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

522 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Example18.8 Calculate the error term of order 5 due to y(x) = e x generating a Taylor polynomial
aboutx = 0.
Solution Heren = 5 and son +1 = 6.Inthisexamplea = 0.We see thaty = e x and so
y ′ (x) =y ′′ (x) = ··· =y (5) (x) =y (6) (x) =e x
and soy (6) (c) = e c . The remainder termoforder 5,R 5
(x), isgiven by
R 5
(x) = ec x 6
6!
forsome numbercbetween 0 andx
Example18.9 (a) Calculatethe fourth-orderTaylor polynomial, p 4
(x), generatedbyy(x) = sin2x
aboutx = 0.
(b) State the fourth-order errorterm,R 4
(x).
(c) Calculate anupper bound forthis errortermgiven |x| < 1.
(d) Comparey(0.5) and p 4
(0.5).
Solution (a) y(x) =sin2x, y(0) = 0
y ′ (x)=2cos2x, y ′ (0) =2
y ′′ (x)=−4sin2x, y ′′ (0) =0
y (3) (x)=−8cos2x,
y (3) (0) = −8
y (4) (x)=16sin2x, y (4) (0) =0
Thefourth-orderTaylor polynomial, p 4
(x), is
p 4
(x) =y(0) +y ′ (0)x +y ′′ (0) x2
2! +y(3) (0) x3
3! +y(4) (0) x4
4!
=0+2x+0− 8x3
6 +0
=2x− 4x3
3
(b) Wenotethaty (5) (x) = 32cos2xandsoy (5) (c) = 32cos(2c).Theerrorterm,R 4
(x),
isgiven by
R 4
(x) =y (5) (c) x5
5!
= 32cos(2c)x5
120
= 4
15 cos(2c)x5 wherecisanumber between 0 andx
(c) In order to calculate an upper bound for this error term we note that |cos(2c)| 1
forany value ofc. Hence an upper bound forR 4
(x) isgiven by
|R 4
(x)|
∣
4x 5
15
∣