Answers to Homework 4, Math 4121 (1) Prove that ∑ (n+ 1)zn has

Answers to Homework 4, Math 4121
P
n
(1) Prove that ∞
n=0 (n + 1)z has radius of convergence 1 and thus
defines a function F (z) on B(0, 1) with F (0) = 1. Find a power
series expansion for 1/F (z) near the origin.
1
The radius of convergence
followsP
from the fact lim(n+1) n =
R
n+1
1.
we see that F (z)dz = ∞
. So, let G(z) =
n=0 z
P∞Thus,
1
n
,
converging
in
B(0,
1).
We
get,
G0 (z) = F (z)
z
=
n=0
1−z
1
1
2
and so F (z) = (1−z)
2 , giving F (z) = 1 − 2z + z .
P∞
(2) Prove that n=1 1+nx2 x2 converges pointwise for any x > 0 and
the convergence is uniform in [r, ∞) for any r > 0. So, we get
a continuous function F (x) for x > 0 defined by this series.
Calculate limx↓0 F (x).
If we prove the uniform convergence for any r > 0, the first
part will follow. But for any x ≥ r, 1+nx2 x2 ≤ n12 r and then by
Weierstrass M-test, we are done.
For calculating the limit, we observe, for any x > 0,
Z n+1
x
xdt
x
≤
≤
.
2
2
2
2
1 + (n + 1) x
1+x t
1 + n 2 x2
n
So, we get adding these
Z ∞
xdt
F (x) ≤
≤ x + F (x).
1 + x2 t2
0
The middle integral is just π/2 (substituting u = xt and then
u = tan θ). So, we get F (x) ≤ π/2 ≤ x + F (x) for all x > 0
and then it is clear that the required limit is just π/2.
(3) Let f be a continuous function on R and assume that Rlimx→+∞ f (x) =
2
A, a finite number. What can you say about limn→∞ 0 f (nx)dx?
R2
R
2n
We have 0 f (nx)dx = n1 0 f (y)dy, by substitution y = nx.
Given any > 0, we have an N ∈ N such that for
R 2nall x ≥ N ,
f (x) ∈ (A−, A+). Thus for any n ≥ N , we get, 0 f (y)dy =
R 2N
R 2n
f
(y)dy
+
f (y)dy. Thus, we have
0
2N
Z 2N
Z 2n
Z 2N
f (y)dy+2(n−N )(A−) ≤
f (y)dy ≤
f (y)dy+2(n−N )(A+).
0
0
0
Dividing by n and letting n go to infinity, we see that
Z 2
2(A − ) ≤ lim
f (nx)dx ≤ 2(A + ).
n→∞
0
Since was arbitrary, we see that the required integral is just
2A.
1
2
√
1
− 12
cos 23 .
n=0 (3n)! = e + 2e
√
− 12 + 23 i, a primitive third root
(4) Prove the formula 3
2πi
P∞
Let ω = e 3 =
x ∈ R, we consider
x
e +e
ωx
+e
ω2 x
=
=
∞
X
xn
n=0
∞
X
n=0
n!
+
∞
X
ω n xn
n=0
n!
+
of unity. For
∞
X
ω 2n xn
n=0
n!
(1 + ω n + ω 2n )xn
n!
Now, 1 + ω n + ω 2n = 0 if n is not a multiple of 3 and equal to
3, when it is a multiple of 3. So, we get,
ex + eωx + eω
2x
=3
∞
X
x3n
.
(3n)!
n=0
Putting x = 1, and using De Moivre’s formula, we get the
required formula.
P
n
(5) Assume f (z) = ∞
n=0 an z is a powerseries (over complex numbers) with radius of convergence R > 0. Prove that the closed
set {a ∈ B(0, R)|f (a) = 0} is discrete (that is, has no accumulation points) unless f (z) = 0 for all z ∈ B(0, R).
Let T ⊂ B(0, R) be the the set of zeros of f which are accumulation points of the set of all zeros of f . We wish to show
that if this set is non-empty, then f is identically zero. By elementary topology, one can easily check that this set is closed
too, since the set of all zeros of f is closed by continuity of
f . We will show that this set is open and by connectedness of
B(0, R), we would be done.
If T is non-empty and not open, it has a boundary point, say
p. (Not all points are interior to T .) Since all points of T are
accumulation points, we can find a sequence pn ∈ T , pn 6= p
and lim pn = p. From what we proved in class, there exists
an open disc, B(p, r) contained in B(0,P
R), where we have a
∞
n
power series representation for f (z) =
n=0 bn (z − p) . We
will show that bn = 0 for all n. Clearly b0 = f (p) = 0. Assume
we have proved that b0 = b1 = · · · = bn−1 = 0. Then f (z) =
bn (z − p)n + bn+1 (z − p)n+1 + · · · and letting g(z) = f (z)/(z −
p)n = bn + bn+1 (z − p) + · · · , we see that g(z) has a convergent
power series expansion in B(p, r). Since lim pn = p, we have
pn ∈ B(p, r) for all large n. But g(pn ) = f (pn )/(pn − p)n = 0
and thus, by continuity of g, g(p) = g(lim pn ) = lim g(pn ) = 0.
3
This implies bn = 0, completing the induction. So, B(p, r) ⊂ T
and we are done.
1
(6) Consider the function f (x) = 1+x
2 on R. Prove that for any
P
point a ∈ R, f has a power series√expansion f (x) = ∞
n=0 an (x−
n
2
a) with radius of convergence a + 1.
The complex picture always tell you more. If you think of
the above function
as a function of a complex variable, the
√
two points ± −1 are where the function is not defined
√ and so
no disc of convergence can contain these points. So, 1 + a2 ,
which is the distance from a to ±i is the maximum possible
radius of convergence.
1
1
1
So, we write f (z) = 1+z
2 = 1+zi × 1−zi and expand these
separately. Letting u = z − a, we have,
1
1
1
1
=
=
×
,
1 + zi
1 + ui + ai
1 + ai 1 + v
1
ui
. The radius of convergence of the series 1+v
=
where
v = 1+ai
P
n n
(−1) v is clearly when |v| <√1 and that is the same as
ui
| 1+ai
| < 1, which is same as |u| < 1 + a2 . The same argument
1
applies to 1−zi
and then Cauchy product finishes the proof.

Download Report