Prepare for your exams
Get points
Guidelines and tips

Sell on Docsity

Log in Sign up

Prepare for your exams

Study with the several resources on Docsity

Earn points to download

Earn points by helping other students or get them with a premium plan

Guidelines and tips

Sell on Docsity

Log in Sign up

Prepare for your exams

Study with the several resources on Docsity

Find documents

Prepare for your exams with the study notes shared by other students like you on Docsity

Search Store documents

The best documents sold by students who completed their studies

Search through all study resources

Docsity AINEW

Summarize your documents, ask them questions, convert them into quizzes and concept maps

Explore questions

Clear up your doubts by reading the answers to questions asked by your fellow students

Earn points to download

Earn points by helping other students or get them with a premium plan

Share documents

20 Points

For each uploaded document

Answer questions

5 Points

For each given answer (max 1 per day)

All the ways to get free points

Get points immediately

Choose a premium plan with all the points you need

Study Opportunities

Choose your next study program

Get in touch with the best universities in the world. Search through thousands of universities and official partners

Community

Ask the community

Ask the community for help and clear up your study doubts

University Rankings

Discover the best universities in your country according to Docsity users

Free resources

Our save-the-student-ebooks!

Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors

From our blog

Exams and Study

Go to the blog

Lecture Notes on Infinite Sequence and Series | MATH 1920, Study notes of Calculus

Pellissippi State Technical Community College Calculus

Material Type: Notes; Class: Calculus II; Subject: Mathematics; University: Pellissippi State Technical Community College; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 08/17/2009

koofers-user-c9u 🇺🇸

10 documents

1 / 2

This page cannot be seen from the preview

Don't miss anything!

INFINITE SEQUENCES AND SERIES

8.6 Representations of Functions as Power Series

Objective: Express a known function as the sum of infinitely many terms

I. This is a useful procedure for

A. Integrating functions that don’t have elementary antiderivatives.

B. Solving differential equations. [Used in Calculus IV]

C. Approximating functions by polynomials.

II. Recall = 1 + x + = | x | < 1

1

x1

−23

xx...

++ n

n0

x

∞

=

∑

III. Example 1: Express as the sum of a power series and find the interval of

2

1

1x

+

convergence.

A. Replace x by in previous equation

2

x−

B. = = = = 1

2

1

1x

+2

1

1(x)

−−

2n

n0

(x)

∞

=

−

∑n2n

n0

(1)x

∞

=

−

∑2468

xxxx...

−+−+−

1. This is a geometric series; converges when 22

|x|1|x|1|x|1.

−<⇒<⇒<

2. | x | < 1 interval of convergence is

⇒

(1,1)−

IV. Differentiation and integration of power series [term-by-term]

A. The sum of a power series is a function f(x) = .

n

n0

c(xa)

∞

=

−

∑

1. Domain is the interval of convergence of the series.

2. Can treat as a polynomial.

B. If the power series has radius of convergence R > 0, then the

n

n0

c(xa)

∞

=

−

∑

function f defined by f(x) =

23

0123

cc(xa)c(xa)c(xa)...

+−+−+−+

= is differentiable (and therefore continuous) on the interval

n

n0

c(xa)

∞

=

−

∑

.(aR,aR)

−+

1. = .

2

123

f(x)c2c(xa)3c(xa)...

′=+−+−+

n1

n

n1

nc(xa)

∞−

=

−

∑

2. = .

23

012

(xa)(xa)

f(x)dxCc(xa)cc...

23

−−

=+−+++

∫

n1

n

n1

(xa)

Cc

n1

∞+

=

−

++

∑

Partial preview of the text

Download Lecture Notes on Infinite Sequence and Series | MATH 1920 and more Study notes Calculus in PDF only on Docsity!

INFINITE SEQUENCES AND SERIES

8.6 Representations of Functions as Power Series

Objective: Express a known function as the sum of infinitely many terms

I. This is a useful procedure for

A. Integrating functions that don’t have elementary antiderivatives.

B. Solving differential equations. [Used in Calculus IV]

C. Approximating functions by polynomials.

II. Recall = 1 + x + = | x | < 1

x − 1

2 3 x + x +... n n 0

x

∞

=

III. Example 1: Express as the sum of a power series and find the interval of

2

1 +x

convergence.

A. Replace x by in previous equation

2 − x

B. = = = = 1

2

1 + x 2

1 − ( −x )

2 n

n 0

( x )

∞

=

∑ − n^ 2n

n 0

( 1) x

∞

=

2 4 6 8 − x + x − x + x −...

1. This is a geometric series; converges when | − x 2 | < 1 ⇒ | x 2 | < 1 ⇒ | x | <1.

2. | x | < 1 ⇒ interval of convergence is( −1, 1)

IV. Differentiation and integration of power series [term-by-term]

A. The sum of a power series is a function f(x) =.

n n n 0

c (x a)

∞

=

∑^ −

1. Domain is the interval of convergence of the series.

2. Can treat as a polynomial.

B. If the power series n n has radius of convergence R > 0, then the

n 0

c (x a)

∞

=

∑^ −

function f defined by f(x) = c 0 + c (x 1 − a) + c 2 (x − a) 2 + c 3 (x − a)^3 +...

= n n is differentiable (and therefore continuous) on the interval

n 0

c (x a)

∞

=

∑^ −

(a − R, a +R).

2 f ( x )′^ = c 1 + 2c 2 (x − a) + 3c 3 (x − a) +... n 1 n n 1

nc (x a)

∞ −

=

∑^ −

2 3 0 1 2

(x a) (x a) f(x)dx C c (x a) c c... 2 3

n 1 n n 1

(x a) C c n 1

∞ (^) +

=

3. The radius of convergence of the derivative and integral are both R.

4. The interval of convergence may change [only at an endpoint].

5. This method only works with power series.

V. Find a power series representation for.

2

(1 −x)

A. Begin with = 1 + x + = , | x | < 1

1 −x

2 3 x + x +... n

n 0

x

∞

=

B. Differentiate each side to obtain = 1 + 2x + = | x | < 1.

2

(1 −x)

2 3x +... n^1 n 0

nx

∞ −

=

1. n^1 = [replacing n with n + 1]

n 0

nx

∞ −

=

n

n 1

(n 1)x

∞

=

∑^ +

2. R still equals 1

VI. Evaluate as a power series

7

dx

∫(1 −x)

A. = 1 + x + =

1 −x

2 3 x + x +... n

n 0

x

∞

=

1. Replace x with

7 − x

7

(1 − x)^7

1 − ( x )−

7 n

n 0

( x )

∞

=

∑ − n^ 7n

n 0

( 1) x

∞

=

7 14 1 − x + x −...

B. = = C + = C + x

7

dx

∫ (1 −x)

n 7n

n 0

( 1) x dx

∞

=

∑^ −

7n 1 n

n 0

x ( 1) 7n 1

∞ (^) +

=

8 15 22 x x x

... 8 15 22

C. The series converges for i.e. for | x | < 1.

7 | −x | < 1 ,

D. See Example 8: =

7 0

dx

∫ (1 −x)

8 15 22

0

x x x x... 8 15 22

 + −^ +^ −^ + 

≈ 0.

Using alternating series estimation theorem with n = 4 gives the answer correct to

within

7

−

Lecture Notes on Infinite Sequence and Series | MATH 1920, Study notes of Calculus

Related documents

Partial preview of the text

Download Lecture Notes on Infinite Sequence and Series | MATH 1920 and more Study notes Calculus in PDF only on Docsity!

INFINITE SEQUENCES AND SERIES

8.6 Representations of Functions as Power Series

Objective: Express a known function as the sum of infinitely many terms

I. This is a useful procedure for

A. Integrating functions that don’t have elementary antiderivatives.

B. Solving differential equations. [Used in Calculus IV]

C. Approximating functions by polynomials.

II. Recall = 1 + x + = | x | < 1

III. Example 1: Express as the sum of a power series and find the interval of

convergence.

A. Replace x by in previous equation

B. = = = = 1

∑ − n^ 2n

1. This is a geometric series; converges when | − x 2 | < 1 ⇒ | x 2 | < 1 ⇒ | x | <1.

2. | x | < 1 ⇒ interval of convergence is( −1, 1)

IV. Differentiation and integration of power series [term-by-term]

A. The sum of a power series is a function f(x) =.

∑^ −

1. Domain is the interval of convergence of the series.

2. Can treat as a polynomial.

B. If the power series n n has radius of convergence R > 0, then the

∑^ −

function f defined by f(x) = c 0 + c (x 1 − a) + c 2 (x − a) 2 + c 3 (x − a)^3 +...

= n n is differentiable (and therefore continuous) on the interval

∑^ −

(a − R, a +R).

∑^ −

3. The radius of convergence of the derivative and integral are both R.

4. The interval of convergence may change [only at an endpoint].

5. This method only works with power series.

V. Find a power series representation for.

A. Begin with = 1 + x + = , | x | < 1

B. Differentiate each side to obtain = 1 + 2x + = | x | < 1.

1. n^1 = [replacing n with n + 1]

∑^ +

2. R still equals 1

VI. Evaluate as a power series

∫(1 −x)

A. = 1 + x + =

1. Replace x with

∑ − n^ 7n

B. = = C + = C + x

∫ (1 −x)

∑^ −

C. The series converges for i.e. for | x | < 1.

D. See Example 8: =

∫ (1 −x)

 + −^ +^ −^ + 

Using alternating series estimation theorem with n = 4 gives the answer correct to

within