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

We know |x| < 1, and so 4x5
15
18.5 Taylor’s formula and the remainder term 523
4
is never greater than . Hence an upper bound for
15
the error term is 4 15 . If we use p 4
(x) to approximatey = sin2x the error will be no
greater than 4 provided |x| < 1.
15
(d) We letx = 0.5.
y(0.5) = sin1 = 0.8415
p 4
(0.5) = 2(0.5) − 4 3 (0.5)3 = 0.8333
Thedifferencebetweeny(0.5)andp 4
(0.5)canneverbegreaterthananupperbound
of the errortermevaluated atx = 0.5. This isverified numerically.
and
y(0.5) −p 4
(0.5) = 0.8415 −0.8333 = 0.0082
|R 4
(0.5)| 4
15 (0.5)5 = 0.0083
EXERCISES18.5
1 Thefunction,y(x),is given byy(x) = sinx.
(a) Calculate the fifth-orderTaylor polynomial
generated byyaboutx = 0.
(b) Find an expression forthe remainder term of
order 5.
(c) Statean upper bound foryour expression in (b).
2 RepeatQuestion1withy(x) = cosx.
3 Thefunctiony(x) = e x maybe approximated by the
quadraticexpression1 +x+ x2
2 .Findanupperbound
forthe error term given |x| < 0.5.
4 (a) Find the third-orderTaylor polynomial generated
byh(t) = 1 aboutt = 2.
t
(b) State the error term.
(c) Find an upper bound forthe errorterm given
1t4.
5 The functiony(x) =x 5 +x 6 isapproximated bya
third-orderTaylor polynomial aboutx = 1.
(a) Find an expression forthe third-ordererror term.
(b) Find an upper boundforthe errorterm given
0x2.
Solutions
1 (a) x− x3
3! + x5
5!
(b) R 5 =− (sinc)x6
6!
whereclies between 0 andx
x 6
(c)
∣6!
∣
2 (a) 1− x2
2! + x4
4!
(b) R 5 =− (cosc)x6
6!
whereclies between 0 andx
x 6
(c)
∣6!
∣