AS COMMONLY INVESTIGATED. 245 



which, by what we have already shown, would give 



z z 



P = Q. 



But if we suppose the development to commence with one of the terms of 

 the expansion, as d° F (x), and thence demand the most general form of the 



terms on the left, we shall observe that the investigation of Lagrange fails in 

 generality, and that the consequence deduced from it, that only a constant can 

 have all its differential coefficients equal to zero for the same value of x, is 

 neither just in itself, nor so great a paradox when violated as Mr. Peacock 

 seems to suppose.* 



Indeed, including in modal functions of the first order of continuity such as 

 have a portion absolutely rectilinear, and which depart thence into a curve by 

 insensible gradations, it is clear that if we can find a numerical representative 

 F X of such a modal function, Taylor's Theorem would hold throughout its 

 whole extent, and that consequently Fx, a function which is not a constant, 

 would yet fulfil the equation 



{d; (F.x = 0} (10) 



^ X = X ... X 



for every value of n. 



The very simple function -. Fx. tan."^ (0) ^~^ is constantly zero from 

 X = — a to X = a, and constantly equal to F x from x = a to x = a; and 

 thus affords an elementary case of a function fulfilling the equation (10) : but 

 it should be remarked that as the transition takes place in an infinitely small 

 interval, this function does not fall under the first order of continuity. 



This example I have taken from the pamphlet before mentioned, and all 

 that it is necessary to say further on the subject, may be taken from the same 

 source. 



"We see therefore," it is remarked, ''that with proper limitations, v/hich, too, 

 are only necessary in regard to functions of a low order of continuity, we may 

 assert that a constant is the only function that can give the first differential 

 zero; and this granted it is easily seen that ax + a' is the only function that 

 can give the first differential a constant. For, let <^ x be any other function 

 that also gives d <|) (x) = a; we have 



X 



d ^(px — (ax + a')| = 0; 



* See what has been already said on the function e ^' 

 VII. — 3 M 



