Example. - Let or 0.0333585. Then the computation for is as follows:-
First term of 0.0621069
Failed to parse (unknown function "\multirow"): {\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \multirow{4}{*}{}\begin{array}{l} N= \\ \log \lambda k \\ \frac{3}{2} \log b \\ \end{array} & & \therefore & 0.0129940 & \log N & . & & . & 8.1137429 \\ \hline & . . & - . & 7.8733658\} & & & & & \\ \hline & \cdot & . . & 0.9030900\} & Difference & \includegraphics[max width=\textwidth]{2024_02_19_0baa176114aa2645ceebg-048} & - & & 6.9702758 \\ \hline & & & & \begin{array}{l} \log t \\ t= \\ \end{array} & & & & \begin{array}{r} 1.1434671 \\ 13.91448 \\ \end{array} \\ \hline \end{array}}
24.
If it has been decided to carry out the calculation with hyperbolic logarithms, it is best to employ the auxiliary quantity which will be determined by equation III., and thence by XI.; the semi-parameter will be computed from the radius vector, or inversely the latter from the former by formula VIII.; the second part of can, if desired, be obtained in two ways, namely, by means of the formula hyp. and by this, hyp. hyp. Moreover it is apparent that here where the quantity