Download MAS221(2015-16) Exam Solutions 1. (i) (a) Given any ϵ > 0 there

MAS221(2015-16) Exam Solutions
1.
(i) (a) Given any > 0 there exists N ∈ N so that for all n >
N, |an − l| < . (2)
(b) Given any two positive real numbers x, y there exists n ∈ N
so that nx > y. (1)
(c) Guess limit is 1/5, then given
√ any > 0, we√must find N ∈ N
so that √
n > N ⇒ |1/5 − 7/ n − 1/5| = 7/ n < . (1)
Now 7/ n < if and only if n > 49/2 (1).
Using the Archimedean property, (with x = 1, y = 49/2 ),
2
there exists N ∈ N so that N > 49/
√ . (1) √
Then for all n > N we have 7/ n < 7/ N < , as required. (1)
(ii) (a) Proof by induction. True for n = 1, assume true for some n,
then 9 ≤ x2n ≤ 16, and so
1
1
1
(9 + 12) ≤ (x2n + 12) ≤ (16 + 12),
7
7
7
i.e. 3 ≤ xn+1 ≤ 4, as required. (2) Hence, by induction, the
required bound holds for all n ∈ N. (1)
(b) For all n ∈ N,
1 2
(x + 12) − xn (1)
7 n
1 2
=
(x − 7xn + 12)
7 n
1
=
(xn − 3)(xn − 4) (1)
7
xn+1 − xn =
But by (a), (xn − 3) ≥ 0 and (xn − 4) ≤ 0, and so
(xn − 3)(xn − 4) ≤ 0. (1) Hence xn+1 ≤ xn , as required (1).
(c) Since the sequence is monotonic decreasing and bounded below, it converges to a limit l = inf n∈N xn (1)
By algebra of limits in (b) (or alternatively, can use (a) and
solve quadratic)
1
0 = lim xn+1 − lim xn = ( lim xn − 3)( lim xn − 4),
n→∞
n→∞
n→∞
7 n→∞
and hence l = 3 or l = 4. (2) But x1 = 3.5 and sequence is
monotonic decreasing, so we must have l = 3. (1)
1
(iii) (a) A sequence (an ) is Cauchy if given any > 0 there exists
N ∈ N so that for all m, n > N we have |am − an | < . (1)
(b) Choose = 1 (say). Then for any n > N , by (a) and the
triangle inequality
|an | ≤ |aN +1 | + |an − aN +1 | ≤ 1 + |aN +1 |, (2)
so the required bound is
K = max{|a1 |, |a2 |, . . . , |aN |, 1 + |aN +1 |}. (1)
(c) If (an ) is Cauchy, then by (b) it is bounded, so there exists
K > 0 so that |an | ≤ K for all n ∈ N. (1)
If (bn ) is null, then given any > 0 there exists N ∈ N so that
for all n > N, |bn | < /K. (1)
Then for the same and N if n > N , we have
|an bn | ≤ K|bn | < ,
as required.
2.
(1)
(i) Using the triangle inequality,
|a| = |a + b − b| (1)
≤ |a| + |a − b| (1)
and so |a| − |b| ≤ |a − b| (1)
Now interchange a and b to get |b| − |a| ≤ |b − a| = |a − b|.
The result follows since ||a| − |b|| = max{|a| − |b|, |b| − |a|}.
(1)
(ii) Given any sequence (xn ) converging to a with xn ∈ Df \ {a} for
all n ∈ N, we have limn→∞ f (xn ) = f (a). (1)1 f continuous at
a if a ∈ Df and limx→a f (x) = f (a). (1)
(iii) (a) For all x ∈ Df \ {a} we have by (i)
0 < ||f (x)| − |l|| ≤ |f (x) − l|, (1)
and the result follows by the sandwich rule (or can use − δ
criterion). (1)
1
This is the definition given in the course, which was later shown to be equivalent to
the − δ criterion. Any student correctly quoting the latter as the definition will also
receive the full mark.
2
(b) If f is continuous at a, l = f (a) and so |l| = |f (a)|. The result
follows. (1)
(c) f (x) = x. (1)
(iv) (a) Since −1 ≤ cos(y) ≤ 1 for all y ∈ R, we have
−|x| ≤ x cos(1/x) ≤ |x| for all x 6= 0. (1) Then by the
sandwich rule, limx→0 x cos(1/x) = 0 (1)
and the required continuous extension of f is
f (x) if x ∈ R \ {0}
f˜(x) =
(1)
0
if x = 0
(b) By product and chain rules, if x 6= 0,
1
f˜0 (x) = f 0 (x) = cos(1/x) + sin(1/x), (1)
x
f˜ is not differentiable at zero. (1) Indeed for x 6= 0,
f˜(x) − f˜(0)
= cos(1/x), (1)
x
which has no limit at x = 0. To see this first consider a
sequence (xn ) for which xn = 1/2nπ for all n ∈ N. Then
limn→∞ xn = 0 and limn→∞ cos(1/xn ) = 1. (1) Now consider
a sequence (yn ) for which yn = 1/(π/2 + 2nπ) for all n ∈ N.
Then again limn→∞ yn = 0, but this time limn→∞ cos(1/yn ) =
0. (1)
(v) (a) Rolle’s Theorem: Let f be continuous on [a, b] and differentiable on (a, b) with f (a) = f (b). Then there exists c ∈ (a, b)
such that f 0 (c) = 0. (1)
(b) By algebra of limits, g is continuous on [a, b] and differentiable on (a, b). (1) We have g(a) = f (a) and g(b) = f (b) −
f (b)−f (a)
.(b − a) = f (a) = g(a). (1) Hence by Rolle’s theb−a
orem, there exists c ∈ (a, b) so that g 0 (c) = 0. (1) But then
f 0 (c) = α and the result follows. (1)
(c) Let a ≤ x < y ≤ b be arbitrary. By the mean value theorem,
there exists c ∈ (x, y) so that
f (y) − f (x) = f 0 (c)(y − x) < 0,
and so f is strictly monotonic decreasing.
3
(2)
3.
(i) (a) B(a, r) = {x = Rk | |x − a| < r}. (1)
We call U ⊆ Rk open if for all a ∈ U there exists r > 0 such
that B(a, r) ⊆ U . (1)
(b) We call f : Rk → Rl continuous if for each a ∈ Rk , for all ε > 0
there exists δ > 0 such that |x−a| < δ implies |f (x)−f (a)| <
ε. (1)
(c) We define the inverse image f −1 [U ] = {x ∈ U such that f (x) ∈
U }. (1)
Let a ∈ f −1 [U ]. Then f (a) ∈ U , so we have ε > 0 where
B(f (a), ε) ⊆ U . (1)
Since f is continuous, we have δ > 0 such that if |x − a| < δ,
then |f (x) − f (x0 )| < ε. (1)
Hence, if x ∈ B(a, δ), then f (x) ∈ B(f (a, ε) ⊆ U , so x ∈
f −1 [U ]. Thus f −1 [U ] is also open. (2)
(ii) The interval (a, ∞) is open. The others listed are not open. (1
each)
(iii) (a) Define a continuous map f : R2 → R by the formula f (x, y) =
x2 + y 2 . Then A = f −1 [(−∞, 1)] which is open since the
interval (−∞, 1) is open. (1) answer, (1) justification
(b) As in (a), B = f −1 [(1, 2)], which is open as (1, 2) is open. (1)
answer, (1) justification
(c) Let (x, y) ∈ C. Then x ≤ 1. Observe (1, 0) ∈ C. For any
r > 0 the ball B((1, 0), r) contains the element (1 + r/2, 0),
which is an element (x, y) where x > 1. So the ball B((1, 0), r)
contains elements not in C. It follows that C is not open. (1)
answer, (1) justification
4.
(i) (a) The sequence (fn ) converges pointwise to f if for each t ∈
[a, b], we have
lim fn (t) = f (t).
n→∞
(1)
The sequence (fn ) converges uniformly to f if for for all ε > 0,
we have N ∈ N such that |fn (t) − f (t)| < ε for all n ≥ N and
t ∈ [a, b]. (1)
(b) Let t0 ∈ [a, b]. We want to prove f is continuous at t0 . Let
ε > 0. Then:
• We have N ∈ N such that |fn (t) − f (t)| <
n ≥ N, t ∈ [a, b]. (uniform convergence)
4
ε
3
whenever
• We have δ > 0 such that |fN (t) − fN (t0 )| <
|t − t0 | < δ. (fN is continuous)
ε
3
whenever
So, let |t−t0 | < δ. Then we want to show that |f (t)−f (t0 )| <
ε follows, and we have shown that f is continuous at the
(arbitary, chosen at the start) point t0 as required. (3)
Using the above two conditions, if |t − t0 | < δ, then:
ε ε ε
|f (t)−f (t0 )| ≤ |f (t)−fN (t)|+|fN (t)−fN (t0 )|+|fN (t0 )−f( t0 )| < + + = ε.
3 3 3
(2)
Since fn and f are continuous, they are integrable. (1)
Let ε > 0. Then we have N ∈ N such that
|fn (t) − f (t)| <
ε
2(b − a)
whenever n ≥ N , for all t ∈ [a, b]. (2)
Hence by linearity and standard inequalities for integrals
Z b
Z b
Z b
(b − a)ε
ε
≤
|fn (t)−f (t)| dt ≤
f
(t)
dt
f
(t)
dt
−
= <ε
n
2(b − a)
2
a
a
a
whenever n ≥ N . (2)
The result now follows.
(c) The sequence fn (t) converges pointwise to the function
0 t<1
f (t) =
1 t=1
which is not continuous, so the convergence is not uniform by
the above. (1) sequence of functions, (1) idea of using
continuity, (1) correct details
(ii) (a) Let fn : [a, b] → R be a sequence of functions. Suppose we
have a sequence of real numbers (Mn ) such that |fn (x)|
P∞≤ Mn
for all n, and all x ∈ [a, b]. such
n=1 Mn
P that the sum
converges. Then the sequence ∞
f
converges
uniformly
n=1 n
(and absolutely). (2)
P
(b) Let |x| ≤ r. Set Mn = rn . Then the series ∞
n=0 Mn converges
as |r| < 1. (1)
P
n
Since xn ≤ rn , by the Weierstrass M -test, the series ∞
n=0 x
converges uniformly. (1)
5
As a geometric series, the limit is
∞
X
1
xn =
1−x
n=0
(1)
(c) By uniform convergence of the series, and the result for uniform convergence and integrals: (1)
Z r
∞ Z r
X
1
dx =
xn dx
1
−
x
−r
n=0 −r
(1)
For n odd, we have (1)
Z
r
xn dx = 0
−r
For n even:
Z
r
Z
n
x dx = 2
−r
0
r
xn dx =
2rn+1
n+1
(1)
Finally.
Z r
−r
1
dx = − log(1 − r) + log(1 + r) = log
1−x
1−r
1+r
(1)
Putting it all together, the result follows.
(iii) (a) Let f : I → R. We say f is uniformly continuous on I if for all
ε > 0 we have δ > 0 such that for all x, y ∈ [a, b], if |x − y| < δ
then |f (x) − f (y)| < ε. (2)
(b) The function g is continuous, with a closed bounded domain,
and is therefore uniformly continuous. (2)
(c) The function h is uniformly continuous. One way to see this
is to use the mean value theorem. Note that, since x ≥ 1,
1
|h0 (x)| = 2 ≤ 1
x
so by the mean value theorem, for all x, y ∈ [1, ∞) we have
|(h(y) − h(x)| < |y − x|
Hence let ε > 0. Taking δ = ε, we see that for x, y ∈ [1, ∞),
if |x − y| < δ then |h(x) − h(y)| < ε, so h is uniformly continuous. (3)
6