[574 
574] 
ON WRONSKl’S THEOREM. 
97 
_ *■» ± p, m, (ffFy i 
1 - 2 * 3 *' e <t>", (py, (if s Fy 1 - 1 - 2 
r, m"> (SfFT 
+ &c., 
where F, f F', &c. denote Fa, fa, F'a, &c. and the accents denote differentiation in 
regard to a; the integral sign / is written instead of f a ; this is introduced for 
symmetry only, and obviously disappears; in fact, we may equally well write 
Fx = F 
~i? /r 
xy yy 
+ 1.2 p 3 
P, pF 
P\ (PF y 
1 
1 
X 3 1 
p » 
(PY, 
PF' 
natics, vol. xii. (1873), 
1.2.3 p 6 
r, 
(PY, 
(,PF') 
(P)"', 
(PF')‘ 
+ &c. 
) the question “ En quoi 
sser par un seul problème, 
t ce problème universel?” 
? de Fonctions Analytiques 
c, a demonstration) in the 
and demonstrated in the 
MX et seq. ; the theorem, 
oédie Mathématique (Paris, 
I stop for a moment to remark that Laplace’s theorem is really equivalent to 
Lagrange’s; viz. in the first mentioned theorem we have x = (f>(a + Xfx), that is 
<h~ l x — a + Xf(f>. (/>~ x x, and then Fx = F<f>. <£ _1 x, viz. by Lagrange’s theorem 
^H+j W -ft+o w' ■ c/w+ &c - 
where on the right hand F(f> and /<£ are each regarded as one symbol, the argument 
being always a and the accents denoting differentiation in regard to a, thus Fp is 
e root of an equation 
d a . F<f>a = F'cfra . pa, &c., 
viz. this is Laplace’s theorem. 
Suppose in Wronski’s theorem (f>x = x — a] that is, let the equation be 
ar case 0 =fx + x x f x x; or x — a + Xcf)x = 0, 
equation then each determinant reduces itself to a single term: thus the determinant of the 
third order is 
{x-a)', {(x — a) 2 }' , f*F’ 
(x-a)", {(x — af\", (pF')' 
(x-a)'", {(x-aYY", (PF')" 
h&c., then expanding the several 
ants of the same order, and we 
where in the first and second columns the accents denote differentiation in regard to 
x, which variable is afterwards put = a; the determinant is thus 
= 1, * , * , 
0, 1.2, * 
0, 0 , (pF’)" 
c. ix. 13