= Reciprocal of a power series
{title2=$b_0=1/a_0,\quad b_n=-a_0^{-1}\sum_{k=1}^na_kb_{n-k}$}
If $f(x)=\sum_{n\geq0}a_nx^n$ has $a_0\ne0$, its reciprocal has a unique <formal power series> $1/f(x)=\sum_{n\geq0}b_nx^n$. Equating coefficients in the <Cauchy product> $f\cdot(1/f)=1$ gives the displayed <recurrence relation>. In particular,
$$
\frac1{f(x)}=\frac1{a_0}-\frac{a_1}{a_0^2}x+\frac{a_1^2-a_0a_2}{a_0^3}x^2+O(x^3).
$$
For a convergent <power series>, continuity and $f(0)\ne0$ give a neighborhood without zeros, where the reciprocal is analytic and this series converges. Its radius need not equal the original radius: zeros of $f$ can introduce closer reciprocal singularities.
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