Let $x$ be a positive real. I want to find a pair of analytic functions $s(x),d(x)$ such that $s(d(x)) = x$ and

$ s(\exp(d(x)))$ ~ $ x + 2 $

More presicely I Also want :

$$ \lim_{x \to \infty} s(\exp(d(x))) - x - 2 = 0$$

Polynomials seem to fail as do polynomials of exp or ln.

Maybe try Lambert-W ?

Or do I want THE impossible ?

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    $\begingroup$ Note that $\lim_n a_n-b_n = 0$ does not mean the same thing as $a_n\sim b_n$. $\endgroup$ – Regret Mar 26 '15 at 13:33
  • $\begingroup$ I want both , so i edited with also $\endgroup$ – mick Mar 26 '15 at 13:40
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    $\begingroup$ Related: mathoverflow.net/questions/4347/… $\endgroup$ – Jack D'Aurizio Mar 26 '15 at 14:45
  • $\begingroup$ @Jack D'Aurizio Im intrested in dynamics and tetration and I welcome your link therefore. However it might be " in the middle , it cannot be very close to a real iteration of exp. In that sense I think the link is not related. $\endgroup$ – mick Mar 26 '15 at 15:04
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    $\begingroup$ @mick: I do not think so. Here we have $d(x+2)=e^{d(x)}$ hence $d(x+1)$ is the "functional square root of the exponential function" applied to $d(x)$. $\endgroup$ – Jack D'Aurizio Mar 26 '15 at 15:06

We have: $$ d(x+2)= e^{d(x)} \tag{1}$$ hence assuming $d(0)=1$ we have that $d$ grows pretty fast: $d(2)=e,d(4)=e^e,d(6)=e^{e^e}$.

Now it is time to look at Anix' answer to this MO question.

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    $\begingroup$ Your correct. IT is tetration. Tommy posted about the same time to tetration forum this fact. We had a bet where the answer would-be first. Hard call :) $\endgroup$ – mick Mar 26 '15 at 16:48
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    $\begingroup$ Multitasking is dangerous. Well Thats my excuse ;) $\endgroup$ – mick Mar 26 '15 at 16:50

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