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52-1. Consider the following system of linear equations over the real numbers, where x, y and z are variables and b is a real constant.
x + 2 y + z = 0 2 x + 4 y + 3 z = 0 x + 3 y + bz = 0
Which of the following statements are true?
I. There exists a value of b for which the system has no solution. II. There exists a value of b for which the system has exactly one solution. III. There exists a value of b for which the system has more than one solution.
(A) II only (B) I and II only (C) I and III only (D) II and III only (E) I, II and III
Solution: Let
M :=
1 3 b
(^) and v :=
x y z
(^) and 0 :=
Then the given system can be written, in matrix form, as M v = 0. Since, for every b ∈ R, x = y = z = 0 is a solution to M v = 0 , it follows that I is false. For every b ∈ R,
the solution x = y = z = 0 is the only solution of M v = 0
iff
det M = 0.
Date: Printout date: September 6, 2015. 1
For every b ∈ R, by expanding det M along the third row of M , we get
det M = 1 · det
[
]
− 3 · det
[
]
[
]
= 1 · 2 − 3 · 1 + b · 0 = − 1.
Therefore, for all b ∈ R, the equation M v = 0 has exactly one solution. So II is true and III is false. Answer: (A)
52-2. Consider the following system of linear equations over the real numbers, where x, y and z are variables and b is a real constant.
x + 2 y + z = 0 2 x + 4 y + 3 z = 0 3 x + 6 y + bz = 0
Which of the following statements are true?
I. There exists a value of b for which the system has no solution. II. There exists a value of b for which the system has exactly one solution. III. There exists a value of b for which the system has more than one solution.
(A) II only (B) I and II only (C) I and III only (D) II and III only (E) III only
Solution: Let
M :=
3 6 b
(^) and v :=
x y z
(^) and 0 :=
Then the given system can be written, in matrix form, as M v = 0. Since, for every b ∈ R, x = y = z = 0 is a solution to M v = 0 , it follows that I is false. For every b ∈ R,
the solution x = y = z = 0 is the only solution of M v = 0
iff
det M = 0.
Parametrization of C′^ and computation (remembering the negative ori- entation of C′) gives ∫
C′
(z + 3)^2
dz = 0 and
C′
z + 3
dz = − 2 πi.
We have c 1 = g′(−3) =
(− 3 − 1)^2
. Then ∫
C′
c 0 (z + 3)^2
dz = 0 and
C′
c 1 z + 3
dz = − 2 πic 1 =
πi 8
Then ∫
C
g(z) (z + 3)^2
dz =
C′
c 0 (z + 3)^2
dz
C′
c 1 z + 3
dz
πi 8
πi 8
Alternate solution: The integrand is holomorphic inside C, except at z = −3. The winding number of C around z = −3 is equal to −1. The
Taylor expansion of
z − 4
about z = 0 begins
z + · · ·. Chang-
ing z to z + 3, we see that the Taylor expansion of
z − 1
about z = − 3
begins
(z + 3) + · · ·. Then the residue of
(z − 1)(z + 3)^2
at
z = −3 is
. So, by the Residue Theorem, the integral is
[−1]
[
]
[2πi] =
πi 8
53-2. In the complex plane, let C be the circle |z| = 4 with negative
(clockwise) orientation. Compute
C
dz (z − 1)(z + 3)^2
Solution: Define f : C{− 3 , 1 } → C by f (z) =
(z − 1)(z + 3)^2
. We
wish to compute
C
f (z) dz. Let C′^ be a small circle in C about z = 1 on which and inside of which the Taylor expansion of
(z + 3)^2
converges absolutely. Let C′′
be a small circle in C about z = −3 on which and inside of which
the Taylor expansion of
z − 1
converges absolutely. Give C′^ and C′′
negative orientations. By Cauchy’s Theorem, ∫
C
f (z) dz =
[∫
C′
f (z) dz
]
[∫
C′′
f (z) dz
]
The Taylor expansion of
(z + 3)^2
about z = 1 begins
1 (z + 3)^2
Then, on C′, we have
f (z) =
16(z − 1)
where the remainder, denoted · · · , is holomorphic, and so, by Cauchy, integrates to 0 around C′. Remembering that C′^ is oriented negatively, ∫
C′
f (z) dz =
C′
dz 16(z − 1)
dz =
− 2 πi 16
πi 8
The Taylor expansion of
z − 4
about z = 0 begins
z + · · ·.
Changing z to z + 3, we see that the Taylor expansion of
z − 1
about
z = −3 begins
(z + 3) + · · ·. Then, on C′′, we have
f (z) =
4(z + 3)^2
16(z + 3)
where the remainder, denoted · · · , is holomorphic, and so, by Cauchy, integrates to 0 around C′′. Remembering that C′′^ is oriented negatively, ∫
C′′
f (z) dz =
[∫
C′′
−dz 4(z + 3)^2
dz
]
[∫
C′′
−dz 16(z + 3)
dz
]
[
2 πi 16
]
πi 8
Then ∫
C
f (z) dz =
[
πi 8
]
[
πi 8
]
Alternate solution: Let f (z) =
(z − 1)(z + 3)^2
be the integrand, which
is holomorphic inside C, except at z = 1 and z = −3.
WARNING: If we try to use this trick to compute
C
dz z
, we do not
get 0, and, in fact,
C
dz z
= 2πi. Here’s...
THE POINT: Recall that dz changes to
− 16 dw w^2
. Some care is required
to be sure that the transformed integrand (in w) does not have a pole
at w = 0, coming from the pole (of order two) at w = 0 in
w^2
In other words, if we simply transform the original integrand (in z), ignoring the differential dz, then, in the resulting expression (in w), no pole would appear. However, the differential dz itself transforms to a new differential with a pole of order two at w = 0. If we’re lucky, then that “resulting expression (in w)” will have a zero of order (at least) two at w = 0. If so, then, by Cauchy, the transformed integral is 0, and, consequently, so is the original integral.
54-1. Assume that, in a certain two-dimensional world, the wind ve- locity at any point (x, y) is (− 11 x + 10y, − 10 x + 14y). A small particle is simply pushed by the wind. Its position at any time t is given by (f (t), g(t)). Assume that its velocity at time t is
( −11[f (t)] + 10[g(t)] , −10[f (t)] + 14[g(t)] ).
Because its velocity at time t is also given by (f ′(t), g′(t)), its motion will satisfy the equations:
f ′(t) = −11[f (t)] + 10[g(t)], g′(t) = −10[f (t)] + 14[g(t)].
Assume that the initial position of the particle is (f (0), g(0)) = (0, 1). We stand at the origin and watch the particle. Along what slope line
will we look, asymptotically, as t → ∞? That is, compute lim t→∞
g(t) f (t)
Solution: Let M :=
[
]
, and let I :=
[
]
. Define the
function p : R → R^2 ×^1 by p(t) =
[
f (t) g(t)
]
. The characteristic polyno-
mial of M is det (M − λI) = λ^2 − 3 λ − 54 = (λ − 9)(λ + 6), so the
eigenvalues of M are 9 and −6. We have M − 9 I =
[
]
. The
9-eigenspace of M is the kernel of M − 9 I, which is spanned by
[
]
We have M + 6I =
[
]
. The (−6)-eigenspace of M is the
kernel of M + 6I, which is spanned by
[
]
. Let
v :=
[
]
and w :=
[
]
Then M v = 9v and M w = − 6 w. Then, for all t ∈ R, we have
etM^ v = e^9 tv and etM^ w = e−^6 tw.
At any time t, we have p′(t) = M · [p(t)]. Let u :=
[
]
. The initial
position of the particle is p(0) = u. Then the position of the particle, at any time t, is given by p(t) = etM^ u. We calculate v − (1/2)w = (3/2)u, so u = (2/3)v − (1/3)w. Then, at any time t, we have
p(t) = (2/3)e^9 tv − (1/3)e−^6 tw,
so (^) [ f (t) g(t)
]
= p(t) =
[
(2/3)e^9 t^ − (2/3)e−^6 t (4/3)e^9 t^ − (1/3)e−^6 t
]
so g(t) f (t)
(4/3)e^9 t^ − (1/3)e−^6 t (2/3)e^9 t^ − (2/3)e−^6 t^
(4/3) − (1/3)e−^15 t (2/3) − (2/3)e−^15 t^
Then
tlim→∞
g(t) f (t)
54-2. Assume that, in a certain two-dimensional world, the wind ve- locity at any point (x, y) is (−y, x). A small particle is simply pushed by the wind. Its position at any time t is given by (f (t), g(t)). Assume that its velocity at time t is (−[g(t)], f (t)). Because its velocity at time t is also given by (f ′(t), g′(t)), its motion will satisfy the equations:
f ′(t) = −[g(t)], g′(t) = f (t).
Assume that the initial position of the particle is (f (0), g(0)) = (2, 0). Find its position (f (t), g(t)) at any time t.
We will draw this picture, using the following 16 points:
(1, 0), (0, 1), (− 1 , 0), (0, −1), (3, 0), (0, 3), (− 3 , 0), (0, −3), (2, 0), (0, 2), (− 2 , 0), (0, −2), (
Note that the last eight points are on the circle x^2 + y^2 = 4, which is represented in blue in the picture below. The last arrow starts at (
2), runs −(−
2, and rises
- It therefore ends at (
2 , 0). Thus the last arrow starts on the blue circle x^2 + y^2 = 4 and ends on the x-axis, slightly to the left of (3, 0). This arrow, together with the other 15, results in the 16 red arrows appearing in the picture below.
Note that each red arrow is perpendicular to
the line from its footpoint (i.e., starting point) to the origin.
Moreover, any radius from a point on a circle to the circle’s center is perpendicular to the circle’s tangent at that same point. From these observations we see that, if a particle, at a certain time t, is on
a circle centered at the origin,
then it must be travelling tangent to that circle, and so the instanta- neous rate of change in
the distance from the particle to the origin
is equal to zero. That is, the distance from the particle to the origin has derivative 0 at any time. That is, the distance from the particle to the origin is constant. The particle we wish to study in this question starts at (2, 0), which is on the blue circle, and so it must travel around that circle. Moreover, every arrow footed on the blue circle will have length 2, and so our particle travels with speed 2 at all times. The distance around the circle is 4π, so the particle revolves once around the origin (i.e., covers 2π radians) every 2π units of time. Thus, between time 0 and a time t, it has traveled through t radians. It also travels counterclockwise. Therefore, by trigonometry, we see that its position at any time t is given by (f (t), g(t)) = (2 cos t, 2 sin t).
55-1. Let f : R → R be differentiable. True or False: If f ′(0) = 0, then f (x) has a local extremum at x = 0.
Solution: False. Counterexample: Define f : R → R by f (x) = x^3. Then f ′(0) = 0, but f is increasing on R, so f has no local extrema.
55-2. Let f : R → R be differentiable. True or False: If f (x) has a local extremum at x = 0, then f ′(0) = 0.
Solution: True. This is Fermat’s Theorem (a.k.a. the Interior Ex- tremum Theorem).
55-3. Let f : R → R be differentiable. True or False: If f ′(x) has a local extremum at x = 0, then f (x) has a point of inflection at x = 0.
Solution: True. Proof: Assume that f ′(x) has a local extremum at x = 0. We wish to show that f (x) has a point of inflection at x = 0. We will assume that f ′(x) has a local maximum at x = 0; the proof for local minimum is similar. Choose δ > 0 such that f ′(x) is increasing on −δ < x < 0 and such that f ′(x) is decreasing on 0 < x < δ. Then f (x) is concave up on −δ < x < 0 and concave down on 0 < x < δ. Thus f (x) has a point of inflection at x = 0.
56-1. True or false: For any metric d on R, there is a norm ‖ • ‖ on R such that, for all x, y ∈ R, d(x, y) = ‖x − y‖.
Solution: False. Counterexample: Let | • | denote absolute value on R and define a metric d on R by d(x, y) = min{|x − y|, 1 }. Let ‖ • ‖ be a norm on on R, and assume: for all x, y ∈ R, d(x, y) = ‖x − y‖. We aim for a contradiction. Let a := ‖ 1 ‖. Then a > 0. Let x := 2/a and let y := 0. Then, by assumption, d(x, y) = ‖x − y‖. Since x > 0, we get |x| = x = 2/a. Then |x| · ‖ 1 ‖ = xa = 2. We have d(x, y) = d(x, 0) = min{|x|, 1 } ≤ 1. Also, we have ‖x − y‖ = ‖x − 0 ‖ = ‖x‖ = |x| · ‖ 1 ‖ = 2. Then d(x, y) ≤ 1 < 2 = ‖x − y‖. Contradiction.
NOTE: Every norm on R is a positive multiple of absolute value, so the collection of norms on R is very restricted. (For n ≥ 2, norms on Rn^ are much more plentiful.) On the other hand, there are lots of metrics on R, and there are many ways to manipulate one metric to get another. For example, if δ is a metric on a set S, then we can form another metric δ′^ on S by defining δ′(x, y) = min{δ(x, y), 1 }. In fact, if we define f : [0, ∞) → [0, ∞) by f (t) = min{t, 1 }, then f is semi- increasing and semi-concave down and, moreover, we have f (0) = 0. It turns out that because of these three properties of f , we can prove that f ◦ δ is a metric. Since δ′^ = f ◦ δ, we get that δ′^ is a metric. If you are aware of all of these facts, then it’s easy to find a metric on R that doesn’t come from a norm.
56-2. True or false: For every norm ‖•‖ on R, there is an inner product 〈•, •〉 on R such that, for all x ∈ R, we have ‖x‖^2 = 〈x, x〉.
Solution: True. Proof: Let ‖ • ‖ be a norm on R. We wish to show that there is an inner product 〈•, • , 〉 on R such that, for all x ∈ R, we have ‖x‖^2 = 〈x, x〉. Let a := ‖ 1 ‖. Then, for all x ∈ R, we have ‖x‖ = |x| · ‖ 1 ‖ = a · |x|. Define an inner product 〈•, •〉 on R by 〈x, y〉 = a^2 xy. Given x ∈ R. We wish to prove that ‖x‖^2 = 〈x, x〉. We have |x|^2 = x^2 , so ‖x‖^2 = (a · |x|)^2 = a^2 x^2 = 〈x, x〉.
NOTE: Every norm on R is a positive multiple of absolute value, so the collection of norms on R is very restricted. Every inner product on R is
a positive multiple of multiplication, so the collection of inner products on R^2 is similarly restricted. So, since the absolute value norm comes from the multiplication inner product, it follows that every norm comes from an inner product.
56-3. True or false: For every norm ‖ • ‖ on R^2 , there is an inner product 〈•, •〉 on R^2 such that, for all v ∈ R^2 , we have ‖v‖^2 = 〈v, v〉.
Solution: False. Counterexample: Let | • | denote the absolute value on R. Define a norm ‖ • ‖ on R^2 by ‖(x, y)‖ = |x| + |y|. Let 〈•, •〉 be a metric on R^2. Assume, for all v ∈ R^2 , that ‖v‖^2 = 〈v, v〉. We aim for a contradiction. Let p := (1, 0) and q := (0, 1). We have 〈p, p〉 = ‖p‖^2 = [| 1 | + | 0 |]^2 = 1
and 〈q, q〉 = ‖q‖^2 = [| 0 | + | 1 |]^2 = 1
and 〈p + q, p + q〉 = ‖p + q‖^2 = [| 1 | + | 1 |]^2 = 4.
Then
4 = 〈p + q, p + q〉 = 〈p, p〉 + 2〈p, q〉 + 〈q, q〉 = 1 + 2〈p, q〉 + 1,
so 2 = 2〈p, q〉, so 1 = 〈p, q〉. Then
〈p − q, p − q〉 = 〈p, p〉 − 2 〈p, q〉 + 〈q, q〉 = 1 − 2 · 1 + 1 = 0,
so ‖p − q‖^2 = 〈p − q, p − q〉 = 0. On the other hand, we calculate ‖p − q‖^2 = [| 1 | + | − 1 |]^2 = 4. Contradiction.
NOTE: A norm is always determined by its unit level set. Thus ques- tions about norms can be rephrased in geometric terms. For a norm coming from an inner product on R^2 , the unit level set of the norm will be an ellipse; it will have no corners. It’s easy to make a norm on R^2 whose unit level set has corners. One example is ‖(x, y)‖ = |x| + |y|; the unit level set of this norm is a diamond with corners at (1, 0), (0, 1), (− 1 , 0) and (0, −1). Once you realize that this norm, for geometric reasons, cannot possibly come from an inner product, you know that the answer is false. The argument above is simply an effort to find an algebraic argument to confirm that geometric reasoning. By polarization, when a norm does come from an inner product, that inner product is determined
58-1. Let f : R → R be continuous and injective. Let U be an open subset of R. True or false: f (U ) is necessarily an open subset of R.
Solution: True. Proof: Any continuous injective function R → R is either increasing or decreasing. We will assume that f is increasing; the proof for decreasing f is similar. Let V be a set of bounded open intervals such that ∪V = U. Then f (U ) = ∪{f (V ) | V ∈ V}, so it suffices to prove, for all V ∈ V, that f (V ) is an open subset of R. Given V ∈ V. We wish to show that f (V ) is open. Let a := inf V and let b := sup V. Then V = (a, b). So, since f is increasing f (V ) = (f (a), f (b)). Then f (V ) is a bounded open interval, and, in particular, is open.
58-2. Let f : R → R be continuous. Let U be an open subset of R. True or false: f (U ) is necessarily an open subset of R.
Solution: False. Counterexample: Define f : R → R by f (x) = x^2 , and let U := (− 1 , 1). Then f (U ) = [0, 1), so f (U ) is not open.
58-3. Let f : R → R be continuous. Let U be an open subset of R. True or false: f −^1 (U ) is necessarily an open subset of R.
Solution: True, because the preimage of an open set under a continuous function is always open.
58-4. Let f : R → R be continuous. Let B be a bounded subset of R. True or false: f (B) is necessarily a bounded subset of R.
Solution: True. Proof: Since B is bounded, choose a compact inteval C in R such that B ⊆ C. Then f (B) ⊆ f (C). The image of a compact set under a continuous map is compact, so f (C) is compact. Then f (C) is bounded. So, since f (B) ⊆ f (C), we see that f (B) is bounded.
58-5. Let f : R → R be continuous. Let B be a bounded subset of R. True or false: f −^1 (B) is necessarily a bounded subset of R.
Solution: False. Counterexample: Define the function f : R → R by f (x) = 1/(x^2 +1), and let B := (0, 1]. For all x ∈ R, we have x^2 +1 ≥ 1, so f (x) ≤ 1. For all x ∈ R, we have x^2 + 1 > 0, so f (x) > 0. Then, for all x ∈ R, we have 0 < f (x) ≤ 1. Then f (R) ⊆ (0, 1] = B. Then R ⊆ f −^1 (B). Thus f −^1 (B) is bounded.
NOTE: The function f defined above is not injective, so leaves open the question of whether an injective counterexample exists. It turns out that it’s not possible to find an injective rational counterexample.
However, the agebraic function f : R → R defined by f (x) =
x √ x^2 + 1 is injective and continuous and satisfies f (R) = (− 1 , 1). So, setting B := (− 1 , 1), we have f −^1 (B) = R, and so f −^1 (B) is bounded.
58-6. Let f : (0, 1) → R be continuous. Let B be a bounded subset of R. Assume that B ⊆ (0, 1). True or false: f (B) is necessarily a bounded subset of R.
Solution: False: Counterexample: Define f : (0, 1) → R by f (x) = 1/x and let B := (0, 1). For every integer n > 1, we have 1/n ∈ B, so n = f (1/n) ∈ f (B). Thus { 2 , 3 , 4 ,.. .} ⊆ f (B). So, since { 2 , 3 , 4 ,.. .} is unbounded, it follows that f (B) is unbounded.
