082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

18.6 Taylor and Maclaurin series 527
Substitutingz = 2x we obtain
(b) We note that
e 2x =1+2x+ (2x)2
2!
+ (2x)3
3!
+ (2x)4
4!
=1+2x+2x 2 + 4x3
3 + 2x4
3 +···
sinz=z− z3
3! + z5
5! − z7
7! +···
By putting z = 3x we obtain
(c) We note that
sin3x=3x− (3x)3
3!
+ (3x)5
5!
− (3x)7
7!
+···
+···
=3x− 9 2 x3 + 81
40 x5 − 243
560 x7 +···
cosz=1− z2
2! + z4
4! − z6
6! +···
Puttingz = x we obtain
2
( x
cos = 1 −
2) (x/2)2 + (x/2)4 − (x/2)6 +···
2! 4! 6!
= 1 − x2
8 + x4
384 − x6
46080 +···
Example18.14 Determinethe Maclaurinseries fory(x) = 1
1 +x .
Solution The value ofyand its derivatives atx = 0 arefound.
y(x) = 1
1 +x , y(0)=1
y ′ (x) =
−1
(1 +x) 2, y′ (0)=−1
y ′′ 2!
(x) =
(1 +x) 3, y′′ (0) =2!
y ′′′ (x) = −3!
(1 +x) 4, y′′′ (0) = −3!
.
.
y (n) (x) = (−1)n n!
(1 +x) n+1, y(n) (0) = (−1) n n!
Hence usingthe formulafor the Maclaurin series wefind
p(x) = 1 −1(x) +2! x2
2! −3!x3 3! +···+ (−1)n n! xn
n! +···
=1−x+x 2 −x 3 +x 4 −···+(−1) n x n +···