Mr. W. H. L, Russell on the Calculus of Symbols. 441 

 As a last example, we will take the equation 



or putting 



d 



^ 



we can write^the equation (since p=e ^) 



{ p^-p (^^^^^^) ('+ 1) } «=FW- 



Applying the method of divisors, we see that if the symbolical portion of 

 the first member admit of an internal factor, it must be either /o— tt or 

 p-(7r+l). ^ 

 Now in this case 



Hence 



Wherefore the equation 



becomes, if we take the factor /o— tt, and put 



(tt + 1) - (^^^ - 07r + 07r = ; 



an identical equation if we put for the symbol 6 its equivalent as given 

 above. 



Hence p — tt is an internal factor of the symbolical portion of the first 

 member. Effecting the internal division, we have 



(p-(7r+l))(p-7r)/(^) = F(^). 



Let 



(p-t)/'(^)=x(^)' 



and the equation resolves itself into the two, 

 (p-(7r+l))x(^)=F(^) 



and 



(p-7r)/(^) = x(^); 



or 



x(^')-(^+i)xW=FW 



and 



2 k2 



