# Is it possible to get the coefficients of the power series

If we have a function $f(s)$ with this form:

$$f(s) = \sum_{i=0}^{\infty} p_i s^i$$

We also know that:

$$f(1) = 1$$

and

$$p_i \ge 0 \quad \text {for all  i \ge 0}$$

Assume we can calculate $f(s)$ for any $s$, is it possible that with all the info we know, we would be able to get $p_n$ for any n?

(Actually $p_i$ is the probability that $[Z=i]$ where Z is a random variable.)

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This is not a polynomial (unless all but finitely many $p_n$'s vanish); the degree of the polynomial has to be finite. – Srivatsan Oct 1 '11 at 23:22
Coefficients of a power series, you mean. If you can evaluate $f$ for any complex $s$, then sure... – J. M. Oct 1 '11 at 23:26
Your series will necessarily be the Taylor series of $f(s)$ about $s = 0$. Then ${\displaystyle p_i = {f^{(i)}(0) \over i!}}$ for all $i$, where $f^{(i)}(0)$ denotes the $i$th derivative of $f(s)$ at $s = 0$. – Zarrax Oct 1 '11 at 23:30
In any event... – J. M. Oct 1 '11 at 23:34
Could you answer my question? Can you, or can you not, evaluate $f(s)$ for complex $s$, or are you restricted to evaluating for real $s$? – J. M. Oct 1 '11 at 23:53

This is the discrete version of the moment problem or the infinite version of a Vandermonde matrix. One approach is that $p_0=f(0),\quad p_1=\left.\dfrac{\mathrm df(s)}{\mathrm ds}\right|_{s=0}$ and the higher $p$'s are higher derivatives at $0$. Of course, this is rather unstable numerically.
The thing is, according to my understanding of the problem. What I can get is only $f(s)$, not $f'(s)$. In fact, I'm trying to construct a random variable with the same distribution as $Z$, but without knowing $p_0, p_1, p_2, \ldots$, it seems impossible . That's why I'm trying to get all $p_i$'s. – ablmf Oct 1 '11 at 23:42
You can take a numeric derivative. Take $s$ smaller and smaller and check $\frac{f(s}-f(0)}{s}$. That is where the instability comes from-you are subtracting two nearly equal quantities. – Ross Millikan Oct 2 '11 at 0:03
@ablmf: you can do Richardson extrapolation in conjunction with Ross's "take $s$ smaller and smaller" strategem. It doesn't completely cure the numerical instability, but you might manage to squeeze out a few more digits of accuracy as long as you don't shrink $s$ too much. – J. M. Oct 2 '11 at 0:17