1890 - 91 .] Dr Sang on the Extension of Brouncker’s Method. 343 
On using the first of these variations, that is, making 
C = 2.D + l.E + F, 
the entire scheme becomes this : — 
1.5874011 = A = 1.B +0.C +D 
1.2599210 = B =1.C + 0.D + E 
1.0000000 = 0 = 2.D +1.E +F 
.3274801 = D = 1.E + 0.F +G 
.2599210 = E = 2.F +1.G +H 
851188 = F = 1.G + 0.H + 1 
675591 = G =2.H + 1.1 +K 
221243 = H = l.X +0.K+L 
175597 = 1 = 2.K + l.L +M 
57508 = K = l.L + 0.M + H 
45646 = L = 2.M + l.H + 0 
14935 = M=l.H+0.O +P 
11862 = N =2.0 +1.P + Q 
3914 = 0 =1.P +0.Q + R, 
3073 = P = 2.Q + 1.R +S 
961 = Q = l.R +0.S +T 
841 = R =2.S +1.T +17 
310 = S = l.T +1.U + Y 
120 = T 
101=U 
89 = Y 
in which the pair of groups of quotients 
is repeated almost to the end, ceasing only when the accumulation 
of the last-place inaccuracies may have interfered. Hence we are 
led to assume that this recurrence ought to continue for ever. The 
soundness of this inference is confirmed when we operate on the 
symbolical representatives of the three quantities. 
Beginning with the three values, A = £/4, B = ®/2, C = 1, and mak- 
ing A=1.B + 0.C + D, B = 1.C + 0.D. +E, we find D= ^4- $2, 
E = 4/2 - 1, and have now to operate on the three quantities C, D, E. 
