Math 500, Final
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HOMEWORK 7
SHUANGLIN SHAO
1. P143, Ex. 1
b). cn = n. Then
cn+1
= 1.
cn
1
Then the radius of convergence is = 1 = 1. So the series cn xn converges
when |x| < 1. For x = 1, 1, the series diverges because |cn xn | = |cn | does
not have a limit.
lim
n
h). cn
HOMEWORK 9
SHUANGLIN SHAO
1. P 175, Ex. 1
Proof. For any x and h,
(x + h)n = xn + nxn1 h +
n n2 2
n
nn
x
h + +
xhn1 +
h.
2
n1
n
Then
n n2
n
n n1
(x + h)n xn
= nxn1 +
x
h + +
xhn2 +
h
.
h
2
n1
n
Hence
(x + h)n xn
= nxn1 .
h0
h
This shows that f (x) = nxn1
HOMEWORK 10
SHUANGLIN SHAO
1. P221. Ex. 1
Proof. If c = 0, for any > 0, there exists =
|c | ,
then for |x y | ,
|f (x) f (y )| = |c| |x y | |c| = .
If c = 0, f 0.
We conclude that f is uniformly continuous in both cases.
2. P221. Ex. 2
Proof. From |f | <
MIDTERM REVIEW FOR MATH 500
SHUANGLIN SHAO
1. The limit
Dene limn an = A: For any > 0, there exists N N such that for
any n N ,
|an A| < .
The key in this denition is to realize that the choice of N depends on .
We explain it by using an example.
lim
n
2n
Mathematical Induction
Example: Consider the sum of
n=1
n=2
n=3
n=4
the rst n natural numbers:
sum = 1
sum = 1 + 2 = 3
sum = 1 + 2 + 3 = 6 = (3)(4)/2
sum = 10
= (4)(5)/2
Based on the above, one might guess that for n = k, 1 + 2 + 3 + + k = k (k + 1)/2, an
Equivalence relations
Definition: Let X be a set. A relation on X is a subset R of the product X X. If
(x, y) R, then we say that x is related to y.
Youre already familiar with one example of a relation:
Definition: R X X is a function if (x, y), (x, z) R
Decimal expansions
Denition: A decimal expansion is an expression of the form N.a1 a2 a3 an where
(a) N Z, (b) there are a countably innite number of an s, and (c) each an is a decimal
digit - that is an cfw_0, 1, 2, . . . , 9. (We could dene binary, or t
Differentiability and some of its consequences
Denition: A function f : (a, b) R is dierentiable at a point x0 (a, b) if
lim
xx0
f (x) f (x0 )
x x0
exists. If the limit exists for all x0 (a, b) then f is said to be dierentiable in the interval.
It is dier
Conditional convergence and rearrangements of infinite series
Denition: A series which converges but does not converge absolutely is said to be conditionally convergent.
n
n=1 (1) /n
Examples: The alternating harmonic series,
is conditionally convergent.
The Complex Numbers
The set of complex numbers, denoted C, is obtained from the real numbers R by adjoining
the single element i, which has the property that i2 = 1 (so i is not a real number).
The basic properties of complex numbers are deduced by assumi
Limits of functions
We now begin the study of functions or maps of the form f : D R, with D R.
Denition: A function from a set D X to another set Y is a rule that associates to each
element x D (called the domain of f ) a unique element y Y . If x is asso
Newtons proof of Keplers second law
The general version of Keplers second law states that the regions swept out by the radius
vector of a mass point moving in a central force eld in equal time intervals have equal areas.
Newton also shows that the motion
Cardinal numbers
Denition: A set X is nite if for some n N theres a bijection f : X cfw_1, 2, . . . , n.
We say that the cardinal number of X is n and write Card(X ) = n.
Denition: A set X is innite if its not nite. (For any natural number n, no bijection
Sets and Mathematical Notation
A set X is a well-dened collection of objects. In this course, the objects will be (mostly)
real numbers. To say that X is well-dened means that there must be an unambiguous test
for determining whether or not something belo
HOMEWORK 6
SHUANGLIN SHAO
1. P129. Ex. 1
b). The series is convergent by Proposition 9.3.8.
d ). The series is convergent by Proposition 9.3.8.
f ). The series is divergent because
lim
n2
= 1.
+1
n n2
h). The series is divergent because
n3
= .
n n2 + 1
li
HOMEWORK 5
SHUANGLIN SHAO
1. P119. Ex. 1
(b).
3
( )n .
4
This series converges as | 3 | < 1 and in fact
4
1
1
4
3
=
=.
( )n =
4
1 (3/4)
7/4
7
Note that the series begins with a0 = (3/4)0 = 1.
1
(f ). The series
5( 2 )n + 12( 1 )n converges as each of the
Math 500, Final
(last)
Name (Print): (rst)
Signature:
There are a total of 100+20 points on this 2 hours and 30 minutes exam. This contains 12 pages
(including this cover page) and 10+ 1 problems. Check to see if any page is missing. Enter all
requested i
HOMEWORK 1
SHUANGLIN SHAO
1. P 6. Ex. 1
(a) The series 1 +
sum is
1
4
+
1
16
+ is a geometric series with ratio 1 . Then the
4
1
1
1
4
4
=.
3
(b) The series
1
1
1
1
1
1
1
+
= 1 + ( ) + ( )2 + ( )3
4 16 64
4
4
4
is a geometric series with ratio 1 . Then
HOMEWORK 2
SHUANGLIN SHAO
1. P63, Ex. 1
Proof. We prove it by contradiction. Assume that there exists a rational
number r such that r2 = 3 and r > 0. Since r is a rational number, then
there exists r = p such that (p, q ) = 1, where the notation (a, b) =
HOMEWORK 3
SHUANGLIN SHAO
Abstract. Please send me an email if you nd mistakes. Thanks.
1. P91. Exercise 7.1
For convenience of writing, we will replace n by N in the reasoning below.
(b) cfw_ 2n1 0.
+3
Given any > 0. We need to nd N > 0 such that for an
HOMEWORK 4
SHUANGLIN SHAO
1. P108. Ex 1
Proof. 1. The sequence diverges. If the sequence converges, assume that
an A for some A R. Then taking limits on both sides of an+1 =
1 + a2 , we see that
n
A=
1 + A2 , A2 = 1 + A2 .
This is absurd. So the sequence
HOMEWORK 5
SHUANGLIN SHAO
1. P119. Ex. 1
(b).
3
( )n .
4
This series converges as | 3 | < 1 and in fact
4
1
1
4
3
=
=.
( )n =
4
1 (3/4)
7/4
7
Note that the series begins with a0 = (3/4)0 = 1.
1
(f ). The series
5( 2 )n + 12( 1 )n converges as each of the
HOMEWORK 6
SHUANGLIN SHAO
1. P129. Ex. 1
b). The series is convergent by Proposition 9.3.8.
d ). The series is convergent by Proposition 9.3.8.
f ). The series is divergent because
lim
n2
= 1.
+1
n n2
h). The series is divergent because
n3
= .
n n2 + 1
li
HOMEWORK 7
SHUANGLIN SHAO
1. P143, Ex. 1
b). cn = n. Then
cn+1
= 1.
cn
1
Then the radius of convergence is = 1 = 1. So the series cn xn converges
when |x| < 1. For x = 1, 1, the series diverges because |cn xn | = |cn | does
not have a limit.
lim
n
h). cn
HOMEWORK 9
SHUANGLIN SHAO
1. P 175, Ex. 1
Proof. For any x and h,
(x + h)n = xn + nxn1 h +
n n2 2
n
nn
x
h + +
xhn1 +
h.
2
n1
n
Then
n n2
n
n n1
(x + h)n xn
= nxn1 +
x
h + +
xhn2 +
h
.
h
2
n1
n
Hence
(x + h)n xn
= nxn1 .
h0
h
This shows that f (x) = nxn1
HOMEWORK 10
SHUANGLIN SHAO
1. P221. Ex. 1
Proof. If c = 0, for any > 0, there exists =
|c | ,
then for |x y | ,
|f (x) f (y )| = |c| |x y | |c| = .
If c = 0, f 0.
We conclude that f is uniformly continuous in both cases.
2. P221. Ex. 2
Proof. From |f | <