Given $1<\beta<2$ and positive integer $n\geq2013$, then can we find a non-zero vector $(a_n\,,a_{n-1}\cdots\,a_1\,a_0 )$ where all $a_i\in\{-1\,,0\,,1\}$, such that $\sum\limits_{k=0}^{n-1}a_k\beta^k=0$
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3You mean a non-zero vector. – Ehsan M. Kermani May 06 '13 at 12:05
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yes, of course. – Tao May 06 '13 at 12:07
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2$\beta = \pi/2$ or $\beta = e/2$ – JiminP May 06 '13 at 12:08
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2As suggested by JiminP, the answer is NO when $\beta$ is transcendental, and the interval $(1,2)$ contains uncountably many such numbers. – Ewan Delanoy May 06 '13 at 12:15
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May I ask where this question comes from? The "2013" makes it look like a math competition problem, however, competition problems have the tendency to be correct, while this one is not. – Elmar Zander May 06 '13 at 12:44
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I want to make this question funny, it is from fractal – Tao May 06 '13 at 12:49
2 Answers
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If $\beta\in\mathbb{Q}$ then doesn't exists.
Indeed, suppose $(a_n,\cdots,a_0)\ne0$ such that
$$
a_n\beta^n+\cdots+a_1\beta+a_0=0
$$
Let $m=\max\{i:a_i\ne0\}$. Then, we have
$$
a_m\beta^m+\cdots+a_0=0
$$
and we can suppose $a_0\ne0$, otherwise, if $r=\min\{i:a_i\ne0\}$, then $a_m\beta^m+\cdots+a_0=a_m\beta^m+\cdots+a_r\beta^r=\beta^r(a_m\beta^{m-r}\cdots+a_r)=0$, and since $\beta^r\ne0$, then $a_m\beta^{m-r}+\cdots+a_r=0$.
Multiplying by $-1$, if necessary, we can suppose $a_m=1$. Now, writing $\beta=\frac ab$, with $a,b\in\mathbb{Z}$ and $\gcd(a,b)=1$, and multiplying the equation by $b^m$, we obtain
$$
a^m+ba_{m-1}\left(\frac ab\right)^{m-1} + \cdots + b^{m}a_m=0
$$
letting $b$ in evidence, we obtain
$$
a^m=bq
$$
hence, since $\gcd(a,b)=1$, we have that $b$ divides $a$, but this implies $\beta=\frac ab\in\mathbb{Z}$, contradiction.
Yuki
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I think you could have used the rational root theorem directly (http://en.wikipedia.org/wiki/Rational_root_theorem), which tells you that $a|a_0$ and $b|a_m$. – Elmar Zander May 06 '13 at 12:32
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Even if $\beta \in \mathbb{Q}$, there are infinitely many rational numbers $\in (1, 2)$, and set of all $\beta$s are finite (there are $3^{n+1}$ such polynomials and maximum # of solutions is much less than $(n-1)3^n$).
JiminP
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The question says $n\geq2013$, so the set of roots of the set of polynomials is countably infinite, too. However, the restriction $n\geq2013$ looks pretty weird to me, since it doesn't really restrict anything, unless you don't say something like $a_n\not=0$ also. – Elmar Zander May 06 '13 at 12:36
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@JiminP Yes, you're right. I first thought $n$ may be chosen depending on the given $\beta$, but if it may be chosen independently, then the set of roots is of course finite. – Elmar Zander May 06 '13 at 12:43