LaTeX What's Causing the Display Issue with My LaTex Code?

bomba923 · Apr 6, 2005

What's wrong with my code? Why would LaTex show my work?

\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}

What does it mean;--what's "invalid" here?

bomba923 · Apr 6, 2005

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!)
\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}

dextercioby · Apr 6, 2005

1.It's in the wrong forum;
2.Don't abuse the latex code and mess up page layout.I don't want to 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 · Apr 6, 2005

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 · Apr 6, 2005

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

Daniel.

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

LaTeX What's Causing the Display Issue with My LaTex Code?

Similar threads

Hot Threads

Simple problems that Wolfram can't do

How do I cite a chapter from lecture notes?

Wrapfig leaves text permanently curtailed thereafter

GNU Octave vs Wolfram Alpha

Mathematica NDSolve with Quantities

Recent Insights

Insights Quantum Entanglement is a Kinematic Fact, not a Dynamical Effect

Insights What Exactly is Dirac’s Delta Function? - Insight

Insights Relativator (Circular Slide-Rule): Simulated with Desmos - Insight

Insights Fixing Things Which Can Go Wrong With Complex Numbers

Insights Fermat's Last Theorem

Insights Why Vector Spaces Explain The World: A Historical Perspective