1SI5.] Solution of a Prollem of Col. Silas Titus. 59 



By substituting these values of the exponential in the equations 

 (31), (32), (33), they take the following forms : — 



q: V . 2 TT ^/ — 1 :p \\ 2 tt V — 1 = ± 2 tt V^^ x 16 . .(40) 



+ V-2t V^n q: V— TT \/ - 1 = ± 2:r ^/^HT x 17..(41) 



q: V • 2 7r \/ — I z^ ^-£ . 2 iv \/ - 1 = ± 2 tt \/ - 1 X IS . ,(42) 



Now 2 IT expresses the circumference, of which one half is on 

 the positive side, and the other half on the negative side : in 

 respect to the same diameter, — 2 tt expresses the same things 

 taken in contrary directions ; consequently ± 2 t and qp 2 tt are 

 two different expressions of the savie circumference, when the 

 diameters are the same ; they then represent the sjme rings ; and 

 the last three equations give by reduction the following : — 



± 2 T V~I71^Hi^*j = ± 2 ,r n/^^ X V = ± 2 ,r .^^=^ 

 X 16 (^i3) 



± 2 TT ^'^^ [^^^) = ± 2 TT ^r^^\ X %^ = ±2^ ^/"^ 

 X 17 (44) 



± 2 ,r ^"^ {^^^) =±277 ^~^^i X V = ± 2 TT ^/^^l 



X 18 (15) 



These last equations are evidently identical, and the problem 

 proposed is now completely resolved by the roots (25), {-IG), (27), 

 which, expressed aiithmetically, are 



«= V =H (16) 



b= ^^ =4\ (47) 



c = V = H (48) 



Unitij being the area of any circle, which is an essential remark ; 

 for if unity were an abstract number, tliis solution would be absurd. 



It is only true when unity is expressed by c'* ~ '^ ^ ~ (= the area 

 of any ring or central circle.) 



Remark. 



Wallis, who devoted much time and attention to the proposed 

 problem, resolved the equations (1), (2), (3), by aj)proximation. 

 (Wallis's Algebra, Chap. G2.) His roots are, 



a = 2,525,.513,'JSG,744,158 (49) 



b = 2.yG9, 1. "52,708,6 1«),84S (50) 



c = 3,240,580,681,617,174 (51) 



Comparing tliese roots with the roots (46), (17), (48), It is easily 

 perceived that they have not the least relation to each other. In 

 order to prove how nearly the roots (41>), (50), (51), verify the 



