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

514 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series
Hence,
Finally,
p 2
(x) =y(a) −ay ′ (a) + a2
2 y′′ (a)
+{y ′ (a) −ay ′′ (a)}x + y′′ (a)
2 x2
p 2
(x) =y(a) +y ′ (a)(x −a) +y ′′ (x −a)2
(a)
2
p 2
(x) isthesecond-order Taylor polynomialgenerated byyaboutx =a.
Example18.3 Giveny(1) = 0,y ′ (1) = 1,y ′′ (1) = −2,estimate
(a) y(1.5)
(b) y(2)
(c) y(0.5)
usingthe second-order Taylor polynomial.
Solution Thesecond-order Taylor polynomialis p 2
(x):
p 2
(x) =y(1) +y ′ (1)(x −1) +y ′′ (x −1)2
(1)
2
(x −1)2
=x−1−2 =x−1−(x−1) 2 =−x 2 +3x−2
2
We use p 2
(x) as an approximation toy(x).
(a) The value ofy(1.5) isapproximated by p 2
(1.5):
y(1.5) ≈ p 2
(1.5) = 0.25
(b) The value ofy(2) isapproximated by p 2
(2):
y(2) ≈ p 2
(2) =0
(c) The value ofy(0.5) isapproximated by p 2
(0.5):
y(0.5) ≈ p 2
(0.5) = −0.75
Example18.4 (a) Calculatethe second-order Taylor polynomial, p 2
(x), generatedby
y(x)=x 3 +x 2 −6aboutx=2.
(b) Verifythaty(2) = p 2
(2),y ′ (2) = p ′ 2 (2)andy′′ (2) = p ′′ 2 (2).
(c) Comparey(2.1) and p 2
(2.1).
Solution (a) We needtocalculatey(2),y ′ (2) andy ′′ (2). Now
and so
y(x)=x 3 +x 2 −6,y ′ (x)=3x 2 +2x,y ′′ (x)=6x+2
y(2) =6,y ′ (2) =16,y ′′ (2) =14