Question

Hello

If $f(x)=\sum _{n=0}^{\infty } b_nx^n$, and $\frac{1}{f(x)}=\sum _{n=0}^{\infty } d_nx^n$. Then the coefficients of the reprical of f 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 was a general recursive (prefferably not ofcourse) formula for the coefficients of the reciprocal fuction?? That is given an arbitrary $n$, 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

Answer 1

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 $p(n)$).

Answer 2

Without loss of generality we can take $b_0$ to be 1, since \begin{equation*}\sum_{n=0}^\infty b_n x^n = b_0\biggl( 1+\sum_{n=1}^\infty (b_n/b_0)x^n\biggr). \end{equation*} Then for $b_0=1$ we have \begin{equation*} \frac1{f(x)} = \biggl( 1+\sum_{n=1}^\infty b_n x^n\biggr)^{-1}\ =\sum_{m=0}^\infty (-1)^m\biggl( \sum_{n=1}^\infty b_n x^n\biggr)^m. \end{equation*} Expanding by the multinomial theorem and extracting the coefficient of $x^n$ gives \begin{equation*} \frac1{f(x)} = \sum_{n=0}^\infty \kern 3pt x^n \kern -5pt \sum_{m_1+2m_2+3m_3+\cdots = n} (-1)^{m_1+m_2+\cdots} \binom{m_1+m_2+\cdots}{m_1, m_2, \ldots} b_1^{m_1} b_2^{m_2}\cdots.\end{equation*}