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Let $h>0$ be a positive odd integer. Let $n=4h^2.$

Let $R(t)=r_0+r_1t+ \cdots + r_{n-1}t^{n-1}$

be a polynomial with integer coefficients in $\{-1,1\}$ such that

$R(\omega)$ is a nonzero integer for all complex $\omega \notin ${-1,1} such that

(a)

$$ \omega^n=1 $$

and

(b)

$$ R(1) = 2h $$

Can we deduce that all these integers $R(\omega)$ have the same sign ???

reason: R(t) is the ``representer" polynomial of a circulant $-1,1$ matrix $C$ of order $n$ with first row $(r_0, \ldots,r_{n-1})$. I am trying to understand what happens when all the eigenvalues of $C$ (i.e., the $R(\omega)$) are real.

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2  
Why are you interested in these polynomials? This question is extremely similar to your previous one, and this question is probably too localized for MO. –  J.C. Ottem Dec 30 '10 at 22:48
    
thanks again, have you un example with n >4 ? –  Luis H Gallardo Dec 30 '10 at 23:00
    
I have not looked at the question properly, but on first glance agree with J.C. Ottem. Without better motivation, these questions just seem somewhat random, and generated on a whim rather than for a coherent purpose. –  Yemon Choi Dec 30 '10 at 23:11
1  
reason added to question –  Luis H Gallardo Dec 31 '10 at 0:03
1  
I agree with you wadim... –  Luis H Gallardo Dec 31 '10 at 10:21

1 Answer 1

Following the advice of the unknown: $$ R(t)=\frac{1-x^{16}}{1-x}-12x^8 $$ is a counter-example to your sign expectation. You can easily generalize this to any degree $n=4h^2$.

Edit. Igor brings to my attention the fact that the coefficients of $R(t)$ has to be all $\pm1$. For this situation one (including the author) could simply use brute force: if $n=16$, the polynomial of the desired form is a linear combination of the polynomials $(1-x^{16})/(1-x^j)$, $(1-x^8)/(1-x^k)$ and $x^8$. Just pick all possible integer combinations with the $\pm1$ restriction on the coefficients and check their values at the 16th roots of unity. This would also suggest how to proceed for the general $n$.

I turn my "answer" to the community wiki mode, so please feel free (if necessary) to add further details.

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I believe the question asks for $\pm 1$ coefficients, so this is not a counterexample. –  Igor Rivin Dec 31 '10 at 2:38
1  
I finally realized that there was a pre-question to this question already answered by the (same) unknown. And I would join the above remark of the unknown: it is a strange question without sufficient motivation... –  Wadim Zudilin Dec 31 '10 at 3:51
    
$16 = 4 \cdot 2^2$ i.e., $h=2$ that is not odd –  Luis H Gallardo Dec 31 '10 at 7:54
2  
Luis, you are kidding! There was already a post which has recognized 2 to be the oddest integer. :-) Too many constraints. Please try yourself to modify everything for $4\cdot3^2$. –  Wadim Zudilin Dec 31 '10 at 9:56

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