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

520 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
3 Giveny(x) = sinx,obtain the third-, fourth-and
fifth-orderTaylor polynomials generated byy(x)
aboutx = 0.
4 Giveny(x) = cosx, obtain the third-, fourth- and
fifth-orderTaylor polynomials generated byy(x)
aboutx = 0.
5 (a) Giveny(x) = sin(kx),k a constant, obtain the
third-, fourth-and fifth-orderTaylor polynomials
generated byy(x)aboutx = 0.
(b) Writedown the third-, fourth-and fifth-order
Taylor polynomials generated byy = sin(−x)
aboutx = 0.
6 (a) Giveny(x) = cos(kx),k aconstant, obtain the
third-, fourth-and fifth-orderTaylor polynomials
generated byy(x)aboutx = 0.
(b) Writedown the third-, fourth-and fifth-order
Taylor polynomials generated byy = cos(2x)
aboutx = 0.
7 If p n (x) isthenth-order Taylor polynomial generated
byy(x) aboutx = 0, showthat p n (kx) is thenth-order
Taylor polynomial generated byy(kx) aboutx = 0.
8 Thefunction,y(x),satisfiesthe equation
y ′′ =y+3x 2 y(1) =1,y ′ (1) =2
(a) Obtain athird-orderTaylor polynomial generated
byyaboutx = 1.
(b) Estimatey(1.3)using(a).
(c) Obtain afourth-orderTaylor polynomial
generated byyaboutx = 1.
(d) Estimatey(1.3)using(c).
9 Afunction,y(x),satisfiesthe equation
y ′′ +y 2 =x 3
y(0) =1,y ′ (0) = −1
(a) Estimatey(0.25)usingathird-orderTaylor
polynomial.
(b) Estimatey(0.25)usingafourth-orderTaylor
polynomial.
10 Afunction,y(x),hasy(1) = 3,y ′ (1) = 6,y ′′ (1) = 1
andy (3) (1) = −1.
(a) Estimatey(1.2)usingathird-orderTaylor
polynomial.
(b) Estimatey ′ (1.2)usingan appropriate
second-order Taylor polynomial.
[Hint: define anew variable,z,given byz =y ′ .]
Solutions
1 (a) 3+x− x2
2 + x3
3
2 (a)
t 4 8 − 5t3
6 +t2 +6t− 31
3
(b) 0.1589
3 p 3 (x)=x− x3
3! ,
p 4 (x) =x− x3
3! ,
p 5 (x) =x− x3
3! + x5
5!
4 p 3 (x)=1− x2
2! ,
p 4 (x) =1− x2
2! + x4
4! ,
p 5 (x) =1− x2
2! + x4
4!
(b) 3.1827
5 (a) p 3 (x) =kx− k3 x 3
3! ,
p 4 (x) =kx− k3 x 3
3! ,
p 5 (x) =kx− k3 x 3
3!
(b) p 3 (x) = −x + x3
3! ,
p 4 (x) = −x+ x3
3! ,
+ k5 x 5
5!
p 5 (x) = −x+ x3
3! − x5
5!
6 (a) p 3 (x) =1− k2 x 2
2! ,
p 4 (x) =1− k2 x 2
2!
p 5 (x) =1− k2 x 2
2!
+ k4 x 4
4! ,
+ k4 x 4
4!