1818.] Mr Adams's Solution of an Equation. 275 



Article VI. 



.4 Solution of the Equation A" (<p x) = (e T * - 1)" <p x . n being 

 a whole Number, <p x a Function of x, and e a Number whose 

 Naperian Logarithm is Unity. By James Adams, Esq. 



(To Dr. Thomson.) 



SIR, Nov. 11, 1817. 



From the theory of increments, we have 

 A" (<p x) = (<p x + n d <px) — n {f x + (« — 1) d <p x} + 



"-It^i?* + (n-2)d?x}-bc (I.) 



But by Taylor's theorem, and separating <p x from its symbol 

 d, we get 



« r -I- nd<DX-n +— + "'**' + — - + &C.}« X 



I 



n , fl (n - 1) d , (n - 1)' * , (« - I) 3 <*» 



f x + (n — 1) d * * = {1 + — — — + l . 2djfl + 1.2.3d*' 



M * 



■f &c.} <p x = e dz x <p x. 



on^ ci ("- 2 W ■ (" - 2 )' * (" ~ 2 ) 3 <* 3 



f> X + (» -2)rf^X = {1 + jj— + , 2 d ^ + 1-2-3 d^ 



d 



(n — a) j- 



•4- &c.} <p x= e dz x f x. 

 &c. &c. 



By substituting in equation (I). 



A- (, x) = (e*- - n e (n -"^ + =i^ e*"**' - &c.} f a 



= (e*' 1 — 1)" f x. 



See Translation of Lacroix's Differential and Integral Calculus, 

 p. 488. 



I trust, Sir, that the importance of the theorem and the con- 

 ciseness of the demonstration, will, in some degree, be an 

 apology for my sending them to you. 



I am, your most obedient servant, 



James Adams. 



s 2 



