1820.] of Taylor's Theorem, S^c. 105 



+ -Afi- (28 + 87 tan.^ x + 60 tan.'' x) tan. .r sec. x 



+ ^•''i (28 + 317 tan.^ x + 648 tan.* x + 360 



tan.® x) sec. x 

 &c ••—•••• 



Example 20. — To find the increment of cosec. x (radius unity). 

 By example 17, 

 A (cosec. X) = — A X cot. x cosec. x 



+ —^ (1 + 2 cot." x) cosec. x 



— ^-^ (2 + 3 cot.- x) cot. X cosec. x 



+ ^ (2 + 13 cot.- X + 12 cot.* x) cosec. x 



i±l (28 + 87 cot.- X + 60 cot." x) cot. r 



cosec. X 

 -f _±fl^ (28 +317 cot."- X + 648 cot.* x + 



3ti0 cot." x) cosec. x 

 Example 21 . — To find the second increment of log. x. 



By example 11, A (log. ^0 = ^-2^.+ ^73-774+ ^^• 

 Where 7v = d x = A x, a constant quantity ; put z = A {I x) 



„, .„ , «,' . 4.c3 I.U.* . 48 Ki 240 «,« „ 



Then m\\dz= --^ + -^--^r + —. 7- + &C. 



ri't «:3 6 10* , 24 tt)S 120 u)« „ 

 = 4. — 4 — \- &C. 



(Pi a^ 8 u)^ 40 U)6 „ 



T= - li +-^ --i^ + 8^c- 



£15= + J^ _l!^+&c. 



i!^= - 4- +scc. 



120 a.^ 



TU C A. /I N Ax"^ 5A^3 19 Aj:4 81 Aj> 



Therefore A^ (log. x) = - -^ + —^ ^^ + —^ 



^•'^^' + &c. 



X' 



This series may be continued at pleasure, as the coefficients 

 observe the following law, viz. 

 2x1+3=5, 3x5 + 4= 19, 4x19 + 5=81, 5x81 + 6=411, &c. 



In finding the increments of examples 17, 18, 19, 20, it was 

 necessary to use the differentials of the powers of the sines, 

 cosines, tangents, secants, &.c. I begleave to insert the follow- 

 ing table which I formed for similar purposes : 



