New Investigation of Taylor's Theorem. 273 
d? Pee Hons aah 
2. B=$ 5 h-+02,h).h, which substituted in 3”) gives 4qe= 
d? ox 
atta = +62, h). h?, (3); and since the form of 4A 
must be ns to that of aa; when we use A instead of gz, 
d?A 
we may denote 4A by pedi h+4 dar 4-6/(0, hy. h?, (5), 
oe, h) denoting a fanetion tr x and h, Kasai that it =0 when 
(G2) 
dx? 
ant and if (according to the usual method) we denote — 
d7A 
3m 
by © aie » we get 7.9 esti ; and we may here observe that we 
d*-'ox 
J a a=" | dor 
shall denote any expression of the form ——j>—— by G3 
where n is supposed to bea positive integer preater than zero. In- 
dA 
stead of using the equation that remains after putting 2B,— 7 =9, 
we shall use (a’) in what follows ; and since the forms fo B and 
B’ are to be similar to that of 4A, we may by (5”) represent 
them by B=Bh+B,h?+0"(2, h). h?, and B’=B,2h+B, (2h)? 
in 2h).(2h)?, (x, h) being the same function of « and h 
that 6”(a, aay is of x and sec each of Reese functions being, =0 
doz 
mee hi and since si. a? Bat ae a “8 we get AtB= 7, 
ri sts e h+B, h? 40" (a, h). he; hence substituting the values 
of B, B, and A+ Bin. (oi beopmes2(8 Ge ah + 3B, he + 
d*ox 
(2, 2h). (2h)? =0"(m h). “3 na (HE 4a Sh Byh + 
Ox, h). it) =0, fe") a 
If we rae 4 (= og: ee eh + &c. * by (3) we e get 
dor. 
a(F- Be chs 25 6 ee _ ~ ht Ag h? +&c. which being 
substituted in (a’”’), rejecting the terms which destroy each other, 
dividing by 4?, then putting the terms which are independent 
Vol. xxv, No. 2.—July-Sept. 1843, 35 
