162 PROFESSOR KELLAND, ON THE 



Adopting the general expression, we have now to find the value of 

 fsin rmx\ 2 (0 - fl" 1 ) 2 (&* ~ •" 



, . so (smrmx\ 2 (0 - 0-') 2 [9* - 0- r "Y 



( sin *) llhTTFJ or i— U^H • 



Assume 



/firm __ *J-"n\ 2 



fL. — ^_j = 0iK— 1) + ^©trc— »> + + p-MtV", 



.-. 2 ™- 2 + 0-2™ = ^™ + « 2 2 '' ( '"- l) + a30 2r( '"- 2) + ... 



+ 2r(m - 2) + ... 

 which by equating coefficients, gives 



ch - 2 = 0, 



fl 3 — 2« 2 + 1 = 0, 



a 4 — 2a 3 + « 2 = 0, 



#m + 1 — 2 Cf m + fl! m _ 1 = — 2, 



=0: 



hence a 2 — 2, 03= 3, a 4 = 4, &c. 



and a m+ i = 2»a - (m — 1) — 2 

 = »a — 1 



that is, the coefficients form an arithmetic series, increasing up to m, 

 and then diminishing down to 1. 



By multiplying by the factor (0 — 9~ l f, we obtain 



02r(m-l)+2 1 2Q 2r(m-2)+2 X.,.4 /ftg 2 + . . . 4. 0-2r(m- l)+2 



+ 0- { *^ ui) + ZQ-V'"^** + ...+ mQ- 2 + ...+ d*'^-" ~ 2 



= 2 {cos (2/-/W - 1 + 2)# + 2cos(2rw - 2 + 2)#+... 

 + m cos 2a; + (m — 1) cos (2r — 2) x +...+ cos [2r(?« - 1) - 2]x\ 

 — 4 {cos 2r(m — l)x + 2 cos 2r(m - 2)x +...+ (m ~ 1) cos 2rx] 



- 2m = K. 



