Is $a_1\cdots a_n + \frac1{a_1\cdots a_n}$ a polynomial in $a_i + \frac1{a_{i+1}}$?

Question

Question

Let $n$ be a positive integer, and let $a_1,\ \dots,\ a_n$ be formal variables. Is $$ P = a_1\cdots a_n + \frac1{a_1\cdots a_n} $$ a polynomial in the $n$ expressions $a_1 + \frac1{a_2},\ a_2 + \frac1{a_3},\ \dots,\ a_{n-1} + \frac1{a_n},\ a_n + \frac1{a_1}$?

As examples, letting $b_i = a_i + \frac1{a_{i+1}}$ for convenience, with indices taken modulo $n$,

1. for $n = 1$, we have $P = b_1$.
2. for $n = 2$, we have $P = b_1b_2 - 2$.
3. for $n = 3$, we have $P = b_1b_2b_3 - b_1 - b_2 - b_3$.
4. for $n = 4$, we have $P = b_1b_2b_3b_4 - b_1b_2 - b_2b_3 - b_3b_4 - b_4b_1 + 2$.

Motivation

The motivation for this question comes from the specialization $a_1 = \dots = a_n = a$, for which the answer is known to be yes, with the Chebyshev polynomials of the first kind (involving a different normalization) serving as suitable polynomials. Indeed, letting $b = a + \frac1a$, the four example polynomials above become $$b,\quad b^2 - 2,\quad b^3 - 3b,\quad b^4 - 4b^2 + 2,$$ which match the polynomials $2T_\bullet(\frac b2)$.

Thus, these polynomials could be interpreted as multivariate generalizations of Chebyshev polynomials. However, I'm not aware of any such polynomials in the literature.

Possible Progress

A polynomial for $P$, if it exists, should be unique. This is because the monomials in the $b_\bullet$ are linearly independent, as (interpreted as Laurent polynomials in the $a_\bullet$) each monomial has exactly one highest-degree term, and these highest-degree terms are all different.

The polynomial for $P$ seems to have a reasonable description, as follows:

• The polynomial is multilinear in the $b_\bullet$.
• Given a nonempty index set $S \subseteq \mathbb Z/n\mathbb Z$, the term $\prod_{i \in S} b_i$ is included if and only if the following condition is satisfied: as we travel along $\mathbb Z/n\mathbb Z$ in the natural way, the distance between any two consecutive elements of $S$ is odd.
  • Note that this automatically forces $|S| \equiv n \pmod 2$.
  • The coefficient corresponding to this term is $(-1)^\frac{n - |S|}2$; i.e., the sign we'd expect from the Chebyshev polynomial.
• The empty set $\varnothing$ obeys similar rules. It is included (i.e. the polynomial has a constant term) if and only if $n$ is even; this matches the parity condition above.
  • Its coefficient is doubled compared to what the previous rule would suggest. That is, it appears with a coefficient of $2(-1)^\frac n2$, matching the constant term of the corresponding Chebyshev polynomial.

The following (inefficient) Sage code carries out this construction. I've used it to verify the construction for $n \le 16$, though it gets slow very quickly.

from math import prod
from itertools import combinations, pairwise

def valid(l, N):
return len(l) % 2 == N % 2 and all(u != v for u, v in pairwise(x % 2 for x in l))
N = 10
a = [var(f'a{i}') for i in range(N)]
b = [a[i] + 1/a[(i+1)%N] for i in range(N)]
expr = 0
for m in range(N+1):
for p in combinations(range(N), m):
if valid(p, N):
expr += (2 if m == 0 else 1) * (-1) ** ((N - m) // 2) * prod(b[i] for i in p)
print(expand(expr))

Answer 1

Finally, this becomes an answer. Possible closed-form expression through the determinant of 3-diagonal matrices with two extra $i=\sqrt{-1}$ elements $$ P_n=\det\begin{pmatrix} b_1 & 1 & 0 & ... & i \\ 1 & b_2 & 1 & ... & 0 \\ 0 & 1 & b_3 & ... & 0 \\ ... & ... & ... & ... & ... \\ -i & 0 & 0 & ... & b_n \end{pmatrix}. $$ A few first identities $$ P_1=\det(b_1),\ \ P_2=\det\begin{pmatrix}b_1 & 1+i \\ 1-i & b_2 \end{pmatrix},\ \ P_3=\det\begin{pmatrix}b_1 & 1 & i \\ 1 & b_2 & 1 \\ -i & 1 & b_3 \end{pmatrix}, \ \ P_4=\det\begin{pmatrix}b_1 & 1 & 0 & i \\ 1 & b_2 & 1 & 0\\ 0 & 1 & b_3 & 1 \\ -i & 0 & 1 & b_4\end{pmatrix}. $$ Another formula that follows from this one - trace of the product of $2\times 2$ matrices - $$ P_n={\rm tr}\begin{pmatrix} b_n & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} b_{n-1} & 1 \\ -1 & 0 \end{pmatrix}...\begin{pmatrix} b_1 & 1 \\ -1 & 0 \end{pmatrix}. $$ See the comments below - thanks to Darij Grinberg. I repeat it there. Each matrix can be factorized as $$ \begin{pmatrix} b_j & 1 \\ -1 & 0 \end{pmatrix}=\begin{pmatrix} \frac1{a_{j+1}} & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ a_j & 1 \end{pmatrix} $$ and we have $$ P_n={\rm tr}\begin{pmatrix} \frac1{a_{1}} & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ a_n & 1 \end{pmatrix}...\begin{pmatrix} \frac1{a_{2}} & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ a_1 & 1 \end{pmatrix} $$ $$ ={\rm tr}\begin{pmatrix} 1 & 0 \\ a_1 & 1 \end{pmatrix}\begin{pmatrix} \frac1{a_{1}} & 1 \\ -1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ a_n & 1 \end{pmatrix}\begin{pmatrix} \frac1{a_{n}} & 1 \\ -1 & 0 \end{pmatrix}...\begin{pmatrix} 1 & 0 \\ a_2 & 1 \end{pmatrix}\begin{pmatrix} \frac1{a_{2}} & 1 \\ -1 & 0 \end{pmatrix} $$ $$ ={\rm tr}\begin{pmatrix} \frac1{a_1} & 1 \\ 0 & a_1 \end{pmatrix}\begin{pmatrix} \frac1{a_n} & 1 \\ 0 & a_n \end{pmatrix}...\begin{pmatrix} \frac1{a_2} & 1 \\ 0 & a_2 \end{pmatrix}={\rm tr}\begin{pmatrix} \frac1{a_1...a_n} & * \\ 0 & a_1...a_n \end{pmatrix}. $$

Answer 2

Here is an alternate proof inspired by AAK's great answer above, but that doesn't use any matrices. Instead we use a combinatorial argument with a "sign-reversing involution".

Consider the set $\mathcal{S}_n$ of all possible sets of nonoverlapping pairs $\{i, i+1\} \subseteq \mathbb{Z}/ n \mathbb{Z}$. For a given $S \in \mathcal{S}_n$, let $w(S) = (-1)^{|S|} \prod_{i \notin \bigcup S} b_i$.

Let $P_n(b_1, \ldots, b_n) = \sum_{S \in \mathcal{S}_n } w(S)$.

We could think of this as choosing an arbitrary set of non-overlapping pairs $b_i b_{i+1}$ in the word $b_1 b_2 \cdots b_n$ and replacing each pair with $-1$, and then adding them up.

For example, with $n=4$, we get

$$\begin{align*} \emptyset & \to b_1b_2b_3b_4,\\ \{12\} &\to (-1)b_3 b_4 \\ \{23\} &\to b_1 (-1)b_4 \\ \{34\} &\to b_1 b_2 (-1) \\ \{41\} &\to b_2b_3(-1) \\ \{12, 34\} & \to (-1)(-1) \\ \{23, 41\} & \to (-1)(-1) \\ \end{align*}$$

and so $P_4 = b_1 b_2 b_3 b_4 - b_3 b_4 - b_1 b_4 - b_1b_2 - b_2b_3 + 2$, matching OP's expression above.

Claim:

$$P_n(a_1 + 1/a_2, a_2 + 1/a_3, \ldots, a_n + 1/a_{1}) = a_1 \cdot \cdots \cdot a_n + \frac{1}{a_1 \cdot \cdots \cdot a_n}$$

(This is another way of stating OP's claim in the "Possible Progress" section.)

Proof:

Identify $\mathcal{S_n}$ with the set of words formed by starting with $b_1 \cdots b_n$, choosing arbitrary sets of adjacent pairs and replacing each pair with $-1$. Form the set $\mathcal{T}_n$ given by taking any word in $\mathcal{S}_n$ and replacing an arbitrary set of $b_i$ by $a_i$, and the other $b_i$ by $a_{i+1}^{-1}$. By expanding out all the factors $a_i + 1/a_{i+1}$, we see that

$$P_n(a_1 + 1/a_2, \ldots, a_n + 1/a_{1}) = \sum_{x \in \mathcal{T}_n} x.$$

Now, the words in $\mathcal{T}_n$ come in pairs: For each word we can take the first appearance of either a pair $a_i^{-1} a_i$ or a $-1$, and replace it with a $-1$ or a pair $a_i^{-1} a_i$. The exception are the words that have no $-1$ or $a_i^{-1} a_i$, namely $a_1 \cdots a_n$ and $a_2^{-1} a_3^{-1}\cdots a_n^{-1} a_1^{-1}$, which have no partner under this pairing. All the others except for these two will cancel. For example for $n=3$ we have $$ \mathcal{T}_3=\left\{ \begin{array}{ll} a_1 a_2 a_3, & \\[3pt] a_2^{-1} a_2 a_3 ,& (-1)a_3,\\[3pt] a_1 a_3^{-1} a_3 ,& a_1(-1),\\[3pt] a_1 a_2 a_1^{-1} ,& a_2(-1),\\[3pt] a_2^{-1} a_3^{-1} a_3 ,& a_2^{-1}(-1),\\[3pt] a_1 a_3^{-1} a_1^{-1} ,& a_3^{-1}(-1),\\[3pt] a_2^{-1} a_2 a_1^{-1} ,& (-1)a_1^{-1},\\[3pt] a_2^{-1} a_3^{-1} a_1^{-1}& \end{array}\right\} $$

and it's clear how when we take the sum they all cancel but the first and the last.