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Why won't LaTex display this?

  1. Apr 6, 2005 #1
    What's wrong with my code? Why would LaTex show my work?

    [tex] \begin{array}{l}
    r = \frac{1}{{k - 1}}\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n - \bar x}}{{S_x }}} \right)\left( {\frac{{y_n - \bar y}}{{S_y }}} \right)} \right]} = \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}{k}}}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)\left( {\frac{{y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}{k}}}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)} \right]} \Rightarrow \left( \begin{array}{l}
    {\rm multiply both fractions} \\
    {\rm by }\frac{{\left( {{\raise0.7ex\hbox{$1$} \!\mathord{\left/
    {\vphantom {1 k}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$k$}}} \right)}}{{\left( {{\raise0.7ex\hbox{$1$} \!\mathord{\left/
    {\vphantom {1 k}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$k$}}} \right)}} = \frac{{k^{ - 1} }}{{k^{ - 1} }} = \frac{k}{k} = 1 \\
    \end{array} \right) \Rightarrow \\
    \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)} \right]} \Rightarrow \left( \begin{array}{l}
    {\rm rewrite the bottoms sums over} \\
    {\rm a common denominator,}\;\left( k \right) \\
    \end{array} \right) \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{k}} \right)^2 } }}{{k - 1}}} }}} \right)} \right]} \\
    \Rightarrow \left( \begin{array}{l}
    {\rm Expand the bottom sums to factor } \\
    {\rm out }\left( k \right).{\rm Because we have a sum of } \\
    squares{\rm , we will factor out a }\left( {k^2 } \right) \\
    \end{array} \right) \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}{{k^2 \left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}{{k^2 \left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \left( \begin{array}{l}
    {\rm From this, we can solve the square root of} \\
    \;\left( {k^2 } \right)\;{\rm to place this value outside the radical } \\
    \end{array} \right) \Rightarrow \\
    \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}{{\left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}{{\left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \left( \begin{array}{l}
    {\rm The }\left( k \right){\rm values will cancel } \\
    {\rm as we multiply:}\;k \cdot \sqrt {\frac{1}{{k^2 }}} = 1 \\
    \end{array} \right) \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}{{\left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}{{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}{{\left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \\
    \Rightarrow \left( \begin{array}{l}
    {\rm We express the bottom radical as a} \\
    {\rm a radical numerator divided by }\sqrt {k - 1} ; \\
    {\rm therefore, we can place }\sqrt {k - 1} {\rm in the } \\
    {\rm numerator of the entire equation } \\
    \end{array} \right) \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left\{ {\left[ {\frac{{\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\sqrt {k - 1} }}{{\sqrt {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } }}} \right]\left[ {\frac{{\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)\sqrt {k - 1} }}{{\sqrt {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } }}} \right]} \right\}} \Rightarrow \left( \begin{array}{l}
    {\rm We can multiply both z - score expressions} \\
    {\rm to simplify the sum and remove }\sqrt {k - 1} \\
    \end{array} \right) \Rightarrow \\
    \Rightarrow \left( {\frac{1}{{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\frac{{\left( {k - 1} \right)\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)}}{{\left( {\sqrt {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } } \right)\left( {\sqrt {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } } \right)}}} \right]} = \sum\limits_{n = 1}^k {\left\{ {\frac{{\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)}}{{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } \right]\left[ {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } \right]} }}} \right\}} = = \frac{{\sum\limits_{n = 1}^k {\left[ {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)} \right]} }}{{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } \right]\left[ {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } \right]} }} \Rightarrow \\
    \Rightarrow \left( {{\rm Substitute}\left\{ \begin{array}{l}
    a = \sum\limits_{n = 1}^k {x_n } \\
    b = \sum\limits_{n = 1}^k {y_n } \\
    \end{array} \right\}} \right) \Rightarrow \frac{{\sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k^2 - k\left( {y_n a + x_n b} \right) + ab} \right]} }}{{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n^2 k^2 - 2ax_n k + a^2 } \right)} } \right]\left[ {\sum\limits_{n = 1}^k {y_n^2 k^2 - 2by_n k + b^2 } } \right]} }} = \Rightarrow \left( \begin{array}{l}
    {\rm We can take out}\left( {ab} \right){\rm from the } \\
    {\rm numerator and the}\left( {a^2 ,b^2 } \right){\rm from the } \\
    {\rm denominator as}\left( {kab} \right){\rm and}\left( {a^2 k,b^2 k} \right) \\
    \end{array} \right) \Rightarrow \frac{{kab + \sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k^2 - k\left( {y_n a + x_n b} \right)} \right]} }}{{\sqrt {\left[ {a^2 k + \sum\limits_{n = 1}^k {\left( {x_n^2 k^2 - 2ax_n k} \right)} } \right]\left[ {b^2 k + \sum\limits_{n = 1}^k {\left( {y_n^2 k^2 - 2by_n k} \right)} } \right]} }} \Rightarrow \\
    \Rightarrow \left( \begin{array}{l}
    {\rm Then, we can factor out the} \\
    \left( k \right){\rm from the sum expressions} \\
    \end{array} \right) \Rightarrow = \frac{{kab + k\sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k - \left( {y_n a + x_n b} \right)} \right]} }}{{\sqrt {\left[ {a^2 k + k\sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 k + k\sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} = = \frac{{k\left\{ {ab + \sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k - \left( {y_n a + x_n b} \right)} \right]} } \right\}}}{{\sqrt {k^2 \left[ {a^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} \Rightarrow \left( \begin{array}{l}
    {\rm The}\left( {k^2 } \right){\rm in the denominator} \\
    {\rm radical can be taken out and} \\
    {\rm will cancel with the }\left( k \right) \\
    {\rm in the equation numerator} \\
    \end{array} \right) \\
    = \frac{{ab + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n b - y_n a} \right)} }}{{\sqrt {\left[ {a^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} = \frac{{ab + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n b - y_n a} \right)} }}{{\sqrt {\left\{ {a^2 + \sum\limits_{n = 1}^k {\left[ {x_n \left( {x_n - 2a} \right)} \right]} } \right\}\left\{ {b^2 + \sum\limits_{n = 1}^k {\left[ {y_n \left( {y_n - 2b} \right)} \right]} } \right\}} }} \Rightarrow \left( \begin{array}{l}
    {\rm Replace variables }\left( {a,b} \right) \\
    {\rm with their original assignments} \\
    \end{array} \right) \Rightarrow \\
    \frac{{\left( {\sum\limits_{n = 1}^k {x_n } } \right)\left( {\sum\limits_{n = 1}^k {y_n } } \right) + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n \sum\limits_{n = 1}^k {y_n } - y_n \sum\limits_{n = 1}^k {x_n } } \right)} }}{{\sqrt {\left[ {\left( {\sum\limits_{n = 1}^k {x_n } } \right)^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n \sum\limits_{n = 1}^k {x_n } } \right)} } \right]\left[ {\left( {\sum\limits_{n = 1}^k {y_n } } \right)^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n \sum\limits_{n = 1}^k {y_n } } \right)} } \right]} }} = \frac{{\left( {\sum\limits_{n = 1}^k {x_n } } \right)\left( {\sum\limits_{n = 1}^k {y_n } } \right) + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n \sum\limits_{n = 1}^k {y_n } - y_n \sum\limits_{n = 1}^k {x_n } } \right)} }}{{\sqrt {\left\{ {\left( {\sum\limits_{n = 1}^k {x_n } } \right)^2 + \sum\limits_{n = 1}^k {\left[ {x_n \left( {x_n - 2\sum\limits_{n = 1}^k {x_n } } \right)} \right]} } \right\}\left\{ {\left( {\sum\limits_{n = 1}^k {y_n } } \right)^2 + \sum\limits_{n = 1}^k {\left[ {y_n \left( {y_n - 2\sum\limits_{n = 1}^k {y_n } } \right)} \right]} } \right\}} }} \\
    \end{array} [/tex]

    What does it mean;--what's "invalid" here?
     
  2. jcsd
  3. Apr 6, 2005 #2
    Hmm---I think I fixed it; how do you know when to press enter and start a new command line, so you can see my WHOLE work? (because browser's don't scroll that far!)
    [tex] \begin{gathered}
    r = \frac{1}
    {{k - 1}}\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n - \bar x}}
    {{S_x }}} \right)\left( {\frac{{y_n - \bar y}}
    {{S_y }}} \right)} \right]} = \left( {\frac{1}
    {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}
    {k}}}
    {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}
    {k}} \right)^2 } }}
    {{k - 1}}} }}} \right)\left( {\frac{{y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}
    {k}}}
    {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}
    {k}} \right)^2 } }}
    {{k - 1}}} }}} \right)} \right]} \Rightarrow \left( \begin{gathered}
    \text{multiply both fractions} \hfill \\
    \text{by }\frac{{\left( {{\raise0.7ex\hbox{$1$} \!\mathord{\left/
    {\vphantom {1 k}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$k$}}} \right)}}
    {{\left( {{\raise0.7ex\hbox{$1$} \!\mathord{\left/
    {\vphantom {1 k}}\right.\kern-\nulldelimiterspace}
    \!\lower0.7ex\hbox{$k$}}} \right)}} = \frac{{k^{ - 1} }}
    {{k^{ - 1} }} = \frac{k}
    {k} = 1 \hfill \\
    \end{gathered} \right) \Rightarrow \hfill \\
    \left( {\frac{1}
    {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}
    {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n - \frac{{\sum\limits_{n = 1}^k {x_n } }}
    {k}} \right)^2 } }}
    {{k - 1}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}
    {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n - \frac{{\sum\limits_{n = 1}^k {y_n } }}
    {k}} \right)^2 } }}
    {{k - 1}}} }}} \right)} \right]} \Rightarrow \left( \begin{gathered}
    \text{rewrite the bottoms sums over} \hfill \\
    \text{a common denominator,}\;\left( k \right) \hfill \\
    \end{gathered} \right) \Rightarrow \left( {\frac{1}
    {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}
    {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}
    {k}} \right)^2 } }}
    {{k - 1}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}
    {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}
    {k}} \right)^2 } }}
    {{k - 1}}} }}} \right)} \right]} \hfill \\
    \Rightarrow \left( \begin{gathered}
    \text{Expand the bottom sums to factor } \hfill \\
    \text{out }\left( k \right).\text{ Because we have a sum of } \hfill \\
    squares\text{, we will factor out a }\left( {k^2 } \right) \hfill \\
    \end{gathered} \right) \Rightarrow \left( {\frac{1}
    {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}
    {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}
    {{k^2 \left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}
    {{k\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}
    {{k^2 \left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \left( \begin{gathered}
    \text{From this, we can solve the square root of} \hfill \\
    \;\left( {k^2 } \right)\;\text{to place this value outside the radical } \hfill \\
    \end{gathered} \right) \Rightarrow \hfill \\
    \Rightarrow \left( {\frac{1}
    {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}
    {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}
    {{\left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}
    {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}
    {{\left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \left( \begin{gathered}
    \text{The }\left( k \right)\text{values will cancel } \hfill \\
    \text{as we multiply:}\;k \cdot \sqrt {\frac{1}
    {{k^2 }}} = 1 \hfill \\
    \end{gathered} \right) \Rightarrow \left( {\frac{1}
    {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\left( {\frac{{x_n k - \sum\limits_{n = 1}^k {x_n } }}
    {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } }}
    {{\left( {k - 1} \right)}}} }}} \right)\left( {\frac{{y_n k - \sum\limits_{n = 1}^k {y_n } }}
    {{\sqrt {\frac{{\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } }}
    {{\left( {k - 1} \right)}}} }}} \right)} \right]} \Rightarrow \hfill \\
    \Rightarrow \left( \begin{gathered}
    \text{We express the bottom radical as a} \hfill \\
    \text{a radical numerator divided by }\sqrt {k - 1} ; \hfill \\
    \text{therefore, we can place }\sqrt {k - 1} \text{ in the } \hfill \\
    \text{ numerator of the entire equation } \hfill \\
    \end{gathered} \right) \Rightarrow \left( {\frac{1}
    {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left\{ {\left[ {\frac{{\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\sqrt {k - 1} }}
    {{\sqrt {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } }}} \right]\left[ {\frac{{\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)\sqrt {k - 1} }}
    {{\sqrt {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } }}} \right]} \right\}} \Rightarrow \left( \begin{gathered}
    \text{We can multiply both z - score expressions} \hfill \\
    \text{to simplify the sum and remove }\sqrt {k - 1} \hfill \\
    \end{gathered} \right) \Rightarrow \hfill \\
    \Rightarrow \left( {\frac{1}
    {{k - 1}}} \right)\sum\limits_{n = 1}^k {\left[ {\frac{{\left( {k - 1} \right)\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)}}
    {{\left( {\sqrt {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } } \right)\left( {\sqrt {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } } \right)}}} \right]} = \sum\limits_{n = 1}^k {\left\{ {\frac{{\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)}}
    {{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } \right]\left[ {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } \right]} }}} \right\}} = = \frac{{\sum\limits_{n = 1}^k {\left[ {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)} \right]} }}
    {{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n k - \sum\limits_{n = 1}^k {x_n } } \right)^2 } } \right]\left[ {\sum\limits_{n = 1}^k {\left( {y_n k - \sum\limits_{n = 1}^k {y_n } } \right)^2 } } \right]} }} \Rightarrow \hfill \\
    \Rightarrow \left( {\text{Substitute}\left\{ \begin{gathered}
    a = \sum\limits_{n = 1}^k {x_n } \hfill \\
    b = \sum\limits_{n = 1}^k {y_n } \hfill \\
    \end{gathered} \right\}} \right) \Rightarrow \frac{{\sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k^2 - k\left( {y_n a + x_n b} \right) + ab} \right]} }}
    {{\sqrt {\left[ {\sum\limits_{n = 1}^k {\left( {x_n^2 k^2 - 2ax_n k + a^2 } \right)} } \right]\left[ {\sum\limits_{n = 1}^k {y_n^2 k^2 - 2by_n k + b^2 } } \right]} }} = \Rightarrow \left( \begin{gathered}
    \text{We can take out}\left( {ab} \right)\text{from the } \hfill \\
    \text{numerator and the}\left( {a^2 ,b^2 } \right)\text{from the } \hfill \\
    \text{denominator as}\left( {kab} \right)\text{and}\left( {a^2 k,b^2 k} \right) \hfill \\
    \end{gathered} \right) \Rightarrow \frac{{kab + \sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k^2 - k\left( {y_n a + x_n b} \right)} \right]} }}
    {{\sqrt {\left[ {a^2 k + \sum\limits_{n = 1}^k {\left( {x_n^2 k^2 - 2ax_n k} \right)} } \right]\left[ {b^2 k + \sum\limits_{n = 1}^k {\left( {y_n^2 k^2 - 2by_n k} \right)} } \right]} }} \Rightarrow \hfill \\
    \Rightarrow \left( \begin{gathered}
    \text{Then, we can factor out the} \hfill \\
    \left( k \right)\text{from the sum expressions} \hfill \\
    \end{gathered} \right) \Rightarrow = \frac{{kab + k\sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k - \left( {y_n a + x_n b} \right)} \right]} }}
    {{\sqrt {\left[ {a^2 k + k\sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 k + k\sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} = = \frac{{k\left\{ {ab + \sum\limits_{n = 1}^k {\left[ {\left( {x_n y_n } \right)k - \left( {y_n a + x_n b} \right)} \right]} } \right\}}}
    {{\sqrt {k^2 \left[ {a^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} \Rightarrow \left( \begin{gathered}
    \text{The}\left( {k^2 } \right)\text{in the denominator} \hfill \\
    \text{radical can be taken out and} \hfill \\
    \text{will cancel with the }\left( k \right) \hfill \\
    \text{in the equation numerator} \hfill \\
    \end{gathered} \right) \hfill \\
    = \frac{{ab + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n b - y_n a} \right)} }}
    {{\sqrt {\left[ {a^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n a} \right)} } \right]\left[ {b^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n b} \right)} } \right]} }} = \frac{{ab + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n b - y_n a} \right)} }}
    {{\sqrt {\left\{ {a^2 + \sum\limits_{n = 1}^k {\left[ {x_n \left( {x_n - 2a} \right)} \right]} } \right\}\left\{ {b^2 + \sum\limits_{n = 1}^k {\left[ {y_n \left( {y_n - 2b} \right)} \right]} } \right\}} }} \Rightarrow \left( \begin{gathered}
    \text{Replace variables }\left( {a,b} \right) \hfill \\
    \text{with their original assignments} \hfill \\
    \end{gathered} \right) \Rightarrow \hfill \\
    \frac{{\left( {\sum\limits_{n = 1}^k {x_n } } \right)\left( {\sum\limits_{n = 1}^k {y_n } } \right) + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n \sum\limits_{n = 1}^k {y_n } - y_n \sum\limits_{n = 1}^k {x_n } } \right)} }}
    {{\sqrt {\left[ {\left( {\sum\limits_{n = 1}^k {x_n } } \right)^2 + \sum\limits_{n = 1}^k {\left( {x_n^2 k - 2x_n \sum\limits_{n = 1}^k {x_n } } \right)} } \right]\left[ {\left( {\sum\limits_{n = 1}^k {y_n } } \right)^2 + \sum\limits_{n = 1}^k {\left( {y_n^2 k - 2y_n \sum\limits_{n = 1}^k {y_n } } \right)} } \right]} }} = \frac{{\left( {\sum\limits_{n = 1}^k {x_n } } \right)\left( {\sum\limits_{n = 1}^k {y_n } } \right) + \sum\limits_{n = 1}^k {\left( {x_n y_n k - x_n \sum\limits_{n = 1}^k {y_n } - y_n \sum\limits_{n = 1}^k {x_n } } \right)} }}
    {{\sqrt {\left\{ {\left( {\sum\limits_{n = 1}^k {x_n } } \right)^2 + \sum\limits_{n = 1}^k {\left[ {x_n \left( {x_n - 2\sum\limits_{n = 1}^k {x_n } } \right)} \right]} } \right\}\left\{ {\left( {\sum\limits_{n = 1}^k {y_n } } \right)^2 + \sum\limits_{n = 1}^k {\left[ {y_n \left( {y_n - 2\sum\limits_{n = 1}^k {y_n } } \right)} \right]} } \right\}} }} \hfill \\
    \end{gathered} [/tex]
     
  4. Apr 6, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    1.It's in the wrong forum;
    2.Don't abuse the latex code and mess up page layout.I don't wanna change the # of pixels only to see your nonsense...
    3.Use \\ or simply break the code (use [ tex ] & [ /tex ] tags more frequently.

    Oh.4.Did i say that your post is a monstruosity which should be deleted?

    Daniel.
     
  5. Apr 6, 2005 #4

    Evo

    User Avatar

    Staff: Mentor

    Be nice!

    But that is the largest I've seen.

    bomba, I suggest you experiment on posts and use the "preview post" feature. Perhaps if you look at some posts by others that use LaTex and hit the "quote" button, as if you were going to reply, you will see how they did the formatting.
     
  6. Apr 6, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Sorry,Evo.I'm not (a) Saint :surprised: I overreact,sometimes...:redface:

    Daniel.

    P.S.The monstrusity part is not real,it's SURREAL :tongue2:
     
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