64 Prof. J. C. Adams on Legendres Coefficients. [Jan. 31, 



By a slight reduction the coefficient of P n becomes 



n(n + l) 



Hence 



3 (* + !)(* + 2) . n(»+l) 

 r 2 r n - 2 ( 2w + i) (272 + 3) r w+2i ~ (2»-l) (2» + 3)^» 



, 3 (n-l)n 

 " 1 "2 (2w-l) (2» + l) r - 



Again, putting 72=2 in our original formula, we have 



PA=| A J? t P.-|p i P. 



_5 (72 + 1) (w+2) p 5 



~2 (2t2 + 1) (272 + 3)^" +2+ 3 (2t2-1) (2» + 3)^ P - 



5 (n-l)n 2 w + l 2 72 



+ 2 (2*2-1) (2n + l)^" 2 3 2w + r " +1 ""3 2» + 1 



Substitute for ^P B+2 , /*P B and /iP nm2 their equivalents as before 



. P p 5 (72 + I) (t2 + 2) fc + 3) 

 • ' r 3^»- 2 (2,2 + 1) (2t?, + 3) (2t2 + 5) M+3 



(5 (t2 + 1)(t2 + 2) ?i + 2 5 72(72 + !) 72 + 1 _2 72-4-1 1 



12 (2t2 + 1) (2n + 3) 27^ + 5 3(2w-l)(2»+8).2» + l 3 2» + lJ n+l 



J5 77,(72 + 1) ^ ,5 (72 — 1)72- 72 — 1 _2 72 1 



+ l3(272-l)(27i + 3) 2^+1 2(272-1) (2t2 + 1) 2^3 3 2»+T/ 



5 (n-2) (72-1)72 



^2 (2/2-3) (272-1) (27?- + 1)" "- 3 

 By rediiction the coefficient of P n+1 in this expression becomes 



3 72(72 + 1) (fl + 2) 



2 (272-1) (272 + 1) (2/^ + 5) 

 and similarly the coefficient of P n _j becomes 



3 (72-1) 72(72+ 1) 



2 (272-3) (272+1) (272 + 3) 



Hence we have 



