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MA3110
Mathematical Analysis II
AY 2010/2011 Sem 1
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
with credits to Agus Leonardi, Tay Jun Jie
MA3110
Mathematical Analysis II
AY 2010/2011 Sem 1
Question 1
(a) We will first prove supm≥n (am + bm ) ≤ supm≥n am + supm≥n bn .
Let supm≥n am = L and supm≥n bm = M .
By definition, for all m ≥ n, we have am ≤ L and bm ≤ M , which implies am + bm ≤ L + M for
all m ≥ n, hence proving the claim.
By the property of limits, we have
lim sup (am + bm ) ≤ lim sup am + lim sup bn
n→∞ m≥n
n→∞ m≥n
n→∞ m≥n
hence proving the required inequality.
(b) Let an = sin n and bn = − sin n.
Then lim supn→∞ (an + bn ) = 0, while lim supn→∞ an = lim supn→∞ bn = 1.
In this case, we have strict inequality.
Question 2
(a) We will first prove that g 0 (0) exists implies g 0 (0) = 0.
Suppose g 0 (0) exists, but g 0 (0) 6= 0. WLOG, suppose g 0 (0) > 0.
Then there exists δ > 0 such that
g(x) > 0,
g(x) < 0,
∀x ∈ (0, δ)
∀x ∈ (−δ, 0)
However, by definition, g(x) ≡ |f (x)| ≥ 0 for all x. Hence, we have a contradiction. We conclude
that g 0 (0) = 0.
Now, by definition of derivative,
g(x) − g(0)
g(x)
|f (x)|
= lim
= lim
=0
x→0
x→0 x
x→0
x−0
x
g 0 (0) = lim
Moreover, when x > 0, we have the following inequality:
−
|f (x)|
f (x)
|f (x)|
≤
≤
x
x
x
The direction of above inequality is reversed when x < 0.
Hence, by Squeeze Lemma,
f (x)
f 0 (0) = lim
=0
x→0 x
proving the statement.
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AY 2010/2011 Sem 1
(b) Suppose f 0 (0) exists.
One direction has been proven, i.e. if g 0 (0) exists then by part (i), f 0 (0) = 0 and g 0 (0) = 0.
We will prove the other direction.
Now, if f 0 (0) = 0, then by definition
of derivative, given any ε > 0, there exists δ > 0, such that
f (x) < ε.
for all x such that 0 < |x| < δ, x Moreover, we have for all x such that 0 < |x| < δ,
|f (x)|
f (x) x − 0 = x < ε
hence, proving that
g 0 (0) = lim
x→0
g(x)
|f (x)|
= lim
=0
x→0
x
x
by definition.
Question 3
(a) We will first show that f is Riemann-integrable in [0, 1], by showing that f is continuous in [0, 1].
Given ε > 0, take δ = ε2 , then for all |x − y| < δ,
p
√
|f (x) − f (y)| ≤ |y − x| < δ = ε
as required.
Now, we can apply the Riemann sum expression for f (x).
Take the equal-spaced partition of f (with norm n1 ), with the right end-point of each partition
interval as the sample point. Then since f is integrable, and limn→∞ n1 = 0, by the Theorem on
Convergence of Riemann Sums,
Z 1
n
X
i 1
f
=
f (x) dx
n→∞
n n
0
lim
i=1
as required.
Question 4
(a) By Taylor Expansion, we have
sin y = sin 0 +
cos c
y
1!
for some c between 0 and y.
xp xp As | cos c| ≤ 1, we have | sin y| ≤ |y| for all y ∈ R. Therefore, we have sin p ≤ p for all x ∈ R.
n
n
To prove convergence of the required series, we will use Weierstrass M-test. Note that if p > 1,
∞ p
∞
X
X
x 1
= |xp |
np np
n=1
n=1
converges for all x ∈ R by the p-series test.
Hence,
∞
X
n=1
sin
xp
converges uniformly on [−r, r] for any r > 0 by Weierstrass M-test.
np
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MA3110
Mathematical Analysis II
AY 2010/2011 Sem 1
(b) By Taylor Expansion, we have
sin y = sin 0 +
cos 0
sin c 2
y−
y
1!
2!
for some c between 0 and y.
1
As −1 ≤ sin c ≤ 1, we have sin y ≥ y − y 2 for all y ∈ R.
2
Therefore, we have
xp
xp
1 x2p
sin p ≥ p −
n
n
2 n2p
Now, we will show that the series
X
∞
∞ p
X
1 x2p
1 xp
x
1
p
−
=
−
x
np 2 n2p
np 2 n2p
n=1
n=1
diverges for any x 6= 0.
X
∞ ∞
X
1
np − k
k
=
−
, where k = 12 xp .
np n2p
n2p
n=1
n=1
np − k
1
Let an =
and bn = p .
n2p
n
Consider the series
If p ∈ (0, 1], we have
lim
n→∞
an
k
= lim 1 − p = 1 > 0
n→∞
bn
n
Clearly, if p = 0, the series diverges.
∞
X
Moreover, if p ∈ [0, 1], the series
bn diverges by p-series test.
n=1
Hence, by Limit Comparison Test, if p ∈ [0, 1], the series
∞
X
sin
n=1
xp
is divergent at any x 6= 0.
np
Question 5
1
x
sin √ .
n
n
Clearly, fn (x) are differentiable for all x ∈ R.
∞
X
Moreover,
fn (1) converges. This is because (by Question 4)
(a) Let fn (x) =
n=1
1
1 sin √1 ≤ √
n
n n n
∞ X
1 and the series
32 converges by the p-series test.
n
n=1
∞
X
Therefore,
fn (1) converges by Comparison Test.
n=1
Furthermore,
∞
X
n=1
fn0
=
∞
X
1
x
cos √
n
n=1 n
3
2
converges uniformly. This is because
1
x
3 cos √ ≤ 13
2
n n2
n
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and
Mathematical Analysis II
∞
X
1
converges by p-series test.
3
n=1
AY 2010/2011 Sem 1
n2
Hence, the convergence of
∞
X
fn0 follows by Weierstrass M-test.
n=1
∞
X
0
In this case, we have f (x) is uniformly convergent on R and f (x) =
fn0
=
n=1
∞
X
1
x
cos √
n
n=1 n
3
2
as required.
Question 6
√
∞
∞
X
cos nx X 1
√
√ diverges. Furthermore, since cosine is even, it suffices to consider
(a) For x = 0,
=
n
n
n=1
n=1
√
x > 0. Let x > 0 be given. Now, we note that cos nx ≥ 12 whenever
π
π √
≤ nx ≤ 2kπ + ,
3
3
2kπ −
that is,
(6k + 1)π 2
(6k − 1)π 2
≤n≤
.
3x
3x
2
2
2
(6k+1)π
Let lk = (6k−1)π
. Since `k → ∞ as
,
L
=
, Ik = [lk , Lk ] and `k = Lk − lk = 8kπ
k
3x
3x
3x2
k → ∞, we deduce that ∃N1 ∈ N such that Ik ∩ N 6= ∅ for every k ≥ N1 . In addition, let b·c, d·e
and #Ik denote the floor function, ceiling function and number of integers in Ik respectively. Then
#Ik = bLk c − dlk e + 1 > Lk − lk − 1 = `k − 1. Now, for every k ≥ N1 ,
√ √
X cos nx X cos nx
1 X 1
1 X 1
≥
√
√
√ ≥
√
≥
2
2
n n
n
Lk
n∈I
n∈I
n∈I
n∈I
k
k
k
k
1 #Ik
1 `k − 1
= √
> √
2 Lk
2 Lk
Now, limk→∞
√1
Lk
= 0 and
8kπ 2
`k
3x
4π
= lim
lim √
·
=
.
k→∞ 3x2
k→∞
(6k + 1)π
3x
Lk
`√k −1 4π 2π
−1
= 4π
.
Thus
∃N
∈
N
such
that
−
Hence, limk→∞ `√k L
2
3x
3x < 3x whenever k ≥ N2 . Therefore,
Lk
k
for every k ≥ max(N1 , N2 ), we have
√ X
π
cos nx √
>
.
n 3x
n∈Ik
√
∞
X
cos nx
√
In conclusion,
fails the Cauchy Criterion and thus diverges.
n
n=1
(b) For x = 0,
√
∞
X
sin nx
n=1
n
= 0.
Furthermore, since sine is an odd function, we have −
√
∞
X
sin nx
n=1
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n
=
√
∞
X
sin (− nx)
n=1
n
, hence it
NUS Mathematics Society
MA3110
Mathematical Analysis II
AY 2010/2011 Sem 1
suffices to consider the case where x > 0.
Let x > 0 be given and let yn =
√
nx for each n ∈ N. Note that
∞
[
[yn , yn+1 ) forms a parti-
n=1
2
tion on [x, ∞). Moreover, limn→∞ yn+1 − yn = 0 and x2 = yn+1
− yn2 for all n ∈ N. Therefore,
√
∞
X
sin nx
n=1
n
= x2
sin yn
sin yn
= (yn+1 + yn ) 2 (yn+1 − yn ).
2
yn
yn
√
∞
X
sin nx
Since limn→∞ yn+1 −yn = 0, heuristically yn+1 ≈ yn for large n. Hence, the tail sum of
n=1
Z
2 sin y
is the Riemann sum of
dy over some appropriate interval I.
y
I
√
Z ∞
∞
X
sin t
sin nx
Since
dt converges, we conclude that
converges.
t
n
0
n
n=1
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