# Roots of sparse “quadratic-like” polynomial.

So I know about this question and I've seen papers like this and this. But the former isn't exactly what I want and the latter two papers are too deep and I'm lazy and I wanna quick-and-easy answer and I'm sure you legitimate mathematicians are happy to supply this lazy electrical engineer with a quick-and-easy answer, right?

So, in general, an $$N$$th-order polynomial with real coefficients is:

$$f(z) = \sum\limits_{n=0}^{N} a_n \, z^n \qquad \qquad \text{where }N\in\mathbb{Z}\ge0,\ a_n\in\mathbb{R},\ z\in\mathbb{C}$$

\begin{align} w^2 + bw + c &= w^2 (1 + bw^{-1} + cw^{-2})\\ &= (w-r_1)(w-r_2) \\ \end{align}

again $$b,c\in\mathbb{R}$$ and $$w,r_1,r_2\in\mathbb{C}$$ . We know that

\begin{align} r_1 &= -\tfrac{b}{2} + \sqrt{\left(\tfrac{b}{2}\right)^2 - c} \\ r_2 &= -\tfrac{b}{2} - \sqrt{\left(\tfrac{b}{2}\right)^2 - c} \\ \end{align}

if $$b^2 \ge 4c$$ and

\begin{align} r_1 &= -\tfrac{b}{2} + i\sqrt{c - \left(\tfrac{b}{2}\right)^2} \\ r_2 &= -\tfrac{b}{2} - i\sqrt{c - \left(\tfrac{b}{2}\right)^2} \\ \end{align}

if $$b^2 < 4c$$

In the latter complex-conjugate case, we know that

$$|r_1| = |r_2| = c$$

which is both simple and handy.

Now, whether the roots are real or complex-conjugate, suppose we have chosen $$b$$ and $$c$$ so that both

\begin{align} |r_1| &< 1 \\ |r_2| &< 1 \\ \end{align}

Fine, now let's return to the sparse polynomial $$f(z)$$. Suppose $$N$$ is even and all coefficients $$a_n$$ are zero except:

\begin{align} a_0 &= 1 \\ a_{N/2} &= b \\ a_N &= c \\ \end{align}

What are the roots of $$f(z)$$? Suppose $$N$$ is pretty big (and even), say $$N\approx 1000$$. If I can guarantee that $$b$$ and $$c$$ are chosen to insure that $$|r_1|$$ and $$|r_2|$$ are less than $$1$$, can I rely on all $$N$$ roots of $$f(z)$$ also being less than $$1$$?

I think that I can. I've been fiddling with the substitution of

$$w = z^{N/2}$$

and I know about the Nth-roots-of-unity, but am I guaranteed that all of the roots of the $$N$$th-order polynomial have magnitude less than $$1$$?

• I guess if $|w|<1$, we know that $\big| z^{N/2} \big| < 1$ and $|z|<1^{2/N}$ and $|z|<1$. So, although I dunno all $N$ roots of $f(z)$, I think I can say they are all inside the unit circle and that is the most important thing. (But I would like to know what the values of the roots are, anyway.) – robert bristow-johnson Dec 14 '18 at 1:55

So you have the equation: $$cz^{2n}+bz^n+1 = 0$$, where it's guaranteed that $$z^n_{1,2}$$ have modulus less than $$1?$$ If this is the case, then your intuition is correct because if: $$|z^n| = |z|^n = |\omega| <1,$$ it must necessarily follow that: $$|z|<1$$ since $$|z|$$ is a positive real number.