Why won't LaTex display this?

bomba923

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?

bomba923

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]

dextercioby

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.

Evo

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.

dextercioby

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

Daniel.

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