# Description of region

• Mar 17th 2011, 06:13 AM
leoemil
Description of region
Hi there,

Given the expression (for a region):

$\displaystyle \left| \frac{-x \pm \sqrt{x^2 - 4y}}{2} \right|< 1$

Can anybody help me reducing this, to a simpler description of the region. I know it can somehow be reduced to a triangular region.

Thank you so much!
• Mar 17th 2011, 07:36 AM
HallsofIvy
Quote:

Originally Posted by leoemil
Hi there,

Given the expression (for a region):

$\displaystyle \left| \frac{-x \pm \sqrt{x^2 - 4y}}{2} \right|< 1$

Can anybody help me reducing this, to a simpler description of the region. I know it can somehow be reduced to a triangular region.

Thank you so much!

The simplest way to handle this is to look at the equation
$\displaystyle \frac{-x\pm sqrt{x^2- 4y}}{2}= 1$
which can be rewritten
$\displaystyle \pm\sqrt{x^2- 4y}= x+ 2$

Squaring both sides, $\displaystyle x^2- 4y= x^2+ 4x+ 4$
or $\displaystyle y= -x- 1$.

That is the straight line through (0, -1) and (-1, 0). But in order that the left side be a real number (so it can be compared to 1) we must have $\displaystyle x^2- 4y> 0$ or $\displaystyle 4y< x^2$. $\displaystyle 4y= x^2$ is a parabola. $\displaystyle 4y< x^2$ for (x, y) below that parabola.

If x= 0, y= 0, the original inequality is 0< 1 which is true. That means that all (x, y) on the same side of y= -x- 1 as (0, 0) (above and to the right) satisfy the inequality provided they also lie below the parabola. The line and parabola do NOT intersect so that is not anything like a "triangle"- it is and unbounded region.
• Mar 17th 2011, 07:59 AM
leoemil
Thanks, but.....
Thank you so much for the answer,

However, what if you include the region for which the squareroot is negative. Then you can take modulus, in order to compare it with 1 - remember the | | sign, i have included in the original post. I think it is suppose to end up being a triangular region. What do you think?
• Mar 17th 2011, 09:36 AM
FernandoRevilla
Quote:

Originally Posted by leoemil
However, what if you include the region for which the squareroot is negative. Then you can take modulus, in order to compare it with 1 - remember the | | sign, i have included in the original post. I think it is suppose to end up being a triangular region. What do you think?

Taking into account that $\displaystyle (-x\pm \sqrt{x^2-4y})/2$ are the roots of $\displaystyle p(t)=t^2+xt+y\in\mathbb{R}[t]$ and their modulus must be less than one, you can use a well known bound:

If $\displaystyle p(t)=a_0t^d+\ldots+a_{d-1}t+a_d\in\mathbb{R}[t]$ where $\displaystyle a_0\neq 0$ and $\displaystyle z\in\mathbb{C}$ is a root of $\displaystyle p$ then,

$\displaystyle |c|\leq \max_{i=1,\ldots ,d} (d|a_i/a_0|)^{1/i}=M$

In our case,

$\displaystyle M=\max\{2|x|,\sqrt{2|y|}\}$ .

P.S. Of course I don't know if you have covered that theorem.