Mr. W. H. L. Russell on the Calculus of Symbols. 427 



more of the symbols Y, and taking the aggregate of the terms so formed. 

 The conditions of division will he, as before, 



Y„F = 0, V,_iR,=0, V„_2E2=0 



Let us now investigate the conditions that F may be externally divisible 



We see at once that F, as before, must take the form Z°F+y„Z'„F, and 

 also that Z'„F can contain neither y,^ nor Hence we shall have 



VIF=0, and also V„_iV,,F=0. 



Now 



^(tJ,_,Z;F)=^{Y„_2+y,_iV„_,}U„_,Z'„F 



=XY,,_.oU,.2Z'„F+y„_iXZ'„F + y,Z;F. 

 Hence we shall have 



Rx=Z«F-XY„_,U,-2Z'»F-y„_iXZ'„F, 

 when we must introduce the conditions 



V2_,E, = 0, and V„_2V„_iRi=0 ; 

 consequently we shall have 



Ili=Z°_,Z«F- ZL,XY„_,U„-2Z;F + 



(Z'„_iZ°F-Z'„_iXY„_,U„_.oZ'„F-ZLiXZ'„F)y„_i. 



Now 



#,U„_3(Z',,_iZ«F-Z'„_,XY,_,U„_,Z',.F 

 dx 



-ZLiXZ'„F) = 

 XY„_3U„_3Z'„_,Z«F - XY„_3lJ„_3Z'„_,XY„_,U„-2Z'nF 

 -XY,,_3U._3Z«_,XZ'„F+ (XZ'„_,Z«F- 



XZ',._iXY,_oU„-2Z'„F - XZLiXZ',F)y„_2 + 

 (Z'„_,Z?.F-Z'„_,XY„_XT,_,Z'„F-Z°_,XZ'„F)y,_,. 



Hence 



E2= ZLiZ«F - ZLxXY„_A.-2Z'„F 



- XY„_3U„_3Z'„_xZ?.F + XY,_3LT„_3Z'„_iXY„_,U,._2Z',F 

 + XY,_3U,,_3Z«_,XZ'„F + (XZ',_,XY„_2U,-2Z'.F 

 + XZLiXZ'„F-XZ'„_iZ?.F)y„_2. 

 Introducing the conditions 



VL.oR^^o, v„.3Y„_2R2=o, 



we find 



R2=Z«-2ZLiZ«F- 



ZL2Z^xXY„_oU,-2Z'.F-Zt,XY„_3U,_3Z'„_iZ«„F 



+ ZL2XY,_3U„_3Z'„_iXY„_,U„_2Z'„F 



+ ZL,XY„_3U,_3Z»_,XZ',.F + (Z'„_,ZL,Z°F 



-Z'„_2Z«_iXY„_A,_2Z'„F-Z',_2XY„_3U„_3Z'„-iZf.F 



+ Z'„_2XY„_3U„_3Z',_iXY„_JJ„_oZ',F 



+ Z'„_2XY„_3U„_3ZriXZ'„F + Z°_2XZ'„_iXY„.2U,..2Z;F 



+ Z°_2XZ?,_iXZ',F-Z?._2XZ',._iZ°F)y„.2. 



