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

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516 Chapter 18 Taylor polynomials, Taylor series and Maclaurin seriessop 2(V) =I s(e 40V a −1) +40Is e 40V a (V −Va ) +1600I se 40V (V −Va a) 22Thecoefficientsneedtobecalculatedonlyonce.Afterthatthecalculationofacurrentvalue only involves evaluating a quadratic.EXERCISES18.31 (a) Obtainthe second-order Taylor polynomial,p 2 (x),generatedbyy(x) = 3x 4 +1aboutx = 2.(b) Verifythaty(2) = p 2 (2),y ′ (2) = p ′ 2 (2)andy ′′ (2) = p ′′ 2 (2).(c) Evaluate p 2 (1.8)andy(1.8).2 (a) Calculate the second-order Taylor polynomial,p 2 (x),generated byy(x) = sinx aboutx = 0.(b) Calculate the second-order Taylor polynomial,p 2 (x),generated byy(x) = cosx aboutx = 0.(c) Compare your resultsfrom (a) and(b)with thesmall-angle approximations given in Section6.5.3 Afunction,y(x),issuchthaty(−1) = 3,y ′ (−1) = 2andy ′′ (−1) = −2.(a) Statethe second-order Taylor polynomialgenerated byyaboutx = −1.(b) Estimatey(−0.9).4 Afunction,y(x),satisfiesthe equationy ′ =y 2 +xy(1)=2(a) Estimatey(1.3)usingafirst-orderTaylorpolynomial.(b) Bydifferentiatingthe equationwith respect tox,obtain an expression fory ′′ .Hence evaluatey ′′ (1).(c) Estimatey(1.3)usingasecond-order Taylorpolynomial.5 Afunction,x(t),satisfiesthe equationẋ=x+ √ t+1x(0)=2(a) Estimatex(0.2)usingafirst-orderTaylorpolynomial.(b) Differentiate the equation w.r.t.t and henceobtain an expression forẍ.(c) Estimatex(0.2)usingasecond-order Taylorpolynomial.6 Afunction,h(t),isdefined byh(t) = sin2t +cos3tObtainthe second-order Taylor polynomial generatedbyh(t)aboutt = 0.7 Thefunctionsy 1 (x),y 2 (x)andy 3 (x)are defined byy 1 (x) =Ae x ,y 2 (x) =Bx 3 +Cx,y 3 (x) =y 1 (x) +y 2 (x)(a) Obtainasecond-order Taylor polynomial fory 1 (x)aboutx = 0.(b) Obtainasecond-order Taylor polynomial fory 2 (x)aboutx = 0.(c) Obtainasecond-order Taylor polynomial fory 3 (x)aboutx = 0.(d) Canyou draw any conclusionsfrom youranswers to (a),(b)and(c)?Solutions1 (a) p 2 (x) = 72x 2 −192x +145(c) p 2 (1.8) = 32.68,y(1.8) = 32.49282 (a) p 2 (x) =x (b) p 2 (x) =1− x223 (a) p 2 (x) = −x 2 +4(b) p 2 (−0.9) = 3.19.This isan approximation toy(−0.9).4 (a) 3.5(b) y ′′ = 2yy ′ +1,y ′′ (1) = 21(c) 4.4455 (a) 2.61(b) ẍ=ẋ +2 √ t +1(c) 2.67

18.4 Taylor polynomials of the nth order 5176 p 2 (t) =1+2t −4.5t 2( )7 (a)A1 +x+ x22(b) Cx(c) A+(A+C)x+ Ax22TechnicalComputingExercises18.3Many technicalcomputing languageshave thecapability ofproducingaTaylor polynomial ofafunction.Insome languagesthisis available asadefaultwhereasin others, such asMATLAB ® ,thisisoffered via a toolboxfunctionwhichmayneedto beloaded orpurchased separately. Ifthe software youare usingdoes not have the capability to produce theTaylor seriesautomaticallyyou maychoose to dothispartmanuallyandthenplot the resultsusingthetechnicalcomputing language.1 (a) Calculate the second-order Taylor polynomial,p 2 (x),generatedbyy(x) =x 3 aboutx = 0.(b) Drawy(x)andp 2 (x)for −2 x2.2 (a) Calculate the second-order Taylor polynomial,p 2 (x),generated byy(x) = sinx aboutx = 0.(b) Drawy(x)andp 2 (x)for −2 x 2.3 (a) Calculate the second-order Taylor polynomial,( )1p 2 (x),generated byy(x) = sin aboutx = 3.x(b) Drawy(x)andp 2 (x)for1 x 5.4 (a) Calculate the second-order Taylor polynomial,p 2 (x),generated byy(x) = e cosx aboutx = 0.(b) Drawy(x)andp 2 (x)for −2 x 2.18.4 TAYLORPOLYNOMIALSOFTHEnTHORDERIf we know y and its first n derivatives evaluated at x = a, that is y(a), y ′ (a),y ′′ (a), ...,y (n) (a), thenthenth-orderTaylorpolynomial, p n(x), maybewrittenasp n(x)=y(a) +y ′ (a)(x −a) +y ′′ (x −a)2(a) +y (3) (x −a)3(a)2! 3!+···+y (n) (x −a)n(a)n!This provides an approximation toy(x). The polynomial and its firstnderivatives evaluatedatx = a match the values ofy(x) and its firstnderivatives evaluated atx = a,thatisp n(a) =y(a)p ′ n (a) =y′ (a)p ′′ n (a) =y′′ (a).p (n)n (a) =y(n) (a)

516 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series

so

p 2

(V) =I s

(e 40V a −1) +40Is e 40V a (V −Va ) +1600I s

e 40V (V −V

a a

) 2

2

Thecoefficientsneedtobecalculatedonlyonce.Afterthatthecalculationofacurrent

value only involves evaluating a quadratic.

EXERCISES18.3

1 (a) Obtainthe second-order Taylor polynomial,

p 2 (x),generatedbyy(x) = 3x 4 +1aboutx = 2.

(b) Verifythaty(2) = p 2 (2),y ′ (2) = p ′ 2 (2)and

y ′′ (2) = p ′′ 2 (2).

(c) Evaluate p 2 (1.8)andy(1.8).

2 (a) Calculate the second-order Taylor polynomial,

p 2 (x),generated byy(x) = sinx aboutx = 0.

(b) Calculate the second-order Taylor polynomial,

p 2 (x),generated byy(x) = cosx aboutx = 0.

(c) Compare your resultsfrom (a) and(b)with the

small-angle approximations given in Section6.5.

3 Afunction,y(x),issuchthaty(−1) = 3,y ′ (−1) = 2

andy ′′ (−1) = −2.

(a) Statethe second-order Taylor polynomial

generated byyaboutx = −1.

(b) Estimatey(−0.9).

4 Afunction,y(x),satisfiesthe equation

y ′ =y 2 +x

y(1)=2

(a) Estimatey(1.3)usingafirst-orderTaylor

polynomial.

(b) Bydifferentiatingthe equationwith respect tox,

obtain an expression fory ′′ .Hence evaluate

y ′′ (1).

(c) Estimatey(1.3)usingasecond-order Taylor

polynomial.

5 Afunction,x(t),satisfiesthe equation

ẋ=x+ √ t+1

x(0)=2

(a) Estimatex(0.2)usingafirst-orderTaylor

polynomial.

(b) Differentiate the equation w.r.t.t and hence

obtain an expression forẍ.

(c) Estimatex(0.2)usingasecond-order Taylor

polynomial.

6 Afunction,h(t),isdefined by

h(t) = sin2t +cos3t

Obtainthe second-order Taylor polynomial generated

byh(t)aboutt = 0.

7 Thefunctionsy 1 (x),y 2 (x)andy 3 (x)are defined by

y 1 (x) =Ae x ,y 2 (x) =Bx 3 +Cx,

y 3 (x) =y 1 (x) +y 2 (x)

(a) Obtainasecond-order Taylor polynomial for

y 1 (x)aboutx = 0.

(b) Obtainasecond-order Taylor polynomial for

y 2 (x)aboutx = 0.

(c) Obtainasecond-order Taylor polynomial for

y 3 (x)aboutx = 0.

(d) Canyou draw any conclusionsfrom your

answers to (a),(b)and(c)?

Solutions

1 (a) p 2 (x) = 72x 2 −192x +145

(c) p 2 (1.8) = 32.68,y(1.8) = 32.4928

2 (a) p 2 (x) =x (b) p 2 (x) =1− x2

2

3 (a) p 2 (x) = −x 2 +4

(b) p 2 (−0.9) = 3.19.This isan approximation to

y(−0.9).

4 (a) 3.5

(b) y ′′ = 2yy ′ +1,y ′′ (1) = 21

(c) 4.445

5 (a) 2.6

1

(b) ẍ=ẋ +

2 √ t +1

(c) 2.67

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