174 



Mr Mayall, On Current Sheets, especially [Feb. 26, 



The last line is got by finding the coefficient of sin nO in 

 the preceding line and using (30), and the expression holds for 

 all values of n down to unity provided we consider A,^ = yu,^ = 0. 

 The left-hand side of (33) is then 



C^ln [ii' + ^) - 2 ('^' + ^*^ + 4 ) /*"+! 



^SSpiaOf Zx 



R 



^SHpiadf Z^ 



1 / 3\ 



sin nd 



+ n (fln-i - fln+i) - 2 ^\-^ 



sin 72^ ...(34). 



For the right-hand side of (33) suppose fl^ is of the form 



J-o (ad) Q.,n (cosh a) sin m^, 



then we have 



d d ' 



j^ 26„ (ae) Q„ sin 72^ + ^ -^0 (o"^) Qm sin m(9 



{aey 



^K\1 A Q. + (<r0)QA sin n0 



{adf L » [2 {ad) 



■\-A, 



(1 ^ 



Q» + M)Q«/ si^^^ 



2(o-^)» 



" ]2 ((7^) 



S [K m,. + 2CQ;) - 6„_,Q'„_, - 6„,.Q'„,J sin n^ 



4- ^0 {(^Q,„ -f 2CQ,,/) sin md - (sin m + 1 ^ 4- sin 7w - 1 ^) Q,«'| 

 (33) written in full is therefore 



R 



2SHp 



hn^+^y2Gf.,^-f.,^_,-^,^J 



sin nd 



= 2 {K {SQ,^ + ^CQJ - K_,Q'^_, - K,, Q'^J sin ne 



+ ^0 {m.n + 2(7Q,;)sin m^ __ 



- (sin m -h 1 ^ + sin 7w - 1 6') Q,„'] . . .(35). 



By equating to zero the coefficients of sin nd in this equation 

 we arrive at a system of equations which in conjunction with (30) 

 furnish the complete solution of the problem. 



