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508 Chapter 18 Taylor polynomials, Taylor series and Maclaurin series

18.2 LINEARIZATIONUSINGFIRST-ORDERTAYLOR

POLYNOMIALS

Suppose we know that y is a function of x and we know the values of y and y ′ when

x =a,thatisy(a)andy ′ (a)areknown.Wecanusey(a)andy ′ (a)todeterminealinear

polynomial which approximates toy(x). Let this polynomial be

p 1

(x) =c 0

+c 1

x

We choose the constantsc 0

andc 1

so that

p 1

(a) =y(a)

p ′ 1 (a) =y′ (a)

that is, the values of p 1

and its first derivative evaluated atx = a match the values ofy

and itsfirst derivative evaluated atx =a. Then,

p 1

(a) =y(a) =c 0

+c 1

a

p ′ 1 (a) =y′ (a) =c 1

Solving forc 0

andc 1

yields

Thus,

c 0

=y(a) −ay ′ (a)

c 1

=y ′ (a)

p 1

(x) =y(a) −ay ′ (a) +y ′ (a)x

p 1

(x) =y(a) +y ′ (a)(x −a)

p 1

(x) isthefirst-order Taylor polynomialgenerated byyatx =a.

y

O

Q

Figure18.1

Graphical representation

ofafirst-orderTaylor

polynomial.

a

x

The function,y(x), isoften referred toas thegenerating function. Note that p 1

(x) and

its first derivative evaluated atx =a agree withy(x) and its first derivative evaluated at

x=a.

First-order Taylor polynomials can also be viewed from a graphical perspective.

Figure 18.1 shows the function, y(x), and a tangent at Q where x = a. Let the equation

ofthe tangent atx =abe

p(x)=mx+c

The gradient of the tangent is, by definition, the derivative ofyatx = a, that isy ′ (a).

So,

p(x) =y ′ (a)x +c

The tangent passes through the point (a,y(a)), and so

y(a) =y ′ (a)a +c

that isc =y(a) −y ′ (a)a. The equation of the tangent isthus

p(x) =y ′ (a)x +y(a) −y ′ (a)a

p(x) =y(a) +y ′ (a)(x −a)

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