Post Closed as "too localized" by Willie Wong, Gjergji Zaimi, Andres Caicedo, Qiaochu Yuan, Gerry Myerson
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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

2 added 6 characters in body

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:

$d_0 = \frac{a_0}{b_0}$, frac{1}{b_0}$,$d_1 = \frac{a_1 b_0-a_0 b_1}{b_0^2}$,-\frac{b_1}{b_0^2}$,

$d_2 = \frac{a_2 b_0^2-a_1 frac{b_1^2-b_0 b_2}{b_0^3}$

$d_3 = -\frac{b_1^3-2 b_0 b_1+a_0 \left(b_1^2-b_0 b_2\right)}{b_0^3}$ 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

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