135 
hy a Process of mere Inspection. 
give as the sum of the diagonal products 
I hmc-{-alnc-\-a mb 71 -\-lacn 
be i.elhmc-\- a mhn+2acln. 
Thus the result of eliminating betvveen ax'^+bx-\-c-=0 
I -\-m X a •=. 0 
ought to, and is 
n^+ l^c^ — 2 acln-\- Ib^n + am^ c — Ibmc—amb n = 0. 
Rule for finding the prime derivative of the l5^ degree vohich is 
of the form A a;— B. 
1. Begin as before, only attach one zero less to each pro- 
gression ; we shall thus obtain 7iot a square, but an oblong 
broader than it is deep, containing (7n-\- n — 2) rows, and 
{m-\- 7 i — \) terms in each row: in a word, (m + w — 2) rows, 
and {7U -{-71—1) columns. 
To find (A) reject the column at the extreme right, we 
thus recover a square arrangement + w — 2) terms, broad 
and deep. 
Proceed with this new square as with the former one ; the 
difference between the sums of the positive and negative diago- 
nal products will give A. 
To find B, do just the same thing, with the exception of 
striking off not the last column, but the last but one. 
Rule for finding the prime derivative of any degree^ say the 
rth, viz. Ay. — Ar— i x^—^ + + Aq . 
Begin with adding zeros as before, but the number to be 
added to the (a) progression is {m — r') and to the (5) pro- 
gression {n — r). 
There will thus be formed an oblong containing (m + w — 2r) 
rows, and {7n-\-7i—r) terms in each row, and therefore the 
same number of columns. 
To find any coefficient as Ag , strike off all the last (r + 1) 
columns except that which is ( 5 ) places distant from the ex- 
treme right, and proceed with the resulting squares as before. 
Through the well-known ingenuity and kindly preferred 
help of a distinguished friend, 1 trust to be able to get a ma- 
chine made for working Sturm’s theorem, and indeed all pro- 
blems of derivation, after the method here expounded ; on 
which subject I have a great deal more yet to say, than can 
be inferred from this or my preceding papers. 
University College, London, Jan. 16, 1840. 
[To be continued.] 
