MATH 3334
Homework 3, selected solutions
1(a). Let A = cfw_Bk (0) : k N = cfw_B1 (0), B2 (0), B3 (0), . . . .
1(b). Let A = cfw_B
k
k+1
(0) : k N = cfw_B 1 (0), B 2 (0), B 3 (0), . . . .
2
3
4
2(a). Let A be an open cover of K S . Then A covers K and S .
MATH 3334
Homework 2, selected solutions
1. To show that
p
p
m i r ri
2
2
m i ri c
=
+m rc
2
i=1
i=1
or, equivalently,
p
m i ( r ri
2
ri c 2 ) = m r c 2 ,
i=1
we note that
x
2
y
2
= (x + y) (x y),
x, y Rn .
Let x = r ri and y = ri c. Then
x + y = r c,
x
MATH 3334
Homework 1, selected solutions
In Rn show that
(a)
2x
2
+2 y
2
(c) 4x y = x + y
Hint: x y
2
= x+y
2
2
xy
+ xy
2
2
(the parallelogram law)
(the polarization identity)
= (x y) (x y) = x
2
2x y + y
2
Use induction on k to prove that if x1 , . .
MATH 3334
Homework 2, selected solutions
1. To show that
p
p
m i r ri
2
2
m i ri c
=
+m rc
2
i=1
i=1
or, equivalently,
p
m i ( r ri
2
ri c 2 ) = m r c 2 ,
i=1
we note that
x
2
y
2
= (x + y) (x y),
x, y Rn .
Let x = r ri and y = ri c. Then
x + y = r c,
x
MATH 3334
Homework 3, selected solutions
1(a). Let A = cfw_Bk (0) : k N = cfw_B1 (0), B2 (0), B3 (0), . . . .
1(b). Let A = cfw_B
k
k+1
(0) : k N = cfw_B 1 (0), B 2 (0), B 3 (0), . . . .
2
3
4
2(a). Let A be an open cover of K S . Then A covers K and S .
MATH 3334
Homework 5, selected solutions
1(a). Let ak = (1/k, 0), bk = (0, 1/k ), k N. Then ak (0, 0), bk (0, 0)
but f (ak ) = (1, 0), f (bk ) = (0, 1). By the main theorem about limits
(Corollary 2) the limit at (0,0) does not exist.
1(b). Recall that ma
MATH 3334
Homework 6, selected solutions
xy if xy 0
. Since |x| and |y | are continuous on R2 ,
xy if xy 0
1(a). f (x, y ) =
so is their product f . Fix any y > 0. Then
f (h,y )0
h
h+0
lim
f (h,y )0
h
h0
= y , lim
f
(x, y )
x
=
= y . Hence, the limit
y if
MATH 3334
Homework 5, selected solutions
1(a). Let ak = (1/k, 0), bk = (0, 1/k ), k N. Then ak (0, 0), bk (0, 0)
but f (ak ) = (1, 0), f (bk ) = (0, 1). By the main theorem about limits
(Corollary 2) the limit at (0,0) does not exist.
1(b). Recall that ma
MATH 3334
Homework 6, selected solutions
xy if xy 0
. Since |x| and |y | are continuous on R2 ,
xy if xy 0
1(a). f (x, y ) =
so is their product f . Fix any y > 0. Then
f (h,y )0
h
h+0
lim
f (h,y )0
h
h0
= y , lim
f
(x, y )
x
=
= y . Hence, the limit
y if
MATH 3334
Homework 7, selected solutions
1. uxx = sin x cos t = utt .
2. By MVT.
3. f (4, 4, 6) = 6. The partial derivatives are:
fx (x, y, z ) =
fy (x, y, z ) =
fz (x, y, z ) =
x
x2 + y 2 + z 2
y
x2 + y 2 + z 2
z
x2 + y 2 + z 2
,
fx (4, 4, 2) =
2
4
=
6
3
MATH 3334 Dmitri Kuzmin
Test 3 April 12, 2010
This test consists of 4 problems that may be solved in any order.
Read the formulations of all problems before you try to solve any.
Compare the numbers of points, estimate the time for solving each
problem
MATH 3334 Dmitri Kuzmin
Test 2 Selected solutions
1. A sequence is Cauchy i it is convergent. Every convergent sequence
is bounded.
2. f is dierentiable on Rn and f (x0 ) = A, x0 Rn since
lim
xx0
Ax Ax0 A(x x0 )
f (x) f (x0 ) A(x x0 )
= lim
=0
xx0
x x0
x
MATH 3334 Dmitri Kuzmin
Test 2 March 12, 2010
This test consists of 3 problems that may be solved in any order.
Read the formulations of all problems before you try to solve any.
If the formulation of a problem is unclear, raise your hand and ask.
Com
MATH 3334 Dmitri Kuzmin
Test 1 Selected solutions
1. See the text (p. 33) or your classnotes.
2. By the Cauchy-Schwarz inequality
x+y
x y (x + y) (x y) =
x
2
y
2
.
3. We will argue as in the proof of the Theorem about open and closed
balls (Br (x0 ) is op
MATH 3334 Dmitri Kuzmin
Test 1 February 12, 2010
This test consists of 4 problems that may be solved in any order.
Read the formulations of all problems before you try to solve any.
If the formulation of a problem is unclear, raise your hand and ask.
Cauchy-Schwarz: |x y | |x| |y |, x, y Rn
TI: |x + y | |x| + |y |, |x| |y | |x y |
Limits of sequences: (xk ) converges to x Rn if
> 0 N s.t. k N
xk x
Cauchy sequences: (xk ) converges i it is Cauchy
> 0 N s.t. m k N |xk xm| <
Limits of Functions: Let
MATH 3334
Homework 10, selected solutions
1. Note that f =
f |f |
,
2
where |f | is integrable by HW-9.
2. By Theorem Integration over Unions of Jordan Regions
f (x)dx =
T
3. |S | =
=
f (x)dx +
f (x)dx
T \S
S
a
0
a
0
b
0
S dydx =
b(1 x )
a
0
1dydx =
f (x
MATH 3334
Homework 9, selected solutions
1. If g is integrable, then > 0 a partition P s.t. U (g, P )L(g, P ) <
U (f, P ) L(f, P ) =
(Mk (f ) mk (f )|Sk | =
k
K
(f (xk ) f (yk )|Sk |
k
K (g (xk ) g (yk )|Sk | K
k
(Mk (g ) mk (g )|Sk |
k
= K (U (g, P ) L(g
MATH 3334
Homework 8, selected solutions
1. First derivative test at interior points: fx = cos x = 0, fy = sin y = 0.
Critical points: ( , ) and ( 32 , ). Function values: f ( , ) = 1 1 = 0
2
2
and f ( 32 , ) = 1 1 = 2. The boundary consists of 4 segments
MATH 3334
Homework 7, selected solutions
1. uxx = sin x cos t = utt .
2. By MVT.
3. f (4, 4, 6) = 6. The partial derivatives are:
fx (x, y, z ) =
fy (x, y, z ) =
fz (x, y, z ) =
x
x2 + y 2 + z 2
y
x2 + y 2 + z 2
z
x2 + y 2 + z 2
,
fx (4, 4, 2) =
2
4
=
6
3
MATH 3334
Homework 8, selected solutions
1. First derivative test at interior points: fx = cos x = 0, fy = sin y = 0.
Critical points: ( , ) and ( 32 , ). Function values: f ( , ) = 1 1 = 0
2
2
and f ( 32 , ) = 1 1 = 2. The boundary consists of 4 segments
Cauchy-Schwarz: |x y | |x| |y |, x, y Rn
TI: |x + y | |x| + |y |, |x| |y | |x y |
Limits of sequences: (xk ) converges to x Rn if
> 0 N s.t. k N
xk x
Cauchy sequences: (xk ) converges i it is Cauchy
> 0 N s.t. m k N |xk xm| <
Limits of Functions: Let
MATH 3334
Homework 1, selected solutions
In Rn show that
(a)
2x
2
+2 y
2
(c) 4x y = x + y
Hint: x y
2
= x+y
2
2
xy
+ xy
2
2
(the parallelogram law)
(the polarization identity)
= (x y) (x y) = x
2
2x y + y
2
Use induction on k to prove that if x1 , . .
MATH 3334
Homework 9, selected solutions
1. If g is integrable, then > 0 a partition P s.t. U (g, P )L(g, P ) <
U (f, P ) L(f, P ) =
(Mk (f ) mk (f )|Sk | =
k
K
(f (xk ) f (yk )|Sk |
k
K (g (xk ) g (yk )|Sk | K
k
(Mk (g ) mk (g )|Sk |
k
= K (U (g, P ) L(g
MATH 3334
Homework 8, selected solutions
1. First derivative test at interior points: fx = cos x = 0, fy = sin y = 0.
Critical points: ( , ) and ( 32 , ). Function values: f ( , ) = 1 1 = 0
2
2
and f ( 32 , ) = 1 1 = 2. The boundary consists of 4 segments
MATH 3334
Homework 7, selected solutions
1. uxx = sin x cos t = utt .
2. By MVT.
3. f (4, 4, 6) = 6. The partial derivatives are:
fx (x, y, z ) =
fy (x, y, z ) =
fz (x, y, z ) =
x
x2 + y 2 + z 2
y
x2 + y 2 + z 2
z
x2 + y 2 + z 2
,
fx (4, 4, 2) =
2
4
=
6
3
MATH 3334
Homework 6, selected solutions
xy if xy 0
. Since |x| and |y | are continuous on R2 ,
xy if xy 0
1(a). f (x, y ) =
so is their product f . Fix any y > 0. Then
f (h,y )0
h
h+0
lim
f (h,y )0
h
h0
= y , lim
f
(x, y )
x
=
= y . Hence, the limit
y if