SECT. II.] GENERAL SOLUTION. 245 



a 3 = (A t sin 2t*, 4 B l cos zty 



- versin ?&amp;lt; 2 



4- (^4 2 sin 2w a 4 1? 3 cos 2w 2 ) e 



- ^ versin */ 3 



4 (J. 3 sin 2?/ 3 4 # 3 cos 2w a ) e f 

 + &c. ; 



a n = (^ sin (n 1)^4- ^ cos (n 1) u t ] e m 



versin w 2 



+ [A a sin (n - 1) w a + B 3 cos (?i - 1) u 9 ] e 



- ?** versin 3 



4 {-4. sin (n 1) w a 4 B a cos (*i 1) iij e 

 4&c. 



265. If we suppose the time nothing, the values a v a 2 , cr 3 , . . . a n 

 must become the same as the initial values a lt a 2 ,a 3 , ... a n . We 

 derive from this n equations, which serve to determine the coeffi 

 cients A v B V -A 2 , B 2 , A y B 3 It will readily be perceived that 



the number of unknowns is always equal to the number of equa 

 tions. In fact, the number of terms which enter into the value 

 of one of these variables depends on the number of different 

 quantities versin u l} versin w 2 , versing, &c., which we find on 

 dividing the circumference 2?r into n equal parts. Now the 



2-7T 2-7T 2-7T 



number of quantities versin , versin 1 , versin 2 , &c., 



n n n 



is very much less than n, if we count only those that are 

 different. Denoting the number n by 2^ 4 1 if it is odd, 

 and by 2i if it is even, i 4 1 always denotes the number 

 of different versed sines. On the other hand, when in the 



, .... . 2?r . n 2-7T . 2-7T p 



series of quantities versin , versin 1 , versm 2 , &c., 



n n n 



9 



we come to a versed sine, versin X , equal to one of the former 



versin V , the two terms of the equations which contain this 



versed sine form only one term ; the two different arcs % and 

 x-, which, have the same versed sine, have also the same cosine, 

 and the sines differ only in sign. It is easy to see that the 

 arcs Ux and u x &amp;gt;, which have the same versed sine, are such that 



