PROFESSOR CAYLEY ON THE SEXTACTIC POINTS OE A PLANE CURVE. 577 
71. For the first line we have 
d 2 U= — d 3 U=-— ~ V gBH, 
and hence 
(m — l) 2 
[m-lf 
first line of BJB 3 U=^^((m-2)HB<l> + <I>BH). 
{m— 1 ) 
72. For the second line, we have 
that is 
V(B 2 U)=VB 2 U + 2(V .B)BU 
=VB 2 U, since V . B=0, and therefore (V . B)BU=0 ; 
V9>U=V(yU)=v(£Y$__|_ sH ) 
or writing 
this is 
whence also 
Similarly 
=^(UVO+$VU)--^ I? (^VH+2^HVa); 
3 
U=0, VU= — H, V^=<1> 
’ m — 1 ’ 
{m — 2)5 
VB 2 U= 
(: m — l) 2 
Hd>— 
s 2 
(m-iy 
VH, 
3(V3-U)=^A|(H3<I.+$3H)- ^B(VH). 
VB 3 U=V(B 3 U) 
mU 
$ 2 
fBO— , 3 so 
m — I ( m — 1 y 
BH 
=^ I (VU34>+UV(3<l>))- fS ^ Tp (a ! V(3H)+2&Va3H); 
or putting 
U=0, VU= V&=<*>, 
1 m — 1 ’ ’ 
and observing also that V(BH), = VBH+(V . B)H is equal to VBH, that is to BVH, 
we obtain 
V3>U= { -^ I )i(mH34>-24.3H)-^ i? BVH; 
and then from the above value of B(VB 2 U), we find 
B(V3 5 TJ)-V3*U=^ T p(-2H34.+m4>BH)+ ( ^~ 1 (-3(VH)+BVH); 
•& 2 
or observing that the term multiplied by ^ m _^2 is = — (B . V)H, we find 
second line of B 2 B 3 U=^ I p(-2HB<I>+m$BH)+ . V)H). 
5 i 
MDCCCLXV. 
