Quadratic roots greater than 1 for $1-\alpha_1 x - \alpha_2 x^2$

78 Views Asked by At

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?

1

There are 1 best solutions below

1
On BEST ANSWER

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 $

===============================================

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:

  1. If ${\alpha_1}^{2} + 4\alpha_2 \geq 0$ and $\alpha_2 \lt 0$ then the required conditions are $\alpha_1 +\alpha_2 \lt 1 $ and $ (\alpha_1 + 2 \alpha_2) (2 \alpha_2) \lt 0 $
  2. If ${\alpha_1}^{2} + 4\alpha_2 \geq 0$ and $\alpha_2 \gt 0$ then the required conditions are $\alpha_1 +\alpha_2 \gt 1 $ and $(\alpha_1 + 2 \alpha_2) (2 \alpha_2) \lt 0 $
  3. If ${\alpha_1}^{2} + 4\alpha_2 \lt 0$ then the required conditions are $-1<\alpha_2<0$