Week #8 Practice Quiz Solutions for Math 412: Convergence of Series and Power Series - Pro, Quizzes of Mathematics

The solutions to problem 1 and problems 2(a) and 2(b) from the week #8 practice quiz of math 412. The problems involve determining the values of complex numbers z for which certain series converge, finding the radii of convergence for power series, and proving the relationship between the radii of convergence for related power series.

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Pre 2010

Uploaded on 08/16/2009

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March 16-20, 2009 Week #8 Practice Quiz Name:
Math 412
SOLUTIONS
Problem 1. (6 points) Determine all values of zsuch that
โˆž
X
n=0 ๎˜’z
1 + z๎˜“n
converges, and for points of convergence, determine the sum.
SOLUTION: This is a geometric series, so convergence hinges upon the modulus of the common
ratio, z
1+z. More precisely, the given series converges if and only if
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
z
1 + z๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
<1โ‡โ‡’ |z|<|1 + z|
โ‡โ‡’ |z|2<|1 + z|2
โ‡โ‡’ zz < (1 + z)(1 + z) = 1 + zz +z+z
โ‡โ‡’ z+z > โˆ’1
โ‡โ‡’ Re(z)>โˆ’1
2.
Thus, the given series converges if and only if Re(z)>โˆ’1
2. The value of the sum is
1
1โˆ’z
1+z
=z+ 1.
2
Problem 2. (4 p oints each) Determine the radius of convergence of each power series
below:
(a):
โˆž
X
n=0 ๎˜’n!
1000n๎˜“zn
SOLUTION: Using the ratio test, we find that
๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
cn+1
cn๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
=๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
(n+ 1)!1000n
n!(1000n+1)๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
=๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œ
n+ 1
1000 ๎˜Œ
๎˜Œ
๎˜Œ
๎˜Œโ†’ โˆž,
as nโ†’ โˆž. Thus, ฯ= 0. 2
(b):
โˆž
X
n=0 ๎˜’n3(n+ 1)n
(3n)n๎˜“(zโˆ’i+ 3)n.
pf2

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Download Week #8 Practice Quiz Solutions for Math 412: Convergence of Series and Power Series - Pro and more Quizzes Mathematics in PDF only on Docsity!

March 16-20, 2009 Week #8 Practice Quiz Name:

Math 412

SOLUTIONS

Problem 1. (6 points) Determine all values of z such that

โˆž โˆ‘

n=

z

1 + z

)n

converges, and for points of convergence, determine the sum.

SOLUTION: This is a geometric series, so convergence hinges upon the modulus of the common

ratio,

z 1+z

. More precisely, the given series converges if and only if

z

1 + z

< 1 โ‡โ‡’ |z| < |1 + z|

โ‡โ‡’ |z|

2 < |1 + z|

2

โ‡โ‡’ zz < (1 + z)(1 + z) = 1 + zz + z + z

โ‡โ‡’ z + z > โˆ’ 1

โ‡โ‡’ Re(z) > โˆ’

Thus, the given series converges if and only if Re(z) > โˆ’

1 2

. The value of the sum is

z 1+z

= z + 1.

Problem 2. (4 points each) Determine the radius of convergence of each power series

below:

(a): โˆž โˆ‘

n=

n!

n

z

n

SOLUTION: Using the ratio test, we find that

cn+

cn

(n + 1)!

n

n!( n+ )

n + 1

as n โ†’ โˆž. Thus, ฯ = 0. 2

(b):

โˆž โˆ‘

n=

n 3 (n + 1) n

(3n) n

(z โˆ’ i + 3)

n .

SOLUTION: To apply the root test, we must examine

|cn|

1 /n

n

3 /n (n + 1)

3 n

n

n 3 | ยท

n + 1

3 n

as n โˆ’โ†’ โˆž, where the limit of

n

n 3 (which was 1) is determined by using logarithms and Lโ€™Hopitalโ€™s

Rule. Thus, L =

1 3

in the root test, and we find that ฯ = 3. 2

Problem 3. (6 points) Assume that the radius of convergence of the power series

โˆž โˆ‘

n=

cnz

n

is ฯ, where 0 < ฯ < โˆž. Prove that the radius of convergence of the power series

โˆž โˆ‘

n=

c

2 n z

n

is ฯ

2 .

SOLUTION: According to the ratio test, we know that

ฯ =

lim nโ†’โˆž

cn+

cn

Now, by the ratio test, the power series โˆž โˆ‘

n=

c

2 nz

n

converges if and only if

lim nโ†’โˆž

c 2 n+ z n+

c 2 nz

n

= |z|

lim nโ†’โˆž

c 2 n+

c 2 n

Hence, its radius of convergence is

lim nโ†’โˆž

c

2 n+

c 2 n

lim nโ†’โˆž

cn+

cn

) 2 =^ ฯ

2 .