How to convert a powerseries into a Dirichlet series?

[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 $s(h)$

• Are you asking how to convert a power series of a function, back into the original function? Say in your example if you're given $ps(x)$, find a function $s(t)$ such that $s(1) = 1.25723$, $s(2) = -0.699161$, etc? – Mitch Mar 9 '11 at 22:06
• @Mitch: I'm 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 $f(x) = \sqrt{2+x}$ at $x=1$, so $s(1)=\sqrt{2+1}$, $s(2)=\sqrt{2 + \sqrt{2+1}}$ and so on. 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 is completely different. – Gottfried Helms Mar 10 '11 at 14:15
• A little more clarification needed...so in this last example where $f(x) = \sqrt{2+x}$, you're defining $s(1) = f(1)$, and $s(t) = f(s(t-1))$ ? And $ps(x)$ is the power series for $s(t)$ or for $f(x)$? (it's hard to tell from the numerical examples you gave; maybe another simpler example where one can see how to calculate the values that correspond) – Mitch Mar 10 '11 at 14:57
• @Mitch: I extend the examples above in the question. – Gottfried Helms Mar 10 '11 at 15:17
• A Dirichlet series is a series of the form $$f(s)=a_1+a_22^{-s}+a_33^{-s}+a_44^{-s}+\dots$$ It has nothing to do with iteration, and your $s(h)$ has nothing to do with Dirichlet series. – Gerry Myerson May 4 '11 at 13:11

Not sure if you found the answer yet, but this link at Math Overflow solves it.

Essentially, if $F(s) = \sum_{n\ge 1} \frac{a_n}{n^s}$ with abscissa of absolute convergence $\sigma_a$, then

\begin{eqnarray*} \lim_{T\to \infty} \frac{1}{2T} \int_{-T}^T F(\sigma + it) m^{it} dt &=& \lim_{T\to \infty} \frac{1}{2T} \sum_{n\ge 1} \int_{-T}^T\frac{a_ndt}{n^{\sigma}(n/m)^{it}}\\ &=& \lim_{T\to \infty} \frac{1}{2T} \sum_{n\ge 1, n\not=m}\frac{a_n}{n^{\sigma}} \frac{(n/m)^{iT} - (n/m)^{-iT}}{i\log (n/m)} + \frac{a_m}{m^\sigma} \end{eqnarray*}

Since $\frac{(n/m)^{iT} - (n/m)^{-iT}}{i\log (n/m)}$ is bounded as $T\to\infty$, everything vanishes except $\frac{a_m}{m^\sigma}$, so

$$\lim_{T\to \infty} \frac{1}{2T} \int_{-T}^T F(\sigma + it) m^{it} dt = \frac{a_m}{m^\sigma}$$

Then to retrieve the dirichlet series from the power series, just apply this transform for each value of $m$.

• Thanks! I'm afraid, this is far beyond my brain at the moment - unfortunately our courses begin and I've nearly no capacity to chew this appropriately. Hope, I can come back to this in a couple of weeks... – Gottfried Helms Oct 22 '12 at 5:46