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

526 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Since the generating function and series are equal forall values ofxwehave
∞∑ (−1) i x 2i+1
sinx= =x − x3
(2i +1)! 3! + x5
5! − x7
7! +···
i=0
Example18.12 Obtain the Maclaurinseries fory(x) = cosx.
Solution y(x)=cosx, y(0)=1
y ′ (x)=−sinx, y ′ (0)=0
y ′′ (x)=−cosx,
y ′′ (0)=−1
y (3) (x)=sinx, y (3) (0)=0
y (4) (x)=cosx, y (4) (0)=1
and soon. Therefore
p(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2! +y(3) (0) x3
3! +y(4) (0) x4
4! +···
= 1 − x2
2! + x4
4! −···
∞∑ (−1) i x 2i
=
(2i)!
i=0
Since the series and the generating function areequal for all values ofxthen
∞∑ (−1) i x 2i
cosx= = 1 − x2
(2i)! 2! + x4
4! − x6
6! +···
i=0
FromExamples18.10, 18.11and 18.12wenotethreeimportantMaclaurinseries:
e x =1+x+ x2
2! + x3
3! + x4
+ ··· forall values ofx
4!
sinx=x− x3
3! + x5
5! − x7
+ ··· forall values ofx
7!
cosx=1− x2
2! + x4
4! − x6
+ ··· forall values ofx
6!
Example18.13 Findthe Maclaurinseries forthe following functions:
( x
(a)y = e 2x (b)y = sin3x (c)y = cos
2)
Solution We use the previously stated series.
(a) We note that
e z =1+z+ z2
2! + z3
3! + z4
4! +···