
PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 291 
Hence, if we equate the coefficients of a” in the two equivalent expressions 
a 
me ET) and si ta ap du eat rh 
l—a(i+a)” ( Sees ee aad) e 
@-a?-$2 aa 4#* 
the result will be 
(ESD; Ss on 
eo (FIIs Tse wanes ha 
(1+a)"u;=(n+1) {1+ 
a? i 
of +i33 Z. 5 5 eat he pewee 
—_ 

— 12 
or tig n= (N+) pee = 


Die A265. } 
n?—l? | 
—n A ae ty 9+ ave. } 
24. Let us apply these formule to examples. 
Ex. 1. Let w,=e°*, then 
a 
Au, w= (et “—1)" e%%= (e%—1)" e** (by A.) 
Ex. 2. Let u,=«, n=4, then, formula (2), 
aaa=V_1 abe So a (a +3) +60. } 

ie ieee 
=/-1e pele 

+ &e.) 
Ex.3. a"e=(-1)"(2-n@+1)+ Levees ) 
= (Te -2"=n(-1)" (1-4 G9) _ ae.) 
=(—1)"2(1—1)"—n(—1)" A—1)"-3 
which is zero when z is greater than 1, finite only when »=1, in which case it is 
1; and infinite when n is less than 1. 
It is evident that this introduction of « may indicate simply that the form 
of the expansion is incorrect: for a © #=0(*#+1)—« «=2+ const. is the analytical 
result of the equation a z=(2+1)—z=02. +const., by dividing both sides by the 
symbol 0. 
