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%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}
}
}
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