# Coefficients of Quotient of two power series

So I have two functions $f$ and $g$ which are holomorphic on some disk $D(a,r)$ and such that $g$ is never zero on that disk.

Where we can represent the two function as power series given by

$f(z)=\sum\limits_{n=0}^\infty a_n(z-a)^n$ and $g(z)=\sum\limits_{m=0}^\infty b_m(z-a)^m$

Let
$f_N=\sum\limits_{n=0}^N a_n(z-a)^n$ and $g_N=\sum\limits_{m=0}^N b_m(z-a)^m$ then by carrying out the long division of the two polynomials we get that

$$f_N=p_Ng_N+r_N$$ where $p_N$ and $r_N$ are two polynomials and $deg(r_N)$ is at most $N-1$

I want to show that for each integer $m$, the coefficient $c_m$ of $(z-a)^m$ is the same in all the polynomials $p_N$ for $N>m$, and that will be the coefficients on $z^m$ in the Taylor series for $f/g$.

I'm not sure how to do that.

Perhaps someone could give me a hint or tell me if I'm going about this the wrong way.

What I have so far is that I know that since $f$ and $g$ are holomorphic on $D(a,r)$ and since $g$ is never zero on $D(a,r)$, $f/g$ will be holomorphic on $D(a,r)$ also. Thus $f/g$ has a power series representation on $|z-a|<r$.

I also know that for every positive integer $m$ and every $s$ such that $0<s<r$,

\begin{align*}c_m&=\frac{1}{2\pi i}\int_{|w-a|=s}\frac{f(w)}{g(w)(w-a)^{m+1}}dw\\ &=\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\frac{f_N(w)}{g_N(w)(w-a)^{m+1}}dw\\ &=\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\frac{p_N(w)g_N(w)+r_N(w)}{g_N(w)(w-a)^{m+1}}dw\\ &=\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\left[\frac{p_N(w)}{(w-a)^{m+1}}+\frac{r_N(w)}{g_N(w)(w-a)^{m+1}}\right]dw\\ &=\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\frac{p_N(w)}{(w-a)^{m+1}}dw+\lim\limits_{N\to\infty}\frac{1}{2\pi i}\int_{|w-a|=s}\frac{r_N(w)}{g_N(w)(w-a)^{m+1}}dw\\ \end{align*}

I think that the first integral gives the coefficients of the polynomial $p$, but I'm not sure what happens with the second integral since $deg(g_N)>deg(r_N)$.

Suppose $f(z)=\sum\limits_{n=0}^\infty a_n(z-a)^n$ and $g(z)=\sum\limits_{m=0}^\infty b_m(z-a)^m$. Then

You can compute the quotient of the infinite series by writing the quotient with a denominator of the form $1-\sum\limits_{m=1}^\infty -b_m(z-a)^m$. Dividing by this denominator is the same as multiplying by a geometric series.

$$\dfrac{\sum\limits_{n=0}^\infty a_n(z-a)^n}{\sum\limits_{m=0}^\infty b_m(z-a)^m}= \dfrac{\sum\limits_{n=0}^\infty a_n(z-a)^\infty}{b_0-\left(\sum\limits_{m=1}^\infty -b_m(z-a)^m\right)}\\ =\frac{1}{b_0}\sum\limits_{n=0}^\infty a_n(z-a)^n\left(1+ \left(\sum\limits_{m=1}^\infty \frac{-b_m}{b_0}(z-a)^m\right)+\left(\sum\limits_{m=1}^\infty \frac{-b_m}{b_0}(z-a)^m\right)^2+\cdots\right)$$ An example might help:

$$\dfrac{e^z} {\frac{2}{2-z^2}}=\dfrac{\sum\limits_{n=0}^\infty \frac{z^n}{n!}} {\sum\limits_{m=0}^\infty \frac{z^{2m}}{2^m}} = \dfrac{\sum\limits_{n=0}^\infty \frac{z^n}{n!}}{{2^0}-\left(\sum\limits_{m=1}^\infty -\frac{z^{2m}}{2^m}\right)}\\ =\frac{1}{2^0}\sum\limits_{n=0}^\infty \frac{z^n}{n!}\left(1 +\left(\sum\limits_{m=1}^\infty -\frac{z^{2m}}{2^{m+0}}\right)+\left(\sum\limits_{m=1}^\infty -\frac{z^{2m}}{2^{m+0}}\right)^2+\cdots\right) \\\left(1+z+\frac{z^2}{2}+\frac{z^3}{6}+\frac{z^4}{24}\cdots\right)\left(1 -\left(\frac{z^2}{2}-\frac{z^4}{4}-\frac{z^6}{8}\cdots\right)+\left(\frac{z^2}{2}-\frac{z^4}{4}-\frac{z^6}{8}\cdots\right)^2+\cdots\right) \\=1+z+\frac{z^2}{2}-\frac{z^2}{2}+\frac{z^3}{6}-\frac{z^3}{2}\cdots=1+z-\frac{z^3}{3}\cdots$$

You can write the coefficients of the quotient series in “closed form” (there will be finite sums in each term) if you google division of power series, but it is nice to see how to do it by hand term by term, as shown here.

• I'm familiar with changing into the product of two series. I can't see how that will help with what I'm trying to show, though. That for each integer $m$, the coefficient $c_m$ of $(z−a)^m$ is the same in all the polynomials $p_N$ for $N>m,$ and that will be the coefficients on $z_m$ in the Taylor series for $f/g.$ – mi986 Mar 1 '15 at 18:19
• Sorry, I’m confused. I don’t think what you want to show is true. Suppose neither of your Taylor series have any nonzero coefficients. Then when you write $f_N=p_Ng_N+r_N$, $p_N$ will be a complex number, and it will never have more than one coefficient at all. (Maybe I’m still confused.) – Steve Kass Mar 1 '15 at 19:20
• I think what it's referring to is if you look at the $m$-th coefficient in any one of the $p_N$'s where $N>m$ it will be the same. For example, if we did $f_4/g_4$ then the third coefficient in $p_4$ would be the same as if we did $f_5/g_5$ and looked at the third coefficient of $p_5$. Also, $g$ must have at least one nonzero coefficient because it is chosen to be not zero on $D(a,r)$. – mi986 Mar 1 '15 at 22:00
• I think this is quite far from being true. Let $f=e^z$ and $g=e^{2z}$. Then ($a=0$), both $f_N$ and $g_N$ are always $N$-th degree polynomials, thus $p_N$ is always a constant. But it’s a different constant for each value of $N$. ($p_N=2^{-N}$). I suspect you can make any of the coefficients of $p_N$ change around just about any way you want. For example, take $g=e^{z^m}$. Then the degree of $p_N$ will cycle from $0$ to $m$, so the $3$-rd coefficient bounces from 0 to nonzero. Typically $p_N$ will have bounded degree, too. Do you an examples where your conjecture holds, or am I still confused? – Steve Kass Mar 1 '15 at 22:51
• I see what you're saying about p_N being constant. Actually this is a problem from a textbook, so perhaps I've screwed up the wording. The exact statement is as follows: Carry out the following discussion in detail. Let $f$ and $g$ be holomorphic on D(a,r), given by the taylor series such that $g$ has no zeros on that disk. Then the taylor series for the holomorphic function $f/g$ can be obtained using long division as follows. Let $f_N$ and $g_N$ be the partial sums of order $N$ and carry out the long division of the two polynomials. We find that $f_N=p_Ng_N+r_N$ where $p_N$ and $r_N$ are... – mi986 Mar 1 '15 at 23:16