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I am stuck on the following problem:

Let $q$ be a real polynomial of real variable $x$ of the form $q(x)=x^n+a_{n-1}x^{n-1}+....+a_1x-1 .\,\,$ Suppose $q$ has no roots in the open unit disc and $q(-1)=0.$
Then which of the following options are correct?

  1. $\lim_{x \to \infty}q(x)=\infty$

  2. $q(3)=0$

  3. $q(2) >0$

  4. $q(1)=0.$

I have also found the following result interesting but I am not sure it has anything to do with the current problem.

enter image description here

I need some detailed clarification to get the desired result.

learner
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  • Oh, I forgot that. Sorry! Then take $x^2-1$ to see that (2) isn't correct. – MichalisN Jun 27 '13 at 12:19
  • Ahh..This sounds ok. But in that case option 4 holds !! So by taking $q(x)=x^2-1$,we get options 1,3,4 correct and only option 2 false. Is it the case? – learner Jun 27 '13 at 12:22
  • See http://math.stackexchange.com/questions/430166/a-problem-on-roots-of-a-real-polynomial-p-that-has-no-roots-in-the-open-unit-d – Isomorphism Jun 27 '13 at 12:39

2 Answers2

2

(1) is correct because the leading coefficient is positive. Let $\alpha_1,\ldots,\alpha_n$ be the roots of $q$. We have $\prod\alpha_i=(-1)^{n+1}$. We see that if one of the roots had modulus $>1$ then another would have modulus $<1$, which contradicts the condition on the roots of $q$. Therefore all the roots lie on the unit circle so the roots split into real roots $\alpha_i=\pm 1$ and complex roots $\beta_j$ on the unit circle and since complex roots come in pairs it is of the form $$ q(X)=(X-1)^k(X+1)^l\prod_{j\in J}|X-\beta_j|^2 $$ So $q(2)=3^l\prod_j|2-\beta_j|^2>0$. Also $q(3)$ can't be zero, which answers $(2)$. The formula $\prod\alpha_i=(-1)^{n+1}$ becomes $q(0)=(-1)^k=-1$, so $k>0$ and odd, which implies $q(1)=0$.

Summing up: (1), (3) and (4) are true and the opposite of (2) is true, i.e. $q(3)\neq 0$. Note also that every polynomial of the above form with $l>0$ satisfies the conditions, so this also gives all possible $q$'s.

Remark: This is not in contrast to Tomas answer. The difference here is that Tomas did not take into consideration the complex roots of $q$, in this answer I do.

MichalisN
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1

(1) is true, because the leading coefficient of the highest monomial is positive.

All others are wrong, to prove this, take a look at the following polynomial: $$(x-2)(x+1)(x^2+\frac{1}{2})=x^4-x^3-\frac{3}{2}x^2-\frac{1}{2}x-1$$ First, it is of the desired form. Further, its only roots are $2$ and $-1$ (since $x^2+\frac{1}{2}$ is irreducible). Hence, there is no root on the open unit disc and $q(-1)=0$, so it meets the preconditions.

But, $q(1)\neq 0, q(3)\neq 0$ and $q(2)=0$ so (2)-(4) are wrong.

Tomas
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  • I think in this problem one should take complex roots into account. That's why he says no roots in the open unit disc and not $(-1,1)$. – MichalisN Jun 27 '13 at 12:48
  • Hmm, probably, but on the other hand, he specifically talks about a polynomial of real variable x. Maybe the OP should clarify this. – Tomas Jun 27 '13 at 12:51