MA3110
Mathematical Analysis II
AY 2009/2010 Sem 1
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
with credits to AP Sun Rongfeng
solutions prepared by Tay Jun Jie
MA3110
Mathematical Analysis II
AY 2009/2010 Sem 1
Question 1
Firstly, we shall rewrite f + as f + (x) = 12 f (x) + 12 |f (x)|. Suppose f + is differentiable at x = 0.
⇒ |f (x)| is differentiable at x = 0
|f (x)|
⇒ lim
exist
x→0
x
Now, since limx→0+
|f (x)|
x
≥ 0 and limx→0−
|f (x)|
x
(1)
(2)
≤ 0. We have
|f (x)|
=0
lim
x f (x) |f (x)| = lim = |0| = 0
⇒ lim x→0
x x→0 x f (x)
∴ f 0 (0) = lim
=0
x→0 x
Conversely, suppose f 0 (0) = 0.
(3)
x→0
(4)
(5)
f (x)
⇒ lim
=0
x→0 x
|f (x)| f (x) = |0| = 0
⇒ lim = lim
x→0
x x→0 x |f (x)|
=0
⇒ lim
x→0
x
⇒ |f (x)| is differentiable at x = 0
(6)
(7)
(8)
(9)
Therefore f + is differentiable at x = 0.
Question 2
Since f is twice-differentiable on [0, 1], ∃c ∈ (0, 1) such that
1
1
f 00 (c)
1 2
0 1
f (x) = f
+f
x−
+
x−
2
2
2
2
2
1
1
1
≥f
+ f0
x−
2
2
2
Z
∴
0
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1
(11)
1
1
1
0 1
f (x) dx ≥
f
+f
x−
dx
2
2
2
0
2
1
1
x
x
0 1
=f
+f
−
2
2
2
2 0
1
=f
2
Z
(10)
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(12)
(13)
(14)
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MA3110
Mathematical Analysis II
AY 2009/2010 Sem 1
Question 3
Define F : [a, b] → R by F (x) :=
[a, b].
Rx
a
f (x) dx. Since f is continuous on [a, b], F is differentiable on
⇒ F 0 (x) = f (x) ∀x ∈ [a, b]
Now, F is continuous on [a, b] and differentiable on (a, b). Hence, ∃c ∈ (a, b) such that
F (b) − F (a)
b−a
Rb
f (x) dx
∴ f (c) = a
b−a
F 0 (c) =
(15)
(16)
Question 4
Z
lim
n→∞ 0
1
cos
h
x i1
1
x
dx = lim n sin
= lim n sin = 1
n→∞
n
n 0 n→∞
n
Alternatively, observe that each cos nx is integrable on [0, 1] and
x
1
sup cos − 1 = 1 − cos → 0 as n → ∞
n
n
x∈[0,1]
Hence cos nx → 1 uniformly on [0, 1].
Z 1
Z 1
Z 1
x
x
∴ lim
cos dx =
lim cos dx =
dx = 1
n→∞ 0
n
n
0 n→∞
0
Question 5
(a)
an+1 xn = 2an xn + 3an−1 xn
∞
∞
∞
X
X
X
⇒
an+1 xn = 2
an xn + 3
an−1 xn
n=0
n=0
(b) Consider the partial fraction of F (x) from Question 5a.
1
1
1
F (x) =
−
4 1 − 3x 1 + x
!
∞
∞
X
1 X
n
n
=
(3x) −
(−x)
4
n=0
=
n=0
Therefore an =
1 n
4 (3
4
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n=0
1
⇒ (F (x) − x) = 2F (x) + 3xF (x)
x
x
⇒ F (x) =
1 − 2x − 3x2
∞
X
1
(17)
(19)
(20)
(21)
(22)
n=0
(3n − (−1)n )xn
(23)
− (−1)n ).
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MA3110
(c)
P∞
n=0 an x
Mathematical Analysis II
n
AY 2009/2010 Sem 1
converges on {x ∈ R : |3x| < 1} ∩ {x ∈ R : |−x| < 1} = x ∈ R : |x| < 13 .
Question 6
Firstly, observe that f is increasing as f (x) = limn→∞ fn (x) ≤ limn→∞ fn (y) = f (y) (by pointwise
convergence of fn to f ) for all 0 ≤ x ≤ y ≤ 1. Now, let ε > 0 be given. Since f is continuous on
[0, 1], f is continuous on [0, 1]. Hence ∃δ > 0 such that ∀x, y ∈ [0, 1],
ε
whenever |x − y| < δ
2
1
k
Let N ∈ N such that N > 1δ . Hence k+1
−
N
N = N < δ.
k+1
k
ε
⇒f
−f
<
∀k = 0, 1, . . . , N − 1
N
N
2
|f (x) − f (y)| <
As fn → f pointwise on [0, 1], for each k = 0, 1, . . . , N , ∃Mk ∈ N such that
fn k − f k < ε whenever n ≥ Mk
N
N 2
m m+1 Let M = max {M1 , . . . , Mk }. Let x ∈ [0, 1], then x ∈ N
, N for some m = 0, 1, . . . , N − 1. For
n ≥ M (note that M is independent of x),
m
m
m+1
ε
m+1
−f
<f
+ −f
<ε
and
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⇒ fn (x) − f (x) ≤ fn
N
N
N
2
N
m
m ε
m+1
m+1
fn (x) − f (x) ≥ fn
−f
>f
− −f
> −ε
(25)
N
N
N
2
N
∴ |fn (x) − f (x)| < ε
(26)
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