Functional "Square Root" of $1/(x+1)$ I've been dabbling in functional iteration lately, but I've run into a problem that's causing me a lot of grief.
First I tried to iterate the softplus function $f(x)=\ln(e^x+1)$, which wasn't hard. By a preexisting theorem, any function of the form $g(h(g^{-1}(x)))$ iterated $n$ times is given by $g(h^n(g^{-1}(x)))$, so the nth iterate of the softplus function is given by $f^n()=\ln(e^x+n)$. I've been mostly focused on functional square roots (a half-iteration of a function), and in this case, the functional square root was $\ln(e^x+\frac{1}{2})$.
Then I tried iterating a very similar function: $f(x)=-\ln(e^x+1)$. This proved to be much more difficult, because in order to put it in the form $g(h(g^{-1}(x)))$, I have to let $h(x)=\frac{1}{x+1}$. Then, to find the functional square root, I would have to find the functional square root of $h(x)$. So far, I haven't been able to do this, but I've convinced myself that the answer will contain imaginary exponents. Can somebody please tell me how to do this (or how to find out if it is possible or not)?
 A: Since $h(x)=\frac{1}{x+1}$ is a fractional linear map, it is enough to devise a matrix square root for $\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$ to have a functional square root for $h(x)$.
If I am not wrong, this approach gives $$ \frac{\sqrt{1-2 i}+x\sqrt{-1+2 i} }{\sqrt{4+2 i}+x\sqrt{1-2 i}} = \color{red}{\frac{1+ix}{(1+i)+x}}$$ as a functional square root of $\frac{1}{1+x}$.
A: With an explorative approach we could first look at rational expressions of the iterates to positive integer heights and try, whether this gives a scheme for a continuous interpolation.
What we get for the first few iterates is
$$ f°^1(x) = {1+0x\over 1+1x} \qquad
 f°^2(x) = {1+1x\over 2+1x} \qquad
 f°^3(x) = {2+1x\over 3+2x} \qquad
 f°^4(x) = {3+2x\over 5+3x} \cdots
$$
and a pattern of Fibonacci-numbers $fib_k$ is much suggestive (and can also be proven)
$$ f°^h(x) = {fib_h+fib_{h-1}x\over fib_{h+1} + fib_h x} \tag 1 $$
Since the Fibonacci-numbers have a generalization to continuous indexes we have the option to define the fractional iterates based on this generalized Fibonacci-numbers.

The most common$\,^{[1]}$ Fibonacci-generalization is that based on the [Binet-formula][1]:
Let $\Phi = {1+\sqrt 5 \over 2} \approx 1.618... $ then the Binet-formula gives
$$ fib_h = {\Phi^h - (1-\Phi)^h\over \sqrt 5} \tag 2$$
If $h=0.5$ as asked here, we have to use $fib_{-0.5}, fib_{0.5},fib_{1.5} $ in formula (1). That three generalized Fibonacci-values are to be inserted
$$ fib_{-0.5} \approx 0.351577584254 + 0.568864481006 î \\
 fib_{ 0.5}  \approx 0.568864481006 - 0.351577584254 î \\
 fib_{ 1.5}  \approx 0.920442065260 + 0.217286896752 î \\
$$
A massive reduction of the explicite formula (1) when that values are inserted, however, can be done when we simply cancel $fib_h$. We have then formally 
$$ f°^h(x) = {1 +{fib_{h-1}\over fib_h}x\over {fib_{h+1} \over fib_h}+  1x} \tag 3 $$
and for $h=0.5$ we get here the handy coefficients ${fib_{h-1}\over fib_h} = 1 î $ and ${fib_{h+1}\over fib_h} = 1 + 1 î $ such that we can write the compacted formula
$$ f°^{0.5}(x) = {1 + 1 î x\over (1 + 1 î )+  1x} \tag 4 $$
which is identical to that already given solution of Jack D'Aurizio.       

The intention of this answer is to show one way, how to generalize that fractional iteration formula to other values than actually $h=1/2$ only.           
 $\,^{[1]}$ There is also a real-to-real-interpolation common (see wikipedia; in an earlier thread here in MSE I've discussed the difference between that two methods); I didn't check what we would get here with that method 

late update: a far more explicited version of this I have in an earlier essay on exactly that subject: see here from mathindex at my homepage
A: I'm not sure about this general approach :
$$f(f(x))=g(x)=\int_0^{f(x)}f'(s)ds+C=\int_0^1f'(f(x)u)f(x)du+C$$
By letting $s=f(x)u$. Then $$\int_0^1f'(f(x)u)f(x)-g(x)du=0\forall x$$
Thus the integrand is orthogonal to the constant 1 function and hence it can be written after orthogonalizing any non constant in u function to 1 via Gram-Schmidt as :
$$f'(f(x)u)f(x)-g(x)+C=h(x,u)-\int_0^1 h(x,s)ds$$
Thus $$f'(f(x))f'(x)=\frac{f'(x)}{f(x)}(g(x)+h(x,1)-\int_0^1h(x,s)ds-C)=g'(x)$$
So that : $$ln(f(x))=D+\int_0^x\frac{g'(u)}{g(u)+h(u,1)-\int_0^1h(u,s)ds-C}$$, giving then f.
