Sckuherf 8 investigation of Kepler- s Problem, 31 



numbers becomes negative, the corresponding coeiBcient, J or N, 

 will be = 0. Therefore, whatever may be the numbers i and ??, 

 there will never be found more than two values of N, sometimes 

 but one, and sometimes none at all ; as will appear by the follow- 

 ing example. 



§ 24?. Let the coefficient of e^° sin 2 ^ be sought. Then 



have 



a 



10, m 



2, s==4, AW 



+ -^(U9),A- = ^(^20), 



I.'2,..l0 



£20 



AW 



1 



2io-M.2....n 



L 2 



N 



2 , N„_i^ /._j[\»»"1 



± 



J 



and the required coefficient A = A^°^ + A^"'^ + A^'>, Let us abridge 

 this notation by writing J for 



JlO— i— » -ikJ j.__ N„_;_2 T^-;., J. _ N«4-t-J 



, N for 



-' , N' for 



N 



and N" for ^S the latter being used only when i is greater than 



m 



or S, in which case N^ disappears, because ^r or n + i 



S 



would be greater than n (§ 23). Therefore, 



A« 



J 



^,,^;rix:^.[±N^±N±N"(-ir]. 



§ 25, The above formulas sjive Aw= + 



10.9.8.7 



1 



I 



a 



1.2.3.4. 1.2.3... lO 



• ■•t 



"Tciv 



2^3^.5 



AC") 



+ 



2.7 8.9 1 



1.5i.3.4 2^° 



+ 



21 



2 



9 



As to the term A% both numbers i and n must be at the same time 



* 



even or odd, because the exponent of N, —5- , must be an integer ; 

 moreover, n cannot be greater than 10 — i, on account of the ex- 



ponent of J, 



10 



n 



2 



Making then, 



3)i 



1, n 

 3, n 



if n 



if n 

 4 



3, w 



4, n 

 3, n 



5, n 



6, n 



6, n 



7, n 



7; 



9: 



»■ 



