on the Equation of the Fifth Order. 213 
Mr. Jerrard’s function W is the sum of two values of his 
function P ; the substitution by which the second is derived from 
the first can only be that which interchanges the two functions 
[«Byde] and [aye8o] ; and hence any symmetrical function of 
[«Pyde] and [wyeGd] is afunction homotypical with Mr. Jerrard’s 
W; such symmetric function is in fact a 6-valued function only. 
Indeed it is easy to see that the twelve pentagons correspond 
together in pairs, either pentagon of a pair being derived from 
the other one by séed/ation, and the six values of the function in 
question corresponding to the six pairs of pentagons respectively. 
Writing with Mr. Cockle and Mr. Harley, 
T = LyXgt Lply + XX3 t+ XjX,+L Ag 
= 2,@y + LyTe+ LM gt Lets tUploy 
then (r+7' is a symmetrical function of all the roots, and it 
must be excluded ; but) (tT—7')? or 77’ are each of them 6-valued 
functions of the form in question, and either of these functions 
is linearly connected with the Resolvent Product. In Lagrange’s 
general theory of the solution of equations, if 
ft=2, +l, 4+ Cr+ Px, + 25, 
then the ee of the equation the roots whereof are 
( ft), (fe?)®, (fe*)?, (fr), and in particular the last coefficient 
(fifi? Joan )5, are peta by an equation of the sixth degree ; 
and this last coefficient is a perfect fifth power, and its fifth root, 
or fi fi*fi* fr’, is the function just referred to as the Resolvent 
Product. 
The conclusion from the foregoing remarks is that 2f the equa- 
tion for W has the above property of the rational expressibility of 
its roots, the equation of the sixth order resulting from Lagrange’s 
general theory has the same property. 
T 
I take the opportunity of adding a simple remark on cubic equa- 
tions. The principle which fur nishes what in a foregoing foot- 
note is called the @ priort demonstration of Lagrange’s theorem is 
that an equation need never contain extrancous roots ; a quantity 
which has only one value will, if the investigation is properly 
conducted, be determined in the first instance “by a linear equa- 
tion ; one ‘which has two values by a quadratic equation, and so 
on ; Pie is always enough, and not more than enough, to 
determine what is required. 
Take Cardan’s solution of the cubic equation #?+qgxe—r=0, 
we have az=a+0, and thence 3ab=—gq, a?+0°=r; and to 
obtain the solution we write 
. a= — 5, a+b=r, 
