[132] PROFESSOR DE MORGAN, ON SOME 



be a complete differential of n variables a, b, c, &c., it follows tbat the partial tests A b = B a , &c. 

 \ n {n — 1) in number, are all contained in one, as follows. Assume b a function of a, and 

 let c, e, &c. be the differential coefficients of 6 with respect to a. Then the condition that 

 Ada + Bdb + ... is integrable, or rather the collection of all the \n(n — 1) conditions, is 

 seen in 



= (A + Be + Ce + ...);, - (A + Be + Ce + ...),' + (A + Be + Ce + ...) e " - ... ; 



and as the above suppositions may be made upon every order in which the letters can be 

 written, we have 1.2.3 ... n modes in which this reduction to one form can be applied ; any 

 one of which is sufficient. 



This one criterion, then, involves many; but as we know that perfectly distinct relations 

 cannot be reduced explicitly to one, we cannot but say we know that V — V + ... = is not 

 merely one relation. The following mode of shewing that it involves others suggested itself to 

 me before I was in possession of the preceding sole equation. 



If we were to suppose any letters beginning with y, say y, p, q, to be constants, so that 

 V is a function of x, r a function of x, and s, t, its first two differential coefficients, it is clear 

 that we should still have an integrable function. For if Ada + Bdb + Sac. be integrable 

 independently of relation between a, b, &c. it is still integrable if we make da = 0, or db = 0, &c. 

 Consequently, on such a supposition, V r - V' s + V" vanishes. That is, if in V r , V s , V t , we 

 write p, a, r, for r, s, t, in every place in which they appear, and then form V r — V' + V t ", 

 allowing r, s, t, to appear whenever they result from y, p, q, but writing a for p &c, — the 

 part containing p, a, t, will vanish identically. 



It appears above that the mode of actually finding fVdx is equally applicable to the 

 determination of f(Ada + Bdb ... ), when a and 6 are unconnected. As cleared from all 

 liability to fail by M. Jacques Binet, it is as follows. 



In V write u + zy for y, u being a function of x only which must be definitely assigned in 

 any convenient way before the operation is completed ; y being an indeterminate function of*; 

 and z a variable independent of x and y. Let p, q, r, &c. be the differential coefficients of y in 

 u + xy. If then we look at V, which is a function of x, u + xy, u + xp, &c, as a function 



of %, say fx, we have/(l) =/(°) + / fxdx, 



or V, = V + ['(Vyy +V p p+V q q+ V r r + V,s+ V t t) dx, 



where V t is the function of x, u + y, u + p, &c. and V the same function of x, u, u, &c, 

 which is a determinate function of x : and V y , V p , &c. have u + xy, u + xp, &c. written for 

 y, p, &c. Consequently 



d * • /(Pj-y + v P p + ••• ) dx 



of which the first term is determinate, while the second term is found by performing / dx 

 upon 



