Full text: The collected mathematical papers of Arthur Cayley, Sc.D., F.R.S., late sadlerian professor of pure mathematics in the University of Cambridge (Vol. 11)

ON THE THEORY OF RECIPROCAL SURFACES. 
230 
[750 
The equation (26) expresses that the surface and its reciprocal have the same deficiency; 
viz. the expression for the deficiency is 
(30) Deficiency = % (n — 1) (n — 2) (n — 3) — (n — 3) (b + c) + ^(q 4- r) 4- 2£ +%/3+%y+i—^6, 
= 1(^-1) ( n ' - 2) (n' - 3) - &c. 
609. The equation (28) (due to Prof. Cayley) is the correct form of an expression 
for /3', first obtained by him (with some errors in the numerical coefficients) from 
independent considerations. But it is best obtained by means of the equation (26); 
and (27) is a relation presenting itself in the investigation. In fact, considering a as 
standing for its value n (n — 1) — 26 — 3c, we have from the first 25 equations 
6 
a 
= 2, 
+ 2 
3 n — c — k 
= 2, 
- 2 
a {n — 2) — k + B — p — 2a — 3a) 
= 2, 
- 4 
b (n - 2) — p — 2/3 — 87 — 31 
= 2, 
- 6 
c (11 — 2) — 2cr — 4¡3 — 7 — 6 — to 
= 2, 
+ 2 
n + k — a — 2G — 4 B — 2j — 3^ — 
3a> = 2, 
- 3 
2q — 2p + j3 +j 
— 2, 
- 2 
3r + c — 5ar — /3 — 4<0 + x — 6) 
= 2; 
multiplying these equations by the numbers set opposite to them respectively, and 
adding, we find 
— 2 n 3 +12 n" + 4 n+b (12?i — 36) + c (12?i — 48) 
— 6q — Qr — 4G — 10B — 41/3 — 30y — 24£ — 7j — 8x+ 26 — 4&> = X, 
and adding hereto (26) we have the equation (27); and from this (28), or by a like 
process, (29), is obtained without much difficulty. As to the 8 X-equations or symmetries, 
observe that the first, third, fourth, and fifth are in fact included among the original 
equations (for an expression which vanishes is in fact = X); we have from them 
moreover 3n — c = 3a' — k, and thence 3n — c—K = 3a' — k — k, which is = X, or we have 
thus the second equation; but the sixth, seventh, and eighth equations have yet to 
be obtained. 
610. The equations (15), (16), (17) give 
n'— a (a — 1) — 28 — 3«, 
c = 3a (a — 2) — 63 — 8k, 
V = \a (a - 2) (a 2 - 9) - (a 2 - a - 6) (28 + 3«) 4- 28 (8 - 1) + 68« + f* (« - 1). 
From (7), (8), (9), we have 
(a- b- c){n- 2) = k - B - 6/3 - 47 - St - 0 + 2o, 
(a-2b-3c) (n-2)(n-3)=2(S- C)-8k-18h-Qbc + 18/3 + 12 7 + 6i -6«;
	        
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