Dividing an infinite power series by another infinite power series

Question

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 $\,\mathrm{H}$.

Answer 1

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.

Answer 2

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.

Answer 3

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 $H(x)$.

Answer 4

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 $B(y)$).