Consider the following:

Now, use the reciprocal function and tangent line to get an approximation.

2

1y

xxf

1)(

4975124378.01.2

1

2

1,2 :Point

Lecture 31 – Approximating Functions

1

1 2 3

)2( :Slope fm

)( 11 xxmyy

2

2

1y

xxf

1)(

4975124378.01.2

1

2

1,2 :Point

2 2.01

)2( :Slope fm 4

1

)2)(2()2()( xffxL

)01.2(01.2

1L

First derivative gave us more information about the function (in particular, the direction).

For values of x near a the linear approximation given by the tangent line should be better than the constant approximation.

3

Second derivative will give us more information (curvature).

For values of x near a the quadratic approximation should be better than the linear approximation.

What quadratic is used as the approximation?

22102 )()()( axcaxccxp

4

Key idea: Need to have quadratic match up with the function and its first and second derivatives at x = a.

5

xxf

1)(

4975124378.01.2

1

2 2.01

)01.2(01.2

12p

Use p2(x) to get a better approximation.

22 )(

2

)())(()()( ax

afaxafafxp

xxf

1)(

2

1)(

xx'f

3

2)(

xx''f

2

1)2( f

4

1)2( 'f

4

1)2( ''f

2

)( xexf Graphical Example at x = 0

6

)2()(2

xexf x

2

2 xxe

)2)(())2()(2()(22

xx exexxf

)24( 22

xe x

1 2 3

)(0 xp

)(1 xp

)(2 xp

What higher degree polynomial is appropriate?

nnn axcaxcaxccxp )()()()( 2

210

7

Key idea: Need to have nth degree polynomial match up with the function and all of its derivatives at x = a.

)( x'pn

)( x''pn

)()( xp nn

The coefficients, ck, for the nth degree Taylor polynomial approximating the function f(x) at x = a have the form:

8

n

k

kkn axcxp

0

)()(

Def: The Taylor polynomial of order n for function f at x = a:

9

The remainder term for using this polynomial:

Lecture 32 – Taylor Polynomials

)()()( xpxfxR nn

n

k

kk

n axk

afxp

0

)(!

)()(

for some c between x and a.

where M provides a bound on

how big the n+1st derivative

could possibly be.

10

Estimate the maximum error in approximating the reciprocal function at x = 2 with an 8th order Taylor polynomial on the interval [2, 3].

xxf

1)(

9)9(

8 )2(! 9

)()( x

cfxR

8)8(

28 )2(

! 8

)2()2(

! 2

)2()2)(2()2()(

x

fx

fxffxp

2

1)(

xx'f

3

2)(

xx''f

4

6)(

xx'''f

5)4( 24

)(x

xf

11

9)9(

8 )2(! 9

)()( x

cfxR

! 9

|2| 9

xM

)!1()(

1

n

axMxR

n

n

12

What is the actual maximum error in approximating the reciprocal function at x = 2 with an 8th order Taylor polynomial on the interval [2, 3]?

xxf

1)(

)3(3

18p

8)8(

28 )2(

! 8

)2()2(

! 2

)2()2)(2()2()(

x

fx

fxffxp

)3(8R

What nth degree polynomial would you need in order to keep the error below .0001?

13

xxf

1)(

1)1(

)2(! )1(

)()(

nn

n xn

cfxR

nn

n xn

fx

fxffxp )2(

!

)2()2(

! 2

)2()2)(2()2()(

)(2

1)( !

)1()( n

nn

x

nxf

2

1)1( ! 1)1()(

nnn

x

nxf

14

1)1(

)2(! )1(

)()(

nn

n xn

cfxR

To keep error below .0001, need to keep Rn below .0001.

The Taylor series centered at x = a:

15

Lecture 33 – Taylor Series

is a power series with

0

)(!

)(

n

nn

axn

af

!

)()(

n

afc

n

n

The Taylor series centered at x = 0 is called a Maclaurin series:

0 !

)0(

n

nn

xn

f

Find the Maclaurin series for f (x) = sin x.

16

xx'f cos)( xx''f sin)( xx'''f cos)( xxf sin)()4( xxf sin)(

0 !

)0(sin

n

nn

xn

fx

kk

xk

fx

fxff

!

)0(

! 2

)0()0()0(

)(2

Example 1

Find the Maclaurin series for f (x) = ex.

17

xex'f )( xex''f )( xn exf )()(xexf )(

0 !

)0(

n

nn

x xn

fe

kk

xk

fx

fxff

!

)0(

! 2

)0()0()0(

)(2

Example 2

18

For what values of x will the last two series converge?

Ratio Test:

0

12

! )12(sin

k

k

k

xx

0 ! k

kx

k

xe

12

32 ! 12

)!32(lim

n

n

n x

n

n

x

Series converges for

n

n

n x

n

n

x !

)!1(lim

1

Series converges for

19

Consider the graphs:

x sin

xy 1

6

3

3

xxy

1206

53

5

xxxy

50401206

753

7

xxxxy

1

! 7! 5! 3

753 xxxx

20

Example 3 Find the Maclaurin series for f (x) = ln(1 + x).

xx'f

1

1)(

21

1)(

xx''f

31

2)(

xx'''f

4)4(

)1(

6)(

xxf

)1ln()( xxf

0 !

)0()1ln(

n

nn

xn

fx

kk

xk

fx

fxff

!

)0(

! 2

)0()0()0(

)(2

21

)1ln( x

1

1432

1432 k

kk

k

xxxxx

n

n

n x

n

n

x

1lim

1

For what values of x will the series converge?

22

2

)( xexf

Creating new series for:

Example 4 ! 4! 3! 2

1432 xxx

xex

0 ! k

k

k

x

Create and use other Taylor series like was done with power series.

23

Lecture 34 – More Taylor Series

! 7! 5! 3

753 xxxx

0

12

! )12(

)1(

k

kk

k

xxsin

xcos

xcos

24

2

2cos2lim

x

xxx

xcos ! 6! 4! 2

132 xxx

Example 1

xcos2

xx 2cos2

2

2cos2

x

xx

25

Example 2 ...432

)1ln(432

xxx

xx

xx

x

31lnlim

26

Example 3

dxx

ex

xe ! 4! 3! 2

1432 xxx

x

0 ! k

k

k

x

27

Example 4

dxx )sin(1

0

2

! 7! 5! 3

753 xxxx

0

12

! )12(

)1(

k

kk

k

xxsin

3102683.

28

Example 5

dtt

t

)1ln(

.2

019080014.

...432

)1ln(432

xxx

xx