106 Demonstration of Taylor's Theorem, ^c. [Feb. 



Differential Formula, 

 d (sin." x) = n cos. x sin."" \idx 

 rf(cos." x) = — n sin. x cos, "'"' x d x 

 d (vers." x) = n sin. x vers." ~ ' x c? r 

 d (covers." x) = — « cos. x covers."" * x d x 

 d (tan." x) = n sec' x tan.""' xd x 

 d (cot." z) = — n cosec.^ .r cot.""' a: d x . 

 d (sec." x) = n tan. x . sec." x dx 

 d (cosec." x) = — n cot. x cosec." x rf x 



Never having seen a general solution to the following problem, 

 I have ventured to give the following, and will thank you to 

 annex it to the foregoing. 



Problem. — To find the «th increment of the log. x. 

 From the principles of incveiiients, we have 

 A (log. x) = log. ( X + lo) — log. X 



= log. ( 1 + ^) x; - log. X = log. (l + ") 



A« (log. x) = log. (x + 2 w) — log. (x + tc) — log. {x + w) + 

 log. X 

 = log. {x + 2w) — 2 log. (x + w) + log. X 



= log.(l+^)-21og.(l+i) 



= "°g-(l + 7f-.)+'«g-('-,-T:.) 

 A» (log. x)= l(x + 3 w) + l(x + w) —2 l(x + 2w) 

 — I {x + 2io) + 21 (x + w) — I X 

 = l{x + 3v::) - 3l{x + 2iv) + 3l(x-\- w)- Ix 



= '(1+t)-3'(i+^")+3/(,+3 



In like manner may be found A'* (log, x), A^ (log. x), &c. 

 Then by induction we conclude that 



A-(iog. I) =;(i + -=f) 



n (n - 1) ^/^ _^ (n - g) >c N 



n(n - n (n-2^ I /i (» - 3) v, \ 



'2 .3 



+ 



n(n- nC -2) fn - 3) , /, (n-4)tB 



iJ . 3 . 4 

 &C 



«j = A X. 



/(I +'i^-) 



