152 VECTOK METHOD IN THE DETEEMINATION OF OEBITS. 



Using an expansion ending with t 2 we can only satisfy one con- 

 dition relating to acceleration, say the second. This will give 



T i T 3 



TTT 



. ^,,. 



T 2 ^'2 ' T 2 



(Gauss uses virtually 



I I T 1 T 3 



h 



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 



which gives, on multiplication by f^x&g and $ 2 X&2, theorems of 

 Olbers and Lambert. 



It is evident that in general the error in I is of the fifth order, in 

 Ila, 116, lie 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 T 2 2 +r 3 2 3x^ = 0, Ilct becomes identical with I, and its 

 error is of the fifth order. 



The same is true of lie in the corresponding case. It follows that 

 when the intervals are nearly as 5:8 we should use Ila or 116 instead 

 of I. This will evidently abbreviate the solution given above as 

 only one of the quantities r lt r 3 is to be used. 



The formula Ila, 116, lie may also be obtained by the following 

 method, which will show their relative accuracy. 



The interpolation formula 



Tnn O tyt O _. M 



2 'I '2 T 2 '3 



has an error evidently of the second order. If we multiply by 



T 2_j_ T 2__3 T 2 



-2 |-r and subtract from I, we get Ila. So if we multiply by 



! frr 2 



- or 



24 24 



we get 116 or lie. 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, Ila and lie become identical with 

 III. There is, therefore, no reason for using Ila or lie when the 

 intervals are nearly equal. 116 is in this case much less accurate 

 than III. 



