390 Mr Dixon, A n example of complex double integration. 
and the double integral = (2t7r) 3 S W > the summation 
/ a yx , y) 
being over all the solutions of the equations U = 0 = V. Since 
the double integral tends to zero as R increases, Jacobi’s theorem 
follows. 
The corresponding theorem for a single variable is of course 
that Xg(x)/f'(x) = 0 if g(x), f(x) are of degrees n — 2, n and the 
summation extends over all the roots of f{x) = 0. If g(x) is of 
higher degree it may be put in the form 
cf>(x)f(x) + c/O) + gfx), 
where c is constant and g 1 (x) of degree n— 2 at most. Then we 
have Sg(x)/f , (x)=nc. There is a similar extension with two 
variables, depending on the following lemma : — 
If W is of degree r, higher than m+n — 3, we can find a 
constant c and quantics A, B, Z in x, y, 1 of degrees r — m, r — n, 
m + n — 3 such that W — AU + BV + cJ + Z where 
J—d(U, V)/d (x, y). 
Assuming this we have 
2 W/J= 2 (A U + BV + cJ + Z)/J, 
where TJ, V vanish in each term and ^Z/J— 0. Since there are 
mn terms %WjJ = mnc. 
To prove the lemma, let u, v, w, j denote the terms of highest 
degree in U, V, W, J and take first the case r = m + n — 1. Then 
we can choose binary quantics <£, \Jr in x, y of degrees n— 1, m — 1 
such that 
U(f) + v-^r — w ; 
for we have here, in fact, m + n linear equations to find the n 
coefficients of <£ and the m coefficients of t/t, and the determinant of 
the left sides of these equations is the eliminant of u, v, which by 
hypothesis does not vanish. Hence </>, ^ are determinate and are 
such that W — U<f> — F/ is of degree m + n— 2 only. 
If W is of higher degree, r, we can put (not uniquely) 
w = af+/3g+ 
where / g ... are quantics of degree m + n— 1, a, ft . . . quantics of 
degree r — m — n + 1, and apply the same method to f g ... ; thus 
<f), -ft may again be found, but not uniquely, so that w = u<p + vy{r 
and the degree of W — U<j) — Vyjr is r — 1 only. 
In this way by successive steps A 1 , B x may be found so that 
W — Aj U — B 1 V is of degree m + n — 2 only, and we have now only 
to discuss the case r = m + n — 2. 
Since here xw is of degree m + n — 1, we have 
xw = u<f> + v\fr, 
