Let $P$ be a polynomial of fixed degree $d$ with integer coefficients of absolute values at most $n$. Assume that $P(\cos 2\pi/n)$ is no zero. Is there a lower bound for $|P(\cos 2\pi/n)|$ ? For instance, is this at least $n^{-C(d) }$ where $C(d)$ is some constant depending on $d$ ?

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    $\begingroup$ Just a commment that this interesting question would be getting more attention if it had a more informative title. $\endgroup$
    – JSE
    May 12, 2011 at 1:45

2 Answers 2


Let $\alpha = 2^dP(\cos 2\pi /n)$. Then $\alpha$ is a non-zero algebraic integer and thus its norm is a positive integer. The norm is the product of the $\phi(n)$ conjugates of $\alpha$, all of whom are of absolute value at most $n2^d$, so $|\alpha| \ge (1/n2^d)^{\phi(n)-1}$. This is not as good as you'd like, especially for large $n$. I think the problem with fixed $d$ and large $n$ is hard. See the following question:

How small can a sum of a few roots of unity be?

By the way, did you mean the two occurrences of $n$ to be the same?

  • $\begingroup$ This is Liouville's inequality. For $d=1$, Roth's theorem gives required bounds. The generalization of Roth theorem for polynomials of higher degree is open. I doubt that the special case of $\alpha$ is significantly easier. $\endgroup$ May 12, 2011 at 0:09
  • $\begingroup$ @Oleg: I am afraid that Roth's theorem wors for any specific polynomial, while $n$ does change... $\endgroup$ May 12, 2011 at 0:16
  • $\begingroup$ Roth's theorem gives bound for $|q\alpha-p|$. This is the linear polynomial. Vanvu need such bounds for polynomials of degree $d$. Conjecturally for algebraic number $\alpha$ and $\epsilon>0$ there are only finitely many polynomials $P$ of degree $d$ with integer coefficients of absolute values at most $N$ with $0<|P(\alpha)|<N^{-d-1-\epsilon}$. $\endgroup$ May 12, 2011 at 0:28
  • $\begingroup$ @Oleg But he seems to want $N$ to be the same as $n$, which is related to the degree of $\alpha$ and so hidden in the ineffective constants in Roth's theorem and conjectural generalizations. $\endgroup$ May 12, 2011 at 0:41
  • $\begingroup$ @Felipe Good point. Than even this conjecture (wide open) is too weak for vanvu. $\endgroup$ May 12, 2011 at 0:48

It seems one can get something by just looking at the Taylor expansion of $\cos$. Let me consider $\cos 1/n$ for simplicity. Plugging

$$\cos 1/n = 1- (1/n)^2/2! + (1/n)^4/ 4! -.... $$

into $P= a_d x^d + a_{d-1} x^{d-1}+ ...+ a_1 x + a_0 $ one has

$$(a_d+ ...+ a_0) - \frac{(1/n)^2}{2!} (da_d +...+ a_1) -....$$

If $a_d+ ...+a_0$ is not zero, it dominates everything else, as $|a_i| \le n$. If it is zero, repeat the argument for $da_d+ ...+a_1$ and so on. By linear independency, this cannot go for more than $d$ steps (I guess). So we should have a lower bound like $n^{-2d}$.


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