98 
ON WRONSKIS THEOREM. 
[574 
574] 
viz. it is 
= 1.1.2 (f 3 F')", 
and so in other cases; the formula is thus 
Fx = F- \fF' + ~ OT 3 (f' F T + &e. 
agreeing with Lagrange’s theorem. 
Suppose in general (f>% = (x — a) ifrx, or let the equation be 
(x — a) ifrx + \fx = 0, 
that is, 
x — a + \ = 0 : 
ifrx 
we have then by Lagrange’s theorem 
x ’^'(£)T +&c - 
Fx - F-- F' i- + \F' (/-Vi' - 
l ^ + 1.2rv^y) 1.2.3 
( /A 3V' 
Consider for example the term ji' 7 ' | ; this is 
\ (x-a) 3 (fx) 3 )" 
_ r W~f ’ 
the accents denoting differentiation in regard to x, and x being ultimately put = a; 
or, what is the same thing, it is 
d 
V (n i 63 i/( a + 6r) } 3 
F ( a + 0) 
the accents now denoting differentiation in regard to 6, and this being ultimately put 
= 0. This is 
' dvi >«.+«), i/( ; + * )!; -v 
(</>' a + ^ *"« + ...] 
This may be written , where 
A = f + |0</>" + £0y' + ..., 
it being understood that as regards F'f 3 , which is expressed as a function of a only 
(0 having been therein put = 0), the exterior accents denote differentiations in respect 
to a, whereas in regard to A, = <£' + \0$" + &c., they denote differentiation in regard 
to 0, which is afterwards put =0. And the theorem thus is 
F-*~ I i F 'f-1)+1-2 [rr • ¿y - IT 2.3 ■ i)" +&c - 
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