What's the function before $f_0(x)$ in Fast-growing Hierarchy I was wondering about a function related to the fast-growing hierarchy that would be before $f_0(x) = x + 1$, i.e a function $f_{-1}(x)$ such that:
$f_{-1}^x(x) = x+1$ where, for example $f_{-1}^3(x) = f_{-1}(f_{-1}(f_{-1}(x)))$
Edit: This function isn't technically in the fast-growing hierarchy, because they are functions of integers. This implies that $f_0(x)$ is the slowest function that we can ùake, but we're actually extending the domain right here.
Can we get a formula for such a function (if it exists) ? Can we find more than one ? Is there some good properties ?

Here some precisions and what I already know.

*

*Actually, this is easy to find this solution:
$f_{-1}(x) = x + \frac{1}{\lfloor x \rfloor}$, but it's not (in this problem) a "good" function, i.e it's not continuous or differentiable. But $f_0(x), f_1(x) = 2x$ and $f_2(x) = x2^x$ have these properties. I actually ask the question for these ones.

*$f^0(x) = x$ in general, this is coherent with some properties of the re-iterating functions $f^{-1}(f^1(x)) = x = f^{1-1}(x) = f^0(x)$. It also can be read as a f that isn't applied to x. But here the definition says that $f_{-1}^0(0) = 1$ (which should be false). I think a great domain for $f_{-1}(x)$ is $[1,\infty[$, otherwise there could be contradictions.

*It seems like if $f_{-1}(x)$ can be of the form $x + \sum_{k=1}^{\infty}{\frac{c_k}{x^k}}$ with $c_k \in \mathbb{R}$ (and with a converging serie) then $f_{-1}^x(x) \approx x+1$ for some "big" $x$ (but actually the approximation is in general getting quiet good), i.e $\lim_{x\rightarrow\infty}{\frac{f_{-1}^x(x)}{x+1}} = 1$ But I wasn't able to demonstrate it, though it's intuitive.

*Here is a possible approximation:

$x + \frac{1}{x} + \frac{1}{1.9364832724923593}(\frac{1}{x^2} - \frac{1}{x^3}) + \frac{1}{3.5300211407995237}(\frac{1}{x^4} - \frac{1}{x^5}) + \frac{1}{0.62770748195252}(\frac{1}{x^7} - \frac{1}{x^6})$

Thank you :)
 A: I've recently studied on internet iterated functions and how to solve functional equations like $f(f(x)) = e^x$, which was pretty interesting. There's a method that it is use, which is about finding a solution to the Abel functional-equations:
$\alpha(f(x)) = \alpha(x) + 1$
Because it will follow that
$\alpha(f^t(x)) = \alpha(x) + t
\\ \iff f^t(x) = \alpha^{-1}(\alpha(x) + t)$
Which gives a solution to all iterated functions of f!
We're searching a function $f$ such that $f^n(n) = n+1$, let's try to find it's solution to the Abel functional equation (which is how I solved the problem):
$\alpha(f^t(x)) = \alpha(x) + t
\\ \implies \alpha(f^t(t)) = \alpha(t) + t
\\ \iff \alpha(t+1) = \alpha(t) + t$
This now seems pretty easy, we first try to solve it for t as an integer.
$\alpha(t+n) = \alpha(t) + \sum_{k=0}^{n-1}{t+k}
\\ = \alpha(t) + \sum_{k=0}^{n-1}{t} + \sum_{k=0}^{n-1}{k}
\\ = \alpha(t) + tn + \frac{n(n-1)}{2}$
Choosing $t=0$ we get
$ \alpha(n) = \alpha(0) + \frac{n(n-1)}{2}$
This is not rigourous (as everything that we'll do next), but let's generalise to every real $x$ (just replace $n$ with $x$). We've just found a solution to our original functional equation, with $\alpha(0)$ a constant!
Now, let's try to go further and find $\alpha^{-1}(x)$
$\alpha(x) = \alpha(0) + \frac{x(x-1)}{2}
\\ \iff \alpha(x) = \frac{1}{2}x^2 - \frac{1}{2}x + \alpha(0)
\\ \iff x = \frac{1}{2}\alpha^{-1}(x)^2 - \frac{1}{2}\alpha^{-1}(x) + \alpha(0)
\\ \iff \frac{1}{2}\alpha^{-1}(x)^2 - \frac{1}{2}\alpha^{-1}(x) + \alpha(0) - x = 0$
After using the method for solving second degree equations and simplifying, we get:
$\alpha^{-1}(x) = \frac{1 \pm \sqrt{1 - 8\alpha(0) + 8x}}{2}$
Now we can perhaps find a solution by using this formula.
$f(x) = \alpha^{-1}(\alpha(x) + 1)
\\ = \alpha^{-1}(\frac{1}{2}x^2 - \frac{1}{2}x + \alpha(0) + 1)
\\ = \frac{1 \pm \sqrt{1 - 8\alpha(0) + 8(\frac{1}{2}x^2 - \frac{1}{2}x + \alpha(0) + 1)}}{2}$
And then after simplification, we get rid of $\alpha(0)$!
$f(x) = \frac{1 \pm \sqrt{4x^2 - 4x + 9}}{2}$
If we choose the sign to be positive, we get a function that actually works!
Well I didn't prove it (yet, but I don't now if it's difficult), but this is essential since I've made a lot of assumptions.
Also, I don't know if we can construct others solutions, which I think might be really interesting! I was also surprised that there was such a simple formula for the problem, I'm sure that somebody has found this before.
For answering the question, yes such a continuous and derivable function exists, and we have:
$f_{-1}(x) = \frac{1 + \sqrt{4x^2 - 4x + 9}}{2}$
