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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.
Typology: Quizzes
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Problem 1. (6 points) Determine all values of z such that
โ โ
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 โโ |z| < |1 + z|
โโ |z|
2 < |1 + z|
2
โโ zz < (1 + z)(1 + z) = 1 + zz + z + z
โโ z + z > โ 1
โโ Re(z) > โ
Thus, the given series converges if and only if Re(z) > โ
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): โ โ
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 โ. Thus, ฯ = 0. 2
(b):
โ โ
n=
n 3 (n + 1) n
(3n) n
(z โ i + 3)
n .
SOLUTION: To apply the root test, we must examine
|cn|
n
3 /n (n + 1)
3 n
n
n 3 | ยท
n + 1
3 n
as n โโ โ, where the limit of
n
n 3 (which was 1) is determined by using logarithms and LโHopitalโs
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
โ โ
n=
cnz
n
is ฯ, where 0 < ฯ < โ. Prove that the radius of convergence of the power series
โ โ
n=
c
2 n z
n
is ฯ
2 .
SOLUTION: According to the ratio test, we know that
ฯ =
lim nโโ
cn+
cn
Now, by the ratio test, the power series โ โ
n=
c
2 nz
n
converges if and only if
lim nโโ
c 2 n+ z n+
c 2 nz
n
= |z|
lim nโโ
c 2 n+
c 2 n
Hence, its radius of convergence is
lim nโโ
c
2 n+
c 2 n
lim nโโ
cn+
cn
) 2 =^ ฯ
2 .