U MD Dept. of Mathematics
MATH437 F inal
by Michael Thomas
(4-33)
a. Let z0 be an arbitrary element of A. For f (z ) = z ,
f (z ) f (z0)
z z0
z z0
z z0
= lim
z z0 z z0
=1
f (z0) = lim
1 is continuous,
U MD Dept. of Mathematics
MATH437 HW#11
by Michael Thomas
(4-22)
Since f and g are nonzero at nitely many points, their sum is also nonzero at nitely many
points; more specically, it is nonzero at no
U MD Dept. of Mathematics
MATH437 HW#10
by Michael Thomas
(4.13)
a. For the rst part of the problem:
( g f )(v p) : =
=
=
=
=
(D( g f )( p)(v )( g f )(p)
(Dg ( f ( p) Df ( p)(v ) g( f ( p)
g( f ( p)(
U MD Dept. of Mathematics
MATH437 HW#9
by Michael Thomas
(4-1)
a.
i1 i2
i1 i2
ik ei1,
k!
Alt( i1 i2 ik)
1! 1! 1!
= k ! Alt( i1 i2 ik) ei1, , eik
ik =
, eik
= k!
1
k!
k
sgn
i j e j
j =1
Sk
k
=
U MD Dept. of Mathematics
MATH437 HW#8
by Michael Thomas
(3-35)
a. Let U = [a1, b1] [a2, b2]
case specied,
1
0
g = 0
0
[a j , b j ]
0
1
0
a
0
[an , bn] be an n-dimensional rectangle. In the rst
0
U MD Dept. of Mathematics
MATH437 HW#7
by Michael Thomas
(3-29)
Let J denote the Jordan-measurable set in the yz -plane. We partition J using closed subrectangles, which is permissible along the bound
U MD Dept. of Mathematics
MATH437 HW#6
by Michael Thomas
(3-8)
According to Theorem 3-5, if [a, b] R and a < b, then [a, b] does not have content 0. This is
to say that there exists an > 0 such that t
U MD Dept. of Mathematics
MATH437 HW#5
by Michael Thomas
(3-1)
1
1
Dene P to be the partition formed by 0, 2 , 1 0, 2 , 1 , Then m
and m
1
,1
2
[0, 1]
=M
1
,1
2
[0, 1]
= 1. Therefore, L( f , P ) = U
U MD Dept. of Mathematics
MATH437 HW#4
by Michael Thomas
(2-25)
The only point in f (x) where there could be a discontinuity is at x = 0 , but the hint to this
exercise implies that f (0) exists. We m
U MD Dept. of Mathematics
MATH437 HW#3
by Michael Thomas
(1-25)
A linear transformation T : Rn
Rm may be described in terms of component functions. In
other words, we may write T (x) = (t1(x), t2(x),