MATH 300 ASSIGNMENT 7: SOLUTION
(1) Using ratio or comparison test, show that the following series converge:
P∞ 2j
a)
j=3 j!
P∞ sin(2j)
b)
j=1
3j
Sol:
a) use ratio test: set aj = 2j /j!, then
aj+1 /aj = 2/(j + 1) → 0,
as j → ∞.
therefore the series converges.
b) use comparison test: set aj = sin(2j)/3j , then
|aj | ≤ 3−j
P∞
which converges since j=1 3−j converges.
(2) Consider the sequence of functions Fn (x) = xn on (0,1) which converges
pointwise to F (x) = 0. Show that for 0 < x < 1,
1
2
when and only when n > Log 2/ Log(1/x). This problem shows that the
sequence Fn (x) does not converge uniformly on (0, 1).
Sol: xn < 21 if and only if n Log x < − Log 2 or simply n > Log 2/ Log(1/x).
|Fn (x) − F (x)| <
(3) Verify the following Taylor series by finding a general formula for f (j) (z0 ):
cosh z =
z
∞
X
z 2j
,
(2j)!
j=0
z0 = 0.
−z
, f 2j−1 (z) = sinh z, f (2j) (z) = cosh z, for
Sol: f (z) = cosh z = e +e
2
(2j−1)
j ≥ 1. Therefore f
(0) = 0, f (2j) (0) = 1 and the Taylor series follows.
(4) By using the Cauchy product, find first three nonzero terms in the MacLaurin expansion of the function ez sin z.
Sol: Since
z2
z3
+
+ ··· ,
ez = 1 + z +
2
6
z5
z3
+
+ ··· ,
sin z = z −
6
120
we get by using Cauchy product
1 1
ez = z + z 2 + z 3 (− + ) + · · ·
6 2
z3
2
=z+z + .
3
1
2
MATH 300 ASSIGNMENT 7: SOLUTION
(5) Find and state the convergence properties of the Taylor series of the function f (z) = z/(1 − z) around the point z0 = i.
Sol1: the function f (z) = z/(1 − z) has
√ a singularity at z = 1, therefore
it is analytic
inside
the
disk
|z
−
z
|
<
2 (note that the distance between
0
√
1
, and
1 and i is 2). Now f (z) = −1 + 1−z
f (j) (z) = j!(1 − z)−j−1 ,
f (j) (i) = j!(1 − i)−j−1 .
Therefore
∞
f (z) =
X
i
+
(1 − i)−j−1 (z − i)j ,
1 − i j=1
√
valid for |z − i| < 2.
Sol2: the function f (z) = z/(1 − z) has
√ a singularity at z = 1, therefore
2 (note that the distance between
it is analytic
inside
the
disk
|z
−
z
|
<
0
√
√
1 and i is 2). Now for |z − i| < 2, we have |z − i|/|1 − i| < 1, therefore
1
1
1
z − i −1
=
=
· (1 −
)
1−z
1 − i − (z − i)
1−i
1−i
∞
1 X z−i j
=
(
) .
1 − i j=0 1 − i
Therefore
∞
f (z) = (z − i + i) ·
1 X
(z − i)j (1 − i)−j
1 − i j=0
∞
∞
=
X
1 X
(z − i)j+1 (1 − i)−j +
(z − i)j · i(1 − i)−j
1 − i j=0
j=0
=
∞
X
1 i+
(z − i)j (1 − i)−j ,
1−i
j=1
valid for |z − i| <
√
2.