Mathematics 3205
Final with solutions
May 7th, 2013
Partial credit will be awarded for your answers, so it is to your advantage to explain your
reasoning and what theorems you are using when you write your solutions. Please answer
the questions in the space provided and show your computations.
Good luck!
I
II
III
IV
V
VI
VII
VIII
Total
Name:
1
I. (15 points) Compute the lim inf and lim sup of the following sequence
nπ (−1)n
+
,
xn = sin
2
n
n∈N
Solutions:
Let us assume that n = 2k, then
x2k = sin
2kπ
2
+
= sin (kπ) +
=
1
2k
(−1)2k
2k
1
2k
since
sin (kπ) = 0.
On the other side if n = 2k + 1, then
(2k + 1)π
(−1)2k+1
x2k+1 = sin
+
2
2k + 1
1
= sin (kπ + π/2) −
2k + 1
1
= (−1)k −
2k + 1
1
=1−
if k = 2j, and
4j + 1
−1−
1
4j − 1
if
k = 2j − 1,
Hence, the set of all subsequences is given by (x4j+1 )j , (x4j−1 )j , (x2j )j , j ∈ N, where
x4j+1 = 1 −
1
4j + 1
x4j−1 = −1 −
x2j =
Hence
sup
and
inf
j = 0, 1, . . . .
1
4j − 1
1
2j
j = 1, 2, . . . .
j = 0, 1, . . . .
1
1
1
, 1−
, −1 −
,
2j
4j + 1
4j − 1
1
1
1
, 1−
, −1 −
,
2j
4j + 1
4j − 1
j ∈ N = +1,
j ∈ N = −4/3.
Hence, we conclude that
lim inf xn = −4/3,
2
lim sup xn = +1
II. (15 points) Define f : R −→ R as follows:
5x,
if x is rational
f (x) =
2
x + 6, if x is irrational
1. Prove that f is discontinuous at 1 and continuous at 2.
2. Are there other points besides 2 at which f is continuous?
Solutions:
1. Since 1 is rational, then f (1) = 5(1) = 5. On the other hand, limx−→1 f (x) =
limx−→1 x2 + 6 = 7 6= f (1) = 5. Hence, f is discontinuous at 1.
Again, 2 is rational, this implies that f (2) = 5(2) = 10. On the other hand, limx−→2 f (x) =
limx−→2 x2 + 6 = 10 6= f (2). Hence, f is continuous at 2.
2. More generally, f is continuous at a rational number r0 if
lim f (x) = lim x2 + 6 = 5r0
x−→r0
x−→r0
Hence, solving the equation r02 + 6 = 5r0 leads to the points 2 and 3. Hence, besides 2
f is continuous at 3.
3
III. (10 points) Let f : R −→ R be a continuous function and let k ∈ R. Prove that the set
f −1 ({k})
is closed.
Hint: Recall that f −1 ({k}) = {x ∈ R : f (x) = k}
Solutions:
A point x is a limit point for a set C if there exists a sequence xn ∈ C such that xn −→ x
and x ∈ C.
Now, let us assume that there is sequence xn ∈ f −1 ({k}) such that xn −→ x. Our goal is to
prove that x ∈ f −1 ({k}). Using the definition of the set f −1 ({k})
xn ∈ f −1 ({k}) =⇒ f (xn ) = k
=⇒ lim f (xn ) = k
Since f is continuous, this means that
xn −→ x =⇒ f (xn ) −→ f (x).
Now, using the uniqueness of the limit, this implies that f (x) = k. Hence x ∈ f −1 ({k}).
And this complete the proof.
4
VI. (10 points) Using the − δ property, prove that the function
f (x) =
x−1
,
x+1
is uniformly continuous on [0, ∞).
Solutions:
f is uniformly continuous on [0, ∞) if
∀ > 0, ∃δ > 0, ∀x, y ∈ [0, ∞),
|x − y| < δ =⇒ |f (x) − f (y)| < .
x − 1 y − 1
|f (x) − f (y)| = −
x + 1 y + 1
(x − 1)(y + 1) − (x + 1)(y − 1) = (x + 1)(y + 1)
2|x − y|
=
|(x + 1)(y + 1)|
Since
x ≥ 0 =⇒ x + 1 ≥ 1
1
≤ 1.
=⇒
x+1
and
y ≥ 0 =⇒ y + 1 ≥ 1
1
=⇒
≤ 1.
y+1
Hence,
|f (x) − f (y)| ≤ 2|x − y|
<
Thus, it is enough to choose δ = /2.
5
V. (10 points) Use the mean value theorem to establish the following inequality
| cos x − cos y| ≤ |x − y|,
for x, y ∈ R.
Solutions:
The function x −→ cos x is continuous and differentiable on R, hence in particular it is
continuous on [x, y] and differentiable on (x, y) for arbitrary x, y ∈ R. Hence, using the
Mean value theorem there exists c ∈ (x, y) such that
− sin c =
cos x − cos y
.
x−y
Hence
| cos x − cos y|
= | sin c|
|x − y|
≤ 1.
=⇒ | cos x − cos y| ≤ |x − y|,
This completes the proof.
6
x, y ∈ R.
VI. (10 points) Determine whether the following series converges or diverges. Justify your
answer.
X
2n e−n
Solutions:
Here an = 2n e−n . We can use the ratio test
lim
Hence,
P
2n+1 e−n−1
an+1
= lim
an
2n e−n
2
= <1
e
2n e−n is convergent.
7
VII. (10 points) Find the values of x for which the following power series is convergent
X
(n3−n )(x − 2)n
Solutions:
Here, an = n3−n .
an+1
|
an
(n + 1)3−n−1
= lim |
|
n3−n
1n+1
=
3 n
1
=
3
α : = lim |
=⇒ R = 3. Hence,
P −n
(n3 )(x − 2)n is convergent for |x − 2| < 3, hence for x ∈ (−1, 5).
8
x ∈ (−1, 1).
VIII. (20 points) Let f (x) = ln(1 + x),
1. Write out the Taylor polynomial of order n for f at the point x0 = 0
2. What is the Taylor remainder Rn (x)?
3. Prove that f can be written in terms of a power series on (−1, 1).
Solutions:
1.
f (x) = ln(1 + x) =⇒ f (0) = ln(1) = 1
1
=⇒ f 0 (0) = 1
1+x
−1
f 00 (x) =
=⇒ f 00 (0) = −1
(1 + x)2
f 0 (x) =
f (3) (x) =
2
=⇒ f (3) (0) = 2
(1 + x)3
−2 · 3
=⇒ f (4) (0) = −3!
4
(1 + x)
...
n−1
(−1) (n − 1)!
=⇒ f (n) (0) = (−1)n−1 (n − 1)!
f (n) (x) =
(1 + x)n
f (4) (x) =
Hence,
pn (x) =
n
X
f (k) (0)
k=1
=
k!
xk
n
X
(−1)k−1
k=1
k
xk
2. For 0 < c < x, the Taylor remainder is given by
f (n+1) (c) n+1
x
(n + 1)!
xn+1
(−1)n
=
(n + 1) (1 + c)n+1
Rn (x) =
9
3. It is enough to prove that Rn (x) converges to 0 when n goes to ∞. In fact, since
|x| ≤ 1, we have that
|x|n+1
|Rn (x)| =
(1 + n)|1 + c|n+1
|x|n+1
≤
(1 + n)
1
≤
−→n→∞ 0
(1 + n)
Hence for |x| ≤ 1,
ln(1 + x) =
∞
X
(−1)n−1
n=1
10
n
xn .