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

25.6 Taylor polynomials and Taylor series in two variables 837
(c) VerifythatthefirstpartialderivativesoftheTaylorpolynomialand f haveidentical
values at(1, 2).
(d) Estimate f(1.1,1.9)usingtheTaylorpolynomial.Comparethiswiththetruevalue.
Solution (a) We aregiven f =x 2 +xy −y 3 and so
∂f
∂x =2x+y
∂f
∂y =x−3y2
Evaluating these atthe point(1, 2) gives
f(1,2)=−5
∂f
∂x (1,2)=4
∂f
(1,2) = −11
∂y
Puttinga = 1 andb = 2 in the formula for p 1
(x,y) we are able to write down the
Taylor polynomial:
p 1
(x,y) = f(1,2)+(x−1) ∂f
∂x (1,2)+(y−2)∂f ∂y (1,2)
= −5 + (x −1)4 + (y −2)(−11)
=4x−11y+13
The first-order Taylor polynomial is p 1
(x,y) = 4x −11y +13.
(b) We can evaluate the Taylor polynomial and the function at(1, 2).
p 1
(1,2)=4−22+13=−5
f(1,2)=−5
Hence p 1
(1,2) = f (1,2); that is, the Taylor polynomial and the function have
identical values at(1, 2).
(c) The first partialderivatives of p 1
(x,y) are found.
∂p 1
∂x = 4 ∂p 1
∂y = −11
∂f
∂x (1,2)=4
∂f
(1,2) = −11
∂y
Hence the first partialderivatives of p 1
and f are identical at(1, 2).
(d) p 1
(1.1,1.9) = 4(1.1) −11(1.9) +13 = −3.5
f(1.1,1.9) = (1.1) 2 + (1.1)(1.9) − (1.9) 3 = −3.559
ThevaluesoftheTaylorpolynomialandthefunctionareincloseagreementnearto
the point(1, 2).
25.6.2 Second-orderTaylorpolynomialintwovariables
Given a function, f, and its first and second partial derivatives at (a,b) we can write
down the second-order Taylor polynomial, p 2
(x,y).
p 2
(x,y) = f(a,b)+(x−a) ∂f
∂x (a,b)+(y−b)∂f ∂y (a,b)
+ 1 (
)
(x−a) 2∂2 f
2! ∂x (a,b)+2(x−a)(y−b) ∂2 f f
2 ∂x∂y (a,b)+(y−b)2∂2 ∂y (a,b) 2