PROFESSOR KELLAND ON GENERAL DIFFERENTIATION. 


280 
+ae* {2s Ot elo 2} 
rate {go 4D} y = 2 af (a)e-* 
which being reduced gives 


2(D+1)(D+2) =a 

=a O41) — 
1 
Or ~ +3) O+H WF5)7 
1 i 
2 g2e-24 he 26 
+2a’e (D+2)(D ay? Tae ee 
Now y=/(a+z), but since a is itself a function of 2, we cannot proceed fur- 
ther with the reduction of this equation by division, but must proceed to obtain a 
relation between « and R or a and z. 
To do this we shall expand / (a +z) by Taytor’s Theorem. 
The result is 
y= tL = = which being substituted in the reduced equation, cives by (A) 
In 
ey a ee 
(D+ 2) (D+ 3) (D+4) 





d™ f(a) en? =A 
a es d a =) erarese (n+ 9) 
ah at e274 \-2 —26 
*Gib@sa Gab. “Gamat sis eo 
But a=a—R= a—5: hence 
f(@)_@fa w*tya. 
ia? is ioe 




which being substituted for a the sum being taken for n and p, we get 
pt eal aires le Se 
mP QP  da®*P [n+] fp+1 { (m+) (w+ 4) (wn +5) 
emg sea 2a tP 2 \ ae fa 
+ (1 +2) (n+8) +4) * 42) (n43)$ —3 
every integer value of n and p being taken from 0 to 0. 
When x=0, p=0, the left-hand side gives 
—fa. adfa 
60° Se 







