24b Mr. Knight’s note respecting the demonstration, &c. 
equation, we see immediately that A =0, and that A ' and A f/ 
are arbitrary. Then, to find the law of the coefficients, in 
this and other similar cases, where there are any number of 
independent quantities, oc, x ' , x" x"- n , transpose all the 
equation to one side, and find the coefficient of the first power 
of x"- n , then the coefficient of the first power of x” " n — 1 in 
the former coefficient, then again, in this last, the coefficient 
of the first power of x"' n ~ z : and having arrived in this 
manner at the coefficient of x’ , it will have the form a-\-bx-\- 
cx 1 + ...... rx n - f-, and the equation r = o will give the law 
sought for. 
In the present case, putting p — x + (1 ^x)z, <r= 1 -\~x-\- 
(i-{-x)z, we find, by equalling to 0 the coefficient of the first 
power of y, 
A / 4-2A /, '| +SA //, ’S 
(x+x) (x’+Z*) 
+ A' J +2A" J 
— A V — - 2 A"trp — 3 A n, <rp a — — (« -J- 1 ( A"--( n + l ) crp n — 
If in this we equal to 0 the coefficient of the first power of %, 
there arises 0 = 2 A" — A'x — 2 A" (1 + 3^ + 2x9 ) 
— 3 A'" (2^4-5X 2 + 3o: 3 ) 
— 4A ,//, (3^ 2 + 7x 3 + 4a: 4 ) 
nA"— n (n-~ 1 . x n ~~ z -f — 1 . x n ~ l -f- nx n ) 
— &c 
whence we find for the general law of the coefficients («>2), 
n(n — i)A' ,n + 2 n — %.n — j . A ,/ '*'^” 1 )+ ( n — 2) 2 A "••(«— 2 ) 
= ° •(*) 
From which let us suppose that we have calculated a few of 
the coefficients, and arrived at the result of Mr. Spence, viz. 
