Solution of the Electromagnetic Relations. 263 



I£ fi = cos 0, the equation for B t becomes 



Now i z _ m « 



/tP? = ^j (« - »+ 1 ■ *Cn + « + m • PT-.) • 



Also 



1 — /A" . -j— = W + TO . JT „_x — W/*r M 



= 5 z \n+l . n + m . r n „i- n . n — m + lr n+1 J-. 



zw + 1 



Thus on reduction 

 2;i+l . 4~ • (l-/* 2 ■ P ") = nr-1 . w + to . P^-n + 2 . n-m + lPr+i. 

 The substitution 



5 <in 0=pp;T-i+£Pr+i 



is therefore suggested, giving 



i ■ 1 -^i. (Sl Si ° *> - I^? • 8l Sin »=-» • »-l :PP~' 



— n + 1 . n + *2 . yP„+i. 

 The equation (15) is therefore satisfied if 



n + m _ n— ra + 1 



P ~ n.2n + V q ~~n + 1.2n + V 



and thus, 



x — 1 f n+m p Wl n — m + 1 ^ m 1 



1 ~ 2n+ 1 . sin 6 \ n »-i n + 1 »+i / 



sinfl dP% 



it . 7Z+1 ^yU, 



(19) 



on reduction. 

 Thus 



b = b 1 + b 2 



a sin e d 1T dp;r 



-j- . r*o . — =— cos (ma) + e) 



rn .n + lar a/j, 



V n . ?z + 1 . sm 



