Upload
others
View
11
Download
0
Embed Size (px)
Citation preview
Definitions Theorems
Cauchy and Montone Sequences
Dr. Lance Nielsen
Creighton University Department of Mathematics
January 31, 2007
Definitions Theorems
Outline
1 DefinitionsCauchy SequencesMonotonic Sequences
2 TheoremsComments and CautionA Corollary - The Contraction Principle
Definitions Theorems
Outline
1 DefinitionsCauchy SequencesMonotonic Sequences
2 TheoremsComments and CautionA Corollary - The Contraction Principle
Definitions Theorems
Outline
1 DefinitionsCauchy SequencesMonotonic Sequences
2 TheoremsComments and CautionA Corollary - The Contraction Principle
Definitions Theorems
Definition of Cauchy Sequence
DefinitionLet {an}∞n=1 be a sequence of real numbers. We say that{an}∞n=1 is a Cauchy sequence if, for every ε > 0, there is anN ∈ N such that |am − an| < ε whenever m, n > N.
There are sequences that are not Cauchy. For example,an =
∑nk=1
1k is not a Cauchy sequence.
The idea of Cauchy sequence is very closely related to theidea of completeness.
Definitions Theorems
Definition of Cauchy Sequence
DefinitionLet {an}∞n=1 be a sequence of real numbers. We say that{an}∞n=1 is a Cauchy sequence if, for every ε > 0, there is anN ∈ N such that |am − an| < ε whenever m, n > N.
There are sequences that are not Cauchy. For example,an =
∑nk=1
1k is not a Cauchy sequence.
The idea of Cauchy sequence is very closely related to theidea of completeness.
Definitions Theorems
Definition of Cauchy Sequence
DefinitionLet {an}∞n=1 be a sequence of real numbers. We say that{an}∞n=1 is a Cauchy sequence if, for every ε > 0, there is anN ∈ N such that |am − an| < ε whenever m, n > N.
There are sequences that are not Cauchy. For example,an =
∑nk=1
1k is not a Cauchy sequence.
The idea of Cauchy sequence is very closely related to theidea of completeness.
Definitions Theorems
Outline
1 DefinitionsCauchy SequencesMonotonic Sequences
2 TheoremsComments and CautionA Corollary - The Contraction Principle
Definitions Theorems
Definition of Monotonic Sequence
DefinitionLet {an}∞n=1 be a sequence of real numbers. We say that{an}∞n=1 is a monotonic increasing sequence if an < an+1 for alln ∈ N. (It is possible for a sequence to be monotonic increasingfor n ≥ N0 for some N0; i.e. it becomes monotonic for largeenough n.) We say that {an}∞n=1 is monotonic decreasing ifan > an+1 for all n ∈ N. (As before, it is possible for a sequenceto become monotonic decreasing for large enough n.)If we replace the strict inequalities above with a "≤" or "≥", wethen say that the sequence is nondecreasing or nonincreasing,respectively.
Definitions Theorems
Examples of Monotonic Sequences
an = nn+1 : We show that this sequence is monotonic
increasing. Since nn+1 < n+1
n+2 is equivalent to 0 < 1(n+1)(n+2)
and this inequality is certainly true for all n ∈ N.bn = n+1
2n+1 : We show that this sequence is monotonicdecreasing. Since n+1
2n+1 > n+12n+2 is equivalent to n
2n+2 > 0 andthis last inequality is true for all n, the sequence ismonotonic decreasing.
cn = n7−n2+1n+e2n : This sequence is not monotonic initially but
is eventually monotonic. Can you figure out if it’s eventuallyincreasing or decreasing? How?
Definitions Theorems
Examples of Monotonic Sequences
an = nn+1 : We show that this sequence is monotonic
increasing. Since nn+1 < n+1
n+2 is equivalent to 0 < 1(n+1)(n+2)
and this inequality is certainly true for all n ∈ N.bn = n+1
2n+1 : We show that this sequence is monotonicdecreasing. Since n+1
2n+1 > n+12n+2 is equivalent to n
2n+2 > 0 andthis last inequality is true for all n, the sequence ismonotonic decreasing.
cn = n7−n2+1n+e2n : This sequence is not monotonic initially but
is eventually monotonic. Can you figure out if it’s eventuallyincreasing or decreasing? How?
Definitions Theorems
Examples of Monotonic Sequences
an = nn+1 : We show that this sequence is monotonic
increasing. Since nn+1 < n+1
n+2 is equivalent to 0 < 1(n+1)(n+2)
and this inequality is certainly true for all n ∈ N.bn = n+1
2n+1 : We show that this sequence is monotonicdecreasing. Since n+1
2n+1 > n+12n+2 is equivalent to n
2n+2 > 0 andthis last inequality is true for all n, the sequence ismonotonic decreasing.
cn = n7−n2+1n+e2n : This sequence is not monotonic initially but
is eventually monotonic. Can you figure out if it’s eventuallyincreasing or decreasing? How?
Definitions Theorems
A Theorem on Monotonic Sequences
TheoremSuppose that {an}∞n=1 is a monotonic increasing sequence thatis bounded above. Then there is a real number L such thatlimn→∞ an = L. Similarly, if {bn}∞n=1 is a monotonic decreasingsequence that is bounded below, then there is a real number Msuch that limn→∞ bn = M.
Proof.We assume, without loss of generality, that {an} is monotonicdecreasing and bounded below. Let a = inf {an : n ∈ N}. (Whydoes a exist?) Given ε > 0, note that a + ε is not a lower boundfor the sequence. Hence there is an n0 such that an0 < a + ε.But, for any n ≥ n0, an ≤ an0 < a + ε. Moreover, a is a lowerbound so a− ε < a ≤ an < a + ε which implies that |an − a| < εfor all n ≥ n0.
Definitions Theorems
Example
Define an recursively by an+1 = 12
(a2
n + 1), a1 = 0.
Claim: {an} is an increasing sequence.Proof: We calculate as follows.
an+1 − an =12(a2
n + 1)− an =12
(an − 1)2 ≥ 0
Claim: {an} is bounded.Proof: We proceed by induction. If |an| ≤ 1, then|an+1| ≤ 1
2 · 2 = 1 and also, |a1| ≤ 1.We have proved that an is an increasing and boundedsequence. Therefore it has a limit. What is this limit?
Definitions Theorems
Example
Define an recursively by an+1 = 12
(a2
n + 1), a1 = 0.
Claim: {an} is an increasing sequence.Proof: We calculate as follows.
an+1 − an =12(a2
n + 1)− an =12
(an − 1)2 ≥ 0
Claim: {an} is bounded.Proof: We proceed by induction. If |an| ≤ 1, then|an+1| ≤ 1
2 · 2 = 1 and also, |a1| ≤ 1.We have proved that an is an increasing and boundedsequence. Therefore it has a limit. What is this limit?
Definitions Theorems
A Theorem on Cauchy Sequences
TheoremLet {an}∞n=1 be a sequence of real numbers. The sequence isCauchy if and only if it converges to some limit a ∈ R.
Proof:(Convergent implies Cauchy) Assume that the sequenceconverges to a limit a. Take any ε > 0. Then there is a positiveinteger N such that if n ≥ N, then |an − a| < ε/2. Hence
|aj − ak | ≤ |aj − a|+ |a− ak | < ε/2 + ε/2 = ε
if j , k ≥ N. It follows that the sequence is Cauchy. We addressthe implication Cauchy implies convergence on the next slides.
Definitions Theorems
Proof, Continued
Conversely, suppose that the sequence is Cauchy. Define theset S = {x ∈ R : x < aj except for finitely many j}.Since thesequence is bounded (why?), say by M > 0, we know thatevery term in the sequence is bigger than −M. Therefore−M ∈ S. Also, every term of the sequence is smaller than M,so that S is bounded by M. Hence we see that S is a boundedand nonempty subset of real numbers and so it has a leastupper bound, say a. We claim that an → a as n →∞. Let ε > 0be given. Choose N so that
|aj − ak | <ε
2
if j , k ≥ N. Then, in particular, |aj − aN+1| < ε2 . It follows that,
for j ≥ N,aj > aN+1 −
ε
2for j ≥ N.
Definitions Theorems
Proof, Continued
Thus, aN+1 − ε2 ∈ S, and we have that a ≥ aN+1 − ε
2 . We alsohave that
aj < aN+1 +ε
2for j ≥ N and therefore aN+1 + ε
2 /∈ S. Thus a ≤ aN+1 + ε2 . But
now, looking at the inequalities above, we have that|a− aN+1| < ε
2 . We finally obtain∣∣a− aj∣∣ ≤ |a− aN+1|+
∣∣aN+1 − aj∣∣ <
ε
2+
ε
2
for j ≥ N and the proof is finished.
Definitions Theorems
Outline
1 DefinitionsCauchy SequencesMonotonic Sequences
2 TheoremsComments and CautionA Corollary - The Contraction Principle
Definitions Theorems
Further Comments About Cauchy Sequences
1 It is often difficult to tell whether a sequence is Cauchy.The fact that the convergent sequences are the Cauchysequences is usually used when trying to prove generaltheorems about sequences.
2 Keep in mind that for a sequence to be Cauchy we musthave, for arbitrary ε > 0, |ak − al | < ε for all k , l > N. It isnot sufficient that the difference in successive terms besmall. That is to say, even if |an+1 − an| < ε for all n > N,the sequence may still diverge. Suppose an =
√n. It is
clear that this sequence diverges. However,
|an+1 − an| =(√
n + 1−√
n)(√
n + 1 +√
n)√n + 1 +
√n
<1
2√
n
and so the difference in successive terms can be madearbitrarily small.
Definitions Theorems
Outline
1 DefinitionsCauchy SequencesMonotonic Sequences
2 TheoremsComments and CautionA Corollary - The Contraction Principle
Definitions Theorems
Contraction Principle
There is, however, a result that does give a test forconvergence stated in terms of the size of successivedifferences of successive terms of the sequence.
Corollary
Let {an}∞n=1 be a sequence of real numbers. If r is any realconstant such that 0 < r < 1 and |an+2 − an+1| ≤ r |an+1 − an|,then {an}∞n=1 converges.
Definitions Theorems
Proof of the Contraction Principle
We show that the sequence is Cauchy. It will then follow at oncethat it converges. Note that if n > m, the difference an − am canbe written as a telsecoping sum of successive differences:
an − am =n−1∑k=m
(ak+1 − ak ). (1)
If we apply the hypothesis k − 1 times, we can approximate|ak+1 − ak | in terms of r and the initial difference |a2 − a1|.
|ak+1 − ak | ≤ r |ak − ak−1| ≤ r2|ak−1 − ak−2| ≤ · · ·≤ r k−1|a2 − a1|
(2)
Definitions Theorems
Proof Continued
Therefore, by the triangle inequality,
|am − an| =
∣∣∣∣∣n−1∑k=m
(ak+1 − ak )
∣∣∣∣∣≤
n−1∑k=m
|ak+1 − ak | =n−1∑k=m
r k−1 |a2 − a1| .
(3)
Now, factor rm−1|a2 − a1| out of the sum and adjust the index toobtain
Definitions Theorems
Proof Continued
n−1∑k=m
r k−1|a2 − a1| = rm−1|a2 − a1|n−m−1∑
k=0
r k−1
= |a2 − a1|rm−1 · 1− rn−m
1− r
< |a2 − a1|rm−1 · 11− r
,
(4)
where the inequality follows from Example 4.2.2. But rn → 0(problem 4.2.4) and therefore we can, given ε > 0, find an Nsuch that if m − 1 > N, then rm−1 < ε(1− r)/|a2 − a1|. Usingthis bound for rm−1, it follows that if m − 1 > N, then|an − am| < ε and the proof is done.