# 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 $\,\mathrm{H}$.

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.

• Wouldn't the right-hand side of $(1)$ be the determinant of a square matrix of length $n+1$ instead of $n$, since there are $n+1$ columns? – Mathematically Encrypted Apr 30 at 18:37
• @MathematicallyEncrypted: You are right, but expansion at the bottom left $1$ results in an $(n\times n)$ determinant. I've updated the link where you can find the coresponding statement. – Markus Scheuer Apr 30 at 19:10

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.

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 $H(x)$.

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 technical steps (which I can provide, if you are still interested in this topic, but for now just believe, that it works, or try to prove it yourself) 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.