Thomas – Just another demography site

MathType Translation formats
Formulas were no longer shown, but were repaired by using QuickLatex and adding on top of each page.

Translation into WP Format

$\begin{array}{l}{{\sum\nolimits_{{j=1}}^{m}{M}}_{{ij}}}(x)=\left\{ {{{t}_{i}}(x)} \right\}+\sum\nolimits_{{j\ne i}}^{m}{{-{{e}_{{ij}}}}}(x)\\={{m}_{i}}(x)+\sum\nolimits_{{j\ne i}}^{m}{{{{e}_{{ij}}}}}(x)+\sum\nolimits_{{j\ne i}}^{m}{{-{{e}_{{ij}}}}}(x)\\={{m}_{i}}(x)+\sum\nolimits_{{j\ne i}}^{m}{{{{e}_{{ij}}}}}(x)-\sum\nolimits_{{j\ne i}}^{m}{{{{e}_{{ij}}}}}(x)\\={{m}_{i}}(x)\end{array}$

AMS Latex (line breaks removed)

$\begin{gathered}{\sum\nolimits_{j = 1}^m M _{ij}}(x) = \left\{ {{t_i}(x)} \right\} + \sum\nolimits_{j \ne i}^m { - {e_{ij}}} (x) \\ = {m_i}(x) + \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) + \sum\nolimits_{j \ne i}^m { - {e_{ij}}} (x) \\ = {m_i}(x) + \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) - \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) \\ = {m_i}(x) \\ \end{gathered}$

AMS Tex (Line breaks removed)

$\displaylines{{\sum\nolimits_{j = 1}^m M _{ij}}(x) = \left\{ {{t_i}(x)} \right\} + \sum\nolimits_{j \ne i}^m { - {e_{ij}}} (x) \cr= {m_i}(x) + \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) + \sum\nolimits_{j \ne i}^m { - {e_{ij}}} (x) \cr= {m_i}(x) + \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) - \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) \cr= {m_i}(x) \cr}$

MathJax:Latex

$\begin{array}{c}{\sum\nolimits_{j = 1}^m M _{ij}}(x) = \left\{ {{t_i}(x)} \right\} + \sum\nolimits_{j \ne i}^m { - {e_{ij}}} (x)\\ = {m_i}(x) + \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) + \sum\nolimits_{j \ne i}^m { - {e_{ij}}} (x)\\ = {m_i}(x) + \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) - \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x)\\ = {m_i}(x)\end{array}$

Alignment added

AMS Latex (gathered -> aligned

$\begin{aligned} {\sum\nolimits_{j = 1}^m M _{ij}}(x) &= \left\{ {{t_i}(x)} \right\} + \sum\nolimits_{j \ne i}^m { - {e_{ij}}} (x) \\ &= {m_i}(x) + \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) + \sum\nolimits_{j \ne i}^m { - {e_{ij}}} (x) \\ &= {m_i}(x) + \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) - \sum\nolimits_{j \ne i}^m {{e_{ij}}} (x) \\ &= {m_i}(x) \\ \end{aligned}$

$\displaystyle _{l}^{\downarrow }S(0)=L(0){{\left[ {nl(0)} \right]}^{{-1}}}$

$\displaystyle _{\downarrow }^{\downarrow }S_{\downarrow }^{\downarrow }(0)=L(0){{\left[ {nl(0)} \right]}^{{-1}}}$

Author: Thomas

Formulas

Testing math