Inequality

**niaren** · June 19th 2012, 12:53 PM

I have the following inequality

$-1 \leq \frac{p2R\cos(\phi)+R^2-p-2R\cos(\phi)}{1-pR^2} \leq 3$

where $p\in(-\infty,\infty)$ , $R\in(0,\infty)$ and $\phi\in(0,\pi)$ .

Given a $\phi$ and a $R$ my goal is to find which values of $p$ satisfies the inequality.
I have tried to follow the procedure described in the document here: Solving inequalities
but have not found a way to create a table such as seen in example 5.
Does anyone have an idea how to proceed with this inequality?

**emakarov** · June 19th 2012, 01:10 PM

For given $\phi$ and R, this double inequality has the form $-1\le\frac{ap+b}{cp+d}\le 3$ for some constants a, b, c and d. You need to take the intersection of the sets of solutions of these two inequalities. For example, $\frac{ap+b}{cp+d}\le 3$ <=> $\frac{ap+b}{cp+d}-3\le 0$ <=> $\frac{(a-3c)p+(b-3d)}{cp+d}\le0$ <=> [ $(a-3c)p+(b-3d)\ge0$ and $cp+d< 0$ ] or [ $(a-3c)p+(b-3d)\le0$ and $cp+d> 0$ ].

**niaren** · June 20th 2012, 02:50 AM

Thanks for feedback.

However, it gets rather messy rather quickly.

First case

(a): $1-pR^2\geq0$
and
(b): $p(2R\cos(\phi)-1+3R^2) \leq 2R\cos(\phi)+3-R^2$

(a) is easy, we get $p\leq\frac{1}{R^2}$

(b) is not so easy, we don't know the sign of $2R\cos(\phi)-1+3R^2$ .
If $R\leq1$ then the RHS $2R\cos(\phi)+3-R^2 \geq 0$ , so for $R\leq1$ we must study the sign of the LHS $2R\cos(\phi)-1+3R^2$ . On the other hand, $R\geq1$ the LHS is always positive and we must study the sign of the RHS.

$R\leq1$ : (LHS)

R must reach a minimum value before LHS can be positive. That value is $\frac{1}{3}$ .

For $R\leq\frac{1}{3}$ we have

$p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative.

So for $R\leq\frac{1}{3}$ we can't met equality[/TEX].

For $\frac{1}{3}<R\leq1$

if $R<\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ then $2R\cos(\phi)-1+3R^2 < 0$
if $R>\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ then $2R\cos(\phi)-1+3R^2 > 0$
Then
For $\frac{1}{3}<R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$
$p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative

So for $\frac{1}{3}<R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ we require $\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}<p\leq\frac{1}{R^2}$ .

For $\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R<1$
$p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is positive

For $\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R\leq 1$ we require
$p \leq min(\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2},\frac{1}{R^2})$

$R>1$ : (RHS)

When $R>3$ the RHS is always negative.

For $1<R\leq\3$ we have

if $R<\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ then $2R\cos(\phi)+3-R^2 > 0$
if $R>\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ then $2R\cos(\phi)+3-R^2 < 0$
Then
For $1<R\leq\cos(\phi)+\sqrt{\cos^2(\phi)+3}$
$p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is positive
For $\cos(\phi)+\sqrt{\cos^2(\phi)+3}<R<3$
$p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative

For $R>3$ we have

$p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative

We therefore conclude

For $R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ we require $\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}<p\leq\frac{1}{R^2}$
For $\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R\leq \cos(\phi)+\sqrt{\cos^2(\phi)+3}$ we require $p \leq min(\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2},\frac{1}{R^2})$
For $R\geq\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ we require $p \leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ .

Second case

(a): $1-pR^2\leq 0$
and
(b): $p(2R\cos(\phi)-1+3R^2) \geq 2R\cos(\phi)+3-R^2$

(a) is easy, we get $p\geq\frac{1}{R^2}$

(b)
$R\leq1$ : (LHS)

R must reach a minimum value before LHS can be positive. That value is $\frac{1}{3}$ .

For $R\leq\frac{1}{3}$ we have

$p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative.

So for $R\leq\frac{1}{3}$ we require $p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ .

For $\frac{1}{3}<R\leq1$

if $R<\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ then $2R\cos(\phi)-1+3R^2 < 0$
if $R>\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ then $2R\cos(\phi)-1+3R^2 > 0$
Then
For $\frac{1}{3}<R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$
$p\leq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative
For $\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R<1$
$p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is positive

For $\frac{1}{3}<R\leq \frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ we can't meet equality.

For $\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R<1$ we require $p \geq max(\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2},\frac{1}{R^2})$

$R>1$ : (RHS)

When $R>3$ the RHS is always negative.

For $1<R\leq\3$ we have

if $R<\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ then $2R\cos(\phi)+3-R^2 > 0$
if $R>\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ then $2R\cos(\phi)+3-R^2 < 0$
Then
For $1<R\leq\cos(\phi)+\sqrt{\cos^2(\phi)+3}$
$p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is positive
For $\cos(\phi)+\sqrt{\cos^2(\phi)+3}<R<3$
$p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative

For $R>3$ we have

$p\geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ which is negative

We therefore conclude

For $R\leq\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}$ we can't meet inequality
For $\frac{\sqrt{\cos^2(\phi)+3}-\cos(\phi)}{3}<R\leq \cos(\phi)+\sqrt{\cos^2(\phi)+3}$ we require $p \geq max(\frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2},\frac{1}{R^2})$
For $R\geq\cos(\phi)+\sqrt{\cos^2(\phi)+3}$ we require $p \geq \frac{2R\cos(\phi)+3-R^2}{2R\cos(\phi)-1+3R^2}$ .

For the other inequality
$\frac{ap+b}{cp+d}\geq-1$
$\frac{ap+b}{cp+d}+1\geq0$
$\frac{p(a+c)+b+d}{cp+d}\geq 0$

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