \over 1!} -{a_{n,2} u^2 \over 2!} - ... \tag 6$$ which makes then $$\begin{array} {rrl} W(u,n) &=& u \cdot \exp(-u) + u^n \cdot R(u,n) )\\ B(z,n) &=& \exp(u \cdot \exp(-u) + u^n \cdot R(u,n)) \\ &=& \exp(u \cdot \exp(-u)) \cdot \exp( u^n \cdot R(u,n)) \\ &=& \sqrt 2\cdot \exp( u^n \cdot R(u,n)) \\ \end{array} \tag 7$$

Something on convergence: because the first line in fomula (5) contains an exponential series in $u$ and such a series is entire, the consideration of range of convergence of the whole construct focuses on $R(u,n)$ - other than with $z=2$ and $u=\log(2)$ I did not check it, but for this settings it seems to converge to something $\lim_{n \to \infty} R(u,n) =r_\infty \approx 0.0484903140769$ just by numerically searching (binary search) $b_n=B(2,n)$ and computing backwards. Interestingly the coefficients of the power series of $R(u,n)$ have the same scheme to become constant for the leading part which extends when n increases and has thus also a limiting power series. Pari/GP gives me $$R(u,12)= u - 5 \cdot u^2/2! + 13 \cdot u^3/3! - 19 \cdot u^4/4! + 1 \cdot u^5/5! + 231 \cdot u^6/6! - ...$$
and this is the same for all values n greater than 12, so we might assume this constant also for the case of $n \to \infty$.
I did not find anything related in OEIS yet, perhaps I can later find some pattern...

We see in the last formula in (7) the squareroot occuring, so we get the initial formula as $$\lim_{n \to \infty}\sqrt2 \cdot { \exp(u^n R(u,n)) -1 \over u^n} \tag 8$$ where the exponentialseries (divided by $u^n$) has no constant term and the linear term is $R(u,n)$ and the following terms are powers of it, multiplied by n'th powers of $u=\log(2) \approx 0.693 \lt 1$.

These are so far only some (organized) observations, a real proof of the existence of a finite limit must be provided by analysis of the power series of $R(u,n)$ (but besides the heuristics I don't have the proof yet)

• You mentioned Euler-Summation on your website / tetration forum. It is not clear to me how to do limits of euler summations.
– mick
Nov 18, 2015 at 22:53
• I premultiply the vector of the original partial sums (only a finite number of terms - we'll get always only an approximation, but a "good" one), of say the first 32 or 64 terms with a certain matrix, the ESum-matrix. If the growth-rate of the original terms of the series is at most geometrix, but the signs alternate, then Euler-summation can be made arbitrarily precise by configuring the summation-matrix "ESum". Finally it boils down to have a set of coefficients for each partial sum to make a weighted sum of it. The matrix used is made from (appropriate) powers of the Pascalmatrix. Nov 19, 2015 at 5:42
• You can find an old version of introduction at go.helms-net.de/math/binomial_new/pmatrix.pdf . In chap 3, on page 17 I displayed my procedere. I must confess that at that time when I wrote this I was just beginning with serious work in (amateurish) number theory and the text should not yet fulfill research standards - it was merely a display and explanation of my own heuristical process. But it should give a good explanation anyway... Nov 19, 2015 at 5:51
• Thanks. But why dont you expand at a different point ?
– mick
Nov 20, 2015 at 20:17
• @mick: sorry - what do you mean? Nov 20, 2015 at 20:32

$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$

Notice the resemblance with the Koenigs function

https://en.m.wikipedia.org/wiki/Koenigs_function

In fact it is a Koenigs function with the variable fixed to the value $2$.

Since $1 < 2 < \exp(1/e)$ and the derivative is not $0$ or $1$ , the Koenigs function converges to the correct value ; your limit.

$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$

This limit is only possible if

$$\lim\limits_{n\to\infty}\frac{\sqrt[n+1]2_s - \sqrt 2}{\sqrt[n]2_s - \sqrt 2}= \ln2$$

To show this , use l'hopital

We get with $f(x) = x^{f(x)}$ :

$\frac{D x^{f(x)}} {f ' (x)} = \frac{ \sqrt 2 ^2 2\ln(\sqrt2) }{2} = \ln2$.

Qed

This is part of the answer that justifies the RHS of the limit

$$\mathcal L=\lim\limits_{n\to\infty}\frac{\sqrt[n]2_s-\sqrt2}{(\ln2)^n}\tag1$$

With thanks to Tommy1729 for hints.

• I will start a new question related to this ... Im confused.
– mick
Nov 18, 2015 at 11:55
– mick
Nov 18, 2015 at 19:59
• The confusion is over ! See the answer to the question in the link.
– mick
Nov 20, 2015 at 21:51

I got a message from Tommy1729.

He considered nonzero $T$ such that :

$$T = \lim \frac{A^n}{(f(n,2) - \sqrt 2 - L \ln 2 ^n)}$$.

Where $f(n,2)$ is the nth superroot of $2$ , $L$ is the constant from the Op and $A$ is a constant.

If the limit $T$ does not exist at least the best fitting $A$ is considered. In other words :

$$A^n$$

~

$$\frac{1}{(f(n,2) - \sqrt 2 - L \ln 2 ^n)}$$

Generalizing the Op is easy. And the problem was not new to him.

These generalizations seem even harder and he did not provide formal proofs.

$$L \ln2 = r = 0.0475...$$
Where $r$ is conjectured the radius of Gottfried's answer !