Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

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


Guidelines and tips
Guidelines and tips

Radius and Interval of Convergence for Various Power Series, Schemes and Mind Maps of Signals and Systems

The calculations for determining the radius and interval of convergence for several power series using different tests. The series include p-series, alternating series, and those with exponential and factorial terms.

What you will learn

  • What is the radius of convergence for the series โˆšn.xn?
  • What is the interval of convergence for the series (โˆ’1)nn4nxn?
  • How does the ratio test determine the radius of convergence for a power series?

Typology: Schemes and Mind Maps

2021/2022

Uploaded on 09/12/2022

edmond
edmond ๐Ÿ‡บ๐Ÿ‡ธ

3.8

(10)

251 documents

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
ยง11.8
3-20 Find the radius of convergence and interval of convergence of the series.
3.โˆž
X
n=1
xn
โˆšn.
We will apply the ratio test.
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
xn+1
โˆšn+ 1
โˆšn
xn๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
=๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
xโˆšn
โˆšn+ 1
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œโ†’ |x|as nโ†’ โˆž.
Hence the radius of convergence is 1. For x= 1, the series is a divergent p-series, and for x=โˆ’1, the series
is an alternating series, and since 1
โˆšnis decreasing and converges to zero, the series converges. The interval
of convergence is therefore [โˆ’1,1).
5.โˆž
X
n=1
(โˆ’1)nโˆ’1xn
n3
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
xn+1
(n+ 1)3
n3
xn๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
=๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
xn3
(n+ 1)3๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œโ†’ |x|as nโ†’ โˆž
Hence the radius of convergence is 1. For x=ยฑ1, the series converges absolutely and therefore converges.
Therefore the interval of convergence is [โˆ’1,1].
8.โˆž
X
n=1
nnxn.
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
(n+ 1)n+1xn+1
nnxn๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
=x(n+ 1)n+1
nn
converges if and only if x= 0. Therefore the radius of convergence is 0 and the interval of convergence is
[0,0].
9.โˆž
X
n=1
(โˆ’1)nn4nxn
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
(n+ 1)4n+1xn+1
n4nxn๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
=๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
(4(n+ 1)x
n๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œโ†’ |4x|as nโ†’ โˆž.
1
pf3
pf4

Partial preview of the text

Download Radius and Interval of Convergence for Various Power Series and more Schemes and Mind Maps Signals and Systems in PDF only on Docsity!

3-20 Find the radius of convergence and interval of convergence of the series.

โˆž โˆ‘

n=

x n

โˆš n

We will apply the ratio test.

x

n+

โˆš n + 1

n

x n

x

n โˆš n + 1

โ†’ |x| as n โ†’ โˆž.

Hence the radius of convergence is 1. For x = 1, the series is a divergent p-series, and for x = โˆ’1, the series

is an alternating series, and since โˆš^1 n

is decreasing and converges to zero, the series converges. The interval

of convergence is therefore [โˆ’ 1 , 1).

โˆž โˆ‘

n=

nโˆ’ 1 x n

n 3

x

n+

(n + 1) 3

n

3

x n

xn

3

(n + 1) 3

โ†’ |x| as n โ†’ โˆž

Hence the radius of convergence is 1. For x = ยฑ1, the series converges absolutely and therefore converges.

Therefore the interval of convergence is [โˆ’ 1 , 1].

โˆž โˆ‘

n=

n

n x

n .

(n + 1)

n+ x

n+

nnxn

x(n + 1)

n+

nn

converges if and only if x = 0. Therefore the radius of convergence is 0 and the interval of convergence is

[0, 0].

โˆž โˆ‘

n=

n n 4

n x

n

(n + 1)

n+ x

n+

n 4 n x n

(4(n + 1)x

n

โ†’ | 4 x| as n โ†’ โˆž.

Therefore the radius of convergence is

1 4

. At the end points, x = ยฑ

1 4 , the sequence (โˆ’1)

n n 4

n x

n diverges, so

its sum cannot converge. Therefore the interval of convergence is (โˆ’

1 4

1 4

โˆž โˆ‘

n=

n x n

n ln n

x n+

n+ ln(n + 1)

n ln n

x n

x ln n

4 ln(n + 1)

x

โˆฃ as n โ†’ โˆž.

Therefore the radius of convergence is 4. For x = 4, the sequence

โˆž โˆ‘

n=

n 1

ln n

satisfies the criteria for the alternating series test and hence converges. For x = โˆ’4. the sequence

โˆž โˆ‘

n=

n

n

4 n^ ln n

โˆž โˆ‘

n=

ln n

diverges because for n โ‰ฅ 2,

1 n

1 ln n , and the harmonic series diverges. The interval of convergence is

therefore (โˆ’ 4 , 4].

โˆž โˆ‘

n=

n x 2 n

(2n)!

x 2 n+

(2n + 2)!

(2n)!

x 2 n

x 2

(2n + 1)(2n + 2)

โ†’ 0 as n โ†’ โˆž.

Therefore the radius of convergence is infinity and the interval of convergence is R.

โˆž โˆ‘

n=

n(x โˆ’ 1)

n

n + 1(x โˆ’ 1)

n+

โˆš n(x โˆ’ 1) n

n + 1(x โˆ’ 1) โˆš n

โ†’ |x โˆ’ 1 | as n โ†’ โˆž.

The series converes if |x โˆ’ 1 | < 1, so the radius of convergence is 1. If x = 0 or if x = 2, the series

diverges because

n(x โˆ’ 1)

n does not converge to zero. Therefore the interval of convergence is (0, 2).

โˆž โˆ‘

n=

(3x โˆ’ 2) n

n 3 n

x

2 n+

(n + 1)!(n + 2)!2^2 n+

n!(n + 1)!

2 n+

x^2 n+

x

2

4(n + 1)(n + 2)

โ†’ 0 as n โ†’ โˆž.

Therefore the domain of J 1 is R.

  1. The function A defined by

A(x) = 1 +

x

3

x

6

x

9

is called the Airy function after the English mathematician and astronomer Sir George Airy.

(a) Find the domain of the Airy function.

If we write A(x) =

n= anx 3 n , then we find that

an =

2 ยท 3 ยท 5 ยท 6 ยท... ยท (3n โˆ’ 1) ยท 3 n

Since (^) โˆฃ โˆฃ โˆฃ โˆฃ

x

3 n+ an+

x 3 n an

x

3

(3n โˆ’ 1) ยท 3 n

โ†’ 0 as n โ†’ โˆž,

the series converges for all values of x in R.