maclaurin and taylor 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


• 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.


• 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.


• 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


• 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


• Thus it follows that

• The local quadratic approximation becomes

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

)0(//2

fc

2210)( xcxccxp 2

210)( xcxccxf

2//

/

2

)0()0()0()( x

fxffxf

Example 1

• Find the local linear and local quadratic approximations of ex at x = 0, and graph ex and the two approximations together.

Example 1


• 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

2xxex

Example 1


• 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?


• We are led to consider the following problem:


• 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


• 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


• 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


• This yields the following values for the coefficients of p(x)

!

)0(

.

.!3

)0(

!2

)0(

)0(

)0(

///

3

//

2

/1

0

n

fc

fc

fc

fc

fc

n

n


• 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

memorizen

xxxxp

xxx

fxfxffxp

xx

xfxffxp

xxffxp

fxp

n

n

Example 2

• Why do I need to memorize this?

• They may ask you to find a Maclaurin polynomial to approximate xex. This is what they want you to do:

!;!

...2

1)(2

memorizen

xxxxp

n

n

!...

2

!...

21

132

2

n

xxxxxe

n

xxxe

nx

nx

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.

Example 3


• 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

Example 3


• 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

xxx

fxffxp

xxffxp

fxp 4/1)2(

4/1)2(

2/1)2(

2ln)2(

///

//

/

f

f

f

f

Example 3


• 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

xxn

xfxxxfxfxx

k

xf)(

!

)(...))(()()(

!

)(0

000

/00

0

0

Example 4

• Find the nth Maclaurin polynomials for

(a) (b) (c)xcosxsinx11

Example 4


(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

Example 4


(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

xxp

xxp

xp

1)0(

0)0(

1)0(

0)0(

///

//

/

f

f

f

f

Example 4


(a) (b) (c)

(a) This leads us to:

xcosxsinx11

)!12()1(...

!7!5!3)(

12753

k

xxxxxxp

kk

Example 4


(a) (b) (c)

(a) The graphs are shown.

xcosxsinx11

Example 4


(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

Example 4


(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

Example 4


(a) (b) (c)

(b) This leads us to:

xcosxsinx11

)!2()1(...

!6!4!21)(

2642

k

xxxxxp

kk

Example 4


(a) (b) (c)

(b) The graphs are shown.

xcosxsinx11

Example 4


(a) (b) (c)

You need to memorize these!

xcosxsinx11

)!12()1(...

!7!5!3sin

12753

k

xxxxxx

kk

)!2()1(...

!6!4!21cos

2642

k

xxxxx

kk

Example 4


(a) (b) (c)


xcosxsinx11

)!12()1(...

!7!5!3sin

12753

k

xxxxxx

kk

)!12(

)2()1(...

!7

)2(

!5

)2(

!3

)2(22sin

12753

k

xxxxxx

kk

Example 4


(a) (b) (c)


xcosxsinx11

)!12()1(...

!7!5!3sin

12753

k

xxxxxx

kk

)!12(

)()1(...

!7

)(

!5

)(

!3

)(sin

12272523222

k

xxxxxx

kk

Example 4


(a) (b) (c)

(c) We start with:

xcosxsinx11

24)(;6)(

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

///

xfxf

fff

Example 4


(a) (b) (c)

(c) We start with:

xcosxsinx11

15////

4///

3//

2/

)1(

!)(;

)1(

234)(;

)1(

23)(

;)1(

2)(;

)1(

1)(;

1

1)(

kk

x

kxf

xxf

xxf

xxf

xxf

xxf

24)0(;6)0(

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

///

ff

fff

Example 4


(a) (b) (c)

(c) This leads us to

• You should memorize this as well.

xcosxsinx11

nxxxx

...11

1 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

Example 6


• The nth Maclaurin polynomial for is

• from which we have

xe

!...

!21

!

2

0 k

xxx

k

x kn

k

k

xe

!

1...

!2

111

!

1

0

1

nkee

n

k

k

Example 6


• 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

Example 6


• 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(|

n

MRn

xe

xn exf )(1

Example 6


• 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 obtain

eex

)!1(|)1(|

n

eRn

xe

xe

Example 6


• 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

Example 6


• 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

maclaurin and taylor polynomials

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