If one is given a power series of the form:
$$\sum^\infty_{m=1} a_m x^m = a_0 + a_1x + a_2 x^2 + \dots$$
for known $a_m$'s i.e $a_m = f(m)$ and as $m\rightarrow\infty$, $a_m\rightarrow0$ is there anyway to find out what the function is given by? For example, I know that
$$\sum^\infty_{m=1} x^m = {1\over 1-x}.$$
So is there an expression for for non-unity coefficients? Is there a way t0f doing this? Like inverting the Taylor series or something similar?
There are two questions here, one in the title and one in the body of the text:
(i) Given a power series $$\sum_{k=0}^\infty a_k x^k\tag{1}$$ with $a_0=c\ne0$ in general there is no way to get hold of some functional inverse "in finite terms". Assume that this series defines a monotonic function $f:\ U\to{\mathbb R}$ in some neighborhood $U$ of $0$. The corresponding inverse function would then be defined in the neighborhood of $c$ and maybe have a development of the form $f^{-1}(y)=\sum_{l=0}^\infty b_l(y-c)^l$. Even when this is the case the $b_l$ cannot be obtained from the $a_k$ by some recursive formula involving only finitely many $a_k$ at each step.
If, however, $a_0=0$ and $a_1\ne0$ then the series $(1)$ has a "formal" inverse, as well as a functional inverse. This means the following: There is a sequence $(b_l)_{l\geq1}$ of coefficients $b_l$, defined recursively from the coefficients $a_k$, such that "formally" $$\sum_{l=1}^\infty b_l\left(\sum_{k=1}^\infty a_k x^k\right)^l=x\ .$$ Furthermore, if the series $(1)$ actually converges to some function $f(x)$ in a neighborhood $U$ of $0$, then the series $\sum_{l=1}^\infty b_ly^l$ represents $f^{-1}(y)$ in some neighborhood $V$ of $0$. Mathematica has a command that computes the $b_l$ from the $a_k$, no question of convergence involved.
(ii) A series $(1)$ involves an arbitrary infinite sequence of coefficients $a_k$, whereas any well known special function, as $\exp$, $x\mapsto x/(1-x)^2$ and the like, has a Taylor series whose coefficients follow some rigid law, say, the law of the Fibonacci numbers. Therefore you cannot expect that a "random" series of the form $(1)$ can be identified as the series of a function in your catalogue of "well known functions".
But note that there is some help when the encountered $a_k$ happen to be integers: You can look up an initial segment $(a_0,a_1,\ldots, a_n)$ of the sequence $(a_k)_{k\geq0}$ in the online catalogue OEIS and will maybe find out that it belongs to some "well known function".
Set $b_0 = \frac{1}{a_0}$, that is $a_0b_0 = 1$. Let's try to set the $b_n$ for n > 0 by induction/recursion.
In the base case n = 0 and we can set the $b_n$ for n > 0 trivially (there are none)
So for the induction step (k < n to n) we have set all $b_j$ correctly, so that for all k < n the terms $\sum_{i+j=k} a_ib_j$ yield zero. Now let's just set our coefficient of concern of degree n:
$\sum_{i+j=n} a_ib_j = 0 $
$(\sum_{i+j=k \setminus\{i=0, j=n\}} a_ib_j) + a_0b_n = 0$
$ a_0b_n = -(\sum_{i+j=k \setminus\{i=0, j=n\}} a_ib_j)$
multiply both sides by $b_0$
$ (b_0a_0)b_n = -b_0(\sum_{i+j=n \setminus\{i=0, j=n\}} a_ib_j)$
$b_n = -b_0(\sum_{i+j=n \setminus\{i=0, j=n\}} a_ib_j)$
This gives a recursive algorithm to set the coefficients of the inverse of f(x).
Previous answers said this was not possible, so maybe my answer is wrong or maybe i did something else. I came up with it when trying to solve an excerzise in an algebra book (Aluffi 3.1.16: Prove that a power series $a_0 + a_1x^1 + a_2x^2 + \dots$ is a unit in R[[x]] if and only if $a_0$ is a unit in ring R.), which, if true implies an inverse at least exists.
You cannot "invert" a power series (P.S.) by inverting term by term. If you want to find the inverse of a function of which a powerseries representation is established, you got to kind of start all over. For example, there is a P.S. for sinx. But reciprocating these polynomial terms does not get the P.S. for the cosecant function. Instead you have to find a couple of derivatives of the cosecant and established the coefficients by putting in zero for x every time. Other mechanisms are also possible to find it, but still involves some algebra.
Sometimes you can build an inverse function from a given function. For example, a P.S. representation of arcsine can be found by creating √(1-x) in a denominator (or use Binomial Series) and then replacing x by x² etc.
