Let
C = fa + bi : a; b 2 R g :
We dene addition and multiplication on C for z = a + bi and w = c + di by
z + w = (a + c) + (b + d)i
z w = (ac bd) + (ad + bc)i
We also dene the conjugate of z = a + bi by
z=a
bi:
The complex numbers are a eld, but not an or
Math 575
Problem Set #10 Solutions
1. (p. 107, 1) Consider the function
8
>x
>
<
1
f (x) =
>
>
:
sin(x)
;
cos(x)
x 6= 0
0;
x=0
on the interval ( ; ). First, we claim that f is continuous. As 1
cos(x) 6= 0 and sin(x) 6= 0 for x 2 ( ; ) and x 6= 0, we may a
Math 575
Problem Set #9 Solutions
1. (p. 104, 1b) To nd
lim
x#0
log(1 + x) x
;
sin(x2 )
let f (x) = log(1+x) x, and let g (x) = sin(x2 ). By continuity, limx#0 f (x) =
limx#0 g (x) = 0. Note that g is nonvanishing in (0; 1). Since
1
1
1+x
g 0 (x) = 2x cos
Math 575
Problem Set #8 Solutions
1. (p. 91, problem 1) Let a = f (1) and note that f (n) = na for all integers n.
Moreover, f (1) = a = 2f (1=2) so f (1=2) = a=2, and similarly f (1=2n ) =
Pn
j
a=2n for all positive integers n. It now follows that f
=
j
Math 575
Problem Set #7 Solutions
1. (p. 83, prob.12) Let p be any point in S , and consider the sequence
given by pn = f (pn 1 ) where p0 = p. From the contraction property
we have d(pn ; pn+1 ) r d(pn 1 ; pn ). Letting an = d(pn 1 ; pn ) we then
have 0
Math 575
Problem Set #6 Solutions
1. (p. 78, problem 2) (=) Suppose that p is a limit point of B . Let
r0 = 1. There is a point p1 2 B belonging to Nr0 (p). Let r1 = d(p1 ; p)
and choose r2 = min(1=2; r1 =2). There is a point p2 2 B belonging
to Nr2 (p),
Math 575
Problem Set #5 Solutions
1. (p. 63, prob. 3) We apply the root test with an = an . Since jan j1=n =
an we have
8
0<a<1
<0
1
a=1
lim jan j1=n =
:
n!1
:
1
a>1
P
2
It follows that the radius of convergence of the series 1 an z n is 1
n=0
if 0 < a <
Math 575
Problem Set #4 Solutions
1. Suppose that fan g is aP
sequence with an ! 0, that fsn g is the sequence
of partial sums sn = n=1 aj , and that fs2n g converges to a limit s.
j
We claim that fsn g converges to s. Given " > 0, we can nd N so that
jan
Math 575
Problem Set #3 Solutions
1. We will use the Cauchy criterion to show that fbn g converges. First,
by hypothesis,
jbn+m
bn j
m
X
k=1
m
X
jbn+k
bn+k 1 j
(an+k
an+k 1 )
k1
= an+m
am
Since fan g is convergent and nondecreasing, the right-hand side is
Math 575
Problem Set #2 Solutions
Men pass away, but their deeds abide. Augustin-Louis Cauchy, (17891857)
1. (p. 32, prob. 1) Suppose that fxn g1 is a real sequence with xn ! x,
n=1
and that a
xn
b. For any " > 0 there is a positive integer N so
that jxn
Math 575
Problem Set #1 Solutions
Thanks to Xiaoli Kong, Ryan Curry, and Chad Linkous for ideas used in
the following solutions!
1. (p. 14, prob. 1) Suppose that ja bj < " for every real number " > 0.
If a 6= b, then ja bj > 0 which is impossible, hence a
Math 575
Problem Set #0 Solutions
How does one learn to do proofs? By observation and by practice,
practice, practice.
Richard Beals.
Acknowledgement : Thanks to Kristina Pepe, Qiao Liang, and Steven
Reeves (by way of their homework) for valuable hints!
Exam I Review
October 14, 2011
1. Know the basic properties of the real numbers R and the complex numbers
C.
2. Know what a metric space (S; d) is, and understand how R and C are
metric spaces.
3. Know the least upper bound property for the reals.
4. Know
Alternate Solution of Problem 19 Using the Integral Test
Chad Linkous
October 10, 2011
For what, if any, values of a > 0 do the following series converge?
(a)
1
X
n=2
1
:
n(log n)a
(b)
1
X
n=3
1
:
n log n(log log n)a
Solutions.
Note: We will use the follo
Sequences, Completeness, Sequential
Compactness: An IBL Approach
October 10, 2012
A sequence fpn g from a metric space (S; d) converges to a point p 2 S if for
every " > 0 there is a positive integer N so that d(pn ; p) < " for all n N .
A metric space (S
A linear fractional transformation is a mapping f : C ! C of the form
az + b
cz + d
f (z ) =
where a; b; c; d are complex numbers with ad bc 6= 0. The linear fractional
transformations form a group under composition:
(1) (closure) The composition of any t
Math 575
Problem Set #11 Solutions
Rb
1. (112, prob. 3) Suppose that f is nonnegative on [a; b] and a f = 0.
Suppose there is an x0 with f (x0 ) = c > 0. Taking " = c=2 there is
a > 0 so that jf (y ) f (x)j < " for jx y j < , so that f (y )
c=2
in (x
; x