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Elegant notation?

  1. Sep 3, 2006 #1
    Is the notation
    [tex]\left\{ \begin{gathered}
    {\text{inequality}} \hfill \\
    {\text{inequality}} \hfill \\
    {\text{inequality}} \hfill \\
    \end{gathered} \right\} [/tex]
    generally accepted to denote the solution set (i.e., intersection) of the inequalities?

    If so, then the following (part of a problem I came up with) should be easy to understand:
    [tex]\bigcup\limits_{\begin{subarray}{l}
    \left( {i,j,k} \right) \in \mathbb{N}^3 , \\
    i < j < k \leqslant n
    \end{subarray}} {\left\{ \begin{gathered}
    \left( {y - y_i } \right)\left( {x_i - x_j } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_k } & {y_i - y_k } \\
    {x_i - x_j } & {y_i - y_j } \\

    \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_j } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_k } & {y_i - y_k } \\
    {x_i - x_j } & {y_i - y_j } \\

    \end{array} } \right| \hfill \\
    \left( {y - y_i } \right)\left( {x_i - x_k } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_j } & {y_i - y_j } \\
    {x_i - x_k } & {y_i - y_k } \\

    \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_k } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_j } & {y_i - y_j } \\
    {x_i - x_k } & {y_i - y_k } \\

    \end{array} } \right| \hfill \\
    \left( {y - y_j } \right)\left( {x_j - x_k } \right)\left| {\begin{array}{*{20}c}
    {x_j - x_i } & {y_j - y_i } \\
    {x_j - x_k } & {y_j - y_k } \\

    \end{array} } \right| \geqslant \left( {x - x_j } \right)\left( {y_j - y_k } \right)\left| {\begin{array}{*{20}c}
    {x_j - x_i } & {y_j - y_i } \\
    {x_j - x_k } & {y_j - y_k } \\

    \end{array} } \right| \hfill \\
    \end{gathered} \right\}} [/tex]

    Is there a simpler/more elegant way to express this, or is it fine the way it is?
     
    Last edited: Sep 3, 2006
  2. jcsd
  3. Sep 3, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    That looks like a set of inequalities rather than the solution set to me. Rather like saying that {x2- 4= 0} denotes the set {2, -2}. (The second way is more "elegant"!)
     
  4. Sep 3, 2006 #3
    Well, to be more clear, perhaps I should add intersection symbols:
    [tex]
    \bigcup\limits_{\begin{subarray}{l}
    \left( {i,j,k} \right) \in \mathbb{N}^3 , \\
    i < j < k \leqslant n
    \end{subarray}} {\left( \begin{gathered}
    \left( {y - y_i } \right)\left( {x_i - x_j } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_k } & {y_i - y_k } \\
    {x_i - x_j } & {y_i - y_j } \\

    \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_j } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_k } & {y_i - y_k } \\
    {x_i - x_j } & {y_i - y_j } \\

    \end{array} } \right| \cap \hfill \\
    \left( {y - y_i } \right)\left( {x_i - x_k } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_j } & {y_i - y_j } \\
    {x_i - x_k } & {y_i - y_k } \\

    \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_k } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_j } & {y_i - y_j } \\
    {x_i - x_k } & {y_i - y_k } \\

    \end{array} } \right| \cap \hfill \\
    \left( {y - y_j } \right)\left( {x_j - x_k } \right)\left| {\begin{array}{*{20}c}
    {x_j - x_i } & {y_j - y_i } \\
    {x_j - x_k } & {y_j - y_k } \\

    \end{array} } \right| \geqslant \left( {x - x_j } \right)\left( {y_j - y_k } \right)\left| {\begin{array}{*{20}c}
    {x_j - x_i } & {y_j - y_i } \\
    {x_j - x_k } & {y_j - y_k } \\

    \end{array} } \right| \hfill \\
    \end{gathered} \right)} [/tex]
    Essentially, I wish to denote a union of solution sets :shy:
    Would that be clear/understood from the way I rewrote it here?
     
    Last edited: Sep 3, 2006
  5. Sep 3, 2006 #4

    shmoe

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    Science Advisor
    Homework Helper

    It looks like you are still thinking of an inequality as the set of things that satisfy it, this isn't the usual use of the notation. The set of x's that satisfy x>2 would be written:

    {x| x>2}

    (specifiy some set x is coming from if it's not clear from the context, like x a real number, or integer), not

    x>2
     
  6. Sep 3, 2006 #5
    Thanks shmoe :smile:
    In that case, this would be the more correct way to denote my union of solution sets:
    [tex]\bigcup\limits_{\begin{subarray}{l}
    \left( {i,j,k} \right) \in \mathbb{N}^3 , \\
    i < j < k \leqslant n
    \end{subarray}} {\left\{ {\left( {x,y} \right)\left| \begin{gathered}
    \left( {y - y_i } \right)\left( {x_i - x_j } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_k } & {y_i - y_k } \\
    {x_i - x_j } & {y_i - y_j } \\

    \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_j } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_k } & {y_i - y_k } \\
    {x_i - x_j } & {y_i - y_j } \\

    \end{array} } \right| \hfill \\
    \left( {y - y_i } \right)\left( {x_i - x_k } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_j } & {y_i - y_j } \\
    {x_i - x_k } & {y_i - y_k } \\

    \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_k } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_j } & {y_i - y_j } \\
    {x_i - x_k } & {y_i - y_k } \\

    \end{array} } \right| \hfill \\
    \left( {y - y_j } \right)\left( {x_j - x_k } \right)\left| {\begin{array}{*{20}c}
    {x_j - x_i } & {y_j - y_i } \\
    {x_j - x_k } & {y_j - y_k } \\

    \end{array} } \right| \geqslant \left( {x - x_j } \right)\left( {y_j - y_k } \right)\left| {\begin{array}{*{20}c}
    {x_j - x_i } & {y_j - y_i } \\
    {x_j - x_k } & {y_j - y_k } \\

    \end{array} } \right| \hfill \\
    \end{gathered} \right.} \right\}} [/tex]
    Right?
     
    Last edited: Sep 3, 2006
  7. Sep 3, 2006 #6

    shmoe

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    Science Advisor
    Homework Helper

    Sure, I'd call that more correct, but you'd probably want to put some "or"'s in there.



    edit-spelling, I wasn't calling you "Sue"
     
    Last edited: Sep 3, 2006
  8. Sep 3, 2006 #7
    Some "or's" ? Why so?
     
  9. Sep 3, 2006 #8

    shmoe

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    maybe "and"'s then. You just have a list of inequalities as conditions,

    {x|x>2, x>4, x=65}

    What does this mean? It's ambiguous as it's written.
     
  10. Sep 3, 2006 #9
    Do you mean like this:
    [tex]\bigcup\limits_{\begin{subarray}{l}
    \left( {i,j,k} \right) \in \mathbb{N}^3 , \\
    i < j < k \leqslant n
    \end{subarray}} {\left\{ {\left( {x,y} \right)\left| \begin{gathered}
    \left( {y - y_i } \right)\left( {x_i - x_j } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_k } & {y_i - y_k } \\
    {x_i - x_j } & {y_i - y_j } \\

    \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_j } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_k } & {y_i - y_k } \\
    {x_i - x_j } & {y_i - y_j } \\

    \end{array} } \right| \wedge \hfill \\
    \left( {y - y_i } \right)\left( {x_i - x_k } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_j } & {y_i - y_j } \\
    {x_i - x_k } & {y_i - y_k } \\

    \end{array} } \right| \geqslant \left( {x - x_i } \right)\left( {y_i - y_k } \right)\left| {\begin{array}{*{20}c}
    {x_i - x_j } & {y_i - y_j } \\
    {x_i - x_k } & {y_i - y_k } \\

    \end{array} } \right| \wedge \hfill \\
    \left( {y - y_j } \right)\left( {x_j - x_k } \right)\left| {\begin{array}{*{20}c}
    {x_j - x_i } & {y_j - y_i } \\
    {x_j - x_k } & {y_j - y_k } \\

    \end{array} } \right| \geqslant \left( {x - x_j } \right)\left( {y_j - y_k } \right)\left| {\begin{array}{*{20}c}
    {x_j - x_i } & {y_j - y_i } \\
    {x_j - x_k } & {y_j - y_k } \\

    \end{array} } \right| \hfill \\
    \end{gathered} \right.} \right\}} [/tex]
    Correct?
     
  11. Sep 3, 2006 #10

    AKG

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    Science Advisor
    Homework Helper

    Yes, post #9 is correct.
     
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