New Investigation of Taylor’s Theorem. 273 
d? 
. B=$ Fan b+ Aa, h).h, which substituted in (3’”) gives 4or= 
d d? 
a, btk aw ht +2, h).h?, (3’"); and since the form of 4A 
must be similar to that of 49x, when we use A instead of 9z, 
we may denote 4A by eee h+3 oe i +6(2,h).h?,(5”), 
(x, h) denoting a function of « and h, such that it =O when 
| Te 
h=0; and if (according to the usual method) we denote i 
d* d?A d® 
by a we get 7.2 aS ; and we may here observe that we 
ie) 
“‘dx*-) d" px 
shall denote any expression of the form i by =: ’ 
where 7 is supposed to be a positive integer greater than zero. In- 
d 
stead of using the equation that remains after putting 2B, — irely 
we shall use (a’) in what follows; and since the forms of B and 
B’ are to be similar to that of 4A, we may by (5”) represent 
them by B=BA+B,h?+6(a, h).h?, and B/=B,2h-+B, (2h)? 
+0(x, 2h).(2h)?, 6(x, h) being the same function of x and h 
that 6(x, 2h) is of x and 2h, each of these functions being =0 
2 
when h=0; and since A= ae B,=4 2 we get cee 
dj? 
+3 ae h+B,h? +6’(x, h).h?; hence substituting the values 
dj? 
of B, B’, and A+B in (a’), it becomes a(4 rh + 3B,.h? + 
dyx d? px 
0"(x, 2h) . (2h)? ~0"(a, h). ht) —4 (Fo 44 Fe ht Byhe + 
6(2, h). h | =0, (a’”). 
dx d? x is 
If we develope a(= +4 rasa h+ &e.] by (5”), we get 
d a d*p d? 
A(T +4 aaah a: &e.] == — A+ ce h? +&c. which being 
substituted in (a’”’), rejecting the terms which destroy each other, 
dividing by #2, then putting the terms which are independent 
Vol. xv, No. 2.—July-Sept¥#1843. 35 
