Dividing an infinite power series by another infinite power series Let's say I have two power series
$\,\mathrm{F}\left(x\right) = \sum_{n = 0}^{\infty}\,a_{n}\,x^{n}$ and
$\,\mathrm{G}\left(x\right) = \sum_{n = 0}^{\infty}\,b_{n}\,x^{n}$.
If I define the function
$\displaystyle{\,\mathrm{H}\left(x\right) =
\frac{\mathrm{F}\left(x\right)}{\mathrm{G}\left(x\right)} =
\frac{\sum_{n = 0}^{\infty}\, a_{n}\,x^{n}}{\sum_{n = 0}^{\infty}\, b_{n}\, x^{n}}}$, is there a general way to expand $\,\mathrm{H}$ such that $\,\mathrm{H}\left(x\right) = \sum_{n=0}^{\infty}\,c_{n}\,x^{n}$ ?.
I guess, what i'm asking is if there is a way to get the first few $c_{n}$ coefficients ?. I'm dealing with a physics problem in which I have two such functions $\,\mathrm{F}$, $\,\mathrm{G}$ and I'd like to get the first few terms in the power series $\,\mathrm{H}$.
 A: If we use the geometric series, we end up with
$$\frac1{G(x)}=\frac1{1-(1-G(x))}=\sum_{n=0}^\infty(1-G(x))^n$$
This works out best if $b_0=1$.  If $b_0=b$, then one must rescale as follows:
$$\frac1{G(x)}=\frac{1/b}{1-(1-G(x)/b))}=\frac1b\sum_{n=0}^\infty\left(1-\frac{G(x)}b\right)^n$$
Proceed to foil out and then multiply $F(x)$ in to get the desired $H(x)$.
A: One can also derive the following fast method (which works fast at least for a few coefficients of the expansion). It follows from umbral calculus, since the generating function of the form
$$
\mathrm{F}(x)=\sum_{n=0}^\infty a_n x^n
$$
satisfies umbral differential equation (here $\theta=x\frac{d}{dx} \Rightarrow g(\theta)\cdot x^n = g(n)x^n$):
$$
(1+\theta)^{-1} a_{\theta+1}^{-1} a_{\theta}^{\vphantom{1}} \frac{d}{dx} \cdot \mathrm{F}(x) = \mathrm{F}(x)
$$
Now one needs some trivial steps to obtain the following identity. For given sequence $\{b_n\}_{n=0}^\infty$ consider the operator $\mathfrak{L}_b$ which acts on sequences as
$$
\mathfrak{L}_bf(n) := f(n+1)-\frac{b_{n+1}}{b_0}f(0)
$$
Then
$$
\frac{\sum\limits_{n=0}^\infty a_n x^n}{\sum\limits_{n=0}^\infty b_n x^n}=\sum_{k=0}^\infty x^k \left.\left[\frac{1}{b_0} \mathfrak{L}_b^k \cdot a_n\right] \right|_{n=0}
$$
Indeed
\begin{align*}
&\left.\left[\frac{1}{b_0} \mathfrak{L}_b^0 \cdot a_n\right] \right|_{n=0} = \frac{a_0}{b_0}\\
&\left.\left[\frac{1}{b_0} \mathfrak{L}_b^1 \cdot a_n\right] \right|_{n=0} = \left.\frac{1}{b_0}\left(a_{n+1}-\frac{b_{n+1}}{b_0}a_0\right)\right|_{n=0}=\frac{a_1}{b_0}-\frac{b_1 a_0}{b_0^2}\\
&\left.\left[\frac{1}{b_0} \mathfrak{L}_b^2 \cdot a_n\right] \right|_{n=0} =\\
&=\left.\frac{1}{b_0}\left(a_{n+2}-\frac{b_{n+2}}{b_0}a_0-\frac{b_{n+1}}{b_0}\left(a_1-\frac{b_1 a_0}{b_0} \right)\right)\right|_{n=0}=\\
&=\frac{a_2}{b_0}-\frac{a_0 b_2}{b_0^2}-\frac{a_1 b_1}{b_0^2}+\frac{a_0 b_1^2}{b_0^3}
\end{align*}
I hope it is helpful.
UPD. Similarly, one may use the following method. Suppose that we want to understand what is the expansion of the series
$$
\frac{1}{\theta!} \cdot \frac{R_\theta \cdot A(x)}{H_\theta \cdot B(x)}
$$
Define the $0$-derivative $\mathrm{L}$, an operator that acts on formal power series of $x$ as $\mathrm{L}\cdot f(x)=(f(x)-f(0))/x$. Now write tautologically
\begin{align*}
\frac{1}{\theta!} \cdot \frac{R_\theta \cdot A(x)}{H_\theta \cdot B(x)}&=\exp(x \partial_y)\cdot \frac{1}{\theta_y!} \cdot \frac{R_{\theta_y} \cdot A(y)}{H_{\theta_y} \cdot B(y)}\bigg|_{y=0} = \theta_y !\exp(x \partial_y)\cdot \frac{1}{\theta_y!} \cdot \frac{R_{\theta_y} \cdot A(y)}{H_{\theta_y} \cdot B(y)}\bigg|_{y=0}=\\
&=\exp(x \mathrm{L}_y) \cdot \frac{R_{\theta_y} \cdot A(y)}{H_{\theta_y} \cdot B(y)}\bigg|_{y=0} = (H_\theta\cdot B)(y)\exp(x \mathrm{L}_y) \cdot \frac{R_{\theta_y} \cdot A(y)}{H_{\theta_y} \cdot B(y)}\bigg|_{y=0}=\\
&=\exp(x (H_\theta \cdot B)\mathrm{L}(H_\theta \cdot B)^{-1}) R_\theta \cdot A(y)\bigg|_{y=0}=\\
&=\exp\left(x \frac{H_{\theta+1}}{H_\theta}B\mathrm{L}B^{-1}\right) R_\theta \cdot A(y)\bigg|_{y=0}
\end{align*}
Now if $H_{\theta+1}/H_{\theta}$ is a polynomial in $\theta$, say $p(\theta)$, then we have an action $(H_{\theta+1}/H_{\theta}) \cdot y^n = p(n)y^n$. The latter means that in the exponent we have some reasonable operator, and in certain cases it may be very helpful (for the reasonable choice of the series $B(y)$).
A: The standard way
(in other words,
there is nothing original
in what I am doing here)
to get $H(x)$
is to write
$H(x)G(x) = F(x)$
and get an iteration
for the $c_n$.
$\begin{array}\\
H(x)G(x)
&=\sum_{i=0}^{\infty} c_{i} x^{i} \sum_{j=0}^{\infty} b_{j} x^{j}\\
&=\sum_{i=0}^{\infty}  \sum_{j=0}^{\infty} c_{i}b_{j} x^{i+j}\\
&=\sum_{n=0}^{\infty}  \sum_{i=0}^{n} c_{i}b_{n-i} x^{n}\\
&=\sum_{n=0}^{\infty} x^{n} \sum_{i=0}^{n} c_{i}b_{n-i} \\
\end{array}
$
Since
$H(x)G(x) 
= F(x)
= \sum_{n=0}^{\infty} a_{n} x^{n}
$,
equating coefficients
of $x^n$,
we get
$a_n
=\sum_{i=0}^{n} c_{i}b_{n-i}
$.
If $n=0$,
this is
$a_0 = c_0b_0$
so,
assuming that
$b_0 \ne 0$,
$c_0
=\dfrac{a_0}{b_0}
$.
For $n > 0$,
again assuming that
$b_0 \ne 0$,
$a_n
=\sum_{i=0}^{n} c_{i}b_{n-i}
=c_nb_0+\sum_{i=0}^{n-1} c_{i}b_{n-i}
$
so
$c_n
=\dfrac{a_n-\sum_{i=0}^{n-1} c_{i}b_{n-i}}{b_0}
$.
This is the
standard iteration
for dividing polynomials.
A: Since the multiplication of power series is not that hard we can reduce the task in finding the reciprocal $\frac{1
}{G(x)}$ of a power series
\begin{align*}
G(x)=\sum_{n=0}^\infty b_n x^n
\end{align*}
provided $b_0\ne 0$.

According to H.W. Gould's Combinatorial identities, vol. 4 formula (2.27) the following is valid: Let $b_0\ne 0$, then with
\begin{align*}
\frac{1}{G(x)}=\frac{1}{\sum_{n=0}^\infty b_n x^n}=\sum_{n=0}^\infty B_n x^n
\end{align*}
we obtain
\begin{align*}
B_0&=\frac{1}{b_0}\\
B_n&=\frac{1}{b_0^nn!}\left|
\begin{array}{ccccc}
0&nb_1&nb_2&\cdots&nb_n\\
0&(n-1)b_0&(n-1)b_1&\cdots&(n-1)b_{n-1}\\
0&0&(n-2)b_0&\cdots&(n-2)b_{n-2}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&0&0&\cdots&1\\
\end{array}\tag{1}
\right|
\end{align*}
The right-hand side of (1) is the determinant of an $(n\times n)$-matrix.

