MATH 361
Second Midterm Exam Wednesday, April 5, 2017
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This exam consists of 6 problems. Please write clearly, both in form and substance. Be aware
that a poorly worded sentence might be the difference between a correct and a totally wrong
answer.
Good luck!.
Problem Points
1
6
2
6
3
6
4
6
5
6
6
6
Total
36
Score
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1
Problem 1. Let (K, d) be a compact metric space, and {fk }k∈N a sequence of continuous functions
fk : K → R, such that
(a) fk (x) ≥ 0, for every x ∈ K.
(b) fk (x) ≤ f` (x) whenever k ≥ `.
(c) fk (x) → 0 for every x ∈ K. (fk converges pointwise to 0.)
Prove that fk → 0 uniformly.
Proof. Let ε > 0 be given. For every x ∈ K, there exist Nx ∈ N, such that n ≥ Nx implies
ε
0 ≤ fn (x) < .
2
Since fNx is continuous a x, there exist an open set Ux containing x, such that
ε
|fNx (y) − fNx (x)| < ,
2
for every y ∈ Ux . It follows that if n ≥ Nx and y ∈ Ux , then
0 ≤ fn (y) ≤ fNx (y) < fNx (x) +
ε
< ε.
2
The family {Ux }x∈K , is an open cover of K and since K is compact, we may extract a finite
N
subcover {Uxi }N
i=1 . Let Nε be the maximum of the {Nxi }i=1 . If n ≥ Nε and y ∈ K, then y is in
some Uxi , and since n ≥ Nε ≥ Nxi , it follows that
0 ≤ fn (y) < ε.
In other words fn converges to 0 uniformly.
2
Problem 2. Suppose that f : [0, ∞) → R is continuous, f (0) = 0, f is differentiable on (0, ∞),
and f 0 is nondecreasing (increasing).
Prove that g(x) = f (x)/x is nondecreasing (increasing) for x > 0.
Proof. Since f is differentiable on (0, ∞), it follows that g is differentiable on (0, ∞), and that
dg
f 0 (x)x − f (x)
(x) =
.
dt
x2
On the other hand, the Mean Value Theorem for f on the interval [0, x], implies that
f (x) − f (0) = f 0 (c)(x − 0),
for some c ∈ (0, x). Since f (0) = 0, and f 0 (c) ≤ f 0 (x), it follows that
f (x) ≤ f 0 (x)x
for every x > 0, which implies that the derivative of g is nonnegative at every point x ∈ (0, ∞),
which in turn implies that g is increasing.
3
Problem 3. Prove that the power series
∞
X
zn
n=1
n
converges at every point |z| = 1, except z = 1.
Use Abel’s theorem to find these limits.
(Hint: Use summation by parts.)
Proof. Fix a point z ∈ C \ {1}, |z| = 1. If we denote sn = 1 + z + · · · z n =
that
N
X
zn
n=1
n
=
N
X
sn − sn−1
n=1
n
N
1 − z n+1
, s0 = 0, we see
1−z
N
sN X
1
1
sN X
1
1
=
−
sn−1 ( −
)=
+
sn−1 (
− ).
N
n n−1
N
n−1 n
n=2
n=2
(This is the summation by parts.) Next notice that
2
= M,
|1 − z|
|sn | ≤
(independent of n ∈ N). It follows that
sN
N
∞
X
→ 0 as n → ∞, and that the series
sn−1 (
n=2
is absolutely convergent. So the series
1
1
− )
n−1 n
∞
X
zn
n=1
n
converges. Abel’s theorem now implies that it converges to
lim
∞
X
(zt)n
t→1−
n=1
n
.
For |zt| < 1, this is the series of
− ln(1 − tz),
and so it converges to − ln(1 − z).
If you want to know what this is explicitly when you write z = eiθ , θ ∈ (0, 2π), note that
θ
θ
ei 2 − e−i 2
θ θ−π
1 − e = −2ie
= 2 sin( )ei 2 ,
2i
2
iθ
so that
i θ2
θ
π−θ
− ln(1 − eiθ ) = − ln(2) − ln(sin( )) + i
.
2
2
4
Problem 4. Find the critical points of the function
f (x, y) = sin x + y 2 − 2y + 1
and determine whether they are local maxima, local minima or saddle points.
Proof. Since
∂f
∂f
= cos x,
= 2y − 2,
∂x
∂y
2n + 1
it follows that the critical points are {(
π, 1)}n∈Z . The Hessian is given by the matrix
2
− sin x 0
,
0
2
which computed at the point xn = (
2n + 1
π, 1) is
2
(−1)n+1 0
.
0
2
So the points xn with even n are saddle points, and points xn with odd n are local minima.
5
Problem 5. Determine the second order Taylor polynomial of the function
2
f (x, y) = e(x−1) cos(y)
about the point (x0 , y0 ) = (1, 0).
Proof. The fist derivative is
2
2
Df (x, y) = 2(x − 1)e(x−1) cos(y)dx − e(x−1) sin(y)dy,
whereas the second derivative is
2
2
D2 f (x, y) = (2 − 4(x − 1)2 )e(x−1) cos(y)dxdx − 2(x − 1)e(x−1) sin(y)dydx
2
2
− 2(x − 1)e(x−1) sin(y)dxdy − e(x−1) cos(y)dydy.
Computed at the point (1, 0) they are
Df (1, 0) = 0
and
D2 f (1, 0) = 2dxdx − dydy.
It follows that the second order polynomial is
1
1
T2 (h, k) = f (1, 0) + Df (1, 0)(h, k) + D2 f (1, 0)((h, k, (h, k)) = 1 + h2 − k 2 .
2
2
6
Problem 6. Suppose (E, k k), (F, k k) are Banach spaces, U ⊂ E is an open, connected set, and
f : U → F is a function satisfying
4
kf (x) − f (y)k ≤ kx − yk 3 .
Prove that f is constant.
Proof. Note that the inequality implies that f is differentiable and that Df (x) = 0 at every point
x in U. Indeed:
4
1
kf (x + h) − f (x)k ≤ khk 3 = khkkhk 3 ,
1
and khk 3 → 0 as khk → 0. The Mean Value Theorem implies that the function is locally constant,
and since U is connected, it follows that f is constant.
7