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Consider the even polynomial $$ \Phi(z)=a_6z^6+a_4z^4+a_2z^2+a_0=0\;, $$ where $a_i\in\mathbb{R}$. Given that $a_0<0$ and $a_6>0$, does there exist conditions to guarantee that $\Phi$ has no positive real roots? This seems impossible, but maybe I'm missing something! What I would prefer are good references on this type of problem, or ideas to work with.

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If $z_0$ is a root then $-z_0$ is also a root, so you want all roots of $\phi$ to be non-real. –  Quimey Jan 31 '13 at 15:27
    
Good point. How would you ensure that all the real parts of such roots are in the left half-plane? –  transcendental Feb 1 '13 at 15:32
    
The same argument applies, it can only be possible if all the roots are pure imaginary. –  Quimey Feb 1 '13 at 18:54
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2 Answers 2

Observe $\Phi(0) = a_0 <0$ and $\displaystyle\lim_{t \rightarrow \infty} \Phi(t) = \infty.$ Hence, by the intermediate value theorem $\Phi$ has a positive root.

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Yes, you are correct about the intermediate value theorem stuff. I guess I was hoping one could get a condition (on the coefficients) that somehow negates what you just said; for example, if the relative minimum value of $\Phi$ (at the positive root) is strictly positive. It would really be useful to me if this were possible... –  transcendental Feb 1 '13 at 15:22
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Observe that the complex roots of $\Phi$ come in conjugate pairs because $\Phi(\overline{z})=\overline{\Phi(z)}$. Therefore, if $\Phi$ has no real roots then it can be written as $\Phi(z) = a_6(z-u)(z-\overline{u})(z-v)(z-\overline{v})(z-w)(z-\overline{w})$ for (possibly duplicate) complex numbers $u,v,w$ and you will have $a_0 = \Phi(0) = a_6 |u|^2 |v|^2 |w|^2$. But this would only be possible if $a_6$ and $a_0$ had the same sign.

Of course, as already answered, the intermediate value theorem is a quicker way to see this.

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