176 Mr May all, On Current Sheets, especially [Feb. 26, 



Now the following relations hold between the P's 



.(39); 



^^ p /_ rip p 



2w_l « « «-i 



hence 



4,iSf 

 — p ' — p \ p _ 'onp 



A«2 _-l^n -^ n-i " -^ ,1+1 '^'^■^ n J 



or 



(..^ - ^) (2ap„ - p„_, - p„,j = ^p; (40). 



Again we have also from (39) 



4.SP: = (2n + 1) P„,, - (2n - 1) P„.. - 2CP„ (41), 



and if we substitute in the expression found above from (40) and 

 (41) it is seen to vanish. 



The right-hand side of (35) as far as it depends on the po- 

 tential due to the sheet is 



S [K {8Q. + 2CQ:) - \_,Q\_^ - \^,Q\^;\ sin nd 



tB 



S 



= - ^^ S {SQ,^ + 2CQ: - Q;_, - Q'^J sin nd 



= 0. 



These results do not hold for the cases w = 1 or 2, but by 

 equating the coefficients of sin 6 and sin 26 in (35) to zero we 

 shall have two equations which will determine the values of the 

 constants A, B, in order that (35) may be satisfied by the forms 

 assumed for Xyu. in (36) and (36 a), and we shall thus have found 

 the current function when the external magnetic disturbance is 

 of the form 



The coefficient of sin 6 on the left of (35) is 



- 2^ 1^ (2Cyt., -;.,)-;.,- 1 ^^xj (42), 



now putting for the moment /ji^= — S(APq + BQ^) we may shew 

 as above that 



- {20,., - ^^ - ^^) + ^^ _ ^^ _ I s\ = ^ {ap: + bq:), 



