Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Based on some experiments, I find that the following two statements are correct. But I can not prove this. At the same time, I still can not find the counterexmaples.

Let $p(x)=x^{n}+a_{2}x^{n-2}+a_{3}x^{n-3}+\dots+a_{n-1}x+a_{n}$ be a polynomial with interger coefficients, where $a_{k}\geq0$ for every even $k$ and $a_{k}\leq0$ otherwise. Note here that the coefficient of $x^{n-1}$ is equal to 0. Suppose that there exists some odd number $p$ such that $a_{p}<0$ and $a_{p-1}>0$. Then the following two statements should be true:

  1. $p(x)$ has at most one nonzero real root.

  2. $p(x)$ has no pure imaginary zeros, i.e. $p(x)$ has no zero in the form $\alpha\textrm{i}$, where $\alpha\neq0$ and $\textrm{i}^{2}=-1$.

Thanks for your time.

I am sorry for losing a condition that $a_{2}\geq5$.

share|improve this question
2  
Any other conditions left out? –  Gerry Myerson Mar 1 '11 at 3:34
    
@Gerry Myerson,thanks. The conditions needed are all presented. –  Shunyi Liu Mar 1 '11 at 3:38

3 Answers 3

up vote 7 down vote accepted

$(x^3-1)(x^3-2)(x^2+a^2)=x^8+a^2x^6-3x^5-3a^2x^3+2x^2+2a^2$ has both two different non-zero real roots and $\pm ai$ as roots.

EDIT polynomial adjusted for adjusted conditions. If $a_2$ needs to be at least $b$, set $a$ so $a^2\geq b$. :)

share|improve this answer
    
@Sándor Kovács, thanks for your counterexmaple! I am sorry that I have lost the condition that $a_{2}\geq5$ in my original problem. So, if this condition is added, then the counterexample you given is not appropriate. –  Shunyi Liu Mar 1 '11 at 3:34
1  
check again. :) –  Sándor Kovács Mar 1 '11 at 3:47
    
@Sándor Kovács, thank you very much for your elegant counterexample. –  Shunyi Liu Mar 1 '11 at 4:03

It is not true that $p(x)$ has at most one nonzero real root. For instance, the polynomial $p(x)=x^5+0.001x^3-100x^2+x-0.001$ has three real roots. For the second claim, the polynomial $p(x)=x^5+x^3-x^2-1$ has roots at $\pm i$.

share|improve this answer
    
He specified that $p(x)$ is a polynomial in integer coefficients, so the first counterexample will not work. –  Stanley Yao Xiao Mar 1 '11 at 2:39

With the condition $a_2\ge5$ there's still $x^{16}+5x^{14}+x^{12}+x^8-x^5+x^4-x^3+1$ which vanishes at $i$.

share|improve this answer
    
@Gerry Myerson. Thank you very much for your counterexample. –  Shunyi Liu Mar 1 '11 at 3:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.