Crofton — Applications of the Method of Operative Symbols. 609' 
9. To find the value of e^^^\ wHch will lead to that of J)'f{x\ 
we may proceed as follows : — 
i)r = exp. {\a [x + ^) 0»- ; by (10) 
= ^|«x3 ^ax-d ^axd^ ^iad' ^^^^^^ ^ j ^ . 
Now 
,|ac^3,_^^^ r(r-l)(.-2) ^^^_,^ r(.-l)(.-2)(.-3)(r-4)(r-5) ^,^^., ^ ^^^^ 
3 S.6 
operate on both sides by e<^^^^ ; and 
3 
so that if we write 
e"''' ei'- or = F{r) + _ 3) + '"-■■'r-^' - 6) + . . . 
3 3.0 
If we now operate on both sides by e^^'^^, the effect will be ta 
change 0 everywhere into ax^. Hence if we put 
f{r) = [ax'^y+r {r - 1) ax {ax'^y-^ {r - I) {r- 2) {r - 3) ^^^^ ^^^^y-i^ . . 
or if f{r) = arx'^r + r{r-l) a^'^ x^^-z j^ r{r-\){r- 2){i'- 3) ^^ ^ + &c. ) 
(19) 
If we put = ^ = {^x'Y^D, as 
CF{x^)=^F,{x\ CF{x^)=F2{x^), &c., 
this formula will at once (15) give the value of J)''F{x^), 
10. One or two formulas of some interest are added: — Equa- 
tion (1) gives 
h{D+a^x) ^ -a{x) hD ^a{x) ^ ^a{x+h)-a{x) ^hD, 
thus gH^+^)^^W*f')'-¥\^^=,ih-^hx^hD, 
h^ce the result of any such operator on F{x) is found, as 
e^^F{x) = F{x + h). 
