gg 2 Mr. Knight’s new demonstration of 
liable of course to the same objection. If we multiply a-\-x, 
continually, first by itself, and then by the powers successively 
arising, we easily see that the second term of each succeeding 
product is of the form ntf~~ l x> n being the exponent of the 
- - * * 
power : this does not require a more formal proof, and I as- 
sume it in what follows. Nor is it more difficult to perceive 
that, generally, m being positive or negative, whole or frac- 
tional, the following form may be assumed, 
( a + xy\= a m + 'A a m ~ l x +"Aa m ~ V -f f "Aa m ^x 3 + 
where 'A, "A, "'A, &c. are expressions depending on m alone, 
consequently, (a ^y) m =a m ^'Aa m ~^y -p'Aa ra ~ 2 /-f " A.a m ~ 3 y 3 -{- 
and because \^{a x) (a y) } w == (# 9 + ax + ay -f- xy ) w = 
=a’’(a+x+y+^) m =a"(a+x + vyf l by making ) 
we have also 
j (a- \-x) (a+y) Aan—'^x^ny)-^'' ha m -- 2 (x-\->7ryY -{• 
^oq aloiTw b 9d . "'Aa^^x+Try ) 3 + 1 
or neglecting all the powers of y but the first. 
z=a m [a m + , Aa m - l x + "A a m - z x*-^ '"A a m ~W+ j 
-j- a m y | 'A a m ~ l '7T -J- a "Aa m ~ 2 t tx -J- 3 '"Aa m ~ 3 7rz* -f- j 
4 " 
81 tc 
• • 
Having thus the forms of the series, nothing is required 
but to substitute them in the equation 
{a + x) m x (a +y) m = j(a + x) (a + y) 
and to compare the coefficients of the first power of y on each 
side, and we find 
'A a 2m ~' + 'A 'Aa 2m - 2 x + 'A"Aa 2m ~ 3 x* + 'A'"Aa 2m -*x*+ 
^&a^ l * + * n AJ m ~\x +s'"Aa 2m ~*>irX' + 4'"' A a am ^ 9 rx a + 
