.364 MAJOR P. A. MxcMAHON ON COMBINATORIAL ANALYSIS. 



If no part IT presents itself in the operand, D, causes the monomial function to 

 vanish. Thus 



The compound operation D^D.J),, . . . denotes the successive performance of the 

 operations 



D,, , D, t , D ffj , ... of orders ir-^ , ir. 2 , ir 3 , . . . respectively. 



The law of operation of D,, establishes that the component operations D^.D^D,,, . . . 

 may be performed in any order. Thus 



D.DjTTjTT^o- ) = D^fapo- . . . ) = D.,(TT 2 p<T . . . ) = (pa- . . . ) . 



As the order of operation is immaterial, it is found convenient in most cases to 

 operate with D^D^D^ ... in the order D,,, D,,, D, s , . . . ; this may seem at first 

 sight at variance with the ordinary usage in the Differential Calculus, but there is 

 a convenience in ordering the operator from left to right in agreement with the 

 practice of ordering a partition from left to right. If, further, we note the result 



D.(ir) = 1 



we have a complete account of the operator so far as it is concerned with an operand, 

 which is merely a monomial symmetric function. The operation of D, upon a symmetric 

 function product is of even greater importance in the present theory. It has the 

 effect of erasing a partition of TT from the product, one part from each factor, in all 

 possible ways ; the result of the operation being a sum of products, one product 

 arising from each such erasure of a partition. This has been set forth at length in 

 the papers to which reference has been given, but in deference to the suggestion of 

 one of the Referees appointed by the Royal Society to report upon the present paper, 

 a number of examples are given to familiarise readers with the processes which are so 

 much employed in what follows. 

 Example 1. Consider 



IX(i)". 



The operand consists of n factors, each of which is (1) ; the operator D n is performed 

 through the partition of the number IT which involves TT units ; this partition must 



/ n\ 

 be erased from (l)" = (l)(l)(l). . . ton factors in each of the I j possible ways, and 



the results added. Thus 



As a particular case 



D 2 (l) 4 = (I) (*)(!)(!) + (Z)(1)(D(1) + (*)(!)(!)(*) + (1)00 (!)(!) 

 + (1) (/) (1) (*) + (1) (1) (*) (*) = 6 (I) 2 = (J) (I) 2 . 



