Math 3220-1 Final, December 17, 2015
Solutions
Problem 1. Let x1 , x2 , . . . , xn be a family of mutually orthogonal vectors in Rd . Show that
kx1 + x2 + · · · + xn k2 = kx1 k2 + kx2 k2 + · · · + kxn k2 .
Solution: We prove the result by induction in n. If n = 1 the
statement is obviously true. Assume that n > 1. Then
(x1 + x2 + · · · + xn−1 ) · xn = x1 · xn + x2 · xn + · + xn−1 · xn = 0.
Therefore, x1 + x2 + · · · + xn−1 is orthogonal to xn . By Pythagorean
theorem we have
kx1 + x2 + · · · + xn k2 = kx1 + x2 + · · · + xn−1 k2 + kxn k2 .
Moreover, by the induction assumption, we have
kx1 + x2 + · · · + xn−1 k2 = kx1 k2 + kx2 k2 + · · · + kxn−1 k2 .
This implies the result.
Problem 2. Let K be a compact set. Let Z be a closed subset of K.
Show that Z is compact!
Solution: Let (Ui ; i ∈ I) be an open cover of Z. Then its union
with the open set X − Z is an open cover of K. Since K is compact, it
has a finite subcover of K. This subcover is also a cover of Z. If X − Z
is in this subcover, we can drop it since it is disjoint from Z. In this
way, we get a finite subcover of (Ui ; i ∈ I).
Problem 3. Let f be a continuous function on Rd . Let A be a bounded
subset of Rd . Show that f is uniformly continuous on A.
Solution: Let Ā be the closure of A. Then Ā is closed and bounded
subset of Rd . Therefore, it is compact by Heine-Borel theorem. Hence
f is uniformly continuous on Ā. This immediately implies that it is
also uniformly continuous on A.
Problem 4. Find all points of local maximum, local minimum and all
saddle points of the function
f (x, y) = 3x2 + 2xy + 2x + y 2 + y + 4.
1
2
Solution: The differential of f is
df (x, y) = 6x + 2y + 2 2x + 2y + 1 .
Hence the stationary points are the solutions of linear equations 6x +
2y = −2 and 2x + 2y = −1. By subtracting the second equation from
the first we get 4x = −1. Hence, x = − 41 . From the second equation
we see that 2y = −1 − 2x = −1 + 12 = − 21 . Hence, y = − 14 . It follows
that − 41 , − 14 is the only stationary point of f .
The second partial derivatives of f are
∂ 2f
∂ 2f
∂ 2f
= 2.
=
6,
=
2
and
∂x2
∂y 2
∂x∂y
Since
6 2 2 2 = 12 − 4 = 8
we see that the stationary point is a local minimum.
Problem 5. Let F, G : R2 −→ R2 be the functions given by F (x, y) =
(x − y, x + y) and G(x, y) = (x2 + y 2 , 2xy). Using the chain rule find
the differential dH of the composition H = F ◦ G. Check your result
by direct calculation!
Solution: The differential of F is
1 −1
dF (x, y) =
.
1 1
The differential of G is
2x 2y
dG(x, y) =
.
2y 2x
Hence, by the chain rule, we have
dH(x, y) = dF (G(x, y)) dG(x, y)
1 −1 2x 2y
2x − 2y −2x + 2y
=
=
.
1 1
2y 2x
2x + 2y 2x + 2y
The map H is given by
H(x, y) = F (G(x, y)) = F (x2 + y 2 , 2xy)
= (x2 + y 2 − 2xy, x2 + y 2 + 2xy) = ((x − y)2 , (x + y)2 ).
Hence,we have
2(x − y) −2(x − y)
dH(x, y) =
2(x + y) 2(x + y)
3
what agrees with the original calculation.
Problem 6. Let F (x, y) = (x cos y, x sin y) be a function from R2 into
R2 . Find all points (x, y) ∈ R2 where the function f is locally 1-to-1!
Solution: The differential of the map F is
cos x −x sin y
dF (x, y) =
.
sin y x cos y
Its determinant is
cos x −x sin y = x (cos y)2 + (sin y)2 = x.
det dF (x, y) = sin y x cos y
Hence, the differential of F is invertible at any point (x, y) with x 6= 0.
So, by the inverse function theorem, F is 1-to-1 around any point (x, y)
with x 6= 0. All points (0, y) map into (0, 0), so the map is not locally
1-to-1 there.
Problem 7. Using Fubini’s theorem calculate
Z 1 Z 1
yx
ye dy dx.
0
0
Solution: By Fubini’s theorem we have
Z 1 Z 1
Z 1 Z
yx
ye dy dx =
0
0
0
1
yx
ye dx dy.
0
We calculate first the inner integral. Put u = yx. Then we have
Z
Z
yx
ye dx = eu du = eu + C = eyx + C.
Hence, it follows that
Z
1
yeyx dx = [eyx ]10 = ey − 1.
0
So, we finally have
Z 1 Z 1
Z 1
yx
ye dx dy =
(ey − 1) dy = [ey − y]10 = (e − 1) − 1 = e − 2.
0
0
0
4
Problem 8. Let T be the triangle in R2 bounded by lines x = 1, y = 0
and y = x. Find the integral
Z
(x2 + y 2 ) dV (x, y).
T
Solution: We have
Z
Z
2
2
(x + y ) dV (x, y) =
T
0
1
Z
x
2
2
(x + y ) dy dx.
0
The inner integral is
x
Z x
y3
x3
4x3
2
2
2
(x + y ) dy = x y +
= x3 +
=
.
3 0
3
3
0
Hence, we have
4 1
Z
Z 1 3
4x
x
1
2
2
(x + y ) dV (x, y) =
dx =
= .
3
3 0 3
T
0