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Math Help - Inequality

  1. #1
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    Inequality

    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?
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  2. #2
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    Re: Inequality

    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].
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  3. #3
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    Re: Inequality

    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
    Last edited by niaren; June 20th 2012 at 12:43 PM.
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