Let $f(z)$ satisfy $f(f(z)) = \operatorname{arcsinh}(z/2)$
More precisely, we construct such an $f(z)$ by using the fixpoint at $0$ and the related Koenigs function.
see : https://en.wikipedia.org/wiki/Koenigs_function
Now I wonder about the sign of the $n$ th derivative of $f(x)$ at point $x$ for real $x>0$.
Is there an easy way to compute the sign of $\frac{d^n}{d^n x} f(x)$ at point $x$ ?
Is there maybe a pattern ?
Such as at every fixed value $x>0$ there exists $p_x(x),m_x(x)$ such that $f(x) = p(x) + m(x)$
Where $p_x(x)$ is a real polynomial of $x$ (when expanded around $x$) and $m_x(x)$ is a smooth and completely monotonic function of $x$ at point $x$.
So in other words the tail of the taylor series expanded at some $x>0$ must be a smooth and completely monotonic function of $x$.
Notice the right-plane might not be analytic so the radius of taylor may differ.
Is this pattern true ?
And if not are there other patterns ?
What can we say about the signs of the taylor coefficients ??
I assume $\operatorname{arcsinh}(z/2)$ is a smooth and completely monotonic function for $x > 0$ but I am not sure. The question depends on that.
( it is true that $\operatorname{arcsinh}(z/2)$ is a smooth and completely monotonic function at $x=0$ )
Maybe this fact helps :
A smooth ($C^{\infty}$) and completely monotonic function on an open interval( or semi-axis ) is always analytic on that interval( or semi-axis for the finite part ).
Don't know whether this is what you want. Getting the formal powerseries for $f(z)$ where $f(f(z))=\text{arcsinh}(z/2)$ gives me the first few coefficients as rational numbers scaled by powers of $r=\sqrt 2$:
The rational values are approximations by Pari/GP "bestappr(coeff[j],20)" to $20$ dec digits.
Perhaps a pattern shows up which can then be confirmed by analyzing more of the (real-value) coefficients of $f(z)$.
update We can express the coefficients in polynomials with rational coefficients (and avoid the "bestappr()" q&d-approximations).
The signs of the coefficients seem to alternate (if this is what your are looking for - no proof so far). It should be easy to make strong heuristics up to 256 coefficients or so...
update2 The first 128 coeffs, to be read from left to right and then downwards. The sign-pattern is obvious (computation in Pari/GP with 200 dec digits by default):