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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 $\,\mathrm{H}$.

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    $\begingroup$ Typically you cannot expect the form $\sum_{n={\color{red}1}}^\infty$ $\endgroup$ – Hagen von Eitzen Dec 8 '16 at 21:21
  • $\begingroup$ OOps, that was I typo. I meant from $n=0$ and higher $\endgroup$ – Greg.Paul Dec 8 '16 at 21:23
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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.

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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.

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    $\begingroup$ this is my favourite answer $\endgroup$ – Greg.Paul Dec 11 '16 at 0:39
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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 $H(x)$.

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  • $\begingroup$ This will work well if $G(x)$ has constant term $1$. Otherwise, it's better to rescale $G(x)$ first (making a corresponding rescaling of $F(x)$) so that it does have constant term $1$. Then $1-G(x)$ has vanishing constant term, and so one can ignore $(1-G(x))^n$ if $n$ is large (since it is divisible by $x^n$), if one only wants the first few terms of the quotient. $\endgroup$ – tracing Dec 11 '16 at 0:00
  • $\begingroup$ @tracing Yes, that is correct. Thanks for pointing that out. $\endgroup$ – Simply Beautiful Art Dec 11 '16 at 0:05

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