%This command provides the math for the definition around %pi on first horizontal part of the first column %of page 6. % %This command has one parameter: % 1)The width of the math text \newcommand\TSixPi[1]{% \parbox[t]{#1}{% \TSixPiFontSize \DisplaySpace{\TSixDisplaySpace}{\TSixDisplayShortSpace} \TSixTitle{Wallis' identity:} \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus .5ex}{\SmallChar}{\StyleWithoutNumber} \Fm{\pi = 2 \cdot \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdots}{% 1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdots} } \end{DisplayFormulae} \TSixTitle{Brouncker's continued fraction expansion:} \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus .5ex}{\SmallChar}{\StyleWithoutNumber} \Fm{\tfrac {\pi}{4} = 1 + \frac{1^2}{2 + \frac{3^2}{2 + \frac{5^2}{2 + \frac{7^2}{2 + \cdots}}}} } \end{DisplayFormulae} \TSixTitle{Gregrory's series:} \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus .5ex}{\SmallChar}{\StyleWithoutNumber} \Fm{\tfrac {\pi}{4} = 1 - \tfrac{1}{3} + \tfrac{1}{5} - \tfrac{1}{7} + \tfrac{1}{9} - \cdots} \end{DisplayFormulae} \TSixTitle{Newton's series:} \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus .5ex}{\SmallChar}{\StyleWithoutNumber} \Fm{\tfrac {\pi}{6} = \frac{1}{2} + \frac{1}{2\cdot 3 \cdot 2^3} + \frac{1 \cdot 3}{2\cdot 4 \cdot 5 \cdot 2^5} + \cdots } \end{DisplayFormulae} \TSixTitle{Sharp's series:} \begin{DisplayFormulae}{1}{0pt}{3ex plus 1ex minus .5ex}{\SmallChar}{\StyleWithoutNumber} \Fm{\tfrac {\pi}{6} = \frac{1}{\sqrt{3}} \Big(1 - \frac{1}{3^1 \cdot 3 } + \frac{1}{3^2 \cdot 5 } - \frac{1}{3^3 \cdot 7 } + \cdots \Big) } \end{DisplayFormulae} \TSixTitle{Euler's series:} \begin{DisplayFormulae}{1}{0pt}{4ex plus 1ex minus .5ex}{\SmallChar}{\StyleWithoutNumber} \Fm{\tfrac{\pi^2}{6} = \tfrac{1}{1^2} + \tfrac{1}{2^2} + \tfrac{1}{3^2} + \tfrac{1}{4^2} + \tfrac{1}{5^2} + \cdots } \Fm{\tfrac{\pi^2}{8} = \tfrac{1}{1^2} + \tfrac{1}{3^2} + \tfrac{1}{5^2} + \tfrac{1}{7^2} + \tfrac{1}{9^2} + \cdots } \Fm{\tfrac{\pi^2}{12} = \tfrac{1}{1^2} - \tfrac{1}{2^2} + \tfrac{1}{3^2} - \tfrac{1}{4^2} + \tfrac{1}{5^2} - \cdots } \end{DisplayFormulae} } }