274 New Investigation of Taylor’s Theorem. 
of h (or which do not =0 when h=0) equal to zero, we get 
d*y 1 d’¢ 
2. 3B, — Gz3 = a0, or B wee 77} hence (3’*) becomes dgr= 
dox d? 9x 1 
a h+4 dx? h? +33 
ins ihe 6"(x, h). h’, on and since 
4A has the same form as Jeux, (5) becomes 4A = oaks te 
+55 5 3 = eh oM"(a, h)h?, (8), and as B and B’ must be of 
similar forms, it is evident from what has been done that we may 
put B=BA+B,h? +B wht + BAS + &c., and B’/=B,(2h) + 
B*(2h)? +B,(2h)*+B, hye See dae these values 
dx 
and A+B= ae dat ht 5-3 de 4B, h?+&c. in (a’), 
_ a? 
we get aa dat h+d a8 h24(2'-LB, hs-+&e) 
 .d* ox 1 d*oxr ; 
~4(GttGe h+5-3 Gs h?+B sh? +80.) =0 hc 
‘is under a more convenient form than (a’”). 
d? Led? 
If we develope (44 aa ht53 a ht +é&c. Jove (5”, 
reject the terms which destroy each bse divide by h, then put 
the terms which are manent of h, equal to —_ ise “9 
A \dton _ d* Ge. 
and ssbiitintig oe Faint of B, in eh we at in vibe a 
way find B, = =5 57 syeity 5 me and so on ; sateen hide: 
d? px 1 d*qr 1 do tw 
Fe? Ate 3 3 dx* hi ‘+3734 3.4 dxt = hs pee: and substituting 
doz. 
this in in = Suen we get etic +74; 4 a 
d* ox 
Faeeht ass 3 dae = ht dicen a Ge # Seat8) the ee 
__, continuation being evident. 
(A) is the theorem of Taylor which we vicpceed to “inves 
tigate ; and we have obtained it without making any use 
the binomial theorem. It may not be improper here to ob- 
serve, that although (A) has been found on the supposition 
that @ is i erent i may be applied when a particu- 
