272. New Investigation of Taylor’s Theorem. 
Investigation of Taylor's Theorem. 
We shall now resume (2), 2(B’ —B) — 4A —- 4B=0, (a), which 
we may put under the form 2(B’—B) —4(A+B)=9, (a’), which 
is to be satisfied so as to be an identical equation ; hence since 
B’ is the same function of z and 2h, that B is of x and h, it is 
manifest that B must be of the same form as 4A, also that B’ has 
the same form as 4A, ee that we must use 2/ in B’ where 
we use Ain 4A tes 
If we substitute 9 = for A in 26) it becomes ano = AUB 
(3), which substituted i in (A) ¢ gives Biases h+ Bh, : 
(b); since A is a function of «, the form of A must be similar 
to. oe tg gz when we use A instead of gz, .:. we may PH 
dA 
a h+Ah, (5), where 7 is independent of h, and A’ is 
= Se of z and h, such that it =0 when h=0; since A= 
d =) (ep ; 
diya aa ae] q\ac) dia 
de We get 7 =—7—, or (if we denote © pene Arr 2 an as 
F Pt Bo Re OE ER, " 
1s customary,) 7-=G,21 Which reduces (5) to 4A= Ss 
A’h, (5’) ; hence (from what has been said) we may represent B 
and B’ by B=B,h+(z, h). h, B=B2h+-0(x, 2h). 2h, (6), where 
B, is supposed to be independent of h, and 6(x, h) denotes the _ 
same function of z and h, that (x, 2h) does of « and 2h, these 
functions being such as to =0 when h=0, so that the formsof ~~ 
B and B’ are similar to that of 4A, as they ought to be; and eS 
4B=h[4B,+40(2, h)], (7). By substituting the values of 4A, y 
By B’ and 4B in (a), and rejecting the common factor h, it be- 
comes after a slight reduction 2B SA Aste, 2h) —242, h) 
—4B,—4x, h)=0, which is to be an identical equation, -"+ 
since 2B, and dg 2t independent of h, and since the other terms 
of the equation are not independent of it, i rs of them 
dA d2qt 
=0 when h=0,) we must put 2B, - 7 = or B hae sha 
