The focus of the course will be to understand foundations for
what is the substance of the first year of calculus.
Unlike Math 343 which is an introduction to contemporary abstract
algebra, where the concepts and results are new and eventually
abstract in their nature, Math 316 covers the analysis of real numbers and real valued
functions of one real variable.
The subject matter of real analysis is quite familiar.
Real numbers
have been a topic of study since early courses in algebra and
functions of these numbers were the focus of preparations for
calculus as well as the main subject of calculus. Thus the basic
results of this course are fairly familiar, BUT the proofs for many
of the key theorems of calculus were not proven in the calculus
courses. These results appeared reasonable and so why bother to
spend precious time on proving "the obvious"? Part of this course
will be to develop an appreciation for the need for rigor along with
the rigor itself.
One common approach to the study of real analysis starts with the
foundations: developing first what real numbers are in a rigorous
fashion from the natural numbers or the integers, then the rational
numbers, and finally the real numbers. This involves some vary
careful work in the algebra of equalities and inequalities for
arithmetic.
We will follow an alternative approach.
We will look at some of the
key theorems of calculus and analyze these results to see what is
needed to prove them. Eventually we will delve more deeply into the
nature of the subject arriving finally at a rigorous definition of
what a real number is.
Key theorems we will examine will involve continuity,
differentiability, and integrability. History:The first coherent and
somewhat successful presentation of calculus was by Cauchy in the
early 19th century- about 150 years after the works of Newton and
Leibniz. This work formed a foundation for the new empirical
approach to science as a whole ( in opposition to religion). a major
criticism of the calculus was found in the work of Bishop Berkeley
(The Analyst) . The efforts of the next hundred plus years arrives at a careful definition of the real numbers
comes toward the end of the 19th century with separate works by
Weierstrass and Dedekind. At times I will suggest reading from
original sources to illuminate some of the difficulties and key issues
in the development of a rigorous analysis for the calculus.
We will use aworking definition for what a real number is: a real
number is any number that can be represented as a possibly infinite
decimal (positive or negative) or that can be thought of a the
measure of a length of a line segment either to one side or the
other of a specified point on on a given line an a specific unit for
measurement.
1-21
Organization of the course:
details like tests, homework, etc. as described in the main
course page.
Reference to Polya's :
How to Solve It... available on line:
https://notendur.hi.is/hei2/teaching/Polya_HowToSolveIt.pdf
Polya describes 4 Phases of Problem Solving 1. Understand the problem. 2. See connections to devise a plan. 3. Carry out the plan. 4. Look back. Reflect on the process and results.
It is the first phase that is usually not recognized as being
essential. There is usually more to understand than is apparent.
Next class: Discuss 5 theorems at the heart of calculus.
Mention other texts in Real Analysis:
Analysis WebNotes by John Lindsay Orr
Interactive Real Analysis by Bert G. Wachsmuth
A First Analysis Course by John O'Connor
Basic Analysis: Introduction to Real Analysis by Jiri
Lebl
Introduction to Real Analysis by William Trench
Principles of Mathematical Analysis by Walter
Rudin (1953)
Calculus by Michael Spivak [A calculus book that
proved all the theorems! PDF (with some issues) now on
Moodle.]
On Proofs:
How to Read and Do Proofs by D.Solow
The Keys to Advanced Mathematics by D. Solow
How to Solve It by G. Polya
Watch video "Space Filling Curves"Space
Filling Curves - YouTube
which were introduced in the late 19th Century and were the early
instances of what later came to be called fractals in the late
Twentieth Century. Fractals are now a subject of continuing
research. 1-22
Also you can see this in the library on shelf- please replace
accurately! VIDEO2577
Use of point-wise convergence for a sequence of approximating
curves.
Use of concept of continuity for curves.
Length of a curve.
How does a curve "fill" space?
Week 2: 1-25
Continuation on space filling curves:
To make sense of curves being close and sequence of curves having limits,
we consider the concept of a metric space. - a set where
closeness can be measured using non-negative real numbers for the
measurements. A metric on a set X is a function m: X×X→R+∪{0} where
For all a,b∈X,m(a,b)=m(b,a).
If m(a,b)=0 then a=b.
For all a,b,c∈X,m(a,b)+m(b,c)≥m(ac).
Elementary examples are the real numbers with m(a,b)=|b-a|, and
R2 and R3 with the usual euclidean distance between points
measurement.
Metrics for continuous functions on a closed interval [a,b] are
related to the issues of when functions are "close."
One such metric for continuous functions is m(f,g)=max{|f(x)-g(x)|,x∈[a,b]}.
Another is m(f,g)=∫ba(f(x)-g(x))2dx.
In Linear algebra, a real inner product on vectors,<f,g> gives rise to a norm, ||f||=√<f,f>, and then a metric, m(f,g)=||f-g||.
Mean
Value Theorem [proof based on Rolle's
theorem ]: If f is (i) continuous on [a, b] and (ii) differentiable on (a, b), then
there exists a number c in (a,
b)such thatf′(c)=f(b)-f(a)b-a .
Some theorems that rely on the Mean Value Theorem for
their proof.
If f′(x)=0 for an interval then there is a real
numberK where
f(x)=K for allx in the interval, I.
Two proofs were given: Direct: Pick a∈I. It's enough to show for any b∈I,a<b,thenf(b)=f(a).K will be f(a).
Using differentiability implies continuity, f satisfies the
hypotheses of the MVT, so there is a number c where c∈(a,b) and f′(c)=f(b)-f(a)b-a.
But since c∈I , f′(c)=0, and thus 0=f(b)-f(a)b-a and thus f(b)=f(a). Indirect: Suppose there are a,b∈I with a<b,f(a)<f(b). Using differentiability implies continuity, f
satisfies the hypotheses of the MVT, so there is a number
c where c∈(a,b) and f′(c)=f(b)-f(a)b-a.
But since c∈I , f′(c)=0,thus 0=f(b)-f(a)b-a>0. A contradiction. A similar argument
leads to a contradiction by supposing there are a,b∈I with `a<b,
f(a)> f(b).
1-29
More on the MVT;
Proofs about increasing/decreasing functions: If f′(0)>0
on an interval, then f is an increasing function.
Counter examples to the MVT:
(1) without Continuity: f (x) = x for
x in (0, 1] and f(0) = 1 on the interval
[0,1]
and
(2) without differentiability: f (x) = 1/2
-| x - 1/2| for x in [0,
1].
Also f(x)=(12)13-|(x-12)|13for
x in [0, 1].
Theorem:
Rolle's Theorem [proof based on Extreme Value Theorem and critical
point analysis]: If f is (i) continuous on [a,
b] and (ii) differentiable on
(a, b), (iii) f
(a) = f (b)
then there exists a number c
in (a, b)
such thatf′(c)=0=f(b)-f(a)b-a.
Motivational Question I: What is a number?
Foundational Theorems for Continuity [also about existence].
See the following if you want to see a "simple" approach to
defining real numbers: [This is a main theme for the course!] Vector Calculus, Linear Algebra, and
Differential Forms: A Unified Approach by John
Hubbard and Barbara Burke Hubbard Appendix.A.1
Arithmetic of Real numbers.pdf
Week 3
2-1
Example: algebraic numbers = {z: z is a complex number which is
the root of a polynomial with integer coefficients}
A few more properties of Fields:
Suppose F is a field.
Proposition: If a and b are
in F and ab = 0 then either a =0 or b=0.
Proof: Case 1. If a=0 then we are done.
Case 2. If a is not 0, then a has an inverse... c
where ca=1. Lemma: For any c in F, c*0 = 0.
Then b=1*b= (c*a)*b= c*(a*b)
= c*0 = 0. Thus either a =0 or b=0.
IRMC [I rest my case.]
Proof of Lemma: c*(0+0)=c*0, so c*0+c*0 = c*0.
Let g be the real number where g+c*0=0.
Then g+ (c*0 + c*0) = g+ c*0 = 0,
but g+ (c*0 + c*0=(g+ c*0) + c*0)= 0 + c*0 = c*0. Thus c*0 = 0.
If a, b and c are in F, a+b = a+c impliesb=c.
Proof: Let k be the element of the field where k
+ a = 0 Then b = 0 + b = (k +a)+b =k
+(a+b)= k +(a+c) =(k +a)+c
= 0 + c = c.
If a, b and c are in F with a
not equal to 0, ab = ac impliesb=c.
Proof: Since a is not equal to 0, let k be the
element of the field where k * a = 1 Then
b = 1* b = (k *a)*b =k(a*b)= k *(a*c) =(k *a)*c = 1* c = c.
Other field results: (-1)*a = -a; -(-a)=a;
(-a)*b= -(a*b); (-a)*(-b)= a*b
Interesting: Let n·1 stand
for 1 + 1 + 1 + ... + 1 with n summands.
Either (i) for
all n, n·1 is not 0 in which case we say the field has
"characteristic 0" , or
(ii) for some n, n·1=0.
In the casen·1=0, there is a smallest n for which n·1=0, in which case
we say the field has "characteristic n".
For example: R, Q,
and C all have characteristic 0, while Z2, Zp,
where p is a prime number, and any finite field such as
F4, all have a non zero characteristic.
If the characteristic of F is not zero, then it is a
prime number. Proof: If n is not a prime, n = rs with 1<r,s<n. Let a
= 1+1+...+1 r times and b= 1+1+...+1
s times.
Then ab=n·1=0, so either a = 0
or b = 0. This contradicts the fact that n was supposed to
be the SMALLEST natural number for which n·1=0. IRMC.
Order for the Real Numbers: At the heart of "close".
There is a relation > on R.
(That is, given any pair a, b then a
> b is either true or false).
This can be based on establishing a subset of
the Real Numbers R+ - the "positive real
numbers" with properties:
i) for any number a, either a = 0, a∈R+ , or -a∈R+, but not any two of these are true.
ii) if a,b∈R+ then a+b∈R+ and a⋅b∈R+.
This allows a definition of "<" by saying
a<b is true if (and only if) b-a∈R+.
It follow that 0<a if and only if a∈R+.
The following properties hold for the
relation "<": [Proofs are based on algebra and the
properties of R+.]
a) Trichotomy: For any a, b∈R
exactly one of a > b, a = b,
b > a is true.
b) If a, b > 0 then a
+ b > 0 and a.b > 0
c) If a > b then a
+ c > b + c for any c
Something satisfying the field and order
axioms is called an ordered field.
The field and order axioms for the real numbers can be used to
deduce many simple algebraic or order properties of R.
Examples
The ordering > on R is transitive.
That is, if a > b and b > c
then a > c.
Proof
a > b if and only if a - b
> 0 by definition.
b > c if and only if b - c
> 0
Hence (a - b) + (b- c) > 0
and so a - c > 0 and we have a >
c.
If a > 0 and b > c
then ab > a c.
Proof b > c if and only if b - c
> 0
Hence a (b- c) > 0
and so ab - ac > 0 and we have ab >
ac.
For and real numbers a and b with a<b, a<a+b2<b
Proof: a+a<a+b<b+b so a=12(a+a)<a+b2<12(b+b)=b.
2-2
Remarks on Assignment 2.
Absolute value inequalities: Proofs were best
that did not proceed by cases.
Examples: For any x-|x|≤x≤|x|
so -|a|≤a≤|a| and -|b|≤b≤|b| so -|a|-|b|≤a+b≤|a|+|b|. Thus |a+b|≤|a|+|b|.
Or (|a+b|)2=(a+b)2=a2+2ab+b2≤|a|2+2|a||b|+|b|2=(|a|+|b|)2.
since 0<|a+b| and 0<|a|+|b|,
|a+b|<|a|+|b|. Using inequalities with decimals to prove the IVT:
Outline:
Assume f(a)<P<f(b).
Consider a<a+b2<b and apply the trichotomy property to
f(a+b2) and P. This may give thatf(a+b2)=P
and we have the conclusion needed to prove the theorem.
Otherwise consider a smaller interval depending on
whether f(a+b2)>P or f(a+b2)<P. Continuing with this process on smaller (and smaller) intervals will
lead to a sequence of intervals of lengths b-a2n
and eventually, the estimation of a
decimal number c will arise.
Because of the continuity of f, it must be that f(c)=P. Note: We interpret continuity here informally as "when
x≈c then f(x)≈f(c) " and use that and the
trichotomy law of inequality to understand that f(c) must equal
P. we will clarify this as we evolve a more precise definition of
continuity. An example created with GeoGebra
illustrating this "bisection" argument with a graph and mapping diagram for the
function f(x)=x2-2 on the interval [0,2].
Bisection and Proof Outline of IVT vizualized with GEOGEBRA.
What distinguishes R from Q (and from other subfields of R) is the Completeness Axiom.
Definitions:
An upper bound of a non-empty subset A of R is an element b∈R with b≥a for all a∈A.
An element M∈R is a least upper bound or supremum of A if M is an upper bound of A and if b is an upper bound of A then b≥M.
A lower bound of a non-empty subset A of R is an element b∈R with b≤a for all a∈A.
An element m∈R is a greatest lower bound or infimum of A if m is an lower bound of A and if b is a lower bound of A then b≤m. The Completeness Axiom
If a non-empty set A has an upper bound, it has a least upper bound.
Using decimals to find decimal places of
√3:
Start with 11=1 and 22=4, so the units decimal for
√3 is 1.
Now to find the tenths decimal digit consider 12,1.12,1.22,1.32,...,1.72=2.89,1.82=3.24,1.92,22.
So the tenth digit is 7.
To find the hundredth digit consider 1.702,1.712,1.722,1.732=2.9929,1.742=3.0276,...,1.782,1.792,1.82.
So the hundredth digit is 3.
Continue in this fashion, for each decimal place, examine the
eleven numbers that might have the correct digit, and choose the
digit where the squared value changes from being smaller than 3
to being larger than 3.
Now prove the infinite decimal that results must have its square
equal to 3. ...
Fact: If L and L' are least upper bounds for a set S, then L =
L'.
Proof: Since L' is an upper bound for S and L is a least upper
bound for S, L ≤ L'. Likewise, L' ≤ L. So L = L'.
(trichotomy property) EOP.
Credibility without proof for the Completeness:
We can use decimals, representing real numbers, to help give
credibility to the belief that the Completeness Property
is true for the real numbers.
Discussion: Since S is nonempty, there is an x0∈S.
Suppose
B is an upper bound for S, so B≥x0. If B=x0 then
it is not hard to show that B is the least upper bound
of S.
So,we suppose B>x0. Now there is an integer m0 with
m0>B and an integer n0 with n0<x0. [Why?]
In the interval [n0,m0] there is an integer m1 where
m1 is an upper bound for S but m1-1 is not an upper
bound for S. If m1∈S then m1 is the least upper
bound of S. Otherwise let n1=m1-1 and now consider the
interval [n1,m1].
In the interval [n1,m1] there is a decimal n1+k1⋅0.1
where 0<k1≤10 is an integer and n1+k1⋅0.1 is an upper bound for S but n1+(k1-1)⋅0.1
is not an upper bound for S.
Let m2=n1+k1⋅0.1. If m2∈S then m2 is the least upper bound of S.
Otherwise, let n2=n1+(k1-1)⋅0.1 and
now consider the interval [n2,m2].
In the interval [n2,m2] there is a decimal n2+k2⋅0.01
where 0<k2≤10 is an integer and n2+k2⋅0.01 is an upper bound for S but n2+(k2-1)⋅0.01 is not an upper bound for S.
Let m3=n2+k2⋅0.01. If m3∈S then m3 is the least upper bound of S.
Otherwise, let n3=n2+(k2-1)⋅0.01 and
now consider the interval [n3,m3].
We can continue in this fashion. After each step that continues we obtain a
"decreasing" sequence of terminating decimals mj which are
upper bounds for S and an "increasing" sequence of terminating
decimals nj each of which is just one digit smaller in the
last decimal place than mj and each of which is not an
upper bound for S .
These decimals allow us to determine a number to any decimal
precision which approximates the least upper bound L. L
will be the limit of the two sequences nj and mj.
The indirect proof that L is the least upper bound of S would involve
an argument assuming either that L was not an upper bound for
S or that there was an upper bound of S that was smaller
than L and consideration of how the two sequences were
determined. This argument is omitted here as this argument
is given only to suggest the truth of the completeness
property for real numbers.
Lemma: The set of natural numbers, N, is not a bounded set.
Proof of Lemma: (Indirect) If N is
bounded, there is a real number B that is an upper bound for N,
and thus there is a least upper bound L for the set N.
Then L-1 is not an upper bound for N, so there is a
natural number m∈N where L-1<m<L. But then
L<m+1<L+1 which contradicts L as an upper bound
for N since m+1∈N.
2-5
Another important order property of the real ( and rational) numbers
The Archimedean Property of the real (and rational) numbers:
Proposition: If r is a positive real ( or rational) number, then there is
natural number n′>0 where 0<1n′<r.
Proof: Consider 1r: 1r>0, and by the lemma, there is a natural number
n′>0 where n′>1r>0. Then 0<1n′<r. EOP.
Alternative Proof: [A decimal approach.] Express r as a decimal.
If r≥1 use n′=2. If r<1, express r as a
decimal. Suppose the first nonzero digit of r is in the
kth decimal place. Let n′=10k+1. Then
0<1n′<r.
Greatest Lower Bound Theorem:
If a non-empty set B has a lower bound, it has a greatest lower bound.
Proof: Suppose B is a non-empty set and L is a lower bound for B.
Let A= { x:x=-y for some y∈B}.
Then A is non-empty and -L is an upper bound for A, so by the
completeness axiom, there is a number M where M is the least upper
bound of A. Then -M is the greatest lower bound of B. [Justify
this last statement using the fact that if a≤b then -a≥-b.]
Assume that f is continuous on the interval [a,b]
and f(a)<0<f(b). The theorem concludes that there is a
number c∈(a,b) with f(c)=0.
For the purpose of this "proof" we
interpret "continuous at t " as meaning that when x≈t then f(x)≈f(t).
The Plan: Use the completeness property for a subset
of [a,b] to find a candidate for c and then show that f(c)=0 by showing that f(c) cannot
be positive or negative, leaving f(c)=0 as the only possibility.
The Set: Let X={x∈[a,b]: for all y∈[a,x],f(y)≤0}. X is not empty since f(a)<0, and thus there is some small interval [a,a+ε] where by continuity f is negative for all a≤x≤a+ε. X is bounded above by b, so by the completeness of the real numbers, it has a least upper bound, which we denote c.
From the remark that showed that X is not empty, it also follows that a<c. Because f(b)>0 there is some small
interval [b-ε,b] where by continuity f is positive for all
b-ε≤x≤b. Thusc<b
We now proceed to examine f(c) showing that f(c)<0 and f(c)>0 are both impossible, leading to the conclusion
that f(c)=0.
(i) Suppose f(c)<0. Then by continuity, there is an interval, I, containing c where f(y)<0 for all y∈I.
In I there is some s with s<c, so s is not an upper bound for
X. Thus there is some x* in X with s<x*<c.
Notice that
there is some r with r∈I and c<r.
Considering [a,x*]
and [x*,r] we have that for all y∈[a,r],f(y)≤0. Thus r∈X. This contradicts that c is an upper bound
for X, so f(c)<0 is false.
(ii) Suppose f(c)>0. Then by continuity, there is an interval, J, containing c where f(y)>0 for all y∈J.
But then there is some s with s<c, with
f(s)>0. So if x∈X then x<s and thus s is an
upper bound for X smaller than c. This contradicts that c is the
least upper bound of X, so f(c)>0 is false
Since (i) and (ii) have eliminated f(c)<0 and
f(c)>0 as possibilities, it must be that f(c)
=0`. EOP
2-8
Motivational Question I: What is a number?
Motivational Question II: What does
continuity mean?
Motivational Question III: What
does limit mean? for numbers? for function
values? for functions? for sets?
GeoGebra Applet for visualizing sequences with Mapping Diagrams
and Graphs for a sequence defined by an=f(n). Examples.
GeoGebra Applet for visualizing sequences with Mapping Diagrams and Graphs for a sequence defined by an=f(n).
Provisional Definition: limit
of an is a means "after a while
all an are close to a." 2-9
Def'n: We say limn→∞an=a if
given any ε>0 there is a natural number M so
that
if n>M then |an-a|<ε.
Comment: What kind of number is ε?
Visualizing
the definition of
limn→∞an=a
Example: limn→∞1n=0.
Proof: Suppose ε>0. Then by the Archimedean
Property there is a natural number M where 1M<ε. To show this M works, suppose n>M.
Then we have 1n<1M [Prove this---?] . Thus |1n-0|=1n<1M<ε. EOP.
2-11
Example: Show that limn→∞nn+1is not1.1. Solution: Supose limn→∞nn+1=1.1 . Then let
ε=0.05 and consider any natural number n. Then nn+1<1<1.1 and |nn+1-1.1|>|1-1.1|=0.1>0.05. So for any
natural number M, if n>M then |nn+1-1.1|>0.05 which contradicts the definition of limit which says that for any
ε>0 there is a natural number M so that if n>M then |an-a|=|nn+1-1.1|<ε. Example: Show that limn→∞(-1)n does not exist. Solution: Supose there is a number L where
limn→∞(-1)n=L . Then let ε=0.5. So if
limn→∞(-1)n=L then there is a natural number M, where
if n>M then |(-1)n-L|<0.5 or L-0.5<(-1)n<L+0.5. Consider- if n is even then (-1)n=1 so L-0.5<1<L+0.5 and 0.5<L<1.5. While if n is odd the (-1)n=-1 and L-0.5<-1<L+0.5 so -1.5<L<-.5. But
then L is both positive and negative - which is a contradiction. Proposition: [ "Limits are unique."] If limn→∞an=A and limn→∞an=B then A=B. Proof: Suppose limn→∞an=A and limn→∞an=B.
Let ε=|A-B|3.
By the definition there is a natural number MA where if n>MA then |an-A|<ε.
Likewise there is a natural number MB where if n>MB then |an-B|<ε.
Now let n=MB+MA+1 so that n>MA and n>MB.
Then |an-A|<ε=|A-B|3 or A-|A-B|3<an<A+|A-B|3.
Similiarly B-|A-B|3<an<B+|A-B|3.
If B<A then these inequalities say 2A+B3<an<4A-B3 while 4B-A3<an<2B+A3.
But this means 2A+B3<an<2B+A3which implies2A+B<2B+A orA<B. A contradiction! Similarly B>A also leads to a contradiction. So it must be that A=B. For a few more examples see: Convergence
in the Reals 2-12
Motivation:
Preview of definition for limit of a function and continuity
based on sequence limits.
Suppose a∈domain of f.
We say limx→af(x)=L if for any sequence
an in the domain of f-{a}withlimn→∞an=a, limn→∞f(an)=L.
We say f is continuous on its domain D if
for any a∈D and for any sequence an in the
domain of f where limn→∞an=a,
limn→∞f(an)=f(a) or limx→af(x)=f(a)
Theorems connected to algebra, inequalities
and intervals.
Proof: Suppose {an}→ a and {bn}→ b and we are given ε>0.
[Plan:We need to find a natural number M so that if
n>M, we can show that |an+bn-(a+b)|<ε. We will use
the convergence of an and bn to find M.]
For the sequence {an} there is a natural number
Ma where if n>Ma we have that |an-a|<ε3 or
a-ε3<an<a+ε3 .
Similarly for the sequence {bn} there is a natural number Mb
where if n>Mb we have that |bn-b|<ε3 or b-ε3<bn<b+ε3.
Now let M=Ma+Mb+1.
Then for n>M, n>Ma and n>Mb so a-ε3<an<a+ε3 and b-ε3<bn<b+ε3.
Adding these inequalities gives a+b-2ε3<an+bn<a+b+2ε3 or |an+bn-(a+b)|<2ε3<ε.EOP. 2-15
Class started with discussion of Problem 3 of assignment due Feb. 8. Work
on understanding the problem led to specifying the function as f(x)=x2 and intervals [-12,13] and [1,2] and the set {3,5,√2}.
Discussion considered the graph and mapping diagram and how the extreme
value theorem and intermediate vlue theorems were used to find intervals and number to describe f(D).
Reference was made to the following proof for (i). Alternative Proof: Choose M=max{Ma,Mb}. Then for n>M, n>Ma and n>Mb so |an+bn-(a+b)|=|(an-a)+(bn-b)|≤|an-a|+|bn-b|<ε3+ε3=2ε3<ε.EOP.
(ii) (an- bn)→ a - b
Proof: Leave till after (iii). Then apply i) and iii) to (an+ (-1) bn) → a +(-1) b =a - b. 2-16
(iii) (anbn)→ ab Proof: Suppose {an}→ a and {bn}→ b and we are given ε>0.
[Plan:We need to find a natural number M so that if
n>M, we can show that |anbn-(ab)|<ε. We will use
the convergence of an and bn and the fact that convergent
sequences are bounded to find M. Also we will consider these products
as areas of two rectangular regions: one with sides a and b and and
the other with sides anand bn.]
This visualizes the algebra
that |anbn-ab|=|(an-a)bn+a(bn-b)|.
Let B>0 with B>|bn| for all n. [Explain why this is possible.]
Let A>0 so that A>|a| so 0≤|a|A<1.
For the sequence {an} there is a natural number Ma
where if n>Ma we have that |an-a|<ε3B.
For the sequence {bn} there is a natural number Mb
where if n>Mb we have that |bn-b|<ε3A. Choose M=max{Ma,Mb}. Then for n>M, n>Ma and n>Mb so |anbn-ab|=|(an-a)bn+a(bn-b)|≤|an-a||bn|+|a||bn-b|<ε3BB+|a|ε3A<2ε3<ε.EOP. (iv) (an/bn)→ a/b (provided bn ≠ 0 andb ≠ 0).
Lemma: 1bn→1b (provided bn ≠ 0 andb ≠ 0).
Proof of lemma: Suppose {bn}→ b≠0 and we are given ε>0.
[Plan:We need to find a natural number M so that if
n>M, we can show that |1bn-1b|<ε. We will use
the convergence of bn, bn≠0, b≠0 and the fact that convergent
sequences are bounded to find M.]
Consider that by (iii) the sequence {bnb} converges to b2≠0.
From this it can be shown (an exercise for the reader) that there is a number B so that `|b b_n|> B>0andB/{b b_n} < 1` for all n.
For the sequence {bn} there is a natural number Mb where if
n>Mb we have that |bn-b|<B⋅ε2. Then for n>Mb
we have |1bn-1b|=|b-bnbbn|=|b-bn||bbn|<εB2|bbn|<ε2<ε. EOP.
Application of arithmetic for sequences to limits and continuity of functions:
Example: Let f(x)=x2. Prove that limx→3f(x)=9[ which =f(3)].
Demonstration: Referring to the limit definition for a function
we suppose that we have a sequence {an} with an≠3 and
limn→∞an=3. Then we consider the sequence {f(an)=a2n} and apply (iii) so that limn→∞a2n=limn→∞an⋅an=limn→∞an⋅limn→∞an=3⋅3=9. Notes: 1. This example shows the function f(x)=x2 is continuous at x=3.
It is not difficult to show then that this function f(x)=x2 is continuous for any a, and so is continuous for the domain (-∞,∞).
2. Let g(x)=x2 when x≠3 and g(3)=7. Prove that limx→3g(x)=9 [ which ≠f(3)].
Demonstration: Referring to the limit definition for a function
we suppose that we have a sequence {an} with an≠3 and
limn→∞an=3. Then we consider the sequence {g(an)=a2n} and apply (iii) so that limn→∞a2n=9. Here the limit exists as x
approaches 3, but the limit is not equal to g(3)=7, so g is not
continuous at x=3.
3. Similar arguments can show that if f(x)=b0+b1x+b2x2+...+bnxn and g(x)=c0+c1x+c2x2+...+cmxm
then as long as g(a)≠0, the function h(x)=f(x)g(x) is continuous at x=a.
2-18
For Friday class discussion:
Theorem: Suppose f and g are functions with
domains (-∞,∞), h(x)=g(f(x)) for all x, f is
continuous at x=a and g is continuous at b=f(a). Prove h is continuous at x=a. Here is a GeoGebra mapping diagram to help you visualize the theorem.
The Squeeze Lemma (for sequences): If an≤cn≤bn for
all n and limn→∞an=c and limn→∞bn=c, then limn→∞cn=c. Proof: Suppose the hypotheses and that we are given ε>0. [Plan: We need to find an M so that if n>M, then |cn-c|<ε. Connect |an-c| and |bn-c| with an≤cn≤bn and the limit assumptions |cn-c|]
Since an≤cn≤bn, an-c≤cn-c≤bn-c and this 0≤|cn-c|≤max{|bn-c|,|an-c|}
For the sequence {an} there is a natural number Ma where if n>Ma we have that |an-c|<ε.
For the sequence {bn} there is a natural number Mb where if n>Mb we have that |bn-c|<ε.
Let M=max{Ma,Mb} and then ... [Finish the argument.] Class Exercise: State and prove the squeeze lemma (for functions).
2-19 Theorem: Suppose f and g are functions with
domains (-∞,∞), h(x)=g(f(x)) for all x, f is
continuous at x=a and g is continuous at b=f(a). Prove h is continuous at x=a.
Proof: Suppose {an} with limn→∞an=a. By continuity of
f at x=a we have that limn→∞f(an)=f(a)=b.
By continuity of g at x=b we have that limn→∞g(f(an))=g(f(a))=g(b)=h(a). Thus we have that
limn→∞h(an)=h(a), showing that h is continuous at
x=a. Monotonic sequences Theorem BMC:A bounded
monotonic (increasing) sequence is convergent.
Proof: Suppose {an} is a sequence with an≤an+1 and an<B for all n. By the completeness of the real numbers
let L be the least upper bound of {an}. Claim : limn→∞an=L. Here's why: Suppose ε>0. Then L-ε<L and hence L-ε is not an upper bound for {an}. So
there is a natural number M where L-ε<aM<L and so
for n>M we have L-ε<aM<an<Lor| a_n -
L | < epsilon`.
Comment: This result gives a sufficient condition for a sequence to have
a limit without knowing in advance what that limit is. The proof finds
the limit by using the completeness of the real numbers.
The Bolzano Weierstass Theorem: Every
bounded sequence has a convergent subsequence.
The proof of BWT can be developed using Spivak's
lemma: Lemma: Every (bounded) sequence has a monotone subsequence.
Proof: (Spivak) ... Two cases:
Case i) There are an infinite number of n with the property that if
m>n then am<an. Put these n's in order, so
n1<n2<... and an1>an2>... and we have
found a subsequence that is monotonic decreasing.
Case ii) There is a largest number N with the property that if
m>N then am<aN. Then let n⋅0=N+1>N. Then
there is a number n⋅1>n⋅0 where an⋅1≥an⋅0. We can
continue to create sequence of n⋅'s where an⋅k+1≥an⋅k and we have found a subsequence that is monotonic increasing.
EOP.
Proof of BWT: Suppose {an} is a bounded sequence. Then it has a
monotonic subsequence which is still bounded. This subsequence is
convergent by the BMC theorem. 2-22 Cauchy
criterion for convergence:
Definition
A sequence is called a Cauchy sequence
if the terms of the sequence eventually all become arbitrarily
close to one another.
That is, given
> 0 there exists N such that if m, n
> N then |am- an|
< .
Proposition: If converges, then is a Cauchy
sequence.
Proof: Suppose , then given , there exists N
such that if n > N then |an
- L| < . Thus if then . Theorem (CC): If } is a Cauchy
sequence, then } converges. Proof Plan: 1 . Show } is bounded.
2. Use the B. W. Theorem to give a subsequence of } that
converges to .
3. Show .
2-23
Some "monster" examples of functions and continuity
Note: Between any two rational numbers there is an irrational number. Between any two irrational numbers there is a rational number.
Discussed extensively in class using the decimal estimation of a
irrational numbers and √2 for finding an irrational number between
two rational numbers.
"Density"
Proposition: If a and b are rational numbers with a<b, then there is an irrational number c with a<c<b.
Proof: Using the linear function f(x)=a+(b-a)(x-1) we see that
f(1)=a,f(2)=b, and if 1<x<2 then a<f(x)<b.
Since 1<√2<2, we have c=f(√2)∈(a,b).
But if c is a rational number then c=a+(b-a)(√2-1) is a rational number.
Solving for √2 we have √2=c-ab-a+1, a rational
number. This contradicts the fact that √2 is not a rational
number, so c is not a rational number, i.e., c is an irrational
number.
EOP.
Proposition: If a and b are irrational numbers with a<b, then there is an rational number c with a<c<b.
Proof: Suppose a and b are irrational numbers with a<b.
Treating these as infinite decimals a=na+.a1a2a3... and b=nb+.b1b2b3.... there is some decimal position where these two
numbers differ. If na<nb then since b is not a rational number
a<nb<b and we are done. Id na=nb let k be the smallest
natural number where ak<bk. Then let c=na+.a1a2...bk00000.... Then a<c<b. EOP.
Remark: In fact-- If a and b are real numbers with a<b, then
there is an rational number c0 with a<c0<b and there is
an irrational number c1 with a<c1<b.
Examples:
f(x)=1 for x≤5, f(x)=0 for x>5. f is continuous for all x≠5. f is not continuous at x=5.
Consider an=5+1n;n=1,2,....and¬et^lim_{n to oo}
f(a_n) = 0 ne 1 = f(5)`.
f(x)=1 for x a rational number, f(x)=0 for x an irrational number. f is not continuous at any x=a.
Consider when a is irrational an the nth truncated estimate of
a as a decimal; n = 1,2,.... and note that limn→∞f(an)=1≠0=f(a).
Consider when a is rational an=a+√2n; n = 1,2,.... and note that limn→∞f(an)=0≠1=f(a).
f(x)=1q for x=pq where gcd(p,q)=1, f(x)=0 for x an irrational number.
Let f(x)=x if x is rational and f(x)=0 if x is irrational.
Then f is continuous at x=0 but not continuous at any other real number.
The intermediate value theorem (again!). I V Theorem [0 version]: If is continuous on
and then there is a number
with . Proof: Plan. Create
sequences and where and for any , and .
Then both sequences are monotonic and bounded and converge to
the same number . Using continuity show that .
Here's how to construct the and . [The
basic idea is bisection.]
Let and . Now let
If , we have found the desired for
the theorem... stop here. :)
If , let and .
If , let and .
Then and and .
Continue in the same way to construct and
from and :
Now let
If , we have found the desired for
the theorem... stop here. :)
If , let and
.
If , let and
.
Then and .
Continuation of proof :
Now the sequences and are monotonic and
bounded, so they each have a limit, call them and
.
Noting that it is an
exercise for the reader to show that .
We let and claim that
Since for all , we can say the
limit of is by continuity , and
thus .
Likewise, since for all , we can
say the limit of is by continuity
, and thus .
But since ,
we have and
,
and thus by multiplication,
()
But from the hypothesis, ,
and . If then , a contradiction of ()
Thus and . EOP.
Cor. I V Theorem [general version]: If is continuous
on and is between and then
there is a number with Proof: Consider for all . Then is continuous on and , so satisfies the hypotheses of IV
T[0 -version], so there is a number where Thus . EOP. 2-25 Begin with proof of the extreme value theorem. The
boundedness theorems. 2-26
More on horrible functions from previous class:
f(x)=1q for x=pq where gcd(p,q)=1, f(x)=0 for x an irrational number.
Result: f is not continuous at any rational number, x=pq. Let
an=x+√2 then limn→∞an=x and f(an)=0 so limn→∞f(an)=0≠1q for x=pq,gcd(p.q)=1. f is continuous for all x, where x is irrational.
Let f(x)=x if x is rational and f(x)=0 if x is irrational.
Then f is continuous at x=0 but not continuous at any other real number.
Review of Rational
Numbers Q and Irrational Numbers, R-Q: Def'n. A number x is a rational number
if there exist integers n and m with m≠0 where x=nm.
A number x is a d-rational number if there
exists an integer q and a sequence {di}i=1,2,... where
di∈N, 0≤di≤9
with x=q+.d1d2...=q+∞∑i=1di⋅10-i,
and there exist M and r∈N where if i≥M then
di+r=di.
Examples:35 and -117 are rational numbers.
√3 is not a rational number. 57.4368686868... is a d-rational number, q=57,d1=4,d2=3,d3=6,d4=8,d3+2j=6,d4+2j=8 for j∈N. M=3,r=2. Theorem: x
is a rational number if and only if x is a d-rational number. Proof outline: ⇒: Suppose x is a rational number, x=nm with
m,n∈Z and m>0. Do the "long division". By the
division algorithm for natural numbers, the remainders R∈N
are always 0≤R<m, so eventually the long division
will give repeating blocks of digits and this shows x is an d-
rational number. ⇐: Suppose x is a d- rational number, x=q+.d1d2...=q+∞∑i=1di⋅10-i. Then consider
10rx-x=(10r-1)⋅x a terminating decimal, which shows
that x can be recognized as a rational number by some
simple arithmetic.
More examples of some horrible
functions for limits and
continuity, including the mathematician's sine.
Examples: i. If f(x)=sin(1x),x≠0 then there is no
number L where limx→0f(x)=L.
Explanation: Consider the sequence an=1nπ.
limn→∞an=0 and limn→∞f(an)=limn→∞sin(nπ)=0.
While for the sequence bn=1π2+2nπ.
limn→∞bn=0 and limn→∞f(bn)=limn→∞sin(π2+2nπ)=1.
ii. If g(x)=xsin(1x),x≠0 and
g(0)=0 then g is continuous.
Proof for x=0: Note that |sin(t)|≤1 for all t, so
|g(x)|=|xsin(1x)|≤|x| for all x≠0. So
limx→0g(x)=0 2-29
Theorem: If f is a continuous function f:[a,b]→R then there exist c,d∈R where Rf={f(x):x∈[a.b]}=[c,d].
Proof: Plan. Apply the extreme value theorem to find c and d in
Rf so that c≤f(x)≤d for all x∈[a,b], so Rf⊂[c,d].
The use the intermediate value throrem to show that if c≤z≤d then z∈Rf and so Rf=[c,d]. Derivatives and differentiable functions.
Review of definition for limit of a function and continuity
based on sequence limits.
We say limx→af(x)=L if for any sequence an in the
domain of f-{a} with limn→∞an=a,
limn→∞f(an)=L.
We say f is continuous on its domain D if for any a∈D
and for any sequence an in the domain of f where
limn→∞an=a, limn→∞f(an)=f(a). [We will return to continuity and limits later in the course.]
But now a look at the rest of the first year of calculus-
Derivatives and Integrals! Definition of derivative and differentiable
functions.
Suppose f is defined on an open interval, I, and a∈I. We
say thatf is differentiable at a if there is a
number L so that limx→af(x)-f(a)x-a=L or limx→af(x)-f(a)-L⋅(x-a)x-a=0*. 3-1 Comment: If f is differentiable at a then the
number L in the definition is unique and is denoted f′(a) or Df(a).
Examples: (i) If f(x)=mx+b then f′(x)=m. limh→0f(a+h)-f(a)-m⋅hh=limh→0m(a+h)+b-(ma+b)-m⋅hh=limh→00h=0.
(ii)If f(x)=x2, then f′(3)=6. limh→0(3+h)2-32-6⋅hh=limh→09+6h+h2-9-6⋅hh=limh→0h2h=0.
(iii) If f(x)=|x|, then f is not differentiable at 0.
Consider hn=(-1)nn so limn→∞hn=0.
Now for any m, f(0+hn)-f(0)-m⋅hnhn=|hn|-m⋅hnhn=1n-(-1)nmn(-1)nn={(-1)n-m}=1-m when n is
even and -1-m when n is odd, so there can be no limit.
"Differentiability implies continuity." Theorem (DIC): If f is defined on an interval and f
is differentiable at a, then f is continuous at a.
Proof: By hypothesis, there is a number L so that
limx→af(x)-f(a)x-a=L. Then limx→af(x)=limx→af(a)+f(x)-f(a)x-a⋅(x-a)=f(a).
[Or since limx→af(x)-f(a)-L⋅(x-a)x-a=0, we have
limx→af(x)-f(a)-L⋅(x-a)=0, solimx→af(x)=limx→af(a)-L⋅(x-a)=f(a).]
Thus f is continuous at a. EOP. 3-3
Proof of linearity, product and chain rules.
These proofs use DIC:
Theorem: Suppose f and g are differentiable at a, then α and f⋅g are differentiable at a.
In fact: (α⋅f+g)'(a)=(α⋅f'+g')(a)=α⋅f'(a)+g'(a) [Linearity] and (f⋅g)'(a)=(f'⋅g+g'⋅f)(a)=f'(a)⋅g(a)+g'(a)⋅f(a). [Leibniz Rule].
Proof:
See Differentiable
Functions 6.5 Note on Linearity: We can
consider the set F(R,R)={f:R→R} as a real vector space using function
addition and scalar multiplication. It is a real linear algebra using
function value multiplication.
The subset DIFF(R,R)={f:R→R where f is differentiable for all x∈R} is a vector subspace of F(R,R)
and the transformation D:DIFF(R,R)→F(R,R) is linear and in the language
of linear algebras is described as a derivation because it satisfies the
Leibniz Rule.
Theorem ( Chain Rule): [ Proof in Sensible Calculus.]
Critical Point Theorem. [This completes first proof of MVT!]
Theorem. (The Critical Point Theorem) Suppose that c is a point
in the open interval (a,b) where f(c) is an extreme value for f ,
then
either i) f is not differentiable at c
or ii) f is differentiable at c, and in this case f '(c) = 0. Proof:If f is not differentiable at c then the first alternative is satisfied and the theorem is true.
We may assume then that we are considering a situation where f(c) is
an extreme value for f and f is differentiable at c.
Let's assume that f(c) is actually the minimum value of f and leave
the case when it is the maximum value as an exercise.
Since f(c) is the minimum value for f(x), f(x) \ge f(c) for any
x, so f(x) - f(c)\ge 0 for any x. Consider x_n > c and x_n
\to c. [See Figure].
In this case x_n - c > 0 so \frac{f(x_n) - f(c)} {x-c} \ge 0. But
this means that f '(c) cannot be negative because there are quotients
as close as we want to f '(c) that are non negative.
When we consider x_n<c and x_n \to c then x_n - c<0 so \frac{f(x) - f(c)}{x-c}
\le 0. But this means that f '(c) cannot be positive. [See Figure.]
So the only possibility for f '(c) is that f '(c) = 0, because
we have assumed that f '(c) exists.
EOP. 3-4 The Flashman version of the
brief history of integration... from Euclid, Archimedes, and
Aristotle to Newton, Leibniz, Cauchy,
and Euler.
Introduction of decimals, logarithms, Descartes "analytic
geometry" allows interpretation of higher powers of numbers as
lengths, general solution of the area problems for most
algebraic curves - except hyperbolae.
Galileo connects position and motion to areas in solving the
problem of motion with constant acceleration. The development of
logarithms as tools for solving trigonometric proportions and
the creation of the hyperbolic/natural logarithm from an area
problem.
Barrow's theorem relates area and tangent problems.
Newton considers curves as related to motion and velocity
determines "slope of tangent line".
Leibniz uses infinitesmals to find area of regions determined by
variables that related as functions.
Inability to find simple arithmetic formulae to determine
"integrals" leads to desire for more precision in estimation and
more accurate definitions returning to precision of Euclid and
the method of Exhaustion to justify and clarify the key concepts
used in the calculus of Euler
and Cauchy.
Note: Euler used a partition where Deltax_j =Deltax_k for all
j and k , giving Sigma_{k=1}^nDeltax_k = Sigma_{k=1}^nDeltax_1 = nDeltax_1 = b -
a. So, for Euler Deltax = {b-a}/n. Continuation: How to define the definite integral:
Suppose f is a function, f: [a,b] -> R. We can interpret
f as the height of a curve above the "x" axis or as the
velocity of an object moving on a straight line at time x.
We can define various sums based on an Euler partition of the
interval with n equally sized sub-intervals: L_n : Sigma_{k=1}^{n} f(x_{k-1}) Delta x [A Left Hand Sum] R_n : Sigma_{k=1}^{n} f(x_k) Delta x [A Right Hand Sum]
If we let m_k = {x_{k-1} + x_k}/2 for k = 1,2,... ,n then M_n : Sigma_{k=0}^{n-1} f(m_k) Delta x [A Midpoint Sum].
With these Euler sums we have three sequences determined by the
function f and the interval [a,b].
First provisional attempt to define an integral based on
experience using these sums in Calculus numerical integrals:
If there is a number I where lim_{n->oo} L_n =
lim_{n->oo} R_n =lim_{n->oo} M_n = I
then we say that f is integrable over [a,b] and we denote I
= \int_a^b f.
Comment: In calculus courses it is usually asserted that these
approximations all work for continuous function, as well as
other related estimates. T_n = {L_n + R_n}/2 Trapezoidal estimate. S_n = 2/3 M_n + 1/3 T_n Simpson's (Parabolic) estimate.
It is not hard to see also that lim_{n->oo} T_n =
lim_{n->oo} S_n = I
If we try to use this definition we would want to basic
properties to hold true:
1. If f(x) >= 0 for all x in [a,b] then \int_a^b f
>= 0.
2. If c in (a,b), then \int_a^b f = \int_a^c
f + \int_c^b f .
Example: Consider f(x) = 1 if x is
rational and f(x) = 0 if x is irrational.
On [0,1] we have L_n=R_n=M_n = 1 for all n, so \int_0^1 f
= 1.
However on [1, sqrt{3}] we have L_n= 1/n; R_n=M_n = 0
for all n, so \int_1^sqrt{3} f = 0.
And on [0, sqrt{3}] we have L_n= {sqrt{3}}/n; R_n=M_n =
0 for all n, so \int_0^sqrt{3} f = 0.
But then if this definition of integration also satisfies
property 2 we have \0 = int_0^sqrt{3} f = \int_0^1 f + \int_1^sqrt{3}
f = 1+0=1.
If property 2 is going to hold then this function leads to
a contradiction. This is an example of why we need to be
more careful in defining the definite integral.
3-8
General Euler Integral: Choose a set C with n points
C = {c_1, c_2, ... , c_n} where c_k in [x_{k-1},x_k] [An
Euler set].
Then define S_C = Sigma_{k=1}^{n} f(c_k) Delta x The general
Euler sum that depends on C.
We say that f is Euler integrable over [a,b] if there
is a number I so that as n -> oo, the sums S_C -> I.
If I exists it is called the Euler integral of f over [a,b]
.
Since the size of the intervals Delta x = {b-a}/n -> 0 we
can make this more precise:
We say lim_{n->oo} S_C = I if given any epsilon
>0, there is a natural number M so that if n > M and
C is an Euler set with n points, then |S_C - I | <
epsilon.
Example revisited: Consider f(x) = 1 if x is
rational and f(x) = 0 if x is irrational.
On the interval [0,1] for any n use c_k = x_k. Then
for any n, S_C =1.
But for any n, choose r_k in [x_{k-1},x_k] where r_k is
irrational. Then for C = {r_1,r_2, ..., r_n} we have S_C =
0.
Thus the function f is not "Euler" Integrable.
Discussion of integration and connection to continuity. 3-10&11
Week 10
3-21 through 3- 25
These lectures and notes are found on a separate link:
Integration Notes from Week 10
Discussion of assignment due March 8 problem 5:
Suppose f is continuous on [a,b] and for all x,y in [a,b]
if x<y then f(x) <f(y) .
a. f([a,b]) = [f(a),f(b)].
b. There is a unique function g: [f(a),f(b)] -> [a,b] where
g(f(t)) = t for all t in [a,b] and f(g(s)) = s for all s in
[f(a),f(b)].
c. g is continuous.
Part a - Use definitions and assumptions to prove the set equality.
[IVT used.]
Part b. Show f is one to one.... then review of "inverse
functions" from Math 240. Proof of c.
Consider a sequence {c_n} where lim c_n = c in [f(a),f(b)]. Then
let r_n = g(c_n) in [a,b] so f(r_n) = c_n. Since {r_n} is a
bounded sequence, using BW Theorem we have a convergent subsequence
{s_m} where s_m = r_n for some n>m and lim s_m = s. Thus
f(s_m) = f(r_n) =c_n and by continuity of f, lim f(s_m) = f(s)
= lim c_n = c. Thus g(c)= s. It suffices to show that lim r_n =
s.
Suppose not. Then there is an epsilon >0 where for any natural
number M there is a number j > M where |r_j - s| >
epsilon . Then for all j, s>z = s- epsilon >
r_j or s< w = s+epsilon< r_j. Thus c= f(s) > f(z)
> f(r_j)=c_j or c = f(s)< f(w)< f(r_j) = c_j. But this
contradicts lim c_n = c. Thus lim r_n = lim g(c_n) = s =
g(c) and g is continuous.
FTofC and inverse function theorem connected to Ln and e^X ,
Arctan and tangent functions, Arcsin and sin functions can be
defined through integration and Inverse Function Theorem.
Bounded Function Constraint Result for Integral.
Main properties of the definite integral that we will
prove:
Including: Monotonicity. Linearity. Additivity. Bounded Constraint.
Continuity of Integral Function for integrable functions. Continuous
Functions are integrable. Fundamental Theorem (Derivative form) for
Continuous Functions. FTofC (Evaluation form) for Continuous
Functions. Mean Value Theorem for Integrals for Continuous
Functions.
Additivity. Continuity of Integral Function for Integrable
Functions. Continuous Functions are Integrable.
FTof Calc I and II, MVT for Integrals. Alternative proof of
FTofC using MVT for Integrals.
Discussion of Improper Integral for Discontinuities on bounded
intervals. Sets of Measure Zero introduced. Countable Sets have
measure zero. Continuity, integrability, and measure zero sets
related. The Cantor Set: has measure 0 but is uncountable.
Week 11
3-28 Non-sequence based limits and continuity.
Limits of sequences with intervals:
Note: lim_{n ->oo} a_n = a if for any epsilon >0 there
is a number M in N where if n>M then |a_n -a| <
epsilon.
Rephrase:..... a_n in ( a-epsilon, a+epsilon).
Definition of an open set. Suppose O is a set of
real numbers. O is open if (and only if) for any a in O there
is a epsilon > 0 where ( a-epsilon, a+epsilon) sub
O.
Example: The interval (1,3) is an open set.
Proof: Choose a in (1,3). Let epsilon = 1/2 min{a-1, 3-a}.
Suppose x in ( a-epsilon, a+epsilon). Then 1=
a-(a-1)\le a- min{a-1,3-a}< a-epsilon< x < a+epsilon
< a + min{a-1,3-a} \le a+ (3-a)=3. Thus x in (1,3) , (
a-epsilon, a+epsilon) sub (1,3) and (1,3) is
open. EOP
Example: The interval (1,3] is not an open set.
Proof: Consider a=3 in (1,3]. Choose any epsilon >0.
Then 3 +epsilon/2 in ( a-epsilon, a+epsilon) but 3 +epsilon/2
is not an element of (1,3] so for any epsilon >0, ( 3-epsilon,
3+epsilon) is not a subset of (1,3] and thus (1,3] is not an
open set. EOP
Remark: It can be proven similarly that for any a < b
, (a,b) is an open set.
Note continued: lim_{n ->oo} a_n = a: if for any open set,O, with a in
O, there is a number M in N where if n>M then a_n in
O. Continuity: With sequences:f is continuous on its domain D
if for any a in D and for any sequence a_n in the
domain of f where lim_{n->oo}a_n = a,
lim_{n->oo} f(a_n) = f(a). With sequences and open sets: If a_n
is a sequence in
the domain of f where lim_{n->oo}a_n = a then for any
open set O with f(a) in O, there is an number M in N where
if n>M then f(a_n) in O. epsilon -delta Definition of Limits and Continuity.
The limit of f as x -> a for a in D is L if for any epsilon
>0, there is a real number delta>0 where if x in D and
0<|x-a|< delta , then |f(x)-L|<epsilon. This is written as
lim_{x ->a} f(x) = L. f is
continuous on its domain D if for any a in D and epsilon
>0, there is a real number delta>0 where if x in D
and |x-a|< delta , then |f(x)-f(a)|<epsilon. That is, for any a in D , lim_{x ->a} f(x) = f(a).
See these on-line references:
GeoGebra Mapping Diagram for limits: http://ggbtu.be/mNLaWHLas Continuity
of Real functions Images
of intervals Limits
of functions 3-29
Example: Let f(x) = C, a constant function. Then f is
continuous for the set of real numbers.
The justification of this statement is left as an exercise for
the reader.
Example: Let f(x) = 3x+5, a linear function. Then f is
continuous for the set of real numbers.
Proof: Suppose a in R and epsilon > 0.
[ We worked with inequalities to see that a likely effective
choice for delta was epsilon /3.]
Let delta = epsilon/3. Then suppose x in R and |x-a|
< delta = epsilon/3.
Then |f(x)-f(a)| = |(3x+5) - (3a+5) | = |3(x-a)| = 3|x-a|.
But we assumed |x-a| < epsilon/3 so |f(x)-f(a)| = 3|x-a|
< 3 epsilon/3 = epsilon.
Thus f is continuous for every real number. EOP
For a visualization of epsilon-delta proof of continuity for linear functions see this GeoGebra worksheet: http://ggbtu.be/mfvWrWyBu
Compare the previous proof with the sequence based proof:
Suppose a_n and lim_{n->oo} a_n =a. Then by applying the
results we have about limits of sequences,lim_{n->oo} f(a_n)
= lim_{n->oo} 3a_n+ 5 = 3a +5 = f(a). Theorem: Suppose f is a function with domain D. f is epsilon-delta continuous on D if and only
if f is sequence continuous on D. Proof: (1) "only if" : Suppose f is epsilon-delta
continuous on D. Assume a in D and a_n in D with
lim_{n->oo} a_n =a. Given epsilon > 0, there is a
delta where if x in D and |x-a| < delta then |f(x) -
f(a)| < epsilon. Since lim_{n->oo} a_n =a there
is a number M in N where if n > M then |a_n -a| <
delta and hence |f(a_n)-f(a)| < epsilon. Thus
lim_{n->oo} f(a_n) = f(a) and f is sequence continuous
on D.
(2) "if" : Assume f is sequence continuous
on D. This will be indirect - so we assume f is not
epsilon-delta continuous on D. Then there is an element
of D, a* and a positive real number, epsilon* > 0 where
for any delta > 0 there is an x in D with |x-a*| <
delta while |f(x) -f(a*)| \ge epsilon*.
Now use delta = 1/2 and let x_1 in D be so that |x_1-a*|< 1/2 while |f(x_1) -f(a*)| \ge epsilon*.
Next use delta = 1/4 and let x_2 in D be so
that |x_2-a*|<1/4 while |f(x_2) -f(a*)|
\ge epsilon*.
Continuing use delta = 1/2^n and let x_n in D be so
that |x_n-a*|<1/2^n while |f(x_n)
-f(a*)| \ge epsilon*.
The sequence x_n in D has lim_{n->oo} x_n =a* but
lim_{n->oo} f(x_n) ne f(a*). This contradicts the
assumption that f is sequence continuous on D. EOP
Recall: Definition of an open set. O is open if for
any L in O there is a epsilon > 0 where N(L, epsilon
)sub O.
Review Definitions: Given f : D -> R, and A sub
D
The image of A under f , f (A) = { y in R : y = f
(x) for some x in A}and
the preimage of A under f, f ^{-1}(A) = {x in R :
f (x) in A}.
Example: f(x) = x^2. f ^{-1}\(\[0\,1\)\)= (-1,1) ; f ^{-1}\(\(0\,1\)\)= (-1,0) cup (0,1); f ^{-1}\(\(1,4\)\)= (1,2) cup
(-2,-1).
Continuity and open sets 4-5
Theorem: Given `f
: R ->R` ,
f is epsilon- delta continuous if and only
if whenever O is an open subset of R, f ^{-1}(O)
is also an open set.
["A function is continuous if and only if the inverse image of
an open set is open."]
Proof: (1) => : Suppose f is
epsilon-delta continuous on D.
Suppose O is an open set and that a in f ^{-1}(O). Then
f(a) in O. Since O is an open set, there is an epsilon
>0 where N(f(a),epsilon) sub O. Since f is epsilon-
delta continuous, there is a delta >0 where if x in
N(a,delta) then f(x) in N(a,epsilon). Thus N(a,delta) sub
f ^{-1}(O) and f ^{-1}(O) is an open set.
(2) <= : Suppose that whenever O is an
open subset of R, f ^{-1}(O) is also an open set.
Suppose epsilon >0 and a in R. Then N(f(a),epsilon)
is an open set with a in f ^{-1}(N(f(a),epsilon)). By the
hypothesis f ^{-1}(N(f(a),epsilon)) is an open set, so there
is a number delta>0 where N(a,delta) sub f
^{-1}(N(f(a),epsilon)). Then if x in N(a,delta) then
f(x) in N(a,epsilon) and f is epsilon-delta continuous. EOP.
Prop: (i) O/ and
R are open sets.
(ii) If U and V are open sets, then U nn V is an open
set.
(iii) If O is a family of open sets, then uuu O
= { x in R : x in U for some U in O} is an
open set.
4-7
Proof of (ii)
Suppose a in U nn V. Then a in U and a in V. Since U
is an open set, there are numbers epsilon_U where N(a, epsilon_U)
sub U and epsilon_V where N(a, epsilon_V) sub V. Let epsilon
= min(epsilon_U, epsilon_V). Then it is not hard to show that
N(a,epsilon) sub U nn V and thus U nn V is an open set. EOP
Proof of (iii)
Suppose a in uuu O. Then a in U for some U in O. Since
U is an
open set, there is a number epsilon_U where N(a, epsilon_U) sub U.
Let epsilon = epsilon_U Then it is not hard to show that
N(a,epsilon) sub uuuO and thus uuu O is an open set. EOP Fact:If a in R then (a, oo) and (-oo, a)
are open sets. Fact: {a} is not an open set.
Cor. : If a < b , then (a,b) is an open set.
The family of all open sets of R is sometimes called the
topology of R.
Example: Let U_n = (-1/n,1/n). Then uuu{U_n} = (-1,1) which is open. BUT nnn{U_n} = { 0} which is not open. Continuity
and open sets
Sidenotes: A topological space is a set X together with a family
of subsets O that satisfies the three properties:
(i) O/ in O and X in O.
(ii)
If U , V in O , then U nn V in O.[Closure
under "finite intersection".]
(iii) If {U_{alpha}} is a family of sets with
each U_{alpha} in O, then uuu U_{alpha} = {x in X : x in
U_{alpha} for some U_{alpha} in the family}in O.
[Closure under "arbitrary union".] Definition of a closed set. A set C is closed if (and
only if) R - C is open.
Prop: (i)
O/ and R are closed sets.
(ii) If
a in R then [a, oo ) and (- oo, a] are closed
sets.
Fact: {a} is a closed set. [ R-{a} = (-oo,a)uu(a,oo) which is an open set.] 4-8
Prop: (i) If U and V are closed sets, then U uu V is
a closed set.
(ii) If K is a family of closed sets, then nnn K
= { x in R : x in C for all C in K} is a
closed set.
Cor. : If a < b , then [a,b] is a closed set.
Proof: [a,b]= [a,oo) nn (-oo,b].
Connectedness , Continuity, and Topological
Proof of the Intermediate Value Theorem. Def'n: A subset I of the R is an interval: If
a, b in I and a < x< b, then xinI. Def'n: A subset S of (a metric space) R is disconnected
if there are open sets U and V where
(i) U uu V sup S; (ii) U nn S !=O/
, V nn S !=O/
, AND V nn U = O/. Def'n: A
subset S of (a metric space) R is connected if it is
not disconnected.
Thus, if S is connected with open
sets U and V where
(i) U uu V sup S; and
(ii) U nn S !=O/
, V nn S !=O/,
then V nn U !=O/.
The following result leads immediately to the IVT.
Theorem: A subset S of R is connected if
and only if it is an interval. [Skip to IVT]
Proof: =>: [Indirect-
contrapositive] Suppose S sub R and S is not an
interval. Then there exist a in S and b in S with
a<b and c where a<c<b and c not in S. Let
U = (-oo,c) and V=(c,oo). Then U uu V sup S; a in U
nn S so U cap C!= O/ and b in V nn S so V cap C!=
O/ and certainly V nn U =O/, so S is not connected.
<=: Deferred till 4-11
Suppose S is an interval. To show S is connected,
suppose open sets U and V where
(i) U uu V sup S; and (ii) U nn S !=O/ , V
nn S !=O/ . [We need to show that V
nn U != O/.] Also suppose V cap U = O/.
Choose a in U cap S and b in V cap S. For convenience
assume a<b.
Let R = {t : [a,t] sub U}.
Since a in R, R !=O/. Since b in V , if x in R then
x < b and thus R is nonempty and bounded above, so by
the least upper bound property of the real numbers we can
let z = lub(R). Then a le z le b and since S is an
interval, z in S. So either z in U or z in V. [We will
show that neither of these alternatives are possible.]
(a) Suppose z in U.
Since U is open, there is an epsilon_U>0
where N(z,epsilon_U) sub U. Then z-epsilon_U <z is
not an upper bound for R and there is w in R where
z-epsilon_U < w le z. But then [a,w]sub U and
[a,z+{epsilon_U}/2] = [a,w]cup [w,z+{epsilon_U}/2] sub U.
Thus z+{epsilon_U}/2 in R and z is not an upper
bound for R. So z!in U.
(b) Suppose z in V.
Since V is open, there is an epsilon_V>0
where N(z,epsilon_V) sub V. Then z-epsilon_V <z is
not an upper bound for R and there is w in R where
z-epsilon_V < w le z. But then [a,w]sub U. But then
w in U while w in N(z,epsilon_V) so w in V. This
contradicts U cap V=O/so z!in V.
EOP.
Theorem: If f : R ->
R is continuous and C is a connected set, the f(C) is a connected set. Proof: Suppose not. Then let U and V be open sets and
(i) f(C) sub U uu V ; and (ii) U nn f(C) != O/ , V nn f(C) !=
O/, AND V nnU = O/ . We will show this implies that R is not a
connected set.
Since f is continuous for R, hat U =f ^{-1}(U) and hatV =f ^{-1}(V) are open sets.
Now (i) hatU uu hatV sup R ; and (ii) hatU nn R!= O/ , hatV nn R != O/ , AND hatU nn hatV = O/. [why?]
But this contradicts the fact that R is a connected set. EOP. More
on Connected and disconnected sets
IVT. Theorem: If f : R ->
R is continuous and a<b and v is between
f(a) and f(b) then there is a number c in [a,b] so
that f(c) = v.
Proof: Suppose not. Then let U= (-oo,v)
and V= (v,oo) and
(i) f([a,b]) sub U uu V ; and (ii) U nn f([a,b]) !=
O/ , V nn f([a,b]) != O/, AND V nnU = O/ . So
f([a,b] is not connected. but [a,b] is a connected set, so
f([a,b]) is connected. A contradiction. EOP.
Cor: If a<b
and f : [a,b]->
R is continuous and v is between f(a) and f(b) then
there is a number c in [a,b] so that f(c) = v.
Proof: Extend the function f by letting
f(x) = f(a) for x < a and f(x) =f(b) for x>b.
Then f: R->R and f is continuous so the previous
theorem can be applied. EOP.
4-11 Deferred from 4-8
Theorem: A subset S of R is connected if and only if it is an interval. Proof <=:Suppose S is an interval. To show S is connected, suppose open sets U and V where
(i) U uu V sup S; and (ii) U nn S !=O/ , V nn S !=O/ .
[We need to show that V nn U != O/.] Also suppose V cap U = O/.
Choose a in U cap S and b in V cap S. For convenience assume a<b.
Let R = {t : [a,t] sub U}.
Since a in R, R !=O/. Since b in V , if x in R then x < b
and thus R is nonempty and bounded above, so by the least upper bound
property of the real numbers we can let z = lub(R).
Then a le z le b and since S is an interval, z in S. So either z
in U or z in V. [We will show that neither of these alternatives are
possible.]
(a) Suppose z in U.
Since U is open, there is an epsilon_U>0 where
N(z,epsilon_U) sub U. Then z-epsilon_U <z is not an upper bound
for R and there is w in R where z-epsilon_U < w le z. But then
[a,w]sub U and [a,z+{epsilon_U}/2] = [a,w]cup [w,z+{epsilon_U}/2] sub
U. Thus z+{epsilon_U}/2 in R and z is not an upper bound for
R. So z!in U.
(b) Suppose z in V.
Since V is open, there is an epsilon_V>0 where
N(z,epsilon_V) sub V. Then z-epsilon_V <z is not an upper bound
for R and there is w in R where z-epsilon_V < w le z. But then
[a,w]sub U. But then w in U while w in N(z,epsilon_V) so w in V.
This contradicts U cap V=O/ so z!in V. Continuous
functions and connected sets A topological analysis for the Extreme Value Theorem:
Definitions: Suppose is a family of sets
and . We say
is a cover of if uuuF={x:x∈U for some .
If is a cover of and every is an open
set, then we say is an open cover of . If is a subfamily of , i.e., , and is a cover of , we say that
is a subcover of .
We say that a set is (topologically)
compact if any open cover
of has a subcover that is a finite set,
i.e., for any open cover of , there exist
where .
Examples: i. A finite set is compact.
ii. for is
not compact. [This set is not closed!]
iii. for is compact.[This set is closed.]
iv.
for is not compact. [This set is closed,
but not bounded.]
Outline:
Heine Borel Theorem (simplest version):
is a compact subset of R.
Theorem: A closed subset of a compact set is
compact.
Theorem: If is a compact subset of ,
then is closed and bounded. [HB in R]
Theorem:If
is continuous and is a compact
then is compact.
Extreme Value Theorem (Topological Proof) Suppose
f: [a,b] -> R is continuous. Then there existc_M and c_m in
[a,b] where f(c_m) le f(x) le f(c_M) for all x in [a,b].
Proof: We can suppose that f(x) = f(b) for x > b and f(x) = f(a) for x<a, so f: R -> R is continuous.
Since [a,b] is compact, f([a,b]) is compact.
Since [a,b] is also an interval, f([a,b]) is a closed and bounded interval.
Thus f([a,b]) = [m,M] where for all x in [a,b], m le f(x) le M.
But since m, M in f([a,b]), there exist c_m and c_M in [a,b] where f(c_m) = m and f(c_M) = M. EOP!4-14 Proofs remaining from outline.
Heine Borel Theorem (simplest version):
is a compact subset of R.
Proof: (outline). Suppose O is a family of open
sets where .
Let G= {x
where is covered by a finite subset of the members
of O.}.
Claims: a in G because [a,a] = {a} and since [a,b] subset uuu O, for some U in O, a in U.
Thus G and by its definition G is bounded
by .
Then let = lub of G.
Plan: Show alpha = b and G alpha = b: Suppose alpha < b. Then a le alpha < b. Since O
is an open covervof [a,b], there is an open set U_{alpha} where
alpha in U_{alpha} in O and epsilon_{alpha} > 0 where N(alpha,
epsilon_{alpha}) sub U_{alpha} nn [a,b].
Considering alpha - epsilon_{alpha} < alpha is not
an upper bound for G so there is a number c in G with
alpha-epsilon_{alpha} < c<alpha.
Since c in G, the interval [a,c] is covered by a finite subset of the members of O.
But then add U_{alpha} to this family of members of O and we see
that alpha + epsilon_{alpha}/2 < b is also in G which contradicts
that alpha is an upper bound for G.
So alpha =b.
α∈G: Using U_{alpha} from
above, we have epsilon> 0 where N(alpha, epsilon) sub
U_{alpha}. Following an argument similar to the last one, alpha -
epsilon < alpha is not an upper bound for G so there is a
number c in G with alpha-epsilon < c<alpha.
Since c in G, the interval [a,c] is covered by a finite subset of the members of O.
But then add U_{alpha} to this family of members of O and we see
that alpha = b is also in G. Thus [a,b] is
(topologically) compact. EOP See also : Closed_Bounded_Subset_of_Real_Numbers_is_Compact
Theorem: A closed subset of a compact set is
compact.
Proof: Supposed
is a closed set and is a compact set with .
Now suppose O
is a family of open sets where .
Since is
closed, let U = R-C and .
Then
covers and by compactness a finite subcover of
covers .
But then if needed that finite subcover
less will be a finite subcover of that covers .
So is compact. EOP.
Corollary: If K is a closed and bounded
subset of R, then K is compact.
Proof: K is bounded- so K for some and . But by HB, is compact, so K is a closed
subset of a compact set and thus K is compact. EOP
4-15
Theorem: If is a (topologically) compact subset of ,
then is closed and bounded. [HB in R] Proof:
(i) is bounded: Consider the family of open
sets, . Certainly this
family covers any subset of , so it covers . But a
finite subcover will be bounded by for the largest
of the sets in that finite subcover. Thus is
bounded.
(ii) is closed. Let = R -K, the complement
of K. We will show that is open.
Suppose and .
Then let so and .
Let be the family of open sets .
Then covers .
Because is compact
there a finite number of points where the
family of open sets
covers .
Then let and we have
that for all k.
Hence
Thus and is an open set. EOP.
Compactness and
continuity. Theorem:If
is continuous and is a compact
then is compact.
Proof: To show
is compact, suppose is an open cover of .
Because is continuous, for each , there is an open
set where .
Let . Note that is open
cover of .
Since is compact, has a finite
subcover of .
For these (finite) open sets we have
corresponding so that .
Thus ....
these are a finite subcover of the family . 4-18
Compactness and Uniform Continuity: 4-19 See special page for more complete covereage.
Definition : Suppose f: R -> R and S is an interval. We say that f is uniformly continuous on S if
for any epsilon > 0, there is a real number delta >0 so that
for any a and x in S, if |x-a| < delta then |f(x)-f(a)| < epsilon. Demonstration on wolfram.com of Uniform Continuity
Proposition: If f is uniformly continuous on S, then f is continuous on S.
Example: f(x) = 1/x for x \ne 0, f(0) = 0; S = (0, 1). f is continuous on S, but f is not uniformly continuous on S.
Theorem: Any continuous function on [a,b] is Darboux (or Riemann) integrable.
Proof: [An application of uniform continuity.]
Sequences and series: Review some of
the properties for convergence related primarily to numbers.
Note how geometric series could be thought of as a sequence
of functions. 4-21and 22
Review and prove:
The divergence test:
If sum_{k=0}^{oo} a_k converges then \lim_{n -> oo} a_n = 0.
Geometric series:
If -1<r<1 then sum_{k=0}^{oo} a r^k = a/{1-r} . Otherwise the series diverges.
The harmonic series: sum_{k=1}^{oo} 1/k diverges.
Convergence of positive series: Suppose a_n >0 for all n.
The comparison test: Suppose a_n < b_n for all
n. If sum_{k=0}^{oo} b_k converges, then sum_{k=0}^{oo}
a_k converges.
If sum_{k=0}^{oo} a_k diverges, then sum_{k=0}^{oo} b_k diverges.
The integral test:
Suppose f: [1,oo) -> R is a decreasing continuous function with f(x) > 0 and a_k = f(k). sum_{k=0}^{oo} a_k converges if and only if int_1^{oo} f converges.
p- series:
sum_{k=0}^{oo} 1/{n^p} converges if and only if p> 1.
4-25
Positive Ratio Tests:
If lim_{n->oo} a_{n+1}/{a_n} = R with R<1 then sum_{k=0}^{oo} a_k converges.
If R>1, or there is no limit, then sum_{k=0}^{oo} a_k diverges.
If R=1, there is no conclusion that can be reached from this information alone.
Proof: Plan:
Suppose R<1. Then there is r with R<r<1 and N so that
if n \ge N, a_{n+1}/{a_n}< r. Use this to show that for k \ge
0, a_{N+k}< r^k a_N so that the comparison test can be applied to
compare sum_{k=0}^{oo} a_{N+k} to the geometric series
sum_{k=0}^{oo} r^k a_N.
Since r<1, the geometric series converges, and so then will sum_{k=0}^{oo} a_{N+k} and thus sum_{k=0}^{oo} a_k
For R>1 or no limit, one can find r >1 where for k \ge
0, a_{N+k}> r^k a_N so that lim_{n -> oo} a_n \ne 0. So the
series will diverge.
For R=1 Examples with a_n= 1/n and a_n = 1/{n^2} show that no conclusion is possible from this information alone.
Positive Root Tests:
If lim_{n->oo} \root n {a_n} = R with R<1 then sum_{k=0}^{oo} a_k converges.
If R>1, or there is no limit, then sum_{k=0}^{oo} a_k diverges.
If R=1, there is no conclusion that can be reached from this information alone.
Plan:
Suppose R<1. Then there is r with R<r<1 and N so that
if n \ge N, \root n {a_n} < r. Use this to show that for k
\ge 0, a_{N+k}< r^{N+k} so that the comparison test can be
applied to compare sum_{k=0}^{oo} a_{N+k} to the geometric
series sum_{k=0}^{oo} r^k r^N.
Since r<1, the geometric series converges, and so then will sum_{k=0}^{oo} a_{N+k} and thus sum_{k=0}^{oo} a_k
For R>1 or no limit, one can find r >1 where for k \ge
0, a_{N+k}> r^{N+k}>1 so that lim_{n -> oo} a_n \ne 0. So
the series will diverge.
For R=1 Examples with a_n= 1/n and a_n = 1/{n^2} show that no conclusion is possible from this information alone.
General Series.
Algebra for series convergence: Sums and Scalar Multiples of Series.
Suppose sum_{k=0}^{oo} a_k and sum_{k=0}^{oo} b_k converge to S_a
and S_b and alpha is any real number. Then
sum_{k=0}^{oo} (a_k +b_k) converges to S_a + S_b and
sum_{k=0}^{oo} alpha a_k converges to alpha S_a.
Absolute convergence implies convergence:
If sum_{k=0}^{oo} |a_k| converges then sum_{k=0}^{oo} a_k converges.
Proof: -|a_k| \le a_k \le |a_k| so 0 \le a_k + |a_k| \le 2|a_k|.
Since sum_{k=0}^{oo} |a_k| converges, sum_{k=0}^{oo} 2|a_k| converges.
By the comparison test, sum_{k=0}^{oo}a_k + |a_k| converges.
Also sum_{k=0}^{oo} - |a_k| converges, so sum_{k=0}^{oo}(a_k + |a_k|) - |a_k| =sum_{k=0}^{oo} a_k converges .
Alternating Series Test. If a_k> a_{k+1} > 0 for all k and lim_{k->oo}a_k = 0 then sum_{k=0}^{oo} (-1)^k a_k converges.
Proof: plan: Let S_n = sum_{k=0}^n (-1)^k a_k.
Show {S_{2n}} is decreasing and bounded below by S_1. So {S_{2n}}
will converge. But then {S_{2n+1}} will converge to the same limit. Conditional Convergence
Definition. sum_{k=0}^{oo} a_k converges conditionally if sum_{k=0}^{oo} a_k converges but sum_{k=0}^{oo} |a_k| diverges.
Power
series and Functions defined by series.
Preface:
For x in R and a sequence of real numbers C={c_n: n = 0,1,2,...},
consider an x where the series sum_{k=0}^{oo} c_k x^k converges. For such an x we have a function
f_C(x) = sum_{k=0}^{oo} c_k x^k.
The study of these series and the related functions is the area of analysis described as "power series."
Note: This subject is considered with other fields where it makes sense
to discuss being close, especially the field of complex numbers.
The use of non-negative powers of x is also generalized to allow negative powers.
One can also change the functions used in the series by using
trigonometric functions- sine and cosine and more coefficients to
consider series of the form sum_{k=0}^{oo} c_k cos(kx) + s_k sin(kx) , usually described as Fourier series.
Results on the theory of power series connect to the study of
continuous, differentiable, and integrable functions and differential
equations.
A function f is described as "real analytic on a set S" if there is a power series where f(x)=f_C(x) for all x in S.
Two ways to understand analytic functions:
1. Study functions defined by power series,
2. Study power series defined by functions which leads to Taylor Theory and Fourier Analysis.
For a power series, the partial sums for x are polynomials of degree
n, S_n(x) = sum_{k=0}^{n} c_k x^k. These polynomial functions form a
sequence of functions, and lead to issues about convergence of
sequences of functions.
Pointwise convergence of function sequences.
Uniform convergence of function sequences.
Metrics on functions.
Series and Power Series
Taylor's Theorem. Sensible Calculus : IX B
MacLaurin Polynomials and Taylor
Definition: f:R -> R is called C^{oo} on an interval I if f ^{n}(x) exists for all n= 1,2,3,... and all x in I.
Theorem (Taylor- Integral Remainder): Suppose f:R -> R is C^{oo}
on (-a,a) where a>0. Then for any x in (-a,a) , f(x) =f(0) +f'(0)x + {f''(0)}/2 x^2 + ... + {f^{(n)}(0)}/{n!}x^n + R_n(x)= P_n(x) +R_n(x) where
(i)R_n(x) = 1/{n!}\int_0^x (x-t)^nf^{(n+1)}(t) dt and
(ii) [Lagrange] there exists p between 0 and x where R_n(x) = {f^{(n+1)}(p)}/{(n+1)!} x^{n+1}.
Furthermore: f(x) = sum_{k=0}^{oo} {f^{(k)}(0)}/{n!}x^k if and only if lim_{n-> oo}R_n(x) = 0.
Proof:
(i) Plan: Use induction on n.
Start: n=0: R_{0}(x) = 1/{0!}\int_0^x (x-t)^0f^{(1)}(t) dt = \int_0^x f'(t) dt= f(x)-f(0)
Induction step: Use integration by parts on the definite integral 1/{n!}\int_0^x (x-t)^nf^{(n+1)}(t) dt
So ...
Use dv = (x-t)^n and u =f^{(n+1)}(t) and thus v = - (x-t)^{n+1}/{n+1} and du= f^{(n+2)}(t). 1/{n!}\int_0^x (x-t)^nf^{(n+1)}(t) dt =
x^{n+1}/{(n+1)!} f^{n+1}(0) + 1/{(n+1)!}\int_0^x (x-t)^{n+1}f^{(n+2)}(t) dt
So R_n (x) = x^{n+1}/{(n+1)!} f^{n+1}(0) + R_{n+1}(x)
where R_{n+1}(x) =1/{(n+1)!}\int_0^x (x-t)^{n+1}f^{(n+2)}(t)
dt.
Thus if we assume (i) is true for n, then f(x)= P_{n+1}(x) +R_{n+1}(x) is also true with (i) true for R_{n+1}(x) as well.
(ii) See Sensible Calculus : IX B MacLaurin Polynomials and Taylor. Comment:
These results can be generalized from using 0 in (-a,a) with
a>0 to c in (a,b) to obtain the general Taylor Theorem: Suppose
f:R -> R is C^{oo} on (a,b) and c in (a,b). Then for any x
in (a,b) , f(x) =f(c) +f'(c)(x-c) + {f''(c)}/2 (x-c)^2 + ... +{f^{(n)}(c)}/{n!}(x-c)^n + R_n(x,c)= P_n(x,c) +R_n(x,c) where
(i)R_n(x,c) = 1/{n!}\int_c^x\ (x-t)^nf^{(n+1)}(t) dt and
(ii) [Lagrange] there exists p between c and x where R_n(x) = {f^(n+1)(p)}/{(n+1)!} (x-c)^{n+1}.
Furthermore: f(x) = sum_{k=0}^{oo} {f^{(k)}(c)}/{n!}(x-c)^k if and only if lim_{n-> oo}R_n(x,c) = 0.
For proof see Sensible Calculus : IX D Taylor Polynomials.
or Taylor Series (with remainders -
Integral & Lagrange) Proof
of Taylor Theorem (Integral remainder) Proof
of Taylor Theorem(Lagrange Remainder)4-29
Analysis of Functions Defined with Sequences of Functions and Power Series Example: Consider s_n(x) = x^n for x in [0,1], s_n(x) = 1
for x>1.Discussion: For |x|<1, lim_{n -> oo} s_n(x) = 0.
For x \ge 1, lim_{n->oo}s_n(x)=1. So s(x) = lim_{n->oo}s_n(x)
is not continuous at x=1. Definitions: Suppose D sub R and {f_n: D -> R} is a sequence of functions. We say that f_n converges tof :D -> R if
for every x in R, lim_{n->oo} f_n(x) = f(x).
We say the f_n converges uniformly on D tof:D->R
if for every epsilon >0 there is a natural number N so that
for every x in D and n > N, |f_n(x) - f(x)|< epsilon.
See http://www.mathcs.org/analysis/reals/funseq/uconv.html Example cont'd: Let D = [0,oo) then {s_n} does not converge uniformly to s on D. Example:Consider s_n(x) = x/n for x in [0,100]. Then s_n converges uniformly to s where s(x) = 0 5-2
Theorem: If {f_n: D -> R} is a sequence of continuous functions on D and f_n converges uniformly on D to f:D->R`, then f is continuous on D. Proof: Suppose a in D and epsilon > 0. Since f_n converges uniformly on D to f
:D->R, there is an number N where if n> Nthen for any x in
D and n > N, |f_n(x) - f(x)|< epsilon/3. Applying this to
f_{N+1} which is continuous at a there is a delta where if |x-a|
< delta then |f_{N+1}(x)- f_{N+1}(a)|< epsilon/3. So then
|f(x)-f(x)| \le |f(x) - f_{N+1}(x) | + |f_{N+1}(x)- f_{N+1}(a)| +
|f_{N+1}(a)- f(a)| < epsilon. EOP.
Note: This result is not true if "continuous" is replaced by "differentiable". Theorem: If {f_n: D -> R} is a sequence of C^{oo}functions on [a,b], f_n converges to f:D->R`, and {f_n'} converges uniformly then f is differentiable and
lim_{n->oo} f_n'(x) = f '(x).
Theorem: If {f_n: [a,b] -> R} is a sequence of continuous functions on D and f_n converges uniformly on [a,b] to f:D->R`,
then f is integrable on [a,b] and
Power Series Interval of convergence, differentiability, integrability. Theorem: Suppose C={c_n: n = 0,1,2,...} and
sum_{k=0}^{oo} c_k R^k converges for some R>0. Then
(i) sum_{k=0}^{oo}
c_k x^kconverges absolutely for |x|<R.
Furthermore, (ii) if f_C(x) = sum_{k=0}^{oo} c_k x^k, then f_C is differentiable and for some , . .
See Sensible Calculus XI.A
Proof of (i):
Since the series sum_{k=0}^{oo} c_k R^k converges, we can
apply the divergence test to conclude that c_kR^k ->0
as k -> oo.
Thus there is some number B that is larger
than c_kR^k for any .
Now suppose that
so that for some , . Then
so
.
Now this last
inequality allows us to compare the series of absolute values to a
geometric series:
.
Since , so the geometric series
converges and therefore by comparison the series
converges absolutely for any in the interval .
We will discuss this proof of result (ii) for derivatives further if time permits.
Proof of (ii): As in the proof of (i) for some b>0, ∣x∣<b<R.
Consider h>0 where |x +h| <b. Then we consider that
f_C(x+h) = sum_{k=0}^{oo} c_k (x+h)^k converges absolutely and thus we
can combine and rearrange the terms of the two series {f_C(x+h) -
f_C(x)}/h = sum_{k=0}^{oo} c_k[ (x+h)^k - x^k]/h Theorem: If f is an analytic function with f=f_C, then C is unique and c_n={f^{(n)}(0)}/{n!}. Remark: With this result, the theory of analytic
functions and the Taylor theory coalesce. Unfortunately, there are
functions that are C^{oo} but not analytic. The usual example is given
by f(x) = e^{-1/x^2} for x \ne 0 and f(0)=0. This function has
f^{n}(0) = 0 for all n, so it cannot be equal to a power series
function.`
The Real Numbers!
First - The rational Numbers:
Given the integers as an
ordered integral domain ("commutative ring with unity and no
zero divisors") consider P= {(a,b): a,b integers and b \ne
0} and the equivalence relation on P where (a,b) ~ (c,d) [in
P] if and only if ad= bc.
As a relation on P, ~ is symmetric, reflexive, and
transitive.
Let [a,b] = { (r,s) in P : (a,b) ~ (r,s) } and show this
partitions P, i.e., (i) every (a,b) in P has (a,b) in [a,b], and
(ii) if [a,b] and [c,d] have some element in common, then
[a,b]= [c,d] - as sets.
The rational numbers Q as a set = {[a,b]: a,b are integers
and b \ne 0.} 5-3
Define operations on Q for addition and
multiplication- [a,b]*_Q[c,d] = [ac,bd] and [a,b]+_Q[c,d] = [ad+bc,bd].
Note: If b,d in Z withb \ne 0 and d \ne 0 then bd \ne 0.
Proposition: These operations are "well defined" and
with them, Q is a field.
Proof: in part: well defined- if [a,b]=[a',b'] and [c,d] = [c',d'] then [ac,bd]=[a'c',b'd']. [0,1]+_Q[a,b] =[a,b]; [1,1]*_Q[a,b]= [a,b] [a,b] +_Q[-a,b] = [0,1]
Finally: Characterize the positive rational numbers
and from this define an order relation on Q. Q+ = {[a,b] : a*b > 0 (in Z)].
Prove: Q+ is well defined. Suppose
[a,b] = [c,d]. [a,b] in Q+ if and only if [c,d] in Q+. [a,b]<[c,d] if and only if [c,d] +_Q[-a,b] in Q+.
The result makes Q an ordered field with a 1:1
function from Z to Q defined by a→[a,1]that
preserves the order and algebraic structures of Z. Defining the real numbers:
Consider the ways we describe real numbers- Notice that the decimal notation gives a sequence of rational
numbers.
Note its connection to infinite series using powers of
10 which converge by comparison with geometric series.
Notice other sequences that
characterize real numbers-
In particular e as the
limit of the sums from the Taylor series and
the limit of the powers: .
To define real numbers we need to eliminate the assumptions
of a limit existing for these sequences of rational numbers.
We use the theory that characterizes convergence with the
Cauchy condition.
This is the start of a more general
definition of a real number using "cauchy sequences" of
rational numbers.
5-5
Let QC = { S={a_n} : where a_n is a rational number for
each n and S is a "rational" Cauchy sequence}
Suppose S={a_n} and T =
{} in QC.
Define an equivalence relation on QC , namely S~T: if for any positive rational number
there is a natural number M where for any k >M, .
Proposition: ~ is a reflexive, symmetric, and transitive
relation.
This leads to the partition of QC into equivalence classes
[S]= {T in QC: S~T}. R = {[S]: S in QC} can then be
used to define operations and an order relation [
inherited from Q] so that ` is an ordered field which
satisfies the least upper bound property. Thus is a complete
ordered field- "the real numbers."
Definition: If a,b in R, a = [{a_n}], b= [{b_n}], we define a+b = [{a_n +b_n}] and a*b=[{a_n*b_n}].
Proposition: {a_n +b_n} in QC and {a_n*b_n}in QC. The operations
a+b in R and a*b in R are well defined. R is a field with these
operations. 0 = [{0,0,0,...}], 1=[{1,1,1,...}] are the additive and
multiplicative identities.
Definition: R+ = {a in R, a= [{a_n}] , there is a natural number M where for if k>M, a_k in Q+.}
We will use the convergence of an and bn and the fact that convergent sequences are bounded to find M. Also we will consider these products as areas of two rectangular regions: one with sides a and b and and the other with sides anand bn.]
found a subsequence that is monotonic decreasing.
In this case x_n - c > 0 so \frac{f(x_n) - f(c)} {x-c} \ge 0. But
this means that f '(c) cannot be negative because there are quotients
as close as we want to f '(c) that are non negative. 
Recall: Definition of an open set. O is open if for
any L in O there is a epsilon > 0 where N(L, epsilon
)sub O.
