29] DUE TO SECULAR CHANGE OF THE PLANE OF THE ECLIPTIC. 239 



and hence that 



^1= +0-00798,55 cos (-nt + 2n't) 



+ 1-07986,87 cosnt 



+ 0-00466,54 cos (3nt - 2n't) 



+ 0-00002,98 cos(5nt-4n't), 



and also, as in the paper referred to above, 



^ + ^ = 1 -17804,45 + 0-02523,37 cos 2 (nt - n't) + 0'00025,16 cos 4 (nt - n't), 

 r* T 



Hence the equation to be solved becomes 



'& 

 dt* 



JaSv 



2 +Sz [1-17804,45 + 0-02523,37 cos 2 (nt -n't) + 0'00025, 16 cos 4 (nt-n't)~] 



+ a [0-01597,1 cos (-nt + 2n't) + 2'1 5973,7 cos nt 



+ 0-00933,1 cos (3nt - 2 n't) + O'OOOOG.O x cos (5nt - n'i)~\ = 0. 



Assume 



Sz = ct> [c_ 3 cos ( 3nt + 4n't) + c_ l cos ( nt + 2n't) + c, cos nt 

 + c 3 cos (3nt 2n f t) + c 5 cos (5nt n't}~\, 



then, by substituting for 8z and equating coefficients of similar terms, we 

 have 



- 7 -34339,8c_ 3 + G'01261, 7c_, + 0-00012,60, =0, 



0-01261, 7c_ 3 + 0-33320,60.., + 0-01261,7c,+ 0'00012,6c 3 + 0'01597,lo> =0, 

 0-00012,6c_ 3 + 0-01261, 7c_, + 0-0098 1,00,+ 0-01261,7c. i + 0-00012,6c 5 



+ 2'15973,7w =0, 



+ 0-00012,6c_ 1 + 0-01261,7c, - 8-31358, 5c 3 + 0-01261,7c 5 + 0-00933,l =0, 

 + 0'00012,6c, +0-01261, 7c. ! -24-63698,lc 5 + 0-00006,Ow =0. 



If we find the values of c_ 3 , c. s , and c 6 from these equations in terms 

 of the two remaining coefficients c_ l and c,, which can be advantageously 

 done, since c_ 3 has a large coefficient in the first equation, c 3 in the fourth 

 and c 5 in the fifth equation, we find 



c_ 3 = 0-00171,8c_ 1 + 0-00001,72 c lt 



c 3 =0-00001, Sc^ + O'OOlSl^ ^ + 0-00112,20), 



c, = +0-00000,589c, + 0-00000,3o), 



