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It might be very simple but I need a formal proof for accepting or rejecting the idea below.

Let g be a polynomial of the order of n given below $$ g(L)=1-\theta_1 L-\theta_2 L^2- \ldots -\theta_n L^n, \quad \theta_i \in \mathbb{R} $$ where some of coefficients (not $\theta_n$) are zero.

So I would like to know under condition above, is there any straightforward rules for roots of the polynomial?

For example I have tested $1-x^2$ and found that the roots are the same but the sign is different. Then, knowing one root is enough to know another one.

Thanks you for sharing me your ideas.

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  • $\begingroup$ You can find upper and lower bound for the roots, or only for the positive real roots. Are the coefficients all non-negative? If not, this is an ordinary polynomial of degree $n$. $\endgroup$ – LutzL Jun 18 '14 at 11:36
  • $\begingroup$ Dear @LutzL, thank you for your reply. yes all the coefficients are negative. Can you explain your reply more? $\endgroup$ – TPArrow Jun 19 '14 at 11:15
  • $\begingroup$ Done as answer, would have been too large for a comment. $\endgroup$ – LutzL Jun 19 '14 at 12:13
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The expression $1-x^2$ can be easily factorized by the difference of squares method, which yields $(1-x)(1+x)$. In general $1-ax^2$ is factorizable to $(1-\sqrt a x)(1+ \sqrt a x)$, and that gives an easy way of finding the roots.

In general, polynomials where the only nonzero components are even powers of $x$ are called, not surprisingly, even. For these, the roots are symmetric around the origin, so knowing all positive or negative roots give the other ones. Conversely, polynomials where the only nonzero components are odd powers of $x$ are called odd. Their roots are also symmetric around the origin, and they automatically have a root at $x=0$. All functions (not just polynomials) can be decomposed uniquely in an odd and an even part.

Although this simplifies calculations, there is no way of algebraically finding the roots of all the polynomials of degree 5 and above. This has been shown by Galois.

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  • $\begingroup$ Thank you @JoeyBF. How about a combination of odd and even zero coefficients? $\endgroup$ – TPArrow Jun 15 '14 at 15:37
  • $\begingroup$ @HHM Then nothing can be said, unless you factorize it in the sum of an even and an odd polynomial. $\endgroup$ – JoeyBF Jun 15 '14 at 19:26
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There is a bound named after Cauchy for positive real roots. For any polynomial $p(x)=a_0+a_1x+...+a_nx^n$ assume that $a_0>0$ and set $N=\{k:a_k<0\}$ (and assume $N$ not empty). Then for $x>0$ $$ p(x)\ge a_0-|N|\,\max_{k\in N}\left(|a_k| x^k\right) $$ so that as long as $$ 0\le x<B=\min_{k\in N}\left(\frac{|a_0|}{|N|\,|a_k|}\right)^{\frac1k} $$ you get that $p(x)>0$, i.e., $B$ is a lower bound for the positive real roots of $p$.

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