Convergence of Taylor Series

Convergence of Taylor Series

Objective: To find where a Taylor Series converges to the original

function; approximate trig, exponential and logarithmic functions

The Convergence Problem

• Recall that the nth Taylor polynomial for a function f about x = xo has the property that its value and the values of its first n derivatives match those of f at xo. As n increases, more and more derivatives match up, so it is reasonable to hope that for values of x near xo the values of the Taylor polynomials might converge to the value of f(x); that is

kn

k

k

nxx

k

xfxf )(

!

)(lim)( 0

0

0


• However, the nth Taylor polynomial for f is the nth partial sum of the Taylor series for f, so the formula below is equivalent to stating that the Taylor series for f converges at x, and its sum is f(x).

kn

k

k

nxx

k

xfxf )(

!

)(lim)( 0

0

0


• This leads to the following problem:


• One way to show that this is true is to show that

• However, the difference appearing on the left side of this equation is the nth remainder for the Taylor series. Thus, we have the following result.

0)(!

)()(lim

00

0

n

k

kk

nxx

k

xfxf


• Theorem:

Estimating the nth Remainder

• It is relatively rare that one can prove directly that as . Usually, this is proved

indirectly by finding appropriate bounds on and applying the Squeezing Theorem. The Remainder Estimation Theorem provides a useful bound for this purpose. Recall that this theorem asserts that if M is an upper bound for on an interval I containing xo , then

n0)( xRn|)(| xRn

|)(| 1 xf n

10 ||

)!1(|)(|

n

n xxn

MxR

Example 1

• Show that the Maclaurin series for cosx converges to cosx for all x; that is

),(...;!6!42

1)!2(

)1(cos642

0

2

xxx

k

xx

k

kk

Example 1


• From Theorem 10.9.2, we must show that for all x as . For this purpose let f(x) = cosx, so that for all x, we have or

n0)( xRn

),(...;!6!42

1)!2(

)1(cos642

0

2

xxx

k

xx

k

kk

xxf n cos)(1 xxf n sin)(1

Example 1


• From Theorem 10.9.2, we must show that for all x as . For this purpose let f(x) = cosx, so that for all x, we have or

• In all cases, we have so we will say that M = 1 and xo = 0 to conclude that

n0)( xRn

),(...;!6!42

1)!2(

)1(cos642

0

2

xxx

k

xx

k

kk

xxf n cos)(1 xxf n sin)(1

1|)(| 1 xf n

1||)!1(

1|)(|0

n

n xn

xR

Example 1


• However, it follows that

• So becomes1||)!1(

1|)(|0

n

n xn

xR

),(...;!6!42

1)!2(

)1(cos642

0

2

xxx

k

xx

k

kk

0||)!1(

1lim 1

n

nx

n

0|)(|0 xRn

Example 1


• The following graph illustrates this point.

),(...;!6!42

1)!2(

)1(cos642

0

2

xxx

k

xx

k

kk

Example 2

• Use the Maclaurin series for sinx to approximate sin 3o to five decimal-point accuracy.

Example 2


• In the Maclaurin series

• The angle is assumed to be in radians. Since 3o = /60 it follows that

...!7!5!3)!12(

)1(sin753

0

12

xxxx

k

xx

k

kk

...!7

)60/(

!5

)60/(

!3

)60/(

6060sin3sin

7530

Example 2


• In the Maclaurin series

• We must now determine how many terms in the series are required to achieve five decimal-place accuracy. We have two choices.

...!7

)60/(

!5

)60/(

!3

)60/(

6060sin3sin

7530

Example 2


• The Remainder Estimation Theorem.• For five decimal-place accuracy, we need

• Using M = 1, x = /60, and xo = 0, we have

000005.60

nR

000005.)!1(

)60/( 1

n

n

Example 2


• The Remainder Estimation Theorem.• This happens at n = 3, so we have

05234.!3

)60/(

603sin

30

Example 2


• The Alternating Series Test.• Let sn denote the sum of the terms up to and including

the nth power of /60. Since the exponents in the series are odd integers, the integer n must be odd, and the exponent of the first term not included in the sum sn must be n + 2.

)!2(

)60/(|3sin|

20

ns

n

n

Example 2


• The Alternating Series Test.• This means that for five decimal-place accuracy we

must look for the first positive odd integer n such that

• Again, this happens at n = 3.

000005.)!2(

)60/( 2

n

n

Example 3

• Show that the Maclaurin series for ex converges to ex for all x; that is

),(...;!

...!2

1!

2

0

k

xxx

k

xe

k

k

kx

Example 3


• Let f(x) = ex, so that• We want to show that as for all x.

It will be useful to consider the cases x < 0 and x > 0 separately. If x < 0, then the interval we will look at is [x, 0] and if x > 0, the interval is [0, x].

),(...;!

...!2

1!

2

0

k

xxx

k

xe

k

k

kx

xn exf )(1

0)( xRn n

Example 3


• Let f(x) = ex, so that• Since f n+1 (x) = ex is an increasing function, it follows

that if c is in the interval [x, 0], then• If c is in the interval [0, x] then

),(...;!

...!2

1!

2

0

k

xxx

k

xe

k

k

kx

xn exf )(1

1|)0(||)(| 011 efcf nn

xnn exfcf |)(||)(| 11

Example 3


• We apply the Theorem with M = 1 or M = ex yielding

• In both cases, the limit is 0.

),(...;!

...!2

1!

2

0

k

xxx

k

xe

k

k

kx

0;)!1(

|||)(|0

1

xn

xxR

n

n 0;)!1(

|||)(|0

1

xn

xexR

nx

n

Approximating Logarithms

• The Maclaurin series

is the starting point for the approximation of natural logs. Unfortunately, the usefulness of this series is limited because of its slow convergence and the restriction -1 < x < 1. However, if we replace x with –x in this series, we obtain

)11(...;432

)1ln(432

xxxx

xx

)11(...;432

)1ln(432

xxxx

xx


• The Maclaurin series

taking the top equation minus the bottom gives

)11(...;432

)1ln(432

xxxx

xx

)11(...;432

)1ln(432

xxxx

xx

)11(...;432

)1ln(432

xxxx

xx

11;...753

21

1ln

753

xxxx

xx

x


• This new series can be used to compute the natural log of any positive number y by letting

or equivalently

and noting that -1 < x < 1.

x

xy

1

1

1

1

y

yx


• For example, to compute ln2 we let y = 2 in which yields x = 1/3. Substituting this value in

• gives

...

7

)(

5

)(

3

)(

3

122ln

7315

313

31

1

1

y

yx

11;...753

21

1ln

753

xxxx

xx

x

Binomial Series

• If m is a real number, then the Maclaurin series for (1 + x)m is called the binomial series; it is given by

...!

)1)(1(...

!3

)2)(1(

!2

)1(1 32

kx

k

kmmmx

mmmx

mmmx

Binomial Series


• In the case where m is a nonnegative integer, the function f(x) = (1 + x)m is a polynomial of degree m, so

...!

)1)(1(...

!3

)2)(1(

!2

)1(1 32

kx

k

kmmmx

mmmx

mmmx

0...)0()0()0( 321 mmm fff

Binomial Series


• In the case where m is a nonnegative integer, the function f(x) = (1 + x)m is a polynomial of degree m, so

• The binomial series reduces to the familiar binomial expansion

...!

)1)(1(...

!3

)2)(1(

!2

)1(1 32

kx

k

kmmmx

mmmx

mmmx

0...)0()0()0( 321 mmm fff

mxxmmm

xmm

mx

...!3

)2)(1(

!2

)1(1 32

Binomial Series

• It can be proved that if m is not a nonnegative integer, then the binomial series converges to (1 + x)m if |x| < 1. Thus, for such values of x

or in sigma notation

...!

)1)(1(...

!3

)2)(1(

!2

)1(1)1( 32

km x

k

kmmmx

mmmx

mmmxx

1||;!

)1)...(1(1)1(

1

xxk

kmmmx k

k

m

Example 4

• Find the binomial series for (a) (b)2)1(

1

x x11

Example 4

• Find the binomial series for (a) (b)

(a) Since the general term of the binomial series is complicated, you may find it helpful to write out some of the beginning terms of the series to see developing patterns.

2)1(

1

x x11

Example 4


(a) Substitution m = -2 in the formula yields

2)1(

1

x x11

...!3

)4)(3)(2(

!2

)3)(2()2(1)1(

)1(

1 3222

xxxxx

...!3

!4

!2

!321 32 xxx ...4321 32 xxx

Example 4


(a) Substitution m = -2 in the formula yields

2)1(

1

x x11

...!3

)4)(3)(2(

!2

)3)(2()2(1)1(

)1(

1 3222

xxxxx

...!3

!4

!2

!321 32 xxx ...4321 32 xxx

k

k

k xk )1()1(0

Example 4


(b) Substitution m = -1/2 in the formula yields

2)1(

1

x x11

...!3

)2/5)(2/3)(2/1(

!2

)2/3)(2/1()2/1(1)1(

1

1 322/1

xxxxx

Example 4


(b) Substitution m = -1/2 in the formula yields

2)1(

1

x x11

...!3

)2/5)(2/3)(2/1(

!2

)2/3)(2/1()2/1(1)1(

1

1 322/1

xxxxx

...!32

531

!22

31

2

11 3

32

2

xxx

kk

k

k xk

k

!2

)12(531)1(1

0

Homework

• Pages 702-703

• 1, 3, 5, 9, 11, 17

• Look at page 701. There is a list of several important Maclaurin series. Be familiar with them.

Documents

Convergence of Taylor Series