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This question is about where I made my mistake in the computation of a limit. It relates to An answer I gave that confused me. The question to which I gave the (partial) answer is related to tetration but my mistake is probably a simple general one ( considering tetration as complicated ).

Here is the link to the question and my answer :

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


THE QUESTION

The 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?


MY ANSWER

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

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 used the chain rule to get the derivative of both the numerator and denumerator.

$1 = D x = D g(g^{-1}(x)) = g' (g^{-1}(x)) D g^{-1}(x)$

$=>$

$D g^{-1}(x) = \frac{1}{g ' (g^{-1}(x)) }$

In this case $g$ is the power tower of hight $n$ or $n+1$.

To complete the derivative and limit we introduce a function $f$ and let $x$ go to $g^{-1}(2) = \sqrt 2$.

That is the basic idea.

The problem is with the $f(x)$.

I do not remember how I did it but when I tried again I got a different result.

I got

$\frac{D x^{f(x)} }{f'(x)} = \frac{D exp(ln(x) f(x))}{f ' (x)} = \frac{ x^{f(x)} ( f ' (x) ln(x) + " f(x)/x ")}{f ' (x)} = \frac{ \sqrt 2^2 ( f ' (\sqrt 2) ln(2) + " f(\sqrt 2)/ \sqrt 2 ") } {f ' (\sqrt 2)}$

Everything works out fine .. Apart from the " ... " term. I do not think the " ... " term is $0$ , and neither do I think $ f ' (\sqrt 2) $ is $\infty$ , so nothing vanishes ? .. but I also see no mistake made !?!?

So I am confused and I need help to understand.

Probably a silly mistake. Im already embarrased.

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  • $\begingroup$ Grr i always mess Up the Tex ! Some ) or } probably ! $\endgroup$ – mick Nov 18 '15 at 19:44
  • $\begingroup$ Got it :) finally :) $\endgroup$ – mick Nov 18 '15 at 19:48
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Instead of $x^{f(x)}$ , I should have used $(\sqrt2)^{f(x)}$.

Then we get $ D (\sqrt 2)^{f(x)} / f ' (x) = \ln(\sqrt 2) (\sqrt 2)^{f(\sqrt 2)} f ' (\sqrt 2) / f ' (\sqrt 2) = \ln(2).$

Notice that $ D (\sqrt 2)^{(\sqrt 2)^{f(x)}} / f ' (x) = ( \ln 2)^2 $ as it should.

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