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Let $a_n$ be the number of ways to obtain the amount of $n$ cents, using a supply of 1-cent coins, 3 types of 2-cent coins, and 4-cent coins. Then, $a_n$ is the coefficient of $x^n$ in $$(1+x+x^2+\cdots)(1+x^2+x^4+\cdots)^3(1+x^4+x^8+\cdots),$$ which equals $$G(x):=\sum_{n \ge 0} a_n x^n = \frac{1}{(1-x)(1-x^2)^3 (1-x^4)}.$$

The text I'm reading briefly mentions that $G(x)$ is analytic in $\mathbb{C}$, apart from poles at $\pm 1, \pm i$, and that the asymptotic form of $a_n$ can be obtained by standard analytic arguments. How exactly is the asymptotic form obtained using analytic arguments?

AG.
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    I'm not sure what you mean by asymptotic form. But it seems worth pointing out that you can easily find an exact formula for $a_n$ using partial fractions decomposition. – Cocopuffs Aug 24 '13 at 05:00
  • I'm interested in learning about using techniques from (complex) analysis. Let me double-quote verbatim from the text: the expression is ``analytic in $\mathbb{C}$ apart from poles at $1, -1, i$, and $-i$. So the asymptotic from of $a_n$ can be found by standard analytic arguments.'' How does this method work? – AG. Aug 24 '13 at 05:06
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    There is a nice not easy book by Flajolet, very good. A version is available for free download. I think the title contains the term Analytic Combinatorics. Probably overkill, but very nice. – André Nicolas Aug 24 '13 at 05:15

1 Answers1

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If the pole of smallest modulus has modulus $r$, then asymptotically $a_n=p(n)r^{-n}$ for some polynomial $p$. If the poles are all simple, the polynomial's a constant, and you just get $Cr^{-n}$ for some constant $C$. In the case in hand, the pole at $1$ has multiplicity 5, so the polynomial has degree 4. $a_n$ should be asymptotic to $Cn^4$ for some constant $C$.

The idea is that if the generating function is, as here, a rational function, then every factor $(1-cx)^k$ in the denominator leads to a summand $A/(1-cx)^k$ in the partial fraction decomposition of the rational function. If $k=1$ then expanding the summand (it's just a geometric series) gives a contribution $Ac^n$ to $a_n$. The dominant contribution comes from the largest $c$, which corresponds to the smallest pole $1/c$. When $k\gt1$ the binomial theorem lets you expand $(1-cx)^{-k}$ and the coefficient you get for $x^n$ is of the form $p(n)c^n$ for some polynomial $p(n)$ of degree $k-1$.

Gerry Myerson
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  • I was aware of this method about partial fractions and about the degree of the polynomials, but I was thinking about (i.e. asking more for) the role of (complex) analytic arguments, by which I meant things like derivatives and analytic functions, Cauchy's residue theorem, etc. The book by Flajolet mentioned in the comment above does discuss the role of these concepts, so I'll agree my question is answered. – AG. Aug 25 '13 at 08:05
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    For the particular problem you raised, or any problem where the generating function is rational, Cauchy's residue theorem is overkill. For other problems, that book is a good place to go. – Gerry Myerson Aug 25 '13 at 12:53