Expanding product of binomials $(z^k + z^{-k})$

Question

Suppose $z$ is a complex number, and consider the product

$$f_m(z)=\prod_{k=1}^m \left(z^k + \frac 1 {z^k} \right),$$

for $m = 1,2,\dots$ . Of course, one should be able to expand this into a sum of terms that are either powers of $z$ or of $1/z$. It is easy to see that the highest and lowest order terms will be $z^{1 + 2 + \cdots + m} = z^{\frac 1 2m(m+1)}$ and its reciprocal, respectively.

Here's the situation for low values of $m$:

$$\begin{split} f_1(z) = z &+ \frac 1 z \\ f_2(z) = z^3 + z &+ \frac 1 z + \frac 1 {z^3} \\ f_3(z) = z^6+z^4+z^2+\ &\color{red}2 + \frac 1 {z^2}+ \frac 1 {z^4} + \frac 1 {z^6} \\ f_4(z) = z^{10} + z^8 + z^6 + \color{red}2z^4 + \color{red}2z^2 +\ &\color{red}2 + \frac {\color{red}{2}} {z^2} + \frac {\color{red}2} {z^4} + \frac 1 {z^6} + \frac 1 {z^8} + \frac{1}{z^{10}} \\ f_5(z)=z^{15} + z^{13}+ z^{11} + \color{red}2z^{9} + \color{red}2z^7 + \color{orange}3z^5 + \color{orange}3z^3 + \color{orange}3z &+ \frac{\color{orange}3} z + \frac {\color{orange}3}{z^3} + \frac {\color{orange}3} {z^5} + \frac {\color{red}2} {z^7} + \frac {\color{red}2} {z^9} + \frac 1 {z^{11}} + \frac 1 {z^{13}} + \frac 1 {z^{15}} \end{split}$$

It seems that

1. In each $f_m$ the exponent of $z$ falls from $\tau_m = \frac 1 2 m(m+1)$ down to $-\tau_m$, skipping every other value;
2. For all $m$ the coefficient $a_{m,n}$ of the $n$-th term in the analytic part is equal to the coefficient $a_{m,-n}$ of the $n$-th term in the singular part (which makes sense intuitively).

So:

• Is there a closed-form expression for $a_{m,n}$?
• How can I prove fact 1. above?

Answer 1

Here is a proof for $1$. Suppose the monomial $z^n$ appears in the expanded expression. Then we may write $$ n=\sum_{k=1}^m(-1)^{e_k} k $$ where each $e_k$ is either $0$ or $1$. Let $P$ denote the indices for which $e_k=0$, i.e. the positive terms, and $N$ the indices for which $e_k=1$, i.e. the negative terms. Then $$ n=\sum_{k\in P}k-\sum_{j\in N}j. $$ We can add $0$ as such: $$ n=\sum_{k\in P}k+\sum_{j\in N}j-\sum_{j\in N}j-\sum_{j\in N}j, $$ and now rearrange to get $$ n=\sum_{k\in P\text{ or }N}k-2\sum_{j\in N}j $$ which gives $$ n=\tau_m-2\sum_{j\in N}j. $$ Thus we know that the exponent $n$ must be $\tau_m$ minus an even number, which restricts the possibilities to what you conjectured. To see that all of these are actually possible, consider what happens when $N=\{1\},\{2\},\dots,\{m\},\{1,m\},\{2,m\},\dots,\{m-1,m\}$, etc.

A closed form for $a_{m,n}$ is probably difficult. Here is an idea. If we multiply the entire product by $(z^m)^m$, we obtain $$ f_m(z)=\frac{1}{(z^{m})^m}\prod_{k=1}^m(z^{m+k}+z^{m-k}), $$ and now you are looking almost at partitions of $n$ into distinct parts, althought you add the additional restriction that precisely one of $m+k$ or $m-k$ is used.

Hope this helps.

Answer 2

This is just a start, not a solution, but I can't fit it in a comment. Maybe it's enough for you to answer the question yourself. We can think of the problem this way. Partition $[m] = \{1,2,\dots\}$ into two sets $A$ and $B$ and let $\sum A$ and $\sum B$ be the sum of the elements of $A$ and $B$ respectively. Then $a_m,k= [z^k]f_m(z)$ is the number of such partitions with $\sum A-\sum B = k$. We have $$ a_{m,k}=\cases{ 1,& $m=1,\ |k|=1$\\ 0,& $m=1,\ |k|\neq 1$\\ a_{m-1,k-m}+a_{m-1,k+m},&otherwise } $$

This is because the partitions of $[m]$ arise from adding the element $m$ to one of the sets in a partition of $[m-1].$ Notice that I am assuming that $a_{m,k}$ is defined for all positive integers $m$ and all integers $k.$ Just make $a_{m,k}=0$ when $z^k$ does not appear in $f_m.$

I have no idea whether this can be solved for $a_{m,k}$ in closed form, but at least it makes it easy to calculate.

Answer 3

This is not an answer, but maybe help you to find an answer.

Since the function is same as $$ f_{m}(z) = \frac{1}{z^{m(m+1)/2}}\prod_{k=1}^{m}(1+z^{2k}), $$ it is enough to find an explicit formula for the coefficients of the polynomial $$ g_{m}(z) = \prod_{k=1}^{m}(1+z^{k}). $$ Note that $f_{m}(z) = g_{m}(z^{2})/z^{m(m+1)/2}$. One can observe that the $n$-th coefficient of $g_{m}(z)$ is same as the number of ways to express $n$ as a sum of distinct numbers that are less or equal to $m$. It seems that it is really hard to find the explicit formula for such coefficients, since the explicit formula for $q(n) = \lim_{m\to \infty}[z^{n}]g_{m}(z)$ ($n$-th coefficient of $\prod_{k=1}^{\infty} (1+z^{k})$, which is a number of ways to express $r$ as a sum of distinct numbers, without any condition on the bound of such numbers), closely related to the ordinary partition number $p(n)$: $$ p(n) =\sum_{k=0}^{\lfloor n/2\rfloor} q(n-2k)p(k). $$ Also, it is known that $q(n)$ has a following explicit formula $$ q(n) = \frac{1}{\sqrt{2}}\sum_{k=1}^{n}A(2k-1, n)\frac{\partial J_{0}\left(\frac{\pi i}{2k-1}\sqrt{\frac{1}{3}\left(n+\frac{1}{24}\right)}\right)}{\partial n} $$ where $A(k, n)$ is a generalized Kloosterman sum $$ A(k, n) = \sum_{\substack{1\leq h\leq k \\ \gcd(h, k) = 1}}\exp\left(\pi i \sum_{j=1}^{k=1}\frac{j}{k}\left(\frac{hj}{k} - \lfloor \frac{hj}{k}\rfloor - \frac{1}{2}\right)- \frac{2\pi i n h}{k}\right) $$ and $J_{0}$ is a first kind of Bessel function. This formula seems not to be that helpful, and at least we have a asymptotic formula $$ q(n) \sim \frac{1}{4\sqrt[3]{3}n^{3/4}}\exp\left(\pi\sqrt{\frac{n}{3}}\right). $$ All of these formulas are from here, and sadly I don't know proof of any of these. (Maybe the last one comes from a Hardy-Littlewood's circle method, since similar asymptotic formula for $p(n)$ is proved by such method.)

Answer 4

A proof of 1. could go as follows.

We obtain for integral $m\geq 1$ \begin{align*} \color{blue}{f_m(z)}&=\prod_{k=1}^m\left(z^k+\frac{1}{z^k}\right)\\ &=\prod_{k=1}^m\left(z^{-k}\left(1+z^{2k}\right)\right)\\ &=z^{-\sum_{k=1}^mz^k}\prod_{k=1}^m\left(1+z^{2k}\right)\tag{1}\\ &=z^{-\frac{1}{2}m(m+1)}\sum_{j=0}^{\sum_{k=1}^m k}a_jz^{2j}\tag{2}\\ &\,\,\color{blue}{=z^{-\frac{1}{2}m(m+1)}\sum_{j=0}^{\frac{1}{2}m(m+1)}a_jz^{2j}}\\ \end{align*} In the last line we observe the smallest power of the expression is $-\frac{1}{2}m(m+1)$ while the greatest power is $-\frac{1}{2}m(m+1)+m(m+1)=\frac{1}{2}m(m+1)$.

Comment:

• In (1) we have a finite product where each term is either $1$ or $z^{2k}$. We so obtain a polynomial with smallest power $0$ and greatest power of $z^2$ equal to $\sum_{k=1}^m k$. This polynomial is written more conveniently in (2).