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

18.3 Second-order Taylor polynomials 513
4 (a) Find alinear approximation, p 1 (t),toh(t) =t 3
aboutt = 2.
(b) Evaluateh(2.3)and p 1 (2.3).
5 (a) Findalinearapproximation, p 1 (t),toR(t) = 1 t
aboutt = 0.5.
(b) EvaluateR(0.7)and p 1 (0.7).
Solutions
1 (a) p 1 (x)=x+1
(b) p 1 (x) =e 2 (x −1)
(c) p 1 (x) =e −3 (x +4)
2 (a) p 1 (x) =x
(b) p 1 (x) = 0.5403x +0.3012
(c) p 1 (x) = 0.8776x −0.0406
4 (a) p 1 (t) =12t −16
(b) h(2.3) = 12.167,p 1 (2.3) = 11.6
5 (a) p 1 (t) = −4t +4
(b) R(0.7) = 1.4286,p 1 (0.7) = 1.2
3 (a) p 1 (x) =1
(b) p 1 (x) = −0.8415x +1.3818
(c) p 1 (x) = 0.4794x +1.1173
18.3 SECOND-ORDERTAYLORPOLYNOMIALS
Suppose that in addition to y(a) and y ′ (a), we also have a value of y ′′ (a). With this
informationasecond-orderTaylorpolynomialcanbefound,whichprovidesaquadratic
approximationtoy(x). Let
p 2
(x) =c 0
+c 1
x+c 2
x 2
We require
p 2
(a) =y(a)
p ′ 2 (a) =y′ (a)
p ′′ 2 (a) =y′′ (a)
thatis,thepolynomialanditsfirsttwoderivativesevaluatedatx =amatchthefunction
and its first two derivatives evaluated atx =a. Hence
p 2
(a) =c 0
+c 1
a +c 2
a 2 =y(a) (18.1)
p ′ 2 (a) =c 1 +2c 2 a =y′ (a) (18.2)
p ′′ 2 (a) = 2c 2 =y′′ (a) (18.3)
Solvingforc 0
,c 1
andc 2
yields
c 2
= y′′ (a)
from Equation (18.3)
2
c 1
=y ′ (a) −ay ′′ (a) fromEquation (18.2)
c 0
=y(a) −ay ′ (a) + a2
2 y′′ (a) from Equation (18.1)