Preliminary Exam 2018

Solutions to Morning Exam

Part I.
Solve four of the following five problems.
Problem 1. Consider the series n>2 (n log n)−1 and n>2 (n(log n)2 )−1 . Show
P P
that one converges and one diverges by applying a standard convergence test.
Solution: Use the integral test. An antiderivative for 1/(x log x) is log log x (the
substitution u = log x replaces the integrand by du/u) and consequently
Z ∞
dx
= lim (log log T − log log 2) = ∞.
2 x log x T →inf ty
So n>2 (n log n)−1 diverges. On the other hand, an antiderivative for 1/(x(log x)2 )
P

is −1/ log x (again, use the substitution u = log x), so

Z ∞
dx
2
= lim (−(log T )−1 + (log 2)−1 ) = (log 2)−1 .
2 x(log x) T →inf ty
2 −1
P
Hence n>2 (n(log n) ) converges.
Problem 2. Show that
Z ∞Z ∞ √ Z ∞ Z ∞
2
− x2 +y 2 +y 2 )
e dx dy = 2 e−(x dx dy
−∞ −∞ −∞ −∞
by computing both sides.
Solution: Do the integrals in polar coordinates. On the left-hand side we obtain
Z 2π Z ∞ Z 2π
( e−r r dr) dθ = (−e−r r − e−r |∞
0 ) dθ,
0 0 0
which is (2π)(1) = 2π. On the right-hand side we obtain
Z 2π Z ∞ Z 2π
2 2
( e−r r dr) dθ = (−e−r /2|∞
0 ) dθ,
0 0 0
which is (2π)(1/2) = π.
Problem 3. Prove that if f (x) is sin x or arctan x then |f (b) − f (a)| 6 |b − a|
for all a, b ∈ R and that this inequality also holds for f (x) = log x and a, b > 1.
Solution: By the Mean Value Theorem, f (b) − f (a) = f 0 (c)(b − a) for some c
strictly between a and b, and consequently
|f (b) − f (a)| 6 |f 0 (c)||b − a|.
Since the derivatives of sin x and arctan x are cos x and 1/(1+x2 ) respectively, both
of which are bounded by 1 in absolute value on R, we obtain the stated inequality.
Also, if f (x) = log x then f 0 (x) = 1/x, which is bounded by 1 for x > 1.
Problem 4. Let y be a differentiable function and p a continuous function on
(0, ∞), and suppose that y 0 (t) + p(t)y(t) = p(t) for all t > 0. If p(t) > c/t for some
constant c > 0 prove that limt→∞ y(t) = 1.
Solution: Let P (t) be an antiderivative of p(t) on (0, ∞), say
Z t
P (t) = p(t) dt,
1
1
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and put µ(t) = eP (t) . Multiplying both sides of the differential equation by µ(t),
we obtain (y(t)µ(t))0 = µ0 (t), whence
y(t) = e−P (t) (eP (t) + κ) = 1 + κe−P (t)
for some constant κ. Now for t > 1 the fact that p(t) > c/t implies that
Z t
P (t) = p(t) dt > c log t,
1
−P (t)
whence e < t . Returning to the equation y(t) = 1 + κe−P (t) , we conclude
−c

that limt→∞ y(t) = 1.

Problem 5. Let fn (x) = xn on the interval I = [0, 1] in R. Show that the se-
quence {fn }n>1 does not converge uniformly on I. You may quote general theorems
about uniform convergence.
Solution: Each function fn is continuous on I, but the function f to which
{fn }n>1 is pointwise convergent is not continuous: Indeed f is 0 on [0, 1) and 1 at
1, so f is not continuous at 1. It follows that the convergence is not uniform.

Part II.
Solve three of the following six problems.
Problem 6. Define f : R2 → R by
 xy

if (x, y) 6= (0, 0),
f (x, y) = x2 + y 2
0 if (x, y) = (0, 0).
Show that ∂f /∂x and ∂f /∂y exist at (0, 0) but f is not differentiable at (0, 0). You
may quote general facts about differentiability.
Solution: To show that ∂f /∂x exists at (0, 0) we set x = 0 and attempt to
differentiate with respect to y at 0. Since f (0, y) is identically 0, we see that
∂f /∂x(0, 0) exists and equals 0. Similarly ∂f /∂y(0, 0) exists and equals 0. However,
if a function is differentiable at a point then it is continuous at that point, but our
f is not continuous at (0, 0): Indeed f is identically 1/2 on the line x = y except
at the point (0, 0), where the value is 0. Thus f is not differentiable at (0, 0).
Problem 7. Let I be any interval in R. Show that if f : I → R is uniformly
continuous and {xn } is a Cauchy sequence in I then {f (xn )} is also Cauchy. Is the
assertion still true if we assume merely that f is continuous? Justify your answer.
Solution: Let ε > 0 be given. Since f is uniformly continuous there exists δ > 0
such that if x, x0 ∈ I and |x − x0 | < δ then |f (x) − f (x0 )| < ε. And since {xn } is
Cauchy there exists N such that if m, n > N then |xm − xn | < δ. So if m, n > N
then |f (xm ) − f (xn )| < ε, and we conclude that {f (xn )} is also Cauchy.
Unless I is closed and bounded and therefore compact, the assertion is false
without the assumption that f is uniformly continuous. For example, take I =
(0, ∞), f (x) = log x, and xn = 1/n. Then {xn } is a Cauchy is a sequence in I but
{f (xn )} is not even bounded, because f (xn ) = − log n.
Problem 8. Show that
1 X 1 1 1
= (− + n+1 − )xn
(x − 1)(x − 2)(x − 3) 2 2 2 · 3n+1
n>0

for |x| < 1.

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Solution: By the method of partial fractions,

1 1/2 1 1/2
= − + .
(x − 1)(x − 2)(x − 3) x−1 x−2 x−3
So
1 −1/2 1/2 1/6
= + − .
(x − 1)(x − 2)(x − 3) 1 − x 1 − x/2 1 − x/3
Using the geometric series (1 − r)−1 = n>0 rn , which converges for |r| < 1, we
P
obtain
1 X X X
= (−1/2) xn + (1/2) xn /2n − (1/6) xn /3n .
(x − 1)(x − 2)(x − 3)
n>0 n>0 n>0
n
So the coefficient of x is as asserted.
Problem 9. Let f : R2 → R2 and h : R2 → R2 be the functions
f (x, y) = (e2x−y − ex , e−3x+y − e2y )
and
h(x, y) = (x3 + x + y, y 2 + 2x + 3y).
There is an open neighborhood U of (0, 0) ∈ R2 and a differentiable function g :
U → R2 such that g(0, 0) = (0, 0) and f ◦ g = h. Compute [g 0 (0, 0)], the Jacobian
matrix of g at (0, 0).
Solution: The Jacobian matrix of f is
 2x−y
− ex −e2x−y

0 2e
[f (x, y)] = ,
−3e−3x+y e−3x+y − 2e2y
so
 
0 1 −1
f (0, 0) = ,
−3 −1
which is invertible. Thus by the Inverse Function Theorem, f has a C ∞ inverse on
an open neighborhood of (0, 0), whence g = f −1 ◦ h on an open neighborhood of
(0, 0). The Chain Rule gives
[g 0 (0, 0)] = [f 0 (0, 0)]−1 [h0 (0, 0)].
Now
3x2 + 1
 
0 1
h (x, y) = ,
2 2y + 3
so
 
1 1
h0 (0, 0) = .
2 3
Therefore
 −1       
1 −1 1 1 −1 −1 1 1 1 −1 1 2
[g 0 (0, 0)] = = = .
−3 −1 2 3 4 3 1 2 3 4 5 6

Problem 10. Let P (x, y) = −y/(x2 + y 2 ) and Q(x, y) = x/(x2 + y 2 ).

(a) Compute ∂Q/∂x − ∂P/∂y.
(b) Compute the line integral of P (x, y) dx + Q(x, y) dy around the unit circle
(oriented counterclockwise) x2 + y 2 = 1.
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(c) Explain why (a) and (b) do not contradict Green’s Theorem.
Solution: An easy calculation using the quotient rule shows that
∂Q ∂P
− =0
∂x ∂y
and another easy calculation using the parametrization r(t) = (cos t, sin t) shows
that the line integral in (b) is 2π. But there is no contradiction to Green’s Theorem
Z Z Z
∂Q ∂P
( − ) dx dy = P (x, y) dx + Q(x, y) dy
R ∂x ∂y C
(where R is the unit disk x2 + y 2 6 1) because P and Q are not C 1 functions on
R: Indeed neither is continuous at the origin.
Problem 11. Let C and C 0 be the circles in R3 parametrized by (cos t, sin t, 0)
and (cos t, sin t, 2) respectively (0 6 t 6 2π). Let F(x, y, z) be a C ∞ vector field in
R3 such that ∇ × F = 0. Show that
Z Z
F · dr = F · dr,
C C0
where the integrals on the left and right are the line integrals of F along the oriented
circles C and C 0 respectively.
Solution: Let S be the cylinder x2 + y 2 = 1 for 0 6 z 6 2, oriented so that a
normal vector point outward. Then
Z
(∇ × F) · n dσ = 0,
S
where n is the unit outward normal vector and dσ is the element of surface area on
S. By Stokes’ Theorem, we deduce that
Z Z Z
0= F · dr = F · dr − F · dr,
∂S C C0
and the stated equality follows.
Actually, an even more direct use of Stokes’ Theorem shows that the line integrals
over C and C 0 are equal because both are 0. Indeed, instead of taking S to be the
cylinder, take it to be disk having C as boundary. Since ∇ × F = 0, we deduce
that the integral of ∇ × F over D is 0, whence the line integral of F along C is 0
by Stokes’ Theorem, and similarly for C 0 .

Part III.
Solve one of the following three problems.
Problem 12. For x = (x1 , x2 , . . . , xn ) ∈ Rn , put
q
||x|| = x21 + x22 + · · · + x2n ,
and let S denote the unit sphere ||x|| = 1 in Rn . Let T : Rn → Rn be any linear
transformation. Give a reason why the two sides of the equation
max{x ∈ S : ||T (x)||} = inf{C > 0 : ||T (x)|| 6 C||x|| for all x ∈ Rn }
both exist, and then prove the equation.
Solution: The left-hand side exists because a continuous real-valued function
(such as x 7→ ||T (x)||) on a compact set (like S) attains a maximum value, and the
right-hand side exists because a nonempty set of real numbers which is bounded
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below has a greatest lower bound. Let us denote the left-hand side by M and the
right-hand side by I. Then ||T (x/||x||)|| 6 M for all nonzero x ∈ Rn , and since T
is linear it follows that ||T (x)|| 6 M ||x|| for all x ∈ Rn , including 0. So M is an
element C of the set on the right-hand side, whence
M > I.
On the other hand, any element C of the set on the right-hand side satisfies
||T (x)|| 6 C if x ∈ S, and therefore the maximum of the values ||T (x)||, namely
M , satisfies M 6 C also. In other words, M is a lower bound for the set on the
right-hand side. Therefore M is less than or equal to the greatest lower bound,
namely I, in other words
M 6 I.
We conclude that M = I.
Problem 13. Let X be a metric space with the following property: For every
infinite subset S of X,
inf{d(x, y) : x 6= y, x, y ∈ S} = 0.
Prove that X is totally bounded: In other words, show that for every ε > 0, the
space X can be covered by finitely many open balls of radius ε.
Solution: Suppose the statement is false for a particular ε > 0, and choose
x1 ∈ X. Then there exists x2 ∈ X such that d(x2 , x1 ) > ε. Also, there exists
x3 ∈ X such that d(x3 , x1 ) > ε and d(x3 , x2 ) > ε. Continuing in this way, we
obtain an infinite sequence {xn } with the property that d(xn , xm ) > ε for n 6= m.
Let S = {xn : n > 1}. Then
inf{d(x, y) : x 6= y, x, y ∈ S} > ε,
a contradiction.
Problem 14. Let S be the surface area of the sphere x2 + y 2 + z 2 = 1 and V
the volume of the ball x2 + y 2 + z 2 6 1. Let S 0 be the surface area of the portion
of the sphere x2 + y 2 + z 2 = 1 lying above the plane z = 1/2, and let V 0 be the
volume of the portion of the ball x2 + y 2 + z 2 6 1 lying above the plane z = 1/2.
Show that S 0 = S/4 and V 0 = 5V /32.
Solution: Parametrize the sphere by Γ(θ, ϕ) = (sin ϕ cos θ, sin ϕ sin θ, cos ϕ).
Then
(∂Γ/∂ϕ) × (∂Γ/∂θ) = (− sin2 ϕ cos θ, sin2 ϕ sin θ, cos ϕ sin ϕ)
and consequently the area element is
||(∂Γ/∂ϕ) × (∂Γ/∂θ)|| = sin ϕ dϕ dθ.
Consequently the surface area of S 0 is
Z 2π Z π/3
sin ϕ dϕ dθ = 2π(1 − 1/2) = π.
0 0

(Here recall that ϕ is the angle between the radial vector Γ(ϕ, θ) and the positive
z-axis.) Since the surface area of the whole sphere is 4π (as follows on replacing
the integral from 0 to π/3 by an integral from 0 to π), we do indeed have S 0 = S/4.
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For the volume it is easier to use cylindrical coordinates. Thus

Z 2π Z √3/2 Z √1−r2
0
V = r dz dr dθ.
0 0 1/2
√
The innermost integral is r 1 − r2 − r/2, and then the integral with respect to r is
√
3/2 1 3 1 5
−((1 − r2 )3/2 /3 + r2 /4)|0 = −( + ) + = .
24 16 3 48
After multiplying by 2π we get
5 5 4 5
V0 = π= · π= V
24 32 3 32
as claimed.