Maclaurin and Taylor Polynomials Objective: Improve on the local linear approximation for higher order polynomials

Maclaurin and Taylor Polynomials

Objective: Improve on the local linear approximation for higher

order polynomials.

Local Quadratic Approximations• Remember we defined the local linear approximation

of a function f at x0 as . In this formula, the approximating function

is a first-degree polynomial. The local linear approximation of f at x0 has the property that its value

and the value of its first derivative match those of f at x0.

))(()()( 00/

0 xxxfxfxf

))(()()( 00/

0 xxxfxfxp

Local Quadratic Approximations• If the graph of a function f has a pronounced “bend”

at x0 , then we can expect that the accuracy of the local linear approximation of f at x0 will decrease rapidly as we progress away from x.

Local Quadratic Approximations• One way to deal with this problem is to approximate

the function f at x0 by a polynomial p of degree 2 with the property that the value of p and the values of its first two derivatives match those of f at x0. This ensures that the graphs of f and p not only have the same tangent line at x0 , but they also bend in the same direction at x0. As a result, we can expect that the graph of p will remain close to the graph of f over a larger interval around x0. The polynomial p is called the local quadratic approximation of f at x = x0.

Local Quadratic Approximations• We will try to find a formula for the local quadratic

approximation for a function f at x = 0. This approximation has the form where c0 , c1 , and c2 must be chosen so that the values of

and its first two derivatives match those of f at 0. Thus, we want

2210)( xcxccxp

2210)( xcxccxf

)0()0(),0()0(),0()0( ////// fpfpfp

Local Quadratic Approximations• The values of p(0), p /(0), and p //(0) are as follows:

2210)( xcxccxp 0)0( cp

xccxp 21/ 2)(

2// 2)( cxp

1/ )0( cp

2// 2)0( cp

Local Quadratic Approximations• Thus it follows that

• The local quadratic approximation becomes

)0(0 fc )0(/1 fc 2)0(//

2fc

2210)( xcxccxp 2

210)( xcxccxf

2//

/

2)0()0()0()( xfxffxf

Example 1• Find the local linear and local quadratic

approximations of ex at x = 0, and graph ex and the two approximations together.



• f(x)= ex, so

• The local linear approximation is

• The local quadratic approximation is

1)0( 0/ ef1)0( 0 ef 1)0( 0// ef

xex 1

21

2xxe x



• Here is the graph of the three functions.

Maclaurin Polynomials• It is natural to ask whether one can improve the

accuracy of a local quadratic approximation by using a polynomial of degree 3. Specifically, one might look for a polynomial of degree 3 with the property that its value and the values of its first three derivatives match those of f at a point; and if this provides an improvement in accuracy, why not go on to polynomials of higher degree?

Maclaurin Polynomials• We are led to consider the following problem:

Maclaurin Polynomials• We will begin by solving this problem in the case

where x0 = 0. Thus, we want a polynomial

• such that)0()0(),...,0()0(),0()0(),0()0( ////// nn pfpfpfpf

nnxcxcxcxccxp ...)( 3

32

210

Maclaurin Polynomials• But we know that

nn

nn

nn

nn

nn

cnnnxp

xcnnncxp

xcnnxccxp

xncxcxccxp

xcxcxcxccxp

)...2)(1()(

.

.)2)(1(...23)(

)1(...232)(

...32)(

...)(

33

///

232

//

12321

/

33

2210

Maclaurin Polynomials• But we know that

• Thus we need

33//////

2////

1//

0

!323)0()0(

2)0()0(

)0()0(

)0()0(

ccpf

cpf

cpf

cpf

nn

nn

nn

nn

nn

cnnnxp

xcnnncxp

xcnnxccxp

xncxcxccxp

xcxcxcxccxp

)...2)(1()(

.

.)2)(1(...23)(

)1(...232)(

...32)(

...)(

33

///

232

//

12321

/

33

2210

Maclaurin Polynomials• This yields the following values for the coefficients of

p(x)

!)0(

.

.!3)0(

!2)0(

)0(

)0(

///

3

/

2

/1

0

nfc

fc

fc

fc

fc

n

n

Maclaurin Polynomials• This leads us to the following definition.

Example 2

• Find the Maclaurin Polynomials p0 , p1, p2 , p3, and pn for ex.

Example 2


• We know that

• so

1)0()0()0()0()0( ////// nfffff

!;!

...2

1)(

621

!3)0(

2)0()0()0()(

21

!2)0()0()0()(

1)0()0()(

1)0()(

2

32///2///

3

22///

2

/1

0

memorizenxxxxp

xxxfxfxffxp

xxxfxffxp

xxffxp

fxp

n

n

Example 2


• The graphs of ex and all four approximations are shown.

Taylor Polynomials• Up to now we have focused on approximating a

function f in the vicinity of x = 0. Now we will consider the more general case of approximating f in the vicinity of an arbitrary domain value x0.

Taylor Polynomials• Up to now we have focused on approximating a

function f in the vicinity of x = 0. Now we will consider the more general case of approximating f in the vicinity of an arbitrary domain value x0.

• The basic idea is the same as before; we want to find an nth-degree polynomial p with the property that its value and the values of its first n derivatives match those of f at x0. However, rather than expressing p(x) in powers of x, it will simplify the computation if we express it in powers of x – x0; that is

nn xxcxxcxxccxp )(...)()()( 0

202010

Taylor Polynomial• This leads to the following definition:

Example 3• Find the first four Taylor Polynomials for lnx about

x = 2.


x = 2.• Let f(x) = lnx. Thus

3///

2//

/

/2)(

/1)(

/1)(

ln)(

xxf

xxf

xxf

xxf

4/1)2(

4/1)2(

2/1)2(

2ln)2(

///

//

/

f

f

f

f


x = 2.• This leads us to:

32412

81

21

3

281

21

2///

2

21/

1

0

)2()2()2(2ln)(

)2()2(2ln2)2()2()2)(2()2()(

)2(2ln)2)(2()2()(

2ln)2()(

xxxxp

xxxfxffxp

xxffxp

fxp 4/1)2(

4/1)2(

2/1)2(

2ln)2(

///

//

/

f

f

f

f


x = 2.• The graphs of all five polynomials are shown.

Sigma Notation• Frequently we will want to express a Taylor

Polynomial in sigma notation. To do this, we use the notation f k(x0) to denote the kth derivative of f at x = xo , and we make the connection that f 0(x0) denotes f(x0). This enables us to write

nn

kn

k

k

xxnxfxxxfxfxx

kxf )(

!)(...))(()()(

!)(

00

00/

000

0

Example 4• Find the nth Maclaurin polynomials for

(a) (b) (c)xcosxsinx11


(a) (b) (c)

(a)

xcosxsinx11

xxf

xxf

xxf

xxf

cos)(

sin)(

cos)(

sin)(

///

//

/

1)0(

0)0(

1)0(

0)0(

///

//

/

f

f

f

f


(a) (b) (c)

(a) This leads us to:

xcosxsinx11

!50

!300)(

0!3

00)(

!300)(

00)(0)(0)(

53

5

3

4

3

3

2

1

0

xxxxp

xxxp

xxxp

xxpxxp

xp

1)0(

0)0(

1)0(

0)0(

///

//

/

f

f

f

f


(a) (b) (c)

(a) This leads us to:

xcosxsinx11

)!12()1(...

!7!5!3)(

12753

kxxxxxxpk

k


(a) (b) (c)

(a) The graphs are shown.

xcosxsinx11


(a) (b) (c)

(b) We start with:

xcosxsinx11

xxf

xxf

xxf

xxf

sin)(

cos)(

sin)(

cos)(

///

//

/

0)0(

1)0(

0)0(

1)0(

///

//

/

f

f

f

f


(a) (b) (c)

(b) This leads us to:

xcosxsinx11

!6!4!21)(

!4!21)(

!21)(

1)(

642

6

42

4

2

2

0

xxxxp

xxxp

xxp

xp


(a) (b) (c)

(b) This leads us to:

xcosxsinx11

)!2()1(...

!6!4!21)(

2642

kxxxxxpk

k


(a) (b) (c)

(b) The graphs are shown.

xcosxsinx11


(a) (b) (c)

You need to memorize these!

xcosxsinx11

)!12()1(...

!7!5!3sin

12753

kxxxxxxk

k

)!2()1(...

!6!4!21cos

2642

kxxxxxk

k


(a) (b) (c)


xcosxsinx11

)!12()1(...

!7!5!3sin

12753

kxxxxxxk

k

)!12()2()1(...

!7)2(

!5)2(

!3)2(22sin

12753

kxxxxxx

kk


(a) (b) (c)


xcosxsinx11

)!12()1(...

!7!5!3sin

12753

kxxxxxxk

k

)!12()()1(...

!7)(

!5)(

!3)(sin

12272523222

kxxxxxx

kk


(a) (b) (c)

(c) We start with:

xcosxsinx11

24)(;6)(

;2)0(;1)0(;1)0(///////

///

xfxf

fff


(a) (b) (c)

(c) We start with:

xcosxsinx11

15////

4///

3//

2/

)1(!)(;

)1(234)(;

)1(23)(

;)1(

2)(;)1(

1)(;11)(

kk

xkxf

xxf

xxf

xxf

xxf

xxf

24)0(;6)0(

;2)0(;1)0(;1)0(///////

///

ff

fff


(a) (b) (c)

(c) This leads us to

• You should memorize this as well.

xcosxsinx11

nxxxx

...111 2

The nth Remainder• It will be convenient to have a notation for the error

in the approximation . Accordingly, we will let denote the difference between f(x) and its nth Taylor polynomial: that is

• This can also be written as:

)(xRn

)()( xpxf n

The nth Remainder• The function is called the nth remainder for the

Taylor series of f, and the formula below is called Taylor’s formula with remainder.

)(xRn

The nth Remainder• Finding a bound for gives an indication of the

accuracy of the approximation . The following theorem provides such a bound.

• The bound, M, is called the Lagrange error bound.

)(xRn)()( xfxpn

Example 6• Use an nth Maclaurin polynomial for to

approximate e to five decimal places.

xe


approximate e to five decimal places. • The nth Maclaurin polynomial for is

• from which we have

xe

!...

!21

!

2

0 kxxx

kx kn

k

k

xe

!1...

!2111

!10

1

nkee

n

k

k


approximate e to five decimal places.• Our problem is to determine how many terms to

include in a Maclaurin polynomial for to achieve five decimal-place accuracy; that is, we want to choose n so that the absolute value of the nth remainder at x = 1 satisfies

xe

000005.|)1(| nR

xe


approximate e to five decimal places.• To determine n we use the Remainder Estimation

Theorem with , and I being the interval [0, 1]. In this case it follows that

• where M is an upper bound on the value of for x in the interval [0, 1].

0,1,)( 0 xxexf x

)!1(|)1(|

nMRn

xe

xn exf )(1


approximate e to five decimal places.• However, is an increasing function, so its maximum

value on the interval [0, 1] occurs at x = 1; that is, on this interval. Thus, we can take M = e to

obtaineex

)!1(|)1(|

neRn

xe

xe


approximate e to five decimal places.• Unfortunately, this inequality is not very useful

because it involves e, which is the very quantity we are trying to approximate. However, if we accept that e < 3, then we can use this value. Although less precise, it is more easily applied.

)!1(3|)1(|

n

Rn

xe


approximate e to five decimal places.• Thus, we can achieve five decimal-place accuracy by

choosing n so that or

• This happens when n = 9.

000005.)!1(

3

n

xe

000,600)!1( n

Homework

• Page 684• 1, 3, 7, 9, 11, 15-21 odd• 31, 32

Documents

Maclaurin and Taylor Polynomials Objective: Improve on the local linear approximation for higher order polynomials