300 PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. . 
= X+0 9 X+0 
= TOA) hes Eee 
_A+B.2-" 2- 
1 a+1 
je+1 rce Si a je+1 oe 
Cor. In the same manner may the more general equation 
U,—a(X#+1) UW, 21+6(a4+1) (+2) u,,0+&ce.=X, be solved. 
It is not necesary to solve such equations as u,,,+au,+&c=X, since it 
is evident that, by putting z=4 7, this form of equation is reduced to 
v, +av,,1+&c.=X, which has been already solved. 
We proceed then to the solution of equations involving fractional differences. 
Here we must, at present, confine ourselves to very simple examples. 
Ex. 4. at U, —4U,,=90. 
Since at eme=(em —1)} pme 
It is evident that if m = log (a?+1) the solution of the equation is w,=A e™* 
=A gee’) — Aer 1) 

Ex.. 5. At u,—@ U,=C e** 
Uy, =A POE ge malas ie cm 
( ) (@aiioa 
or, if e*=14+0,u,=A (@ +1)" +5— (841) 
Cor. If 6=a, this solution fails. Put 6=a+@ as in similar cases in Differ- 
ential Equations : 
c 
b—a 
(82 +1)"= © (a? 414208)" =< (a? +1)"+ $2082 (a2+1)"7} 
pB 6 8 
Cy (a24+1)?+2acz(a?+1)"-+ 

then 
A, being an arbitrary constant. 
It may be interesting to verify this solution. 
At (a? +1)*—a. (a? +1)"=0 evidently, 
and At Qacz(a?+1)*-! =2acxa(a?+1)*-!+42ac (a +1) by Prop. 5. 
=2 a? ex (a?+1)*-1+¢ (a? +1)” 
A? .2Qaca(a?+1)*-1-a. 2acx (a? +1)*-1=c (a? +1)"=ce** as it ought. 
Ex. 6. Generally, let a?u,—au,=X 
then Up=A (a2 +1)*+(at—a)-1X 
