Mr. J. Cockle’s Note on the Remarks of Mr. Jerrard. 91 
graphs, are here briefly noticed. Mr. Jerrard has to meet the 
additional objections that— 
1, 2,4. There is no error in Mr. Harley’s processes. The 
error of inferrmg the sextic to be an Abelian I have already 
traced to its source (vol. 3 XIX, up 197 et seq.). 
3. If the functions ,&, .=, 3,5, and ,= are not extraneous, 
let two quintic surds be expelled, as reducible, from them and 
from >=. Then, according to Abel’s theorem, all the roots of 
(ab) are roots of (ac). Denote (ac) and (ab) by 
==0;) e=0 
E=i(Bz—t {Pie .) })- 
But what is the form of €? Certainly not f, for the root & + 
is common to (ab) and (ac). Unless Mr. Jerrard shows that ¢ 
cannot be replaced by f(Be); his conclusion is vitiated by an 
respectively. Then 
error in comparing (ab) and (ac). 
5. The complex process need not supersede the simple method 
of substitutions when we are dealing with symmetric functions 
of @ and ©". Such functions are Mr. Jerrard’s = 5, or 
0; oe 0, “i @;° a ©,, 
and my @, or 
54@, ©, O, O,. 
To take a simple example, 
®n.09(2) t On.a9)(8)= Ons On) BEI): 
6. It can scarcely be said that Mr. Jerrard’s theorem 1s 
ignored by Mr. Harley. That theorem is inapplicable to the 
processes dealt with in Mr. Harley’s paper, into which no 
symbol corresponding to P enters. 
7. Actual calculation shows that & and @ are similar func- 
tions (fonctions semblables). And if in my function y we sub- 
stitute & for @, the result is an equation which, combined with 
(ac), shows that, if the latter be an Abelian, the root of a general 
quintic can be expressed without quintic surds. 
8. If, in expanding a function of two independent quantities, 
we do not obtain an expression symmetric with respect to those 
quantities, and such as to admit of an interchange between them, 
I cannot consider the expansion as algebraically (rigorously) 
true. 
In generalizing the theory of transcendental roots which I 
have sketched, and partially developed, in these pages, we may 
start from 
2” —ne + (n—1)a=0, 
