TITLE: Dividing an infinite power series by another infinite power series
QUESTION [16 upvotes]: 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 $\,\mathrm{H}$.

REPLY [2 votes]: 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 $B(y)$).