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 $d$ ?
2 Answers
Let $\alpha = 2^dP(\cos 2\pi /n)$. Then $\alpha$ is a nonzero 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. $\endgroup$ Commented May 12, 2011 at 0:09

$\begingroup$ @Oleg: I am afraid that Roth's theorem wors for any specific polynomial, while $n$ does change... $\endgroup$ Commented May 12, 2011 at 0:16

$\begingroup$ Roth's theorem gives bound for $q\alphap$. 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 $0<P(\alpha)<N^{d1\epsilon}$. $\endgroup$ Commented 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. $\endgroup$ Commented May 12, 2011 at 0:41

$\begingroup$ @Felipe Good point. Than even this conjecture (wide open) is too weak for vanvu. $\endgroup$ Commented 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_{d1} x^{d1}+ ...+ 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 $n^{2d}$.