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Math Help - inequality

  1. #1
    Super Member Random Variable's Avatar
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    inequality

    Show that  |(Re^{i \theta})^{2} + a^{2} | \ge R^{2} -a^{2} \ \ (R, a >0, \ 0 \le \theta \le \pi)

     |R^{2}e^{2i \theta} + a^{2} | = |R^{2}(cos 2 \theta + i \sin 2 \theta) + a^{2} |= |(R^{2} \cos 2 \theta + a^{2}) + i R^{2} \sin 2 \theta |

     = \sqrt{(R^{2} \cos 2 \theta +a^{2})^{2} + R^{4} \sin^{2} 2 \theta}

     = \sqrt{R^4 \cos^{2} 2 \theta + 2R^{2}a^{2}\cos 2 \theta + a^{4} + R^{4} \sin^{2} \theta}

     = \sqrt{R^{4} + 2R^{2}a^{2} \cos 2 \theta + a^{4}}

     \ge \sqrt{(R^{2}-a^{2})^{2}} ?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Random Variable View Post
    Show that  |(Re^{i \theta})^{2} + a^{2} | \ge R^{2} -a^{2} \ \ (R, a >0, \ 0 \le \theta \le \pi)

     |R^{2}e^{2i \theta} + a^{2} | = |R^{2}(cos 2 \theta + i \sin 2 \theta) + a^{2} |= |(R^{2} \cos 2 \theta + a^{2}) + i R^{2} \sin 2 \theta |

     = \sqrt{(R^{2} \cos 2 \theta +a^{2})^{2} + R^{4} \sin^{2} 2 \theta}

     = \sqrt{R^4 \cos^{2} 2 \theta + 2R^{2}a^{2}\cos 2 \theta + a^{4} + R^{4} \sin^{2} \theta}

     = \sqrt{R^{4} + 2R^{2}a^{2} \cos 2 \theta + a^{4}}

     \ge \sqrt{(R^{2}-a^{2})^{2}} ?
    Apply a variation of the triangle inequality to get

    \begin{aligned}\left|\left(Re^{i\theta}\right)^2+a  ^2\right|\geq\left|\left(Re^{i\theta}\right)^2\rig  ht|-\left|a^2\right| & =\left|Re^{i\theta}\right|^2-a^2\\ & =R^2\left|e^{i\theta}\right|^2-a^2\\ & =R^2\left(\cos^2\theta+\sin^2\theta\right)-a^2\\ &= R^2-a^2\end{aligned}
    Last edited by Chris L T521; August 20th 2009 at 04:33 PM. Reason: fixed some mistakes
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  3. #3
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Apply a variation of the triangle inequality to get

    \begin{aligned}\left|\left(Re^{i\theta}\right)^2+a  ^2\right|\geq\left|\left(Re^{i\theta}\right)^2\rig  ht|-\left|a^2\right| & =\left|Re^{i\theta}\right|^2-a^2\\ & =R^2\left|e^{i\theta}\right|^2-a^2\\ & =R^2\left(\cos^2\theta+\sin^2\theta\right)-a^2\\ &= R^2-a^2\end{aligned}
    Thanks. I don't why I never heard of the reverse triangle inequality (the proof of which is quite simple).
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Random Variable View Post
    Thanks. I don't why I never heard of the reverse triangle inequality (the proof of which is quite simple).
    I'm not sure if this will change things...

    I used the part highlighted in red:

    {\color{red}\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|}\leq\lef  t|a\right|+\left|b\right|.

    So it should technically be \left|\left(Re^{i\theta}\right)^2+a^2\right|\geq\l  eft|R^2-a^2\right|.

    If R\geq a, then it would be ok to leave it as \left|\left(Re^{i\theta}\right)^2+a^2\right|\geq R^2-a^2...
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    I'm not sure if this will change things...

    I used the part highlighted in red:

    {\color{red}\left|\left|a\right|-\left|b\right|\right|\leq\left|a+b\right|}\leq\lef  t|a\right|+\left|b\right|.

    So it should technically be \left|\left(Re^{i\theta}\right)^2+a^2\right|\geq\l  eft|R^2-a^2\right|.

    If R\geq a, then it would be ok to leave it as \left|\left(Re^{i\theta}\right)^2+a^2\right|\geq R^2-a^2...
    I was just about to ask if it matters if a is greater than or less than R. For this particular problem it's less than  R .
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