How to find $s(\exp(d(x)))$ ~ $x + 2$?

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 ?

• Note that $\lim_n a_n-b_n = 0$ does not mean the same thing as $a_n\sim b_n$. – Regret Mar 26 '15 at 13:33
• I want both , so i edited with also – mick Mar 26 '15 at 13:40
• – Jack D'Aurizio Mar 26 '15 at 14:45
• @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. – mick Mar 26 '15 at 15:04
• @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 $d(x)$. – 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: $d(2)=e,d(4)=e^e,d(6)=e^{e^e}$.