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

kxfxf )(!)(lim)( 0

0

0

The Convergence Problem• 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

kxfxf )(!)(lim)( 0

0

0

The Convergence Problem• This leads to the following problem:

The Convergence 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

kxfxf

The Convergence Problem• 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 xxnMxR

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

xxxkxx

k

kk



• 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

xxxkxx

k

kk

xxf n cos)(1 xxf n sin)(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

xxxkxx

k

kk

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

1|)(| 1 xf n

1||)!1(

1|)(|0

n

n xn

xR



• However, it follows that

• So becomes1||)!1(

1|)(|0

n

n xn

xR

),(...;!6!42

1)!2(

)1(cos642

0

2

xxxkxx

k

kk

0||)!1(

1lim 1

n

nx

n

0|)(|0 xRn



• The following graph illustrates this point.

),(...;!6!42

1)!2(

)1(cos642

0

2

xxxkxx

k

kk

Example 2• Use the Maclaurin series for sinx to approximate sin 3o

to five decimal-point accuracy.


to five decimal-point accuracy.• 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

xxxxkxx

k

kk

...!7)60/(

!5)60/(

!3)60/(

6060sin3sin

7530


to five decimal-point accuracy.• 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


to five decimal-point accuracy.• 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


to five decimal-point accuracy.• The Remainder Estimation Theorem.• This happens at n = 3, so we have

05234.!3)60/(

603sin

30


to five decimal-point accuracy.• 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


to five decimal-point accuracy.• 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

kxxx

kxe

k

k

kx


for all x; that is

• 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

kxxx

kxe

k

k

kx

xn exf )(1

0)( xRn n


for all x; that is

• 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

kxxx

kxe

k

k

kx

xn exf )(1

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

xnn exfcf |)(||)(| 11


for all x; that is

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

• In both cases, the limit is 0.

),(...;!

...!2

1!

2

0

kxxx

kxe

k

k

kx

0;)!1(

|||)(|01

xnxxRn

n 0;)!1(

|||)(|01

xnxexRn

xn

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

xxxxxx

)11(...;432

)1ln(432

xxxxxx

Approximating Logarithms• The Maclaurin series

taking the top equation minus the bottom gives

)11(...;432

)1ln(432

xxxxxx

)11(...;432

)1ln(432

xxxxxx

)11(...;432

)1ln(432

xxxxxx

11;...753

211ln

753

xxxxxxx

Approximating Logarithms• 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.

xxy

11

11

yyx

Approximating Logarithms• For example, to compute ln2 we let y = 2 in

which yields x = 1/3. Substituting this value in

• gives

...

7)(

5)(

3)(

3122ln

7315

313

31

11

yyx

11;...753

211ln

753

xxxxxxx

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

kkmmmxmmmxmmmx



• 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

kkmmmxmmmxmmmx

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



• 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

kkmmmxmmmxmmmx

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

mxxmmmxmmmx

...!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

kkmmmxmmmxmmmxx

1||;!

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

xxk

kmmmx k

k

m

Example 4• Find the binomial series for (a) (b)2)1(

1x x1

1

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(1x x1

1


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

2)1(1x x1

1

...!3

)4)(3)(2(!2)3)(2()2(1)1(

)1(1 322

2

xxxxx

...!3!4

!2!321 32 xxx ...4321 32 xxx


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

2)1(1x x1

1

...!3

)4)(3)(2(!2)3)(2()2(1)1(

)1(1 322

2

xxxxx

...!3!4

!2!321 32 xxx ...4321 32 xxx

k

k

k xk )1()1(0


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

2)1(1x x1

1

...!3

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

)2/3)(2/1()2/1(1)1(11 322/1

xxxx

x


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

2)1(1x x1

1

...!3

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

)2/3)(2/1()2/1(1)1(11 322/1

xxxx

x

...!32531

!2231

211 3

32

2

xxx

kk

k

k xkk!2

)12(531)1(10

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.

Convergence of Taylor Series

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