Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/168

Using an expansion ending with $t^{2}$ we can only satisfy one condition relating to acceleration, say the second. This will give

${\frac {\tau _{1}}{\tau _{2}}}{\mathfrak {R}}_{1}-\left(1-{\frac {\tau _{1}\tau _{3}}{2r_{2}^{3}}}\right){\mathfrak {R}}_{2}+{\frac {\tau _{3}}{\tau _{2}}}{\mathfrak {R}}_{3}=0.$

III

(Gauss uses virtually

${\frac {\tau _{1}}{\tau _{2}}}{\mathfrak {R}}_{1}-{\frac {{\mathfrak {R}}_{2}}{1+{\frac {\tau _{1}\tau _{3}}{2r_{2}^{3}}}}}+{\frac {\tau _{3}}{\tau _{2}}}{\mathfrak {R}}_{3}=0,$

which is a little more convenient, but not, I think, generally quite so accurate.)

Writing an equation analogous to III for the earth and subtracting from (7), Mem. Nat. Acad., we have

${\frac {\tau _{1}}{\tau _{2}}}\rho _{2}{\mathfrak {F}}_{1}-\left(1-{\frac {\tau _{1}\tau _{3}}{2r_{2}^{3}}}\right)\rho _{2}{\mathfrak {F}}_{2}+{\frac {\tau _{3}}{\tau _{2}}}\rho _{3}{\mathfrak {F}}_{3}={\frac {\tau _{1}\tau _{3}}{2}}\left({\frac {1}{{\mathfrak {E}}_{2}^{3}}}-{\frac {1}{r_{2}^{3}}}\right){\mathfrak {E}}_{2},$

which gives, on multiplication by ${\mathfrak {F}}_{1}\times {\mathfrak {F}}_{3}$ and ${\mathfrak {F}}_{2}\times {\mathfrak {E}}_{2},$ theorems of Olbers and Lambert.

It is evident that in general the error in I is of the fifth order, in IIa, IIb, IIc of the fourth, and in III of the third. But for equal intervals, the error in I is of the sixth order, and in III of the fourth. And when $\tau _{2}^{2}+\tau _{3}^{2}-3\tau _{1}^{2}=0,$ IIa becomes identical with I, and its error is of the fifth order.

The same is true of IIc in the corresponding case. It follows that when the intervals are nearly as 5:8 we should use IIa or IIb instead of I. This will evidently abbreviate the solution given above as only one of the quantities $r_{1},r_{3},$ is to be used.

The formulæ IIa, IIb, IIc may also be obtained by the following method, which will show their relative accuracy.

The interpolation formula

${\frac {\tau _{1}}{\tau _{2}}}{\frac {{\mathfrak {R}}_{1}}{r_{1}^{3}}}-{\frac {{\mathfrak {R}}_{2}}{r_{2}^{3}}}+{\frac {\tau _{3}}{\tau _{2}}}{\frac {{\mathfrak {R}}_{3}}{r_{3}^{3}}}=0$

has an error evidently of the second order. If we multiply by ${\frac {\tau _{2}^{2}+\tau _{3}^{2}-3\tau _{1}^{2}}{24}}$ and subtract from I, we get IIa. So if we multiply by

${\frac {\tau _{3}^{2}+\tau _{1}^{2}-3\tau _{2}^{2}}{24}}$⁠or⁠${\frac {\tau _{1}^{2}+\tau _{2}^{2}-3\tau _{3}^{2}}{24}}$

we get IIb or IIc. The errors due to using one of these equations instead of I are therefore proportional to these multipliers and very unequal. Again, in case of equal intervals, IIa and IIc become identical with III. There is, therefore, no reason for using IIa or IIc when the intervals are nearly equal. IIb is in this case much less accurate than III.