38 Mr. 0. J. Lodge on some Problems connected 



To determine the value of the product (- — . Q j in the re- 

 sistance-expression (/3) (§ 8), take the images in rows, and find 

 the product for each row separately (cf. § 9). Call the line A B, 



3 











Fig. 



, 



5. 



i 









& - 

 1 





• 





i 













A 





!U ' 









1 









i 









3 — 



i 1 



i 





i 









linages of the pole A in an isosceles right-angled triangle. 



fig. 5, row ; the two rows one on either side of A B call rows 1 ; 

 row 2 will be the row next beyond each of these ; and so on, 

 the images being evidently symmetrical with respect to the 

 line A B. The whole product Q will then equal Q , QJ, Q* ... , 



A.B 



A.A 



where Q^ stands for the product of the -~-^ 's in one row x 

 From the figure we see that 

 1.3.3.5.5.7... 



Q - 2 



l 



q; 



l 2 2 2 + l 2 2 2 + l 2 4 2 + l 2 4 2 + l 2 

 '3 2 + l 2 *3 2 + l 2 *5 2 + l 2 



02= 



q;= 



i 2 +i 2 i 2 +i 2 



l 2 + 2 2 l 2 + 2 2 



*r(0) 



32 +2 2 32 + 2 2 



3 2 



2 2 + 2 2 



2 2 +3 2 

 l 2 + 3 2 'l 2 + 3 2 *3 2 + 3 2 



2 2 + 2 2 

 2 2 + 3 2 



4 2 + 2 a ' # * 



1 



~ 2 V 2 (2)' 



= 3V(3) ; 



and in general 



Q±i =#*<>), 



where the index is to be taken positive when x is odd, negative 

 when x is even ; and where w(#) stands for the product 



2 2 + ^ 4 2 + (2 ,2 6 2 + (2 ,2 



l 2 + x 2 & + x 2 5 2 + x 2 '" ^(oo 2 + x 2 )' 



