274 New Investigation of Taylor’s Theorem. 
of h (or which do not =O when h=0) equal to zero, we get 
Spee EE cues (7 3 
2.3B,—G,2 =9, oo Bi=g 3 Gye ence (3’") becomes 4g7= 
dpx  _ d’ ou 1 d*oxr 
de +2 Wath aL2.3 se h?+ 6”(x, h).h?, (3%); and since 
4A has the same form as dex, (5”) b Lie h Lea 
H gx, (5) becomes = ae +2 Gra 
+ ae Che Loa, h)h®, (5), and as Band B’ must be of 
similar forms, it is evident from what has been done that we may 
put B=BA+B,h?2 +B,h? + B,A* + &c., and B/=B,(2h) + 
B?(2h)?+B,(2h)*+B,(2h)*+&c.; .*. substituting these values 
dpa d? px 1 d'or : it 
and A+B =a +4 ee h-+ 2.3 gs WBA +&e. in (a’), 
d? d? 
we get 22 h+4 Fo H+ (2°-1)B,h+(2!-1)B,ht+&e. | 
dex dx 1 dx bear 
~a(F 18: ae h+5-3 FEES h?+B,h? +&e.)=0, (a’v), which 
is under a more convenient form than (a’”). 
dpe dou 1 dex 
If we develope Debs de @t3.3 AER h?-+60.] by (5”), 
reject the terms which destroy each other, divide by h?, then put 
the terms which are independent of h, equal to zero, we shall 
‘ 1 1 \d‘oux 1 doz 
get 2(2*—1)B, —2( 9349-4) gar =o Of Beg aa ae 
and substituting this value of B, in (a’”), we shall in the same 
1 di px dx 
way find B,=9-3_4.5 eR age and so on; hence A+B=—777-+ 
d? Ids Lbisenatale 
boehts—5 a Tors" h he + &c. and substituting 
this in y(w7+h) = per+(A+B)h, we get o(a+h)=¢x eee + 
digo ly digas ee 
2 h?+5-3 ‘dpe Fong id dee B+ &e., (A), the law of 
continuation being evident. 
(A) is the theorem of Taylor which we proposed to inves- 
tigate; and we have obtained it without making any use of 
the binomial theorem. It may not be improper here to ob- 
serve, that although (A) has been found on the supposition 
that x is indeterminate, yet it may uate when a particu- 
