284 Mr. Adams on [Oct. 



By writing v, v' , v", 8cc. for quantities similar to those in 

 Example 6, and substituting in Problem 2, we have 



A" (Z xT = (l.v xT -n{l . v' xT + 1±^ (/ . ^,// j.y .... 



± (^ ^T 

 By multiplying and dividing by (I a )"' 



A-{ixr = {ixr\(i.vr - nii.vr + "^^'^ /.o- .... 



Example 8. — To find the nth increment of sin. x (radium 

 unity). 



Put ip = sin. X and A x = w, a constant quantity, then will 



tn sin. (j + n jf) sin. j: . cos. nm + cos. j: sin. n to 



-: = : = • — : = COS. 11 W -V 



sm. .r sin. J sin.j: ^^<:>. * w -y- 



sin. n w cot. x 

 g-j^ = cos. {n — 1) w + sin. (« — \) w cot. x 



—^ = cos. (« — 2) w + sin. (/i — 2) M) cot. x 



&c 



Then by substituting in Problem 2, 

 A" (sin. x) = sin. r 5 cos. iiw + sin. m u; cot. a; 



— n (cos. («—])«; + sin. {n — l)wcot. xj 



+ " ~ ■ (cos. (« — 2) w + sin.(« — 2)w cot. a:j 



..'...±,j 



The arc A .r or w being considered as incomparably small, the 

 cos. nw, COS. {n — 1) iv, cos. {n — 2) lo, &c. will be very near 

 unity, and the sin. n w, sin. (« — ])w, sin. (« — 2) w, &c. will be 

 nearly equal to » 7V, (n — 1) ?u, (?« — 2) w, 8cc. on these suppo- 

 sitions, the above equation would become 



A" (sin. x) = sin. .r 5 1 + n w cot. x — n(l + (n — l) w cot. xj 

 + !!i!Lzi} ^1 +(„ _2)t:7cot.x) .... ± Ij 



Example 9. — To find the «th increment of cos. x (radius 

 unity). 



Put ip = cos. X, and A a; = i<7, a constant quantity, then will 



<Pn fs- {x + n to) 



COS. X COS. X 



= COS. n to — sm. n w tan. x 



COS. r 



<r'n— -1 

 COS. X 



&C 



= COS. (n — \)w — sin. (« — l)wtan.a: 

 = cos. (u — 2) w — sin. {n — 2) iv Ian. x 



