Computing a polynomial product over roots of unity

Question

I'm trying to compute the coefficients of the following polynomial, where $\omega$ is a primitive $p$-th root of unity, for $p$ prime:

$$a(x) = \prod_{i=0}^{p-1} f(\omega^ix).$$

It turns out that the $i$-th coefficient is always an integer, and non-zero only when $i$ is a multiple of $p$. So it seems to me like there should be an elementary expression for $a$.

So far I've got this expression for the $i$-th coefficient: $$a_i = x^i\sum_{k_0 + \ldots + k_{p-1} = i}f_{k_0} \cdots f_{k_{p-1}} \omega^{k_1 + 2k_2 + \ldots + (p-1)k_{p-1}}$$

where each $k_i$ is non-negative and bounded by the degree of $f$.

Clearly the roots of unity all cancel out somehow, but I can't figure out how to get a 'nice' expression out. Any suggestions?

Answer 1

Notice that $$ y^p-x^p=\prod_{i=0}^{p-1} (y-\omega^i x)\ . $$ Therefore $$ \prod_{i=0}^{p-1} f(x\omega^i)= Res_y (y^p-x^p, f(y))\ . $$ Where $Res_y$ is the resultant of the polynomials in $y$. The above is just a particular case of the so called Poisson product formula. You can then compute the resultant using for instance Sylvester's determinant formula. See the review http://mate.dm.uba.ar/~alidick/papers/chapter1cd.pdf by Cattani and Dickenstein for a nice introduction to resultants and their properties.

Answer 2

I don't think that $p$ being prime makes any difference. The later thoughts below suggest a Lagrange interpolation method which is perhaps the same as the resultant method mentioned by Abdelmalek Abdesselam.

Let $f(x)=\sum_{j=0}^nc_jx^j. $ One might require $c_n=1$ or $c_0=1$ but it is perhaps nicer not to. Then setting $u=x^n,$ $\prod_{i=0}^{n-1} f(\omega^ix)=F(u)=\sum_{j=0}^nC_j u^j$. One can say that

• $C_j$ is a polynomial of degree $n$ in the coefficients $c_0,\cdots,c_n$ where each term has total degree $n$
• $C_j$ has a term $\pm(c_j)^n$ and no term $c_j^{n-1}$.
• $C_{n-j}$ is $C_j$ with $c_k$ replaced by $c_{n-k}$
• the roots of $F$ are the $n$th powers of the roots of $f$.

Here $c_n$ is a constant and the other $c_i$ are symmetric polynomials of the $n$ roots $\alpha_i$ of $f$. The $\alpha_i$ can be thought of as formal variables. Then $c_0,\cdots,c_{n-1}$ are also a basis for the ring of all symmetric polynomials in those variables ($\frac{1}{c_n}$ times a usual basis) . There are other bases for this ring such as $\sigma_k=\sum_{i=1}^n\alpha_i^k$ and the sum of all the terms $\alpha_1^{m_1}\cdots\alpha_n^{m_n}$ with $m_1+\cdots+m_n=k$. Transforming between these bases (more generally, expressing a given symmetric polynomial in terms of them) is a major topic of invariant theory.

In this case, we want to express the $C_i$, which are certain symmetric polynomials in the $\alpha_i^n,$ in terms of the values $c_i$. This must be a well known case. At any rate here are some results:

For $n=4,$

$$C_0=c_0^4$$

$$C_1=(4c_0^3c_4-2c_0^2c_2^2)-(4c_0^2c_3)c_1+(4c_0c_2)c_1^2-c_1^4$$

$$C_2=(6c_0^2c_4^2-8c_0c_1c_3c_4+2c_1^2c_3^2)+(4c_0c_3^2+4c_1^2c_4)c_2-(4c_0c_4+4c_1c_3)c_2^2+c_2^4$$

$$C_3=(4c_4^3c_0-2c_4^2c_2^2)-(4c_4^2c_1)c_3+4(c_4c_2)c_3^2-c_3^4$$

$$C_4=c_4^4$$

while for $n=5$ we have the following (with the others obtainable by symmetry)

$$C_0=c_0^5$$

$$C_1=5c_0^4c_5-(5c_0^3c_4-5c_0^2c_2^2)c_1+(5c_0^2c_3)c_1^2-5c_0^3c_2c_3-(5c_0c_2)c_1^3+c_1^5$$

$${\small C_2=(10c_0^3c_5^2-15c_0^2c_1c_4c_5+5c_0c_1^2c_4^2+5c_0^2c_3^2c_4+10c_0c_1^2c_3c_5-5c_0c_1c_3^3-5c_1^3c_3c_4)}$$ $${ \small+(5c_0^2c_4^2-15c_0^2c_3c_5+5c_1^2c_3^2-5c_0c_1c_3c_4-5c_1^3c_5)c_2}$$ $${\small +(5c_1^2c_4+10c_0c_1c_5+5c_0c_3^2)c_2^2-(5c_0c_4+5c_1c_3)c_2^3+c_2^5}$$

later thoughts In general one could consider the problem of producing from a polynomial $f(x)=\sum_{j=0}^nc_jx^j$ a polynomial $F(u)=\sum_{j=0}^nC_j u^j$ whose roots are the $m$th powers $\alpha_i^m$ of the $n$ (unknown) roots of $f$. The solution is the polynomial $\prod_{i=0}^{m-1} f(\tau^ix)$ where now $\tau$ is a primitive $m$th root of unity. This process might spread out the roots. In the case $m=2$ one has (with $ q$ repeated application) $\alpha_i^{2^q}$ and the Dandelin–Graeffe method for finding the roots of a univariate polynomial. Splitting $f$ into its even and odd parts speeds up the computation. The method was also discovered by Nikolai Ivanovich Lobachevsky and the linked article suggests that his book Algebra ili Ichislenie Konechnykh Velichin discussed the general product $\prod_{i=0}^{m-1} f(\omega^ix)$. Perhaps the appropriate manipulations of symmetric polynomials are discussed there.

Since a polynomial of degree $n$ is determined by its values at $n$ points (plus its leading coeffcient) one has the following method (which I doubt is new): Let $\zeta$ be a primitive $mn$th root of unity and $\omega$ a primitive $m$th root of unity. Then the desired polynomial $F$ satisfies $F(\zeta^j)=f(\omega^j)$ for $0 \le j \le n-1$. Now Lagrange Interpolation can be used. In this case it might be particularly simple.

Answer 3

The transformation $$ \mathcal{L}_n : \mathbb{C}(x) \longrightarrow \mathbb{C}(x) $$ defined by $$ \mathcal{L}_n(F(x)) = \sum_{\zeta\in\boldsymbol\mu_n} F(\zeta y)\Big|_{y^n\to x} $$ is called a (generalized) Landen transformation. If you take $F(x)=1/f(x)$, where $f(x)$ is a polynomial, then the denominator of $\mathcal{L}_n(F(x))$ is more-or-less the product you are studying. See http://arxiv.org/abs/1308.5355 (especially Section 8, which is really about $\prod_{\zeta\in\boldsymbol\mu_n} F(\zeta y)\Big|_{y^n\to x}$), as well as the references in that paper.

Answer 4

A slightly more general result is shown in Theorem A here

https://www.maths.tcd.ie/pub/ims/bull47/R4701.pdf

which states: If $r$ is relatively prime to $n$ then the coefficient of $x^r$ in $\prod_{i=1}^n f(\omega^i x)$ is 0.

(Here $\omega$ is a primitive $n$-th root of unity, $f$ has rational coefficients.)