# How to expand this generating function

I am new to generating functions / power series and I am struggling with expanding following generating function: $$F(x)=\frac{(x^2-x^{10})(1-x^{10})^3}{(1-x)^4}$$ I tried to pattern my solution on that answer and expand generating function in the same way, so i rewrote $F(x)$ as: $$F(x)=\frac{x^2(1-x^8)(1-x^{10})^3}{(1-x)^4}$$ and in order to obtain coefficient at $x^{20}$ I tried to calculate: $${(20-2)+3 \choose 3}-{(20-2)-8+3 \choose 3}-{(20-2)-10+3 \choose 3} \cdot 3$$ The result of this is $549$ which is smaller by $3$ than proper solution which is $552$. I don't know what I did wrong. In fact, I don't understand this method of expanding coefficients so if you could provide the name of this technique or some resources to read about I would be thankful. I mean, I see the pattern this technique follows, but I don't see why it works.

This generating function gives solution to one problem that I solved first in purely combinatoric way. My calculations were nearly identical to these presented above, but I had to exclude intersection of two sets which was of size $3$. So I understand that there should be $+3$ term in solution above, but I don't know where it would come from in case of expanding generating function.

## 2 Answers

Using negative binomial series,

\begin{align} F(x) &= x^2(1-x^8)(1-3x^{10}+3x^{20}-x^{30})(1-x)^{-4}\\ &= x^2(1-x^8)(1-3x^{10}+3x^{20}-x^{30})\sum_{k=0}^\infty \binom{4+k-1}{k} x^k\\ &= (x^2-x^{10})(1-3x^{10}+3x^{20}-x^{30})\sum_{k=0}^\infty \binom{3+k}{k} x^k\\ &= (x^2-3x^{12}-x^{10}+\color{blue}{3x^{20}}+p(x))\sum_{k=0}^\infty \binom{3+k}{3} x^k\\ \end{align}

where $p(x)$ are higher order terms that are not interesting for the question.

Hence

$$\binom{18+3}{3}-3\binom{11}{3}-\binom{13}{3}+3\binom{3}{3}$$

I suspect you left out the case where we let $k=0$.

• I misunderstood the process of expanding generating function. Looking at the example I linked to, I didn't realize it's author skipped some part of calculations and just gave the final result. This is why I was confused. Thank you for answer. – mrJoe Feb 16 '18 at 18:13

I think that you missed one term. Notice that $$(1-x^8)(1-x^{10})^3=(1-x^8)(1-3x^{10}+O(x^{20}))=1-x^8-3x^{10}+3x^{18}+o(x^{18}).$$ Hence \begin{align} [x^{20}]\frac{x^2(1-x^8)(1-x^{10})^3}{(1-x)^4} &=[x^{18}]\frac{1-x^8-3x^{10}+3x^{18}+o(x^{18})}{(1-x)^4}\\ &=[x^{18}]\frac{1}{(1-x)^4}-[x^{10}]\frac{1}{(1-x)^4}-[x^{8}]\frac{3}{(1-x)^4}+[x^{0}]\frac{3}{(1-x)^4}\\ &=\binom{18+3}{3}-\binom{10+3}{3}-3\binom{8+3}{3}+3=552. \end{align}

• Thank you for explanation. I missed one term because I wrongly deduced the pattern of expanding generating functions. So if there was a term in numerator like (1+x)^123 we would have to calculate it? I thought there is some way to avoid exponentiation. – mrJoe Feb 16 '18 at 18:09
• Could you rephrase your question, please? – Robert Z Feb 16 '18 at 18:24
• I mean, we had to expand numerator and then split function into partial fractions. In my situation there was a value (1+x^10)^3 so it was not hard to expand it. But if the exponent was e.g. 123 we would have a lot of work to do. So I wondered if there is a technique which lets avoid this time-consuming calculation. – mrJoe Feb 16 '18 at 18:54
• No, it suffices to consider $(1-x^{10})^{123}=1-123x^{10}+O(x^{20})$ – Robert Z Feb 16 '18 at 18:59