AY0910 Sem2 - Nus Maths Society

MA2108
Mathematical Analysis I
AY 2009/2010 Sem 2
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
solutions prepared by Wei Boyan, Tay Jun Jie
MA2108 Mathematical Analysis I
AY 2009/2010 Sem 2
Question 1
(a) Let P (n) : xn 6 2. When n = 1, x1 = 1 6 2, so P (1) is true. Suppose P (k) is true, thus xk 6 2.
Then xn+1 = 51 (xk 2 + 6) 6 15 (4 + 6) = 2. By Principle of Mathematical Induction, P (n) is true for
all n ∈ N.
(b) Claim: xn is increasing. Proof:
1
xn+1 − xn = (x2n − 5xn + 6)
5
1
= (xn − 2)(xn − 3)
5
Since xn 6 2, we have xn+1 > xn . Therefore xn is increasing. By Monotone Convergence Theorem,
xn is converges. Let x be the limit of xn .
1
1 2
(xn + 6) = (x2 + 6)
5
5
(x − 2)(x − 3) = 0
x = lim xn+1 = lim
n→∞
n→∞
Thus x = 2 or x = 3. Since xn 6 2, we obtain x = 2. we conclude that xn is convergent and its
limit is 2.
Question 2
(a)
(i) Firstly, observe that
1
√
2n+ n+1
is a decreasing sequence of strictly positive terms with
lim
n→∞
2n +
1
√
n+1
= 0.
Therefore the series converges by Alternating Series Test.
(ii)
2 n1
n2
1 6n ρ = lim n 1 +
n→∞ 3
3n
1
1
1
1 3n
1 3n
lim 1 +
= lim n n lim n n lim 1 +
n→∞
n→∞
n→∞
3 n→∞
3n
3n
1
= e2 > 1
3
Therefore the series diverges by Root Test.
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MA2108
Mathematical Analysis I
(b) Observe that
1
(2n−1)(2n+1)
∞
X
n=1
1
= 12 ( 2n−1
−
1
2n+1 ).
AY 2009/2010 Sem 2
Thus
1 1 1
1
1
1 − + − + ··· +
−
3 3 5
2n − 1 2n + 1
1
1
= lim
1−
n→∞ 2
2n + 1
1
=
2
1
1
= lim
(2n − 1)(2n + 1) n→∞ 2
(c) Since the an , bn > 0 for all n ∈ N, an bn > 0 for all n ∈ N. In addition, bn → 0 as
ρ = lim
n→∞
Since
P
an converges,
P
P
bn converges.
an bn
= lim bn = 0.
n→∞
an
an bn converges by Limit Comparison Test.
Question 3
(a) Given ε > 0, choose δ = min
Since 0 < |x| < 16 , we have
(b)
1 3
6 , 20 ε
. Suppose 0 < |x − 0| < δ,
2
(2x + 1)(x − 2)
2x + 3x + 2 = 3x + 1
3x + 1 |x| |2x + 3|
=
|3x + 1|
δ |2x + 3|
<
|3x + 1|
|2x+3|
|3x+1|
20
3 .
Then,
(2x + 1)(x − 2)
δ |2x + 3|
+ 2 <
3x + 1
|3x + 1|
20
6 δ
3
=ε
6
(i) Let f (x) = (x2 + x + 1) sin( λ3 ). Let xn =
yn → 0.
3
(2n+1)π ,
yn =
3
2nπ .
Then xn 6= 0, xn → 0, yn 6= 0 and
lim f (yn ) = 0
n→∞
lim f (xn ) 6= 0
n→∞
Therefore limn→∞ f (x) does not exist by the Divergent Criterion.
(ii)
6
6
6
−1<
6
x
x
x
x
x 6
3− <
63
∵x>0
2
2 x
Since limx→0+ 3 −
x
2
= limx→0+ 3 = 3, limx→0+
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x
2
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6
x
= 3 by Squeeze Theorem.
NUS Mathematics Society
MA2108
Mathematical Analysis I
AY 2009/2010 Sem 2
Question 4
Let ε > 0 be given. Since limx→a g(x) = 0, ∃δ > 0 such that
|g(x)| <
ε
whenever 0 < |x − a| < δ.
M
Let δ1 = min(δ, h) > 0. If 0 < |x − a| < δ1 , then
|f (x)g(x)| < M ·
ε
= ε.
M
Therefore limx→a f (x)g(x) = 0.
Question 5
Let a ∈ R, take a rational sequence (xn ) and an irrational sequence (yn ) such that xn → a, and
yn → a. Then
f (xn ) = −xn → −a
f (yn ) = 3yn − 8 → 3a − 8.
If f is continuous at x = a, then
−a = 3a − 8
a = 2.
It follows that if a 6= 2, then f is not continuous at x = a. At x = 2, given ε > 0, we choose δ = 3ε ,
then for |x − 2| < δ, we have
| − x + 2| = |x − 2| < δ < ε
|3x − 8 + 2| = 3|x − 2| < 3δ = ε
Therefore, |f (x) − f (2)| < ε, so f is continuous at x = 2.
Question 6
Let ε > 0, since f and g are uniformly continuous on R, there exists δ1 , δ2 > 0 such that
ε
4
ε
x, y ∈ R, |x − y| < δ2 ⇒ |g(x) − g(y)| <
4
x, y ∈ R, |x − y| < δ1 ⇒ |f (x) − f (y)| <
Let δ = min(δ1 , δ2 ), then for x, y ∈ R, with |x − y| < δ, we have
|F (x) − F (y)| = |f (x)g(x) − f (y)g(y)|
= |f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)|
6 |f (x)g(x) − f (x)g(y)| + |f (x)g(y) − f (y)g(y)|
= |f (x)| |g(x) − g(y)| + |g(y)| |f (x) − f (y)|
ε
ε
< |f (x)| + |g(y)|
4
4
1ε
ε
6
+2
24
4
5ε
=
<ε
8
Thus, F is also uniformly continuous.
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NUS Mathematics Society
MA2108
Mathematical Analysis I
AY 2009/2010 Sem 2
Question 7
(a) Let m = lim inf(yn ), M = lim sup(xn ) and ε > 0 be given. Thus ∃K ∈ N such that for n > K,
m − ε < yn
and
xn < M + ε.
Hence M − m > xn − yn for n ≥ K. Let x ∈ C(xn − yn ), so there exist subsequence (xnk − ynk )
such that xnk − ynk → x. Thus ∃K1 ∈ N such that |xnk − ynk − x| < ε whenever k ≥ K1 .
xnk − ynk − ε < x < xnk − ynk + ε
∀k ≥ K1
Now, ∃K2 ∈ N such that K2 ≥ K1 and nk ≥ K whenever k ≥ K2 . Hence,
x < xnk − ynk + ε < M − m + ε
k ≥ K2
Therefore x < M − m + ε for all ε > 0, that is, x ≤ M − m. In conclusion, M − m is an upper
bound of C(xn − yn ) and lim sup(xn − yn ) = sup C(xn − yn ) ≤ M − m.
(b)
(i) Since bn > 0 ∀n ∈ N, Sn > Sn−1 . Then Sn2 > Sn Sn−1 . Therefore,
bn
bn
<
Sn2
Sn Sn−1
Sn − Sn−1
=
Sn Sn−1
1
1
=
−
Sn−1 Sn
(ii) Let Tn =
Pn
bk
k=1 S 2 ,
k
then
b1
1
1
1
1
1
1
+
−
+
−
+ ··· +
−
2
S1 S2 S2 S3
Sn−1 Sn
S1
2
1
=
−
S1 Sn
2
<
S1
P
bn
So (Tn ) is bounded, since Sbn2 > 0, (Tn ) is increasing. Therefore, ∞
n=1 S 2 is convergent.
Tn <
n
n
Question 8
(a) Let ε > 0 be given, ∃µ > 0 such that x > µ implies
|f (x) − L| < ε
Since limn→∞ xn = ∞, ∃K ∈ N such that n ≥ K implies xn > µ. Therefore, n ≥ K implies
|f (xn ) − L| < ε
Therefore, limn→∞ f (xn ) = L.
(b) Let ε > 0 be given. By assumption, ∃M ∈ R such that
|g(x) − g(x0 )| <
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ε
whenever x, x0 > M.
3
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MA2108
Mathematical Analysis I
AY 2009/2010 Sem 2
Let (xn ) be a sequence in R such that xn → ∞. Now, ∃N ∈ N such that xn > M whenever n ≥ N .
Hence
ε
|g(xn ) − g(xm )| < whenever n, m ≥ N.
3
That is, (g(xn )) is Cauchy and whence it converges to some L ∈ R. Let (yn ) be another sequence
in R such that yn → ∞. By the above argument, g(yn ) → L0 for some L0 ∈ R. Now, ∃K1 ∈ N such
that
ε
|g(xn ) − L| < whenever n ≥ K1 .
3
Similarly, ∃K2 ∈ N such that
ε
whenever m ≥ K2 .
3
|g(ym ) − L0 | <
Lastly, ∃K3 ∈ N such that xn , ym > M whenever n, m ≥ K3 . Hence
|g(xn ) − g(ym )| <
ε
whenever n, m ≥ K3 .
3
Let K = max{K1 , K2 , K3 }. If n, m ≥ K,
|L − L0 | ≤ |L − g(xn )| + |g(xn ) − g(ym )| + |g(ym ) − L0 |
ε ε ε
< + +
3 3 3
=ε
Thus |L − L0 | < ε for all ε > 0, that is, L = L0 . In conclusion, for every sequence (zn ) in R such
that zn → ∞, the sequence (g(zn )) converges to L. Therefore limx→∞ g(x) = L.
Question 9
P
(a) Let a = min{x1 , . . . , xn } and b = max{x1 , . . . , xn }. If a = b, then n1 nk=1 f (xk ) = f (x1 ) and we
are done. Suppose a < b, hence [a, b] ⊂ (0, 1) and f is continuous on [a, b]. By Extreme Value
Theorem, ∃c, d ∈ [a, b] such that f (c) ≤ f (x) ≤ f (d) for all x ∈ [a, b].
f (c) ≤ f (xk ) ≤ f (d)
∀k = 1, . . . , n
n
1X
f (c) ≤
f (xk ) ≤ f (d)
n
k=1
P
P
P
If f (c) = n1 nk=1 f (xk ) or f (d) = n1 nk=1 f (xk ) then we are done. Suppose f (c) < n1 nk=1 f (xk ) <
f (d), applying
Intermediate Value Theorem to f on [c, d] or [d, c], ∃e ∈ (c, d) or (d, c) such that
P
f (e) = n1 nk=1 f (xk ).
(b) Firstly, ∃δ > 0 such that for all x, y ∈ [0, ∞),
Now, since kδ
2 −
|g(x) − g(y)| < 1 whenever |x − y| < δ.
(k−1)δ < δ for all k ∈ N, we have
2
g kδ − g (k − 1)δ < 1
∀k ∈ N
2
2
g kδ < 1 + g (k − 1)δ ∀k ∈ N
2
2
g kδ < k
∀k ∈ N
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MA2108
Mathematical Analysis I
AY 2009/2010 Sem 2
h
S
Let C = 2δ > 0. Now, k∈N (k−1)δ
, kδ
forms a partition for [0, ∞). Let x ∈ (0, ∞), then
2
2
h
(m−1)δ mδ
(m−1)δ x∈
,
for
some
m
∈
N.
Furthermore,
x
−
< δ. If m = 1, then
2
2
2
|g(x) − g(0)| < 1
|g(x)| < 1 < 1 + Cx
If m > 1, since
(m−1)δ
2
≤ x, we have
1
x
≤
2
(m−1)δ .
Therefore,
(m
−
1)δ
g(x) − g
<1
2
(m − 1)δ |g(x)| < 1 + g
2
1
|g(x)| < 1 + (m − 1) = 1 + (m − 1)x
x
2
|g(x)| < 1 +
(m − 1)x
(m − 1)δ
2
|g(x)| < 1 + x = 1 + Cx
δ
In conclusion, |g(x)| < 1 + Cx for all x ∈ (0, ∞).
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