Inverse Rule for Formal Power Series I am just really starting to get into formal power series and understanding them.  I'm particularly interested in looking at the coefficients generated by the inverse of a formal power series:
$$\left(\sum_{n\ge 0}a_nx^n\right)^{-1}=\sum_{n\ge 0}b_nx^n$$
I first thought that my approach would be looking at 
$$\frac{1}{\sum_{n\ge 0}a_nx^n}$$
But I'm more thinking that since we know that a series is invertible in the ring if $a_0$ is invertible in the ring of coefficients.  Thus, since if we assume it is, and since the unit series is $\{1,0,0,0,....\}$ then we have
$$\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}b_nx^n\right)=1$$
Thus we know that $a_0b_0=1$ and thus $b_0=\frac1{a_0}$.  And for the remaining terms we are just looking at the convolution generated by the Cauchy Product and so
$$0=\sum_{j=0}^ka_jb_{k-j}$$
$$-a_0b_k=\sum_{j=1}^ka_jb_{k-j}$$
$$b_k=\frac{-1}{a_0}\sum_{j=1}^ka_jb_{k-j}$$
And thus we have a recursive definition.
Is there another approach that defines the numbers $b_k$ without recursive means?  Are you forced to only recursive methods when operating on the ring of formal power series to calculate coefficents?

You are looking for the reciprocal of a power series.
Applying Faà di Bruno's formula (Wikipedia: Bell polynomials - Faà di Bruno's formula) yields the following formula.
$$b_{n}=\frac{1}{n!}\sum_{k=0}^{n}(-1)^{k}k!a_{0}^{-(k+1)}B_{n,k}(1!a_{1},2!a_{2},...,(n-k+1)!a_{n-k+1})$$
$B_{n,k}$: partial exponential Bell polynomial (Wikipedia: Bell polynomials - Exponential Bell polynomials)
This is written e.g. in Singh, M.: nth-order derivatives of certain inverses and the Bell polynomials. J. Phys. A 23 (1990) (12) 2307-2313.
$$b_{n}=\sum_{k=0}^{n}(-1)^{k}a_{0}^{-(k+1)}\hat{B}_{n,k}(a_{1},a_{2},...,a_{n-k+1})$$
$\hat{B}_{n,k}$: Partial ordinary Bell polynomial (Wikipedia: Bell polynomials - Ordinary Bell polynomials)
For certain formal power series, there is a general formula for $B_{n,k}$ and $\hat{B}_{n,k}$.
