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. 12)

NOTE ON ABELS THEOREM. 
39 
[805 
(1883), 
of integrals, 
be stated as 
is a relation 
¡s of a point 
(X, y) on the 
out nodes or 
i intersections 
m_s denote an 
ie theorem is 
>ints of inter 
longing to the 
btain 
805] 
where 8(f> is that part which depends on the variation of the coefficients, of the whole 
variation of <f>; viz. if (f> = ax 11 + bx n ~' i y + then 8(f = x n da + x n ~ Y y db + ...; 8<f> is thus, 
in regard to the coordinates (x, y), a rational and integral function of the order n. 
Writing in this equation 
dx, 
d, J = % dm • 
the equation becomes 
/d(f> df d(f> df\ 
\dx dy dy dx) 
dco + 8(f> = 0, 
or say 
that is, 
— J (/, (f>) dw -j- 8(f> = 0, 
dco = 
8(f) 
J(f, <t>)’ 
and then multiplying each side by the arbitrary function (x, y, l) m ~ 3 , we have 
2 (*, y, 1 )“-* dco = t 8*. 
where 8(f) being of the order n in the variables, the numerator is a rational and 
integral function of (x, y) of the order m + n — 3: hence by a theorem contained in 
Jacobi’s paper “Theoremata nova algebraica circa systema duarum sequationum inter 
duas variabiles,” Crelle, t. xiv. (1835), pp. 281—288, [Ges. Werke, t. III., pp. 285—294], 
the sum on the right-hand side is = 0: hence the required result £ (x, y, ]) m ~ 3 dtu = 0. 
Observing that (x, y, l) m 3 is an arbitrary function, the equation just obtained 
breaks up into the equations 
Xdco = 0, Xxd(o = 0, %y do) = 0,..., Xx m ~ 3 do) = 0,..., %y m ~ 3 do) = 0, 
viz. the number of equations is 
1 + 2 + ... + (m — 2), = ^ (m — 1) (to — 2), 
which is =p, the deficiency of the curve. 
Suppose the fixed curve f (x, y, 1) = 0 is a cubic, m = 3, and we have the single 
relation 2 do = 0, where the summation refers to the 3n points of intersection of the 
cubic and of the variable curve of the order n, (f> (x, y, 1) = 0. 
In particular, if this curve be a line, n — 1, and the equation is do) 1 + do). 2 + d(o 3 = 0 ; 
here the two points (ir x , y)), (x 2) y 2 ) taken at pleasure on the cubic, determine the 
line, and they consequently determine uniquely the third point of intersection (x 3 , y s ); 
there should thus be a single equation giving the displacement d(o 3 in terms of the 
displacements dw 1 , dco 2 ; viz. this is the equation just found 
do) 1 + do) 2 + d(o s = 0.
	        
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