100 Mr. Adams's Demonstration [Feb. 



them be expressed by the fractions -, —7,, -^. Then by substi- 

 tuting in the last equation, we shall have $»„ =f4-7/A^ = (p 



n n (n — l\ n (n - i) (n — ») „ 



By cancelling the function and dividing by ?i, we get 



1 . n-l , (H-\)(n-2) , („- l)(n-8)(n-3) , ^,_ 

 ^ <? = 7/ -1- i—T? H o s -v'" ' o o 4 ,< 77; f- OcC. 



Now since the numerators and denominators of the fractions 

 in the last equation increase together, the former being Jinite 

 whole numbers, and the latter infleftnife/i/, or incomparab/i/ great 

 whole numbers ; no sensible change will be made in the equation, 

 by supposing each of the numerators equal to unity, according 



to which, and restoring the values of -, — , — , Sec. we get, 



A ^ = rt ip + — ^ + ;j-| + <.rs~A + ^^'- which is commonly 



called Taylor's theorem. 



This theorem is as universal in its application in the theory of 

 increments, or differences, as the binomial theorem is, in the 

 expansion of roots and powers. — (Mr. Barlow's Dictionary; 

 Article, Increment.) 



In order to find the increment of any function of a variable 

 quantity, we must take the first, second, third, &c. differentials 

 of the given function, and divide the results by 1, 2, 2 x 3, 

 2 X 3 X 4, &c. respectively, and we shall have the value of 

 A <p ; which will always be finite unless the function be tran- 

 -scendental. 



Example 1. — To find the increment of .r. 



The first differential of a- is dx, and the second differential of 

 X is zero, because d xh supposed constant ; therefore A x = dx^ 



Example 2. — To find the increment of .i'^. 

 dx"^ = 2 xd X, and d (2 x d x) -^ 2 = d x" = A .r^ (example I) ; 

 therefore A (x-) = 2 x A a: + A a- (d x constant). 



Example 3. — To find the increment of .t^ 

 d (1^) -^x"" dx-, dC:^ x^ dx)-^2 = 2x dx""; d (3 x d a°-) ^ 

 Q =dx^; therefore, A (x^) = 3 a* A x + 3 a A z* + A .r' (dx 

 constant). 



Example 4. — To find the increment of a.*". 

 d{x'") - m x'"-' d X, d (m x""-' d x) = m {m - \) x"'-^ d x\ 



d[n (in - 1) x'^-^dx'') = m (m - 1) (m - 2) x'—' d x\ &c. 

 By writing A x for d x, and substituting in the general theorem, 

 we have A (r") = »i a:— ' A x + "'^"."'^ a'"" ^ A x^ + 



^ (,„ _ 1) fa. - 2) g,^_ ,j ^ constant.) 



2.3 ^ 



Example 5. — To find the increment of a- (x + A x). 

 d (x- + X dx)= 2xd X + d a', and d (2 x d x) =2 d a% then 



