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)=0, (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 xz and h, it is 
manifest that B must be of the same form as 4A, also that B’ has 
the same form as 4A, excepting that we must use 2/ in B’ where 
we use # in JA or B. 
dex dya 
If we substitute | for A in (3) it becomes spn = h+Bh 
(3’”), which substituted in (4) gives eth) soe eo h + Bh, 
(b); since A is a function of x, the form of 4A must be similar 
to that of agz when we use A instead of 9x, .*. we may put 
dA dA 
4A= 7, h+ Ath, (5), where 7 1s independent of h, and A’ is 
a function of z and h, such that it =O when h=0; since A= 
ee ae 
dex NN ON des : da d2 yx 
iy Mears de aks Os (Gif we denote ——7— by Ga as 
is customary, ) - Lee , which reduces (5) to A= ee h+ 
A‘h, (5’); hence (from what has been said) we may represent B 
and B/ by B= B,h+ 2, h).h, B/=B2h+ (a, 2h). 2h, (6), where 
B, is supposed to be independent of h, and 6(z, h) denotes the 
same function of x and h, that 9(7, 2h) does of x and 2h, these 
functions being such as to =0 when 4=0, so that the forms of 
B and B’ are similar to that of 4A, as they ought to be; and 
4B=h[4B,+40(2, h)], (7). By substituting the values of 4A, 
B, B’ and 4B in (a), and rejecting the common factor h, it be- 
dA 
comes after a slight reduction 2B,— 7° —A’+-40(z, 2h) — 202, h) 
—4B,—46(x, h)=0, which is to be an identical equation, -’- 
dA 
since 2B, and 7 are independent of h, and since the other terms 
of the equation are not independent of it, (since Ot of them 
dA d? px 
=0 when h=0,) we must put 2B,— 7 =0, or B ae 2 Gas? 
