[Update 3]: I realized, that the series $s(h)$ below is simply a "(general) Dirichlet-series" (1), so after I know the principle how to find the Taylor-series for some function (which may be given by its Dirichlet-series) I can now make my question more precise as:
how to find the Dirichlet-series-representation of a function, given its Taylor-series?
(1) - the attribute "general" means a series $ \sum_{k=1}^{\infty} a_k*e^{- \lambda _k s}$ where the $\lambda _k$ form a strictly monotone sequence of real numbers increasing to $+\infty$ (K. Knopp, "Infinite series...", german ed.)
(I did not change the text below) [end update 3]
In the study of functional iteration and the Schröder function I'm concerned with series of the type:
$$ s(h)= a_0* (u^0)^h + a_1* (u^1)^h + a_2 * (u^2)^h + ... $$
where h is the (iteration-) "height"-parameter.
Usually, if this converges, in the numerical evaluation I change order of computation:
$$ s(h)= a_0*(u^h)^0 + a_1*(u^h)^1 + a_2*(u^h)^2 + ... $$
which means in many cases I handle this as a power series in $u^h$ . But when I considered further analysis, for instance the derivative with respect to h, I differentiate using $h$ as highest exponent. (BTW, what is the name of that type of series?)
Now by accident I asked Pari/GP for the coefficients of $s(x)$ and what I got back was a power series in x. I never thought deeper about such a conversion, except that I remember the case of the conversion of the zeta-series into a power series using the Stieltjes constants.
Example:
s(h) = 2*1^h -1.09662*1/4^h + 0.100215*1/16^h -0.00366327*1/64^h
+ 0.0000717362*1/256^h -0.000000874084*1/1024^h + ... - ...
ps(x) = 1.00000 + 1.25723*x - 0.699161*x^2 + 0.172874*x^3 + 0.0350749*x^4
- 0.0550780*x^5 + 0.0288564*x^6 - 0.00942667*x^7 + O(x^8)
Well, I think, what Pari/GP internally does is to compute the derivatives of $s(h)$ and construct the Taylor series. But now I'm curious (and that's my question): How could I convert a power series ps(x) into a series of the s(h)-type?
[update] I think the comment to Mitch would be a good additional background and explanation for the question, so I copy it to here and extend it a bit
I'm originally interested in the method: how one would do such a conversion of $ps(x)$ to $s(h)$ ?
Note that some values for $s(h)$ are $s(1) \approx 1.73205$ , $s(2) \approx 1.93185$.
My example comes from iteration of the function $$f(x) = \sqrt{2+x}$$ $$f(f(x))=\sqrt{2+\sqrt{2+x}} $$ $$f(f(f(x))) =\sqrt{2+\sqrt{2+\sqrt{2+x}}} $$ and so on.
Now $f(x)$ has a power series in $x$, as well as $f(f(x))$ and for each iterate there is another power series. The method of Schröder functions applied to a recentered version of $f(x)$, which has no constant term (also called "regular iteration"), allows to find a power series $F(x,h)$ in two variables where $h$ means the iteration-"height". In such a function $F(x,h)$ the coefficients at powers of $x$ are polynomials in $\lambda^{ h}$ (where $\lambda$ is an eigenvalue of the function which I usually denote as $u$ for ASCII-readability).
If I set $x=1$ and reorder summation collecting like powers of $\lambda^h $ I get a series in $\lambda^h$ only which I called $S(h)$ This function gives exactly the value of the h'th iterate of $f(x)$ beginning at x=1 simply meaning $ s(h) = F(1,h) $
But $s(h)$ has not the usual form of a power series in $h$ as shown above. When I asked Pari/GP for an evaluation at some h I accidentally typed $s(x)$ and Pari/GP gave back a power series in x, I called it above $ps(x)$. $s()$ and $ps()$ have completely different forms but give the same result: $s(h) = ps(h)$ where both functions converge.
The $s(h)$ as well as the $ps(x)$-form allow also continuous (i.e fractional) iteration giving the same values. Interestingly the behave of the convergence of the two forms is completely different. The limit for $h\to\infty$ is immediately visible in $s(h)$ but for $ps(h)$ that would likely be a divergent series, while for $h=0$ we see immediately that $ps(h)=1$ but would not recognize it by $s(h)$
