1432 day30 inalmus/1432_day30_after.pdf2 math 1432‐ dr. almus the series 1 1 1n n n a is...

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1 Math 1432‐ Dr. Almus Math 1432 DAY 30 Dr. Melahat Almus [email protected] If you email me, please mention the course (1432) in the subject line. Bubble in PS ID and Popper Number very carefully. If you make a bubbling mistake, your scantron can’t be saved in the system. In that case, you will not get credit for the popper even if you turned it in. Check your CASA account for Quiz due dates. Don’t miss any online quizzes! Be considerate of others in class. Respect your friends and do not distract anyone during the lecture. Section 9.6 Alternating Series and Absolute Convergence Alternating Series Test: If an alternating series 1 1 2 3 4 1 1 , 0 n n n n a a a a a a satisfies (i) 1 n n a a for all n (decreasing) AND (ii) lim 0 n n a then the series is convergent.

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1  Math 1432‐ Dr. Almus  

Math 1432 DAY 30

Dr. Melahat Almus

[email protected]

If you email me, please mention the course (1432) in the subject line.

Bubble in PS ID and Popper Number very carefully. If you make a bubbling mistake, your scantron can’t be saved in the system. In that case, you will not get credit for the popper even if you turned it in.

Check your CASA account for Quiz due dates. Don’t miss any online quizzes!

Be considerate of others in class. Respect your friends and do not distract anyone during the lecture.

Section 9.6 Alternating Series and Absolute Convergence

Alternating Series Test:

If an alternating series 1

1 2 3 41

1 , 0n

n nn

a a a a a a

satisfies

(i) 1n na a for all n (decreasing) AND

(ii) lim 0nn

a

then the series is convergent.

2  Math 1432‐ Dr. Almus  

The series 1

1

1n

nn

a

is absolutely convergent if both 1

1

1n

nn

a

and

1

nn

a

are convergent.

The series 1

1

1n

nn

a

is conditionally convergent if 1

1

1n

nn

a

is

convergent but 1

nn

a

is divergent.

POPPER#

Q# 1

2

1

5

n

n n

a. Absolutely Convergent

b. Conditionally Convergent

c. Divergent

Q# 1

1

2

n

n

a. Absolutely Convergent

b. Conditionally Convergent

c. Divergent

3  Math 1432‐ Dr. Almus  

Q# 1

1 2

5

n n

nn

a. Absolutely Convergent

b. Conditionally Convergent

c. Divergent

Q# 1

1 lnk

k

k

a. Absolutely Convergent

b. Conditionally Convergent

c. Divergent

Q# 1

2

1 arctank

k

k

a. Absolutely Convergent

b. Conditionally Convergent

c. Divergent

Q#

2

cos

1

n n

n

a. Absolutely Convergent

b. Conditionally Convergent

c. Divergent

4  Math 1432‐ Dr. Almus  

Q# Estimate the error if 1

2

1n

n n

is approximated using the first 9

terms.

a. Less than 1/100

b. Less than 1/10

c. Less than 1/110

d. Less than 1/90

Q# What is the least n so that Sn approximates 1

3

1

1

n

n

within 0.001?

a. 11 b. 10 c. 9 d. 12

5  Math 1432‐ Dr. Almus  

Section 9.7 Power Series

Definition: A Power Series (centered at x=0) is a series of the form

2 3 40 1 2 3 4

0

...nn

n

c x c c x c x c x c x

where x is a variable and the cn’s are coefficients.

The sum of the series is

2 3 40 1 2 3 4f ... ...n

nx c c x c x c x c x c x

which is a function whose domain is the set of all x for which the series converges.

f (x) resembles a polynomial, but it has infinitely many terms.

The power series may converge for some values of x and diverge for others.

6  Math 1432‐ Dr. Almus  

Examples:

2 3 4

0

( ) 2 2 2 2 2 2 ...n

n

f x x x x x x

2 3

0

1 1 1 1( ) 1 ...

1 2 3 4n

n

g x x x x xn

2 3 4

0

1 1 1 1( ) 1 ...

! 2 6 24n

n

h x x x x x xn

7  Math 1432‐ Dr. Almus  

SPECIAL CASE:

Let cn = 1 for all n; then we get the geometric series, centered at 0,

2 3 4

0

1 ... ...n n

n

x x x x x x

which converges if | x | < 1 and diverges if | x | 1.

That is: 0

n

n

x

is also a power series and

0

1

1n

n xx

when |x|<1

8  Math 1432‐ Dr. Almus  

For example, suppose that 2 3

0

1 ....k

k

P x x x x x

.

By comparing the graphs of 1

1g x

x

and P(x) with more and more terms, you

will see that between 1 and 1 (the interval of convergence), the two graphs converge.

9  Math 1432‐ Dr. Almus  

Using this fact, we can write functions in this form in sigma notation (as a power series):

Example: Write (5

1)f x

x

as a power series:

10  Math 1432‐ Dr. Almus  

Example: Write 2

2( )

4

xg x

x

as a power series:

11  Math 1432‐ Dr. Almus  

A Power Series (centered at x=a) is a series of the form

2 30 1 2 3

0

( ) ( ) ( ) ( ) ...nn

n

c x a c c x a c x a c x a

For notation purposes, (x – a)0 = 1 even when x = a.

When x = a, all the terms are 0 for n 1, so the power series always converges when x = a.

Example:

2 3

0

1 1 1 1( 2) 1 ( 2) ( 2) ( 2) ...

1 2 3 4n

n

x x x xn

is a power series that is centered at x=2.

12  Math 1432‐ Dr. Almus  

Convergence of Power Series

Question: For what values of x is the series convergent?

0

n

n

x

How about?

0

! n

n

n x

13  Math 1432‐ Dr. Almus  

Remark: For a given power series 0

n

nn

c x a

there are only 3 possibilities.

1. The series converges only when x = a.

Example: n nn x converges only at x=0.

2. The series converges absolutely for all x.

Example: !

nx

n converges absolutely for any x.

3. There is a positive number R such that the series converges if

|x – a| R and diverges if |x – a| > R.

Example: nx converges if 1x , diverges if 1x .

In Case 1, we say that the radius of convergence is 0.

In Case 2, we say that the radius of convergence is ∞.

In Case 3, we say that the radius of convergence is R.

14  Math 1432‐ Dr. Almus  

The interval of convergence of a power series is the maximal interval that consists of all values of x for which the series converges. Check endpoints (for regular convergence)!

In Case 1, the interval of convergence is just one point {0} (or {a}).

In Case 2, the interval of convergence is (-∞,∞).

In Case 3, the interval of convergence is (a-R,a+R), or [a-R,a+R], or (a-R,a+R], or [a-R,a+R). One should check the behavior at the end points.

15  Math 1432‐ Dr. Almus  

Example 1: Find the radius of convergence and interval of convergence for

0 3

n

n

x

.