1820.] the Direct Method of Finite Differences. 287 



A" (cosec. X) = cosec. x |i^„„,„,.^ - i + („_",)„ eot. x 

 + "J-^ ....±1? 



2 (l + (n - 2) 10 cot. x\ ^ 



Example 14. — To find the wth increment of a,,. 

 Put (p = x^, then will 



<P. (^J, = (^- + « ^ ^)!,+ n Ay = {X+71AX) + (1/ + uAy) 

 n 



A X = x„ + y„ X Ax, 



n— 1 



X A X, &c. 

 A X and A ?/ supposed constant. 

 By properly arranging the terras, 



A- {x^) = X-, - nx„_, + li!i_li} a:_^ _ &c. 



Therefore by Problem 2, A" (^,) = A" a; + A"y . A a:'. 

 Scholium. — By corollary to Prob. 2, we have 



A" (p = (p (e'' — 1)" = e"''* — /« e'"""'''^ + " ^" ~ ^^ e("-2)rf* — 



&c. (I.) 

 Therefore any of the preceding examples may be solved byfind- 

 ingthe differential of the function <p, and writing its value in the 

 equation marked (I.): take Example 1, where 

 d(f = d (x") =s mx'"~' d X = m x'"~ ' A x = mx'"~' w, 

 Then by substituting in equation (I.) we have 



m — 1 >ii — i n (n 11 w — » 



— &c. 



By taking Example 3, we have 

 d ip = d («') = d X . a" log. a = A x a'' . I a = w a" I a, 



Then by substituting in equation (I.) we have 

 A" (a") = e""""''" — Ke"" (—')«''« + " ^" ~ ^^ ^."("-^ja^n _ g^^,^ 



and so on. 



Example 15. — To find the n\h. increment of a; y. 



By separating the symbols of operation from their quantities, 

 and denoting the symbol of j^ by A, and that ofy by A', we have 



A{x y) = {x + A x) {y -{■ A' y) — xy 

 = (1 + A) (1 '+ A') xy — x'y 

 = xy{{l + A)(l + AO- 1} 

 = X y (A + A' + A AO 

 = x'y {A + A' (1 + A)} 

 Then by operating on the symbols only, we have 



