Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/138

distance of the body from the foot of the perpendicular from the sun upon the line of sight. If we set

$q_{1}=\rho _{1}+({\mathfrak {E}}_{1}.{\mathfrak {F}}_{1}),$$q_{2}=\rho _{2}+({\mathfrak {E}}_{2}.{\mathfrak {F}}_{2}),$$q_{3}=\rho _{3}+({\mathfrak {E}}_{3}.{\mathfrak {F}}_{3}),$

(9)

$p_{1}^{2}={\mathfrak {E}}_{1}^{2}-({\mathfrak {E}}_{1}.{\mathfrak {F}}_{1})^{2},$$p_{2}^{2}={\mathfrak {E}}_{2}^{2}-({\mathfrak {E}}_{2}.{\mathfrak {F}}_{2})^{2},$$p_{3}^{2}={\mathfrak {E}}_{3}^{2}-({\mathfrak {E}}_{3}.{\mathfrak {F}}_{3})^{2},$

(10)

equations (8) become

$r_{1}^{2}=q_{1}^{2}+p_{1}^{2},$$r_{2}^{2}=q_{2}^{2}+p_{2}^{2},$$r_{3}^{2}=q_{3}^{2}+p_{3}^{2}.$

(11)

Let us also set, for brevity,

${\mathfrak {S}}_{1}=A_{1}\left(1+{\frac {B_{1}}{r_{1}^{3}}}\right)({\mathfrak {E}}_{1}+\rho _{1}{\mathfrak {F}}_{1}),$${\mathfrak {E}}_{2}=-\left(1-{\frac {B_{2}}{r_{2}^{3}}}\right)({\mathfrak {E}}_{2}+\rho _{2}{\mathfrak {F}}_{2}),$	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \ \end{matrix}}\right\}\,}$ (12)
${\mathfrak {E}}_{3}=A_{3}\left(1+{\frac {B_{3}}{r_{3}^{3}}}\right)({\mathfrak {E}}_{3}+\rho _{3}{\mathfrak {F}}_{3}).$

Then ${\mathfrak {S}}_{1},{\mathfrak {S}}_{2},{\mathfrak {S}}_{3}$ may be regarded as functions respectively of $\rho _{1},\rho _{2},\rho _{3},$ therefore of $q_{1},q_{2},q_{3},$ and if we set

${\mathfrak {S}}'={\frac {d{\mathfrak {S}}_{1}}{dq_{1}}},$${\mathfrak {S}}''={\frac {d{\mathfrak {S}}_{2}}{dq_{2}}},$${\mathfrak {S}}'''={\frac {d{\mathfrak {S}}_{3}}{dq_{3}}},$

(13)

and

${\mathfrak {S}}={\mathfrak {S}}_{1}+{\mathfrak {S}}_{2}+{\mathfrak {S}}_{3},$

(14)

we shall have

$d{\mathfrak {S}}={\mathfrak {S}}'dq_{1}+{\mathfrak {S}}''dq_{2}+{\mathfrak {S}}'''dq_{3}.$

(15)

To determine the value of ${\mathfrak {S}}',$ we get by differentiation

${\mathfrak {S}}'=A_{1}\left(1+{\frac {B_{1}}{r_{1}^{3}}}\right){\mathfrak {F}}_{1}-A_{1}{\frac {3B_{1}}{r_{1}^{4}}}{\frac {dr_{1}}{dq_{1}}}({\mathfrak {E}}_{1}+\rho _{1}{\mathfrak {F}}_{1}).$

(16)

But by (11)

${\frac {dr_{1}}{dq_{1}}}={\frac {q_{1}}{r_{1}}}\cdot$

(17)

Therefore

${\mathfrak {S}}'=A_{1}\left(1+{\frac {B_{1}}{r_{1}^{3}}}\right){\mathfrak {F}}_{1}-{\frac {3B_{1}q_{1}}{r_{1}^{5}(1+B_{1}r_{1}^{-3})}}{\mathfrak {S}}_{1}$	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \\\ \\\ \ \end{matrix}}\right\}\,}$ (18)
${\mathfrak {S}}''=-\left(1+{\frac {B_{2}}{r_{2}^{3}}}\right){\mathfrak {F}}_{2}+{\frac {3B_{2}q_{2}}{r_{2}^{5}(1+B_{2}r_{2}^{-3})}}{\mathfrak {S}}_{2}$
${\mathfrak {S}}'''=A_{3}\left(1+{\frac {B_{3}}{r_{3}^{3}}}\right){\mathfrak {F}}_{3}-{\frac {3B_{3}q_{3}}{r_{3}^{5}(1+B_{3}r_{3}^{-3})}}{\mathfrak {S}}_{3}$

Now if any values of $q_{1},q_{2},q_{3}$ (either assumed or obtained by a previous approximation) give a certain residual ${\mathfrak {S}}$ (which would be zero if the values of $q_{1},q_{2},q_{3}$ satisfied the fundamental equation), and we wish to find the corrections $\Delta q_{1},\Delta q_{2},\Delta q_{3}$ which must be added to $q_{1},q_{2},q_{3}$ to reduce the residual to zero, we may apply equation (15) to these finite differences, and will have approximately, when these differences are not very large,

$-{\mathfrak {S}}={\mathfrak {S}}'\Delta q_{1}+{\mathfrak {S}}''\Delta q_{2}+{\mathfrak {S}}'''\Delta q_{3}.$

(19)