I am trying to prove the conditions that need to be valid for the below quadratic to have roots outside the unit circle.
$1-\alpha_1 x - \alpha_2 x^2$
and the conditions that need to hold are:
$|\alpha_2| < 1 $ ;
$\alpha_2 + \alpha_1 < 1 $ ;
$\alpha_2 - \alpha_1 < 1 $
so,
$| \frac{\alpha_1 \pm \sqrt{\alpha_1^2 +4 \alpha_2}}{2\alpha_2}| > 1$
EQ1: $\frac{\alpha_1 + \sqrt{\alpha_1^2 +4 \alpha_2}}{2\alpha_2} > 1$
EQ2: $ \frac{\alpha_1 - \sqrt{\alpha_1^2 +4 \alpha_2}}{2\alpha_2} < 1$
using EQ1:
$\sqrt{\alpha_1^2 +4 \alpha_2} > 2\alpha_2 - \alpha_1 $
from EQ2:
$\sqrt{\alpha_1^2 +4 \alpha_2} > -2\alpha_2 + \alpha_1 $
not sure how to proceed further, any pointers?
Case I: Both roots are real
Since both roots are real, ${\alpha_1}^{2} + 4\alpha_2 \geq 0$
Case Ia:
$\alpha_2 \lt 0$ (parabola opens upwards)
Let $$f(x)=1-\alpha_1 x - \alpha_2 {x}^{2}$$
So, $f(1) \gt 0 \implies \alpha_1 +\alpha_2 \lt 1 $
And, lowest point of parabola is also $\gt 1 $
So, $- \frac{\alpha_1}{2 \alpha_2} \gt 1 $
So, $ (\alpha_1 + 2 \alpha_2) (2 \alpha_2) \lt 0 $
Case Ib:
$\alpha_2 \gt 0$ (parabola opens downwards)
$f(1) \lt 0 \implies \alpha_1 +\alpha_2 \gt 1 $
x- co-ordinate of highest point is also greater than 1
So, $- \frac{\alpha_1}{2 \alpha_2} \gt 1 \implies (\alpha_1 + 2 \alpha_2) (2 \alpha_2) \lt 0 $
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Case II: (both roots contain imaginary parts) $\implies$ ${\alpha_1}^{2} + 4\alpha_2 \lt 0$
$$| \frac{\alpha_1 \pm i \sqrt{-({\alpha_1}^{2}+4 \alpha_2})}{-2 \alpha_2}| > 1$$
So, $${(\frac{\alpha_1}{-2 \alpha_2})}^2 - \frac{{\alpha_1}^{2}+4 \alpha_2}{4 {\alpha_2}^{2}} >1$$
$$-4 \alpha_2 > 4 {\alpha_2}^{2}$$
$$-1<\alpha_2<0$$
Summary: