Let $f(x)=\sum \_{n=0}^{\infty } b\_nx^n$ and $\frac{1}{f(x)}=\sum \_{n=0}^{\infty } d\_nx^n$. Then the coefficients of the reciprocal of $f(x)$ can be written down. The first few terms are:


$d\_0 = \frac{1}{b\_0}$, 


$d\_1 = -\frac{b\_1}{b\_0^2}$,


$d\_2 = \frac{b\_1^2-b\_0 b\_2}{b\_0^3}$ 


$d\_3 = -\frac{b\_1^3-2 b\_0 b\_1 b\_2+b\_0^2 b\_3}{b\_0^4}$


...


I was wondering if there is a general recursive (preferably not of course) formula for the coefficients of the reciprocal. If an arbitrary $n$ is given, can I write down a formula for $d\_n$ (recursive or not)?


Regards


//edit: as the comments below suggest I think people are misinterpretating the question. I am not looking for someone to show me how to solve a system of linear equations by substitution... I want a formula for d\_n, Since posting the question, I found such a formula for $d\_n$ at <http://functions.wolfram.com/GeneralIdentities/7/>, see the section on Ratios of the direct function ... if anyone knows of how this formula is derived or any other references to it or similar formulas please let me know... thanks 



power series of the reciprocal... does a recursive formula exist for the coefficients
Assume $b\_0=1$ to simplify things. You want a closed formula for the recursively defined sequence $$d\_0=1$$ $$d\_n=-\sum\_{k=0}^{n-1}d\_kb\_{n-k}. $$
Let $\alpha=(\alpha\_1,\dots,\alpha\_r)\in \mathbb{N}\_ +^\omega$ be a multi-index with length $l(\alpha):=r$ and weight $|\alpha|:=\sum\_{j=1}^r\alpha\_j$. Let's denote $b\_\alpha:=b\_{\alpha\_1}\dots b\_{\alpha\_r}$.


We have (induction)
$$d\_n:=\sum\_{|\alpha|=n}(-1)^{l(\alpha)}b\_\alpha. $$


There are of course several equal terms in the sum, due to the commutativity; summing equal terms, a corresponding smaller set of indices would be the increasing multi-indices (the number of terms in the sum would then be the number of partitions $p(n)$). 


