Math 104 Midterm Notes
1. Natural number
1 belongs to N
if n belongs to N, then its successor n+1 belongs to N
1 is not the successor of any element in N
if n and m in N have the same successor, then n=m
S=A subset of N which contains 1, and which contains n+1 whenever it
contains n
MI, n as the index of the proposition, S is the set which is true
2. Algebraic number, Q<A<R
satisfies a polynomial equation anxn +…a0=0
r=p/q, p, q|a0 and q|an, proof multiply by qn on both sides
49x4-56x2+4=0, possible slns: ±1, 1/7, 1/49, 2, 2/7, 2/49, 4, 4/7, 4/49
3. if x is irrational, x1/a is also irrational, pf by contrapositive
4. Field: a system that has more than 1 element and satisfies the 9 properties
a+(b+c)=(a+b)+c
a+b=b+a
a+0=a
a+(-a)=0
a(bc)=(ab)c
ab=ba
a*1=a
for each a≠0, aa-1=1
a(b+c)=ab+ac
5. Ordering: O
a≤b or b≤a
a≤b and b≤a => a=b
a≤b and b≤c => a≤c
a≤b, a+c≤b+c
a≤b and 0≤c => ac≤bc
6. filed theorems 3.1
a+c=b+c => a=b
a*0=0
(-a)b=-ab
(-a)(-b)=ab
ac=bc and c≠0 => a=b
ab=0 => a=0 or b=0
7. Ordered field theorems: 3.2
a≤b, -b≤-a
a≤b and c≤0 =>bc≤ac
0≤a and 0≤b=> 0≤ab
0≤a2 by cases
0<1
0<a => 0<a-1
0<a<b => 0<b-1<a-1
8. Absolute values:
|a|=a if 0≤a, |a|=-a if a≤0 (3.3)
0 ≤ |a| (3.5→)
|ab| =|a|*|b|
|a+b|≤|a|+|b|, triangle inequality
||a|-|b|| ≤ |a-b| by triangle inequality on |a|≤|a-b|+|b| and |b|≤|b-a|+|a|
9. Bounded
Sup S: least upper bound
Inf S: greatest lower bound
Sup S and inf S need not belong to S
x=Max S x∈S ∧ x=sup S
10. Completeness Axiom:
A = a non-empty subset of S of R that is bounded above has a least upper
bound. i.e: sup A exists ∧ sup A∈R
11. Important consequences
Theorem 11.1 (Archimedean Property): a, x ∈R+, then there exists a positive
integer n0 s.t. n0x>a (pf of Q is easier)
Proof: Let (xn) be n0x. Suppose to the contrary s0 = sup xn≤a. s0 -1/2x is not
the least upper bound of (xn) → ∃ N, s.t. Nx>s0 - 1/2 x → (N+1) x>s0+1/x >s0
∧ (N+1) x ∈(xn) → contradiction
Remark: We can have an arbitrary a, the purpose of the proof is to show that
n0x has no upper bound.
Remark: We always take the advantage of the LEAST upper bound
Theorem 11.2 (Denseness of Q in R): a<b, there exists at least one rational
number r with a<r<b
proof: b-a>0 →∃n st n(b-a)≥2>1 (11.1)
→ pick ℓ∈Q s.t. ℓ > max(|na|, |nb|) ∧ A={na<k≤ℓ}, A≠∅ (ℓ∈A) *
→ m=min A, m-1 not∈A, m-1<na→ na<m≤na+1<nb (n(b-a)>1)
→ a < m/n < b
Remark: We need to show that A is finite and nonempty. We don’t really need
ℓ > max(|na|, |nb|) to guarantee the nonemptyness of A, but this will definitely
work and save more proof time.
Quote: objects are not important for what they are, it’s their relationships that
matter
12. Definition 12.1: Dedekind cut: x^ ={r∈Q: r<x}, then sup x^ =x (extend Q to R)
x^ is a subset of Q but x^≠ ∅, x^≠Q
If r∈ x^, s∈Q and s<r, then s∈ x^ (Ordering: there is no smallest member)
x^ has no largest element, i.e. max x^ does not exist
R={x^: x^ is a Dedekind cut}
x^≤y^ iff x^ ⊂ y^ (the whole set with infinite members is a real number)
x^+y^, x^*y^ …
sup A=∪ x^ (for x^∈A)
Completeness axiom: take a set A∈R and assume that A is bounded above,
there exists a cut M^ in R such that if x^∈A, then x^≤M^ i.e x^⊂M^. indeed
sup A= U{x^:x^∈A}. note that since each x^∈M^, we have sup A⊂M^ and
sup A≠ Q
x^*y^: this can be defined by first considering positive x^, y^ and avoid r*s
with r∈x^, s∈y^ and both r and s negative. We omit the details
-a: let A=negative complement of Dedekind cut a, -a={y: y<x for some x∈A}
Remark: if a is not ∈Q, we don’t have to set y<x.
13. Sequences
Definition 13.1: A sequence xn converges to x, if for every +ve ϵ, there exists
an integer N such that if n>N, then |xn-x|<ϵ, lim xn=x (construct n by ϵ)
Xn =√(n2+n) –n, lim xn =1/2
Quote: 3 tricks: 1. Cauchy Schwarz. 2. Chebyshev. 3. Integration by parts
xn=(-1)n/5, let lim xn=a, ∃1/10 |1/5 –a |<1/10, |-1/5-a|<1/10, triangle inequality
lim xn =x, lim yn=y, lim xn+yn=x+y, (use N and ϵ)
Lemma 13.1: if lim xn=x, then xn is bounded, |xn|≤M for all n.
pf: we can find N such that n≥N =>|xn-x|<1, for these xn, we have -|x|-1≤x1<xn≤x+1≤|x|+1, thus if M=max{|x1|, …, |xn|, |x|+1}→ -M≤xn≤M for all n
lim xn =x, lim yn=y, lim xn*yn=x*y
|xnyn-xy|=|(xn-x)yn +x(yn-y)|= |xn-x||yn| +|x||yn-y| ≤ M|xn-x| +|x||yn-y| (lemma)
if lim xn=x and x≠0 & all xn≠0, then lim 1/xn=1/x →yn/xn =yn* (1/xn)
lemma 13.2: if lim xn=x and x≠0, then there exists λ>0 and N0, such that if
n>N0, then |xn|>λ. (inf |xn| >0 for n>N0)
Proof: there exists n0 such that n≥n0 implies that |xn|(>1/2|x|)>0
→|x|-|xn|≤|xn –x|≤1/2|x|, shown
14. Some useful sequences
Lim an=0 if |a|<1.
Proof: |an|=|a|n=1/(1+z)n →(1+z)n=1+nz+n(n+1)/2z2+…=1+nz (Binom Series)
→ (1+z)n≥nz. In summary, |an|=1/(1+z)n≤1/nz
(strategy, use boundaries)
Lim n1/n=1. Pf: n1/n-1=zn. we need to show that lim zn=0.
From n1/n-1=zn>0 when n>1
We learn that n=(1+zn)n, linear progress ≠ exponential progress
Recall that n=(1+z)n=1+nz+n(n+1)/2z2+…zn>n(n+1)/2z2 *
→ n=(1+zn)n >n(n+1)/2 zn2 → hence zn2<2/(n+1)
Remark: Binomial series largely simplified the power and the later terms
dominate the sum when n is large no matter how small z is (Archimedean).
Lim a1/n=1 if a>0. If a=1 obvious, if a>1, 1<a1/n<n1/n for n>a.
Hence 0<a1/n-1<n1/n-1, 0<a<1, reciprocal of the limit theorem
15. Definition: (an) is non-decreasing if an≤an+1 for all n
Similarly we can have non-increasing sequences.
Non-decreasing + non-increasing= monotonic
Non-decreasing +bounded above ⋀ Non-increasing+ bounded below → limits
Theorem 15.1: For non-increasing sequence A={a1, a2 …} lim A = Sup A
(With some variation, all bounded monotone sequences converge.)
Pf: Sup A exists by the completeness axiom.
Set a=sup A. for n≥N, we want 0≤a-an<ϵ. a- ϵ can’t be an upper bound → ∃
N, a- ϵ<an<a for n>N→ 0≤a-an< ϵ
an=1+1/2!+…1/n! = e
an=(3n+1)/(2n+1) =3/2-(1/2)/(2n+1), an is increasing and bounded above by 3/2
→ it has a limit
16. Definition: lim an=+∞ / -∞ if for any A, ∃ n≥N → an≥A ⋁ ∃ n≤N → an≤A
an=nn, lim an=infinity.
an=(-1)nn has no limit
an=(n+1)2. Given A, if n>√A, then an=(n+1)2>A
17. non-decreasing → bounded above limit = sup{a1, …}, or not bounded limit= +∞
Not bound above→ infinity: Assume an is not bounded above, an is nondecreasing. Pick an A, since an has no bound from above, there exists n0 such
that A<an0. for n≥n0→ A<an0≤an (non-decreasing)
Theorem17.1: Lim an=+∞, then lim(1/an)=0. With an>0, converse is also true
n≥n0 → an>A>0 →
n≥n0 → 0<1/an<1/A <ϵ (Choose ϵ to be 1/A)
an=1+1/1!+1/2!+…1/n! is bounded: an≤1+1+1/2+…1/2n-1
=1+(1-1/2n)/(1-1/2)≤1+2=3
18. Definition: lim sup an, lim lim an represent the extreme behaviors of the sequence
lim sup an = lim sup{an:n>N}
{an^} is a non-increasing sequence (throw more elements away, monotonic)
→ has a limit
lim inf an = lim inf{an:n>N}
{an} is a non-decreasing sequence → has a limit
Remark: when we say lim {an} =∞, we mean {an} diverges to +∞
-∞≤ lim inf an≤lim sup an≤∞
an=(-1)n/n →lim sup an=0, lim inf an=0.
an^=sup{an: n>N}=sup{(-1)N+1/(N+1), (-1)N+2/(N+2) …}
={1st term N if N is odd, 2nd if N is even}
an={1st term N if N is odd, 2nd if N is even}
Theorem 18.1: a=lim an exits iff lim sup an=lim inf an=a
(a) lim an=+∞ → we can find N st n>N → an>A
→ inf{an: n>N}≥A →lim inf an= lim an≥A (non-decreasing)
→ lim inf an=+∞
if lim inf an=+∞ → lim an=+∞ ( Pf: an≥inf {an}>A for n>N )
Similarly we can show that lim sup an=-∞ iff lim an=-∞
(b) Lim an=a is finite → we want lim sup an= lim inf an=a
Pick ϵ>0 and find N such that n>N →|an-a|<ϵ
→ -ϵ<an-a<ϵ→ a-ϵ<an<a+ϵ
→an^=sup{an:n>N}≤a+ϵ → lim an≤a+ϵ (an^ is non-increasing)
→ lim sup an≤a+ϵ
On the other hand → lim inf an≥a-ϵ (an is non-decreasing)
a-ϵ≤lim inf an≤lim sup an≤a+ϵ (pf by contradiction)
(c) Converse: a=lim sup an= lim inf an, a is finite → lim an=a
Pick ϵ>0, for n>N |aN^-a|<ϵ → a≤aN^≤a+ϵ → an≤a+ϵ for n>n0
→ a-ϵ<an≤a for n>n0 (non-decreasing)
→a-ϵ <an<a+ϵ → |an-a|<ϵ
19. Definition: Cauchy: {an} is a Cauchy sequence if for every e>0, we can find N so
that if n, m>N then |an-am|<ϵ
an=1+1/3 …+1/3n
an+m-an=1/3n+1(1+1/3+…+1/3m-1) ≤1/3n+1*3/2=1/2 *1/3n
if m>n → 0≤am-an ≤a/2*1/3n
|an+1-an|≤qn where 0<q<1 → A sequence is Cauchy
|an+m-an|≤|an+m-an+m-1|+|an+m-1-an+m-2| + …+|an+1-an|
≤qn+m-1 + …qn = qn(1-qm)/(1-q)≤qn/(1-q) (note that this is m-free, n only)
We can find N st qn/(1-q)<e since lim qn =0 and 1-q is constant
Remark: Note that (sn)={si=|ai+1-ai|} forms a geometric series
Theorem 19.1: A sequence is Cauchy iff it is convergent.
1. For every ϵ >0, we can find N so that n≥N → |xn-x|<ϵ /2
Now if both n, m≥N, then |xn-xm| ≤ |xn-x|+|x-xm|<ϵ /2+ϵ /2=ϵ
2. Suppose that xn is Cauchy, let us first argue that such a sequence (xn) must
be bounded.
n. m≥n0 → |xn –xm|<1. In particular, |xn-xn0|<1 for n≥n0, or
xn0-1<xn<xn0+1 for n≥n0,
Set A=max{xn0+1, x1, .. xn0-1}, a=min{xn0-1, x1, .. xn0-1},
In summary, a≤xn≤A for all n. This means that (xn) is bounded
To show Cauchy sequence (xn) is convergent, it suffices to show that
x^=lim sup xn = lim inf xn= x. ({xn} is bounded→ x^ and x are finite)
pick ϵ >0 and find N s.t. n, m>N → |xn-xm|<ϵ or – ϵ<xn-xm<ϵ →
xm-ϵ <xn<xm+ϵ ∀ m>N → x^-ϵ≤xn≤ x+ϵ → x^=lim xn = x
Remark: note that the difficulty is that we don’t have a fixed reference
point for lim xn in this case, hence we need to use x^ and x instead.
20. If the limit does not exist, then x<x^. However, we can find a “subsequence” of
xn which converges to x^ and another subsequence which converges to x.
We may have a subsequence which converges to a with x<a<x^
xn=(-1)n2(-1)^n (n) subsequences converge to 0 and ∞
xn= sin(nπ/2): xn^: 1 … xn: -1 …
Definition 20.1: let (xn) be a sequence and let n1<n2< …<nk< …, be an
increasing sequence of +ve integers, yk= xnk is a subsequence of (xn)
xn=(-1)n2(-1)^n (n) nk=2k → yk=x2k=22k is a subsequence
lemma 20.2: let b<c, then there are infinitely many rationals in the interval (b, c)
Proof by constructing a sequence of infinite members where rn+1∈(rn, r^)
Claim 20.3: A sequence (xn) s.t. for every rational r, we can find at least one n
so that xn=r, if a is any real number, then there exists a subsequence (xnk) such
that lim xnk=a
Proof: first pick a rational xn1 in the sequence so that 0<|xn1-a|<1.
For k≥2, let ϵk=min{1/k, |xk-1-a|}>0, Pick xnk after xnk-1 s.t. xnk∈ (a-ϵk, a+ϵk).
By lemma 20.2 we know such xni will definitely exist since ∃ infinite rationals
between any different real number. (Otherwise we can find xN for N finite s.t.
after xN ∃ no rationals in between → number of rationals ≤N → contradiction)
In this way, we can construct a subsequence xn1, xn2, xnk, … so that |xnka|<1/k. Evidently, lim xmk=a
Remark: Here, we make sure both that xmk reaching a and (|xmk| is decreasing
[not necessary in this case]).
note that if xn=sin(πnr) with r∈Q, then xn is periodic. In fact if r=p/q with p∈
Z, q∈N, then xn is periodic of period 2q.
xn=sin(πnr), for r∈R !∈Q, we can find a subsequence xnk so that lim xnk=a,
-1≤a≤1
Theorem 20.4: if lim xn=a and (xnk) is a subsequence, then lim xnk=a
Choose N so that |xn-a|<e for n>N. now if k>N, then nk>N and
as a result |xnk-a|<e, (any subsequence)
Joke: let me give you a break now… some student asks, can we vote for
whether we should have the break? Prof: Yes you can vote, but it has already
been decided…
Theorem 20.5: Every sequence xn has a monotone subsequence *
Pf: xm is a dominant term of the sequence if xm>xn for all n>m.
Case 1: there are infinitely many dominant terms: xn1, xn2, xn3 … with
n1<n2<n3… and xn1>xn2>xn3…
Case 2: ∃ at most finitely many dominant terms: This means that there exists
some n0 so that if n≥n0, then xn can’t be dominant → Hence in particular, xn0
is not dominant → for some n1>n0, xn1≥xn0 → repeat for n1 and onwards
(Bolzano-Weierstarass)Every bounded sequence has a convergent subsequence.
Proof: every bounded monotone subsequence has a limit from (sup or inf)*
Lemma 20.7: let (xn) be any sequence and let (xnk) be a convergent
subsequence. If lim xnk=a, then lim inf xn=x≤a≤x^=lim sup xn.
Pf: it suffices to show that for any subsequence xnk,
x= lim inf xn≤lim inf xnk = lim sup xnk≤ lim sup xn=x^
lim sup xnk= lim sup{xnk: k>N}≤lim sup{xn:n>N} simply k>N →nk>N.
Claim 20.7a: aN≥bN for all N→ lim aN≥lim bN
if lim aN=a, then aN≤a+e for large N. but bN≤aN, so bN≤a+e for large N.
Hence b≤a+e if b=lim bN → we deduce that b≤a
Similarly proofs go with lim inf
Theorem 20.8: if lim sup xn= x^, then ∃ a monotone subsequence (xnk) so that
lim xnk=x^. same for lim inf xn=x
Proof of theorem: (subsequence has limit to the lim sup)
Case 1: we argue that if x^=lim sup xN=-∞, then the conclusion is almost
obvious. → x=-∞ and lim xn=-∞. By our 20.4, we know any subsequence (xnk)
of (xn) necessarily diverges to -∞. So, lim xnk=-∞. (done)
Case 2: assume that x^≠-∞, there are infinitely many dominant terms xn1,
xn2, … with n1<n2< … , indeed if xN^=sup {xn:n>N}, then
xn1^= sup{xn:n≥n1}= xn1, xnk^=xnk for every k. since lim xn^=x^, we deduce
that lim xnk= lim xnk^= x^ (subsequence’s limit=sequence’s limit). we are done
in this case because (xnk) is strictly decreasing.
Remark 1: lim sup xn = x^, hence lim sup xnk = x^
Remark 2: ∃ dominant term → the sequence is bounded above
Case 3: there are at most finitely many dominant terms. Hence there exists n0
so that if n≥n0, then xn is not dominant. We can find m>n0 with xm≥xn0. Let
us pick a sequence (yn) so that yn increases to x^ as n→∞.
If x^ is finite: choose yn to be x^-1/n otherwise set yn=n. Evidently, we can find
m’>n0 so that xm’>yn0. If not, then xm’≤yn0 for all m’>n0. But this means
xn0^=sup{xm’:m’>n0}≤yn0. So, x^≤yn0. (Contradiction).
In summary, we have m and m’>n0 with (xm≥xn0 ^ xm’>yn0)*, choose
n1=max(m, m’) so that n1>n0 and * holds (no more dominant terms, therefore
no overshot)
We continue this. Find n2>n1 so that xn2≥xn1 ^ xn2>yn1
Inductively, we can find n1<n2… so that xnk≥xnk-1 ^ xnk≥ynk-1
Evidently (xnk) is monotone. Moreover lim xnk≤x^.
Finally lim xnk≥lim ynk=x^. thus lim xnk=x^ → lim xnk=x^
Remark: We are not afraid of overshooting because
Quote: 3 steps of mathematics: feel good about it→believe it→prove it
21. Given a sequence (xn), we set S to be the set of the limit points of (xn).
We say a is a limit point of (xn) if there exists a subsequence (xnk) with lim
xnk=a.
If xn is convergent iff S={x} where x=lim xn (convergent = have a limit)
Theorem: S⊂ R U{-∞, +∞}
S ≠ empty, pf: every sequence has a monotone subsequence, which has a limit
x^, x∈ S (might be the same point, subsequence with these limits exist)
S⊂ [x , x^] (by lemma 20.7)
Max S=x^, min S=x (by above 2)
Joke: it’s really a shame to stop at this moment at the middle of the proof.
Theorem 21.1: If sn∈S, sn∉{-∞, ∞} for all n, and lim sn=s. then s∈S
i.e: Limit of limits is the limit (S is closed)
Proof: (yn∈S, yn≠∞ or -∞, yn→y, y∈S) <Diagonal method>
We wish to find n1<n2<…nℓ<… so that lim xnℓ =y
To do so, we find xnℓ so that |xnℓ-yℓ |<1/ℓ * (yℓ = limit of ℓ’s subsequence)
If we manage to find such subsequence, then we are done because given e>0,
we can find N so that if ℓ >N, then |yℓ-y|<e/2 (definition of y).
Now pick ℓ0 so that |xnℓ-yℓ|<1/ℓ<1/ℓ0<e/2, ℓ>ℓ0 (definition of yℓ). Hence for
ℓ >max{ℓ0, N}, |xnℓ-y|=|xnℓ-yℓ+yℓ-y|≤|xnℓ-yℓ| + |yℓ-y|<1/ ℓ+e/2<e/2+e/2=e
we construct xnℓ’s by induction. First claim that there exists n1 so that |xn1yn1|<1. Now suppose n1<n2< … nℓ are chosen satisfying |xnj-yj|<1/j …
Remark: Since we can find such a subsequence which approaches y, its limit
should be included in S.
22. Suppose that lim sup xn=x^ and lim yn=y with 0<y<+∞ → lim sup(xn yn)=yx^
Proof:
Case 1: -∞<x^<∞.
Prove: x^y≤ lim sup(xnyn)
For sure we can find a subsequence xnk so that lim xnk=x^. Now look at (xnkynk)
which is a subsequence of (xnyn). We certainly have lim xnkynk=x^y (limit
theorem)→ lim sup xn*lim sup yn= x^y= lim xnkynk ≤lim sup(xnyn). (by
lemma 22.1) *
∃ N0, y>0 →yn>|y|/2 =y/2 for n>N0
Prove: x^y≥ lim sup(xnyn)
x^=lim sup[(xnyn) (1/yn)]≥lim sup(xnyn) lim 1/yn=lim sup(xnyn) (1/y) (by *)
→ x^y≥ lim sup(xnyn)
Joke: Good time for the break, but it’s a shame… You can vote for this, but by
constitution this is not allowed.
Very active 3 year-old son can spoil everything of mathematics at my home…
Case 2: Assume x^=+∞. We only need to show that lim sup xnyn≥+∞
Again pick xnk with lim xnk=+∞. Obvious that lim xnk ynk=+∞. Give A>0,
choose k0 so that k>k0 → xnk>A. yn is convergent →yn is bounded → for
large n, y/2<yn<3y/2 → for large k, we have xnk>A, y/2<ynk<3y/2 →
xnkynk>A*y/2. A is arbitrary→ we deduce limit is +∞.
Remark: note that lim xnk ynk=+∞ is not part of the limit theorem
Case 3: lim sup xn=-∞→ lim xn=-∞ → lim(-xn)=+∞, lim yn=y>0 →
lim(-xnyn)=+ ∞ → lim(xnyn)=- ∞ → lim sup(xnyn)=-∞
y>0 is necessary
y<0 →lim inf(xnyn)=yx^, lim sup(xnyn)=yx
y=0 →we don’t have enough information.
This is only untrue if x^=+∞ or -∞, since 0*∞ can be anything
Note that if x^ is finite, lim xn yn =0 →lim sup xn yn= lim inf xn yn=0
If we only assume lim sup yn=y with y>0, not valid eg. xn={1, 2, 1, 2…} and
yn={2, 1, 2, 1…}
23. Metric spaces: the notion of convergence for n-tuples (x1, x2 …xn) or even
sequences.
So far we know how to define convergent sequences of numbers. For this, we
needed to check that |xn-x| is small where xn is the n-th member of sequence.
|xn-x| is nothing other than the distance of xn from x.
We can readily generalize the notion of convergence wherever we have a
notion of distance. A “space” equipped with a notion of distance is called a
metric space.
Given a space X, a metric d is a function from X*X into R with the following
properties:
a. d(x, y)≥0, d(x, x)=0, and if d(x, y)=0 →x=y (Positivity)
b. d(x, y)=d(y, x) (commutativity)
c. d(x, z)≤d(x, y)+d(y, z) (triangle inequality)
Examples:
(i)
X=Q and d(x, y)=|x-y|
(ii)
X=R and d(x, y)=|x-y|
(iii) Rk={x*=(x1, x2, x3… xk), xi∈R},
d1(x*, y*) = √(x1-y1)2+…
d2 = |x1-y1| +|x2-y2|
d3= (|x1-y1|p …)1/p
(iv)
X= {(xm) a sequence of R}
d4= ∑ 2-n |xn-yn|/(|xn-yn|+1)
Each element converges to 0 → d4 is finite
(v)
Prove d1 is a metric
Norm d=||x*-y*||, now the triangle inequality means
||x*-z*||≤||x*-y*||+||y*-z*||
Equivalently ||a+b||≤||a||+||b||, square both sides and
Using Cauchy-Schwartz <a, b> ≤||a|| ||b||
||a+b||2≤||a||2+2||a|| ||b|| +||b||2 =(||a||+||b||)2
Proof of Cauchy-Schwartz
t∈R, we have 0≤||a+tb||2= ||a||2+2t<a, b>+t2||b||2
Determinant=C2-AB≤0
24. In a metric space (X, d) we can talk about convergent (Cauchy) sequences:
We say lim xn=x if for every e>0, we can find N so that n>N → d(xn, x)<e
We say xn is Cauchy if for every e>0, we can find N so that n, m>N → d(xn,
xm) <e
We say (X, d) is a complete metric space if every Cauchy sequence in X is
convergent to some element in X. (i.e no gap)
a. (Q, d2) with d2(x, y)=|x-y| is not complete
b. (R, d2) is complete.
c. (R, d1) is complete.
d. (Rk, d1) is complete (k-tuples = k-d Euclidean space)
25. A set A is a bounded subset with ((X, d) a metric space), if ∃ a constant C s.t. d(x,
y)≤C for all x, y∈A
Alternatively, A is bounded if A⊂ {y:d(y, x0)≤C} for some x0∈X and a
positive C.
B(x, r)={y: d(y, x)<r}, D(x, r)={y: d(y, x)≤r}
B(x, r) with d1 →circle; B(x, r) with d2→diamond; B(x, r) with d(x, y) =
max{|x1-y1|, |x2-y2|} →square
Lemma 27.1: A sequence xn is a sequence in Rk→ xn* =(xn1, xn2 …xnk)→
x*=(x1, x2, … xk) iff lim xni=xi for all i (coordinate wise convergent)
xn is Cauchy iff xn1, xn2 … are Cauchy
Proof: max{|x1-y1|, |x2-y2| …}≤d(x*, y*) =√(x1-y1)2+… ≤√k max{|x1-y1|,
2 2
|x -y | …} *
Suppose xn*→x*, for a given e>0, we can find N so that n>N |x*-x*|<e
By *, |xnj-xj|<e, hence xnj→xnj.
Conversely, assume that xnj→xj for j=1…k, choose Nj so that n>Nj →
|xnj-xj|<e/√k, d(xn*, x*)≤√k max{|x1-y1|, |x2-y2| …} ≤√k (e/√k)= e
Theorem 27.2: Rk is complete (every Cauchy sequence has a limit)
Proof: let (xn*) be a Cauchy sequence. By lemma 27.1, each (xnj) is Cauchy.
Hence, by completeness of R, (xnj) converges to a limit xj. Set x*=(x1, … xk).
Then by Lemma 27.1, lim xn*→x*
Theorem 27.3: let xn* be a bounded sequence in Rk. Then xn* has a
convergent subsequence.
Proof: First look at the (xn1), this is bounded → it has a convergent
subsequence (xnj1) with n1<n2<n3… ,
Now look at (xnj2) with index n1, n2 ... This is also bounded→ it has a
convergent subsequence, say (xnjk2) …
Inductively we can find a subsequence in terms of the previous subsequence’s
indices and find the common convergent indices n*r, s.t. all (xn*rj) converges
26. Definition: We say a is an interior point of A if ∃ r>0. So that B(a, r)⊂A.
Ao denotes the set of all interior points. A is open if A= Ao
Let X=R with standard metric → Qo=empty, because any internal (a-r, a+r),
r>0, contains an irrational point (it’s not thick)
Similarly R-Q=I has no interior
Qn1: why?
Ans1: between every pair of different Q there is an I and vice versa
D(a, r)={x, d(a, x)≤r}, D(a, r) o=B(a, r) (remove the boundary)
X={(x1, x2): x1*x2=1}∈R, Xo=empty (thin)
X={(x1, x2): x1, x2>0, x1*x2≥1}, Xo={(x1, x2): x1, x2>0, x1*x2>1}
Empty set is open (trivial) (empty = empty o =empty)
Metric space X is open
{Ui}, i=1..k is a collection of open sets. Union of open set is open
Finite collection of {Ui}, i=1..k, intersection of Ui is open
Remark 1: Above is not true for a infinite collection
Infinite set e.g. 1: Un=(-1/n, 1/n).
∩Un={0} not open (i.e intersection of infinite collection may not be open)
Infinite set e.g. 2: Vn=(-1/n, 1+1/n), ∩Vn=[0, 1] not open
Infinite set e.g. 3: Cantor set (it’s closed)
Remark 2: a closed set in R might be the union of a disjoint sequence of closed
intervals and points: Cantor set: interior=boundary
27. Definition: we say F is closed if X-F{a∈X: a ∉F} is open. We define A~=closure
of A to be the intersection of all closed sets F with A⊂F. A~ =∩{F: F is closed
and A⊂F} (A~ is the most compact closed set of A)
Obviously A~ =A iff A is closed. (A is the tightest F the intersection of all)
Qn2: what is the closure of an open set?
Ans2: the open set ∪ boundary
Quote: If you are an interior creature, you want to see the boundary by
approaching it and estimating it. It’s like a scientific Fiction (but of course you
can’t reach it because you are an interior creature)
My understanding of limit: It’s like human beings trying to figure out what’s
outside earth. They desired to fly, build observatories on the highest mountains,
invent telescopes and shoot satellites. Everything is just to reach out for the
limit though we can never go far beyond a distance from the earth.
Theorem 29.1: the following statements are equivalent (why do we care?)
a. A is closed
b. If xn∈A for each n, and lim xn=x, then x∈A (A’s sequential limits∈A)
(Notion of Closeness)
Proof:
1. a→b: (idea: assume xn∈A^ x∈X-A → xn∈X-A) suppose a. is true
→ let (xn) be a convergent sequence with all xn∈A ^ xn→x
Suppose to the contrary, x is not in A, but X-A is open and x∈X-A.
Hence by openness, we can find r>0, so that B(x, r)⊂X-A. that is B(x,
∅. Since xn→x, we can find N so that n>N→ d(xn, x)<r i.e. xn
r) ∩A=∅
∈B(x, r). This contradicts the fact that all xn belongs to A.
2. b→a: (idea: xn→x∈A ∧ A is not closed→∃ x∉A) suppose b. is true
→ suppose to the contrary, A is not closed→ X-A is not open. → we
can find x∈X-A s.t. B(x, 1/n)!⊂X-A no matter how small r=1/n is (x
has to be on boundary). that is B(x, 1/n) ∩A ≠ ∅. Pick xn∈B(x, 1/n)
∩A for each n. Since xn∈B(x, 1/n), we know that d(xn, x)<1/n. hence
lim xn=x. This contradicts b. because x∉A, xn∈A, xn→x.
Remark: note that we can come up with a definition of convergence with
no direct reference to d: in fact we can say that lim xn=x iff the following
is true. If U is open and x∈U, then there exists N so that n>N →xn∈U
(for any kind of arbitrary tiny U)
Lemma 29.2: Any disk D(a, r)={x: d(a, x)≤r} is closed. (idea: any limit x is
within A) Take a sequence xn in D(a, r). That is d(xn, a)≤r. lim xn=x
Show that d(x, a)≤r
→ d(x, a)≤d(x, xn)+d(xn, a) ≤d(x, xn) + r ∀ n ∧ lim d(x, xn)=0,
→ d(x, a)≤0+r=r
Remark: Take the advantage that xn∈D → d(xn, a)≤r
The set U={x: d(x, a)>r} is open. Simply because this set is X-D(a, r)
Empty is both open and closed → empty U S= S (both S and empty are open)
28. Definition: We say a set C is compact if the following is true: Let {Vn} be a
family of open sets. Assume C⊂∪Vn→ there are finite subcovers n1, n2, …nk so
that C⊂ Vn1 ∪ Vn2 …∪ Vnk (C=∪ Vn for n∈N<∞)
Examples: X=(0, 1), choose Vn=(-1/n, 1-1/n). ∪(Vn for n≥3) =(0, 1). This
has no finite subcover. Vn⊂Vn+1, No matter what k you choose,
U(Vn1…Vvnk) ≠ (0, 1)
Def: A sequence lives in a compact set iff it has a convergent subsequence
Theorem 30.1: If A is compact in a metric space (X, d), then A is closed and
bounded.
Compactness: no matter what cover we come up with, we can find the finite
number of the subcovers *
Remark 1: Compactness (=finiteness) is largely related to closedness. Since
infiniteness is largely related to openness which keeps approaching the
boundary. (However, note that closed ≠ not open)
Proof: Let A be compact, we first show that A is bounded. Pick some x0∈X,
we certainly have A⊂X=U{B(x0, r): r>0}
By compactness, ∃ r1, … rk<∞ s.t. A⊂B(x0, r1) U B(x0, r2) …B(x0, rk)
Now if R=max{r1, …, rk}, then A⊂B(x0, R) and hence bounded.
Closedness: (idea: prove X-A is open by showing that d(x0, ∆)⊂X-A) pick x0!
∈A. Recall that V(x0, r) = {x: d(x, x0)>r} is open. For sure A⊂X-{x0} =
U{V(x0, r): r>0} (because x0∉A)
By compactness, we can find r1, r2 … so that A⊂ V(x0, r1) U…V(x0, rk)
=V(x, ∆) if ∆ =min{r1 … rk}>0 since k<∞. That is A⊂V(x0, ∆)=X-D(x0, ∆)
Hence A∩D(x0, ∆)= empty or D(x0, ∆)⊂X-A. So x0 is an interior point of XA no matter what x0 we choose. Hence X-A is open.
Remark 2: A can be covered by finite V(x0, rk) shows that rk is positive
Remark 3: Note that to show B=X-A is open, we have to show A=X-B is
closed which is implied by finiteness of compactness. Closed or finite means
the sequence will stop at some point, leaving the complement a thick space.
Qn3: why not B(x0, ∆), seems wrong with D here, since it gives closed set
An3: Here D(x0, ∆) does imply B(x0, ∆); since we only need to prove that x0
is of a positive distance from A, we only need to make sure that ∆>0
Remark: If X is “too infinite dimensional”, then there are many closed bounded
sets that are not compact. (the converse is not true)
E.g let X={(xn) sequence in R with x12+…xn2=∑xi2<∞}
Define d((xn), (yn)) = (∑(xi-yi)2)1/2
D(a, r) for r>0 is always non-compact though D(a, r) is closed and bounded.
We can find a sequence in D(a, r) which has no convergent subsequence.
Choose a*=(0, 0 …) and choose r=2. Note that x1*=(1, 0 …), x2*=(0, 1. 0..),
xi*=(0 … 1 0…). Note that the d(xn*, xm*)=√2 <2 if n≠m. This has no
convergent subsequence. (Non-Cauchy)
Remark 1: it needs infinite subcovers to cover the circle in this example …
Remark 2: compactness Any sequence in X has a convergent subsequence.
Quote: If we play with infinite space, there are so many places to hide in
infinite places
Qn4: why is it closed?
Ans4: proved above by lemma 29.2 that any D(a, r) is closed
Lemma 30.3: let Fn be a nested sequence of non-empty bounded closed sets,
then ∩Fn=F1∩F2 … is non-empty. (Nested means Fn+1⊂Fn)
Remark: we really need closedness for this to be true. Here are some examples:
Fn=(0, 1/n), then ∩Fn=empty. (proof by contradiction)
Fn=[0, 1/n], then ∩Fn={0}
Proof of lemma 30.3: pick xn∈Fn. The sequence (xn) is bounded because all
xn are in F1 and F1 is bounded. Hence (xn) has a convergent subsequence (xnk)
(Bolzano-Weierstrass). Let x=lim xnk. Note that xnk∈Fnk⊂ Fℓ , for nk>ℓ.
Here ℓ is fixed. Since xnk∈Fℓ for large k, we deduce that the lim xnk∈Fℓ
(closedness, limit∈space). Since ℓ can be chosen to be arbitrarily large, we
deduce that x∈∩Fℓ for ℓ →∞ →∩Fℓ is non-empty
Remark: Note that we need to find a convergent subsequence (xnk) instead of
any sequence (xn) since the ∩Fn might have many elements
Lemma 30.3: the box [-m, m]k={x*=(x1 … xk) with |xi|≤m} is compact (box
with width 2m)
Remark: the trick is to make the boundary box arbitrarily small to be contained
in an open cover (1 is finite).
Proof: let {Va} be an open cover for A. Suppose to the contrary, this cover has
no finite subcover. Divide the box into 2k sub-boxes of width 2m/2=m. One of
these boxes A1 can’t be covered by finitely many Va’s. Divide A1 into 2k subboxes with width m/2 and choose one of the sub-boxes A2 …
We get a sequence of boxes: … ⊂ An⊂ …⊂ A3⊂ A2⊂ A1, so that the width
of An is 2m/2n and none of An can be covered by finitely many Va’s.
By lemma 30.1 ∩An is not empty. Pick a∈∩An. Pick Va0 so that a∈Va0,
pick r0>0 s.t. B(a, r0)⊂Va0. Since width of An→0, for sure An⊂ B(a, v0)⊂Va0
for large n. (choose n so that √k m 2-n+1<r0) Contradiction.
Remark: note that pick a∈∩An for n→∞ is necessary. This makes sure that
the open cover we are interested in can have an arbitrarily small r since every
box has this point in them. Otherwise with both r and m/2k decreasing, we
can’t assert that one is smaller than the other.
Lemma 30.4: If B is compact and F⊂B with F closed, then F is compact.
Proof of lemma 30.4: Let {Va} be an open cover of A, then B ⊂U{Va}U(X-A).
Now since B is compact, there are a1, a2 … am so that B⊂Va1 U … Vam U
(X-A) (Vai∈{Va}, am<∞)→ hence A⊂ Va1 U … Vam because A is disjoint
from X-A. This completes the proof.
Qn5: Can we add X-A in the open cover?
Ans5: No, X-A is in every cover of A.
Qn6: Can we use the open covers of B to cover A to complete the proof?
Ans6: Open covers of B do not represent every open cover of A, as they are a
subset of it. The formal procedure is to assume for any open covering and
prove its subfamily has finite members.
Theorem 30.2: if A is closed and bounded in Rk, then A is compact.
Proof of theorem 30.2: Assuming the above lemma, for any kind of closed and
bounded set A, we may choose B=[-m, m]k, for large m, so that A⊂B (A is
bounded), then using lemma 30.4 we deduce that A is compact because B is
compact and A is closed.
Quote: Revolution in math: Euclid’s introduction of Axioms and Geometry and
Cantor’s introduction of set and limit, make objects large/small to form a limit.
29. Back to sequences in R. We are interested in convergent sequences. Given a
sequence xn, look at an=xn+1-xn. (Analogue of derivative). a0 +a1+…an-1=x1 +x2x1 +…xn-xn-1=xn (here a0=x1). Lim xn=a0+a1+ … is an infinite series.
Lemma 31.1: We may regard a sequence (xn) as a series xn=a0+a1+ …+an-1.
We are interested in the limiting behavior of xn. If lim xn=x, then for e>0, we
find N so that n>N → |∑am-x|<e. We also use the notion ∑am for the limit x.
For example, the sequence xn=∑am is non-decreasing iff am≥0 for all m. In
this case ∑am is meaningful alw if we allow +∞ as a possible value. Similarly
xn is non-increasing if am≤0 and ∑am well-defined if -∞ is a possible limit.
Remark: bounded monotone sequence is convergent. Note that convergent
sequence has a limit∈R and ±∞!∈R
Lemma 31.2: Since for n>m, ∑a1~n-∑a1~m=∑am~n, the sequence is Cauchy if
for e>0, we can find N so that n>m>N →∑am~n<e. (Cauchy = Convergent)
Remark: to prove a series as convergent, we need to show ∑ak is finite (limit)
or lim∑ am~n →0 when m→∞ ^ n>m (Cauchy)
Claim 31.3: if xn=∑ak is convergent, then lim am=0.
Proof: pick e>0 and let N be as in 31.2. Now choose n=m+1 so that for m>N
→ |am|<e. (Cauchy with difference as 1)
Remark: the converse is not true! We can’t use lim am=0 → xn convergent.
For example if an=1/n, then ∑1/n=∑an=+∞.
(idea: no matter how small m is, 1/m+1/(m+1) …+1/2m≥m(1/2m)=1/2)
Suppose the contrary, ∑1/n=∑an<+∞. n>m>N → 1/m+1/(m+1)+…1/n<e
(Cauchy) So, (n-m)/n =1/(n-1)+ … 1/(n-1) <1/m+1/(m+1)+…1/n<e
Now choose n=2m to get ½<m/2m<m/(2m-1)<e and not possible for 0<e<1/2.
Remark 1: As we will see later ∑1/np <∞ iff p>1.
∑1/ (n (logn)p) <∞, ∑1/ (n logn(log logn)p) <∞, iff p>1
Qn7: How do we prove that ∑1/np <∞ iff p>1
Remark 2: We only need to make sure both n and m are large (>N) in this
proof, the relationship of m and n does not change the assumption
Claim 31.4: suppose that |an|≤bn and ∑bn<∞. Then ∑an is convergent
Proof: we need to show that for e>0, we can find N so that n>n>N →|∑ak|<e.
But |∑ak|≤∑|ak|≤∑bk. So we are done if we choose N so that n>m>N →
∑bk<e.
30. Definition 32.1: We say ∑an is absolutely convergent if ∑|an|<∞
Remark: if a series is absolutely convergent, then it is also convergent. |∑ak|≤
∑|ak|<∞ (Cauchy → convergent).
The converse is not true. For example 1+1/2+ …=∞, but 1-1/2+1/3-1/4+… is
convergent.
Theorem 32.2: let a1≥a2≥ … with lim an=0. (implies everything is +ve).
Then a1-a2+a3-a4 …=∑(-1)n+1an is convergent.
Idea: we need to make |am-am+1+am+2-am+3…±an-1| small for m large ^ n>m.
(Cauchy criterion)
Proof: Case 1: n-m is odd, 0 ≤ (am-am+1)…+(an-2-an-1 )= am-am+1…+an-2-an-1
= am+(-am+1+am+2)…+(-an-3+ an-2)-an-1 ≤ am<e if we choose m>N s.t am<e.
(trick: group in different ways)
Case 2: n-m is odd: 0≤am-am+1…-an-2+an-1≤am
e.g. 1: ∑(-1)n log n /n convergent because log n /n keeps decreasing and the
negation converges to 0 by Theorem 32.2.
e.g. 2: ∑(-1)n 1/√n convergent
e.g. 3: xn=a+aλ+aλ2+…+aλn-1=∑aλk=(1-λn)/(1-λ).
Lim λn=0 |λ|<1→∑a λk=a/(1- λ). For | λ|≥1, ∑a λk=±∞ or not exist
Geometric series or 1/np are the most common tricks for convergent series
e.g. 4: ∑logn/n3<∞, because logn/n3<1/n2
e.g. 5: ∑logn/np<∞ with p>1. For n>n0, log n< nα Hence, logn/np< nα/np<1/np-α.
Choose α>0 ^ a<p-1, so that p-α>1 → ∑logn/np<∑1/np-α <∞
Theorem 32.3: (root test: basically a variation of geometric series) let β=lim
sup |an|1/n. If β<1 then ∑an is absolutely convergent. If β>1, then ∑an is
divergent. If β=1, then we need more information.
Remark: |an|=|a| |λ|n →|an| 1/n =|a| 1/n |λ| →|λ|
e.g. 1: an=n2/3n. lim (an)1/n=1/3, convergent
e.g. 2: an=1/np, lim (an)1/n=1 → no information
Proof:
Case 1: Assume β<1. Recall that β=lim sup |an|1/n=lim sup{|an|1/n :n>N}. ∃ e>0
^ β+e<1, s.t. sup |an|1/n<β+e for n≥N, |an|1/n<β+e or |an|<(β+e)n.
For n≥N, ∑|an|<∑(β+e)n < ∞
Case 2: Assume β>1. β =lim sup |an|1/n=lim sup{|an|1/n :n>N}. ∃ a subsequence
ank s.t. lim |ank|1/nk=β. ∃ e>0 ^ β-e>1 s.t. β-e<|ank|1/nk<β+e for nk>n0→(βe)nk<|ank|. Lim sup|an| > lim sup (β-e)nk =+∞. (Not convergent)
Remark 1: convergent→subsequence has the same real limit subsequence
has no real limit → not convergent
Remark 2: root test is just trying to get our hand on geometric series. Since the
root power itself is the index, the test makes sure that the sequence is bounded
by a geometric series.
Remark 3: lim sup xn gives an upper bound of xn after a sufficient large n
Lemma 32.4: any sequence (an), we have
lim inf |an+1|/|an|≤lim inf |an|1/n≤lim sup |an|1/n≤lim sup |an+1|/|an|
Proof: if λ <lim inf |an+1|/|an|, then λ≤ lim inf |an|1/n (*). This would do the job.
Because if we have the opposite inequality, lim inf |an|1/n< lim inf |an+1|/|an|
we will get a contradiction by finding any λ in between.
Remark: this is a proof by contrapositive. If the opposite of the result leads to
the opposite of (*).
pick u< inf |an+1|/|an|=v. Hence for N large v-e< inf |an+1|/|an|<v+e.
v-e< |an+1|/|an|→ |an|(v-e)< |an+1| → By induction, |an+k|>|an|(v+e)k …
Continue proof in Book 12.7
→ |an| < vn-N|aN| for n>N → α = lim sup |sn|1/n ≤ v
Remark: the limit of sequences with strict inequality will reserve the inequality
but converts to non-strict. e.g. 0<1/n → lim 1/n ≥ 0
Trick: pick a number <the left, prove the number ≤ the right
Mathematical typing:√ (sqrt) ∞ (infty) ∏ (prod) ≡ (equiv) ∘ (circ) ∫ (int) π
(pi) ∮ (oint) σ (sigma) Σ (Sigma) 〖 (begin) 〗 (end) ≈ (approx) ∝
(propto) ∧ (wedge) ∀ (forall) ⇒ (Rightarrow) ⊕ (oplus) ∨ (vee) ∩ (cap) ∪
(cup) ∅ (emptyset) ′ (prime) ≫ (gg) ≪ (ll) ξ (xi) ϵ (e)
Joke: for 19 years, I don’t have a TV, my wife find it’s very strange, but it’s
only a decoration for me.
Claim: lim sup |an+1|/|an|<1 → ∑an is convergent (lemma 32.4 ∧ root test)
Claim: lim inf |an+1|/|an|>1 → ∑an is divergent (lemma 32.4 ∧ root test)
Remark: Note that root test only looks at lim sup|an|1/n whereas ratio test looks
both at both lim sup and lim inf of |an+1|/|an|. Both tests compare to one.
Remark: lim sup |an+1| /|an| >1, does not suffice for the divergence of ∑an.
e.g. : 1, 1/3 a. 2/3 a, 2/32 a, 22/32 a … (2/3)n a, (2/3)n (2a) ...
Both (a2n) and (a2n+1) converges to 0 → 2n√a2n and 2n√a2n+1 =√2 /3 <1
Remark: lim a±1/n =1, lim n±1/n =1 (lim n-1/n = 1/ lim n1/n)
Remark: Theoretically, nth test has a better chance to succeed. However for
some example, the ratio test is easier to apply.
31. ∑ 1/n =∞ →∑ 1/np=∞ (by comparison) for 0<p≤1
We verify ∑1/np<∞ for p>1 by using calculus.
Theorem 33.1: Let f: (0, ∞) → (0, ∞) be a continuous function and assume
that f is non-increasing. Let an=f(n) → ∑an<∞ iff ∫f(x)dx<∞
Remark: here f is a continuous positive non-increasing function with an = f(n)
Proof: an =∫nf(x) dx in [n, n+1]→ min{F(n), F(n+1)}≤an≤max {F(n), F(n+1)}
→∑an+1≤∫f(x) dx≤∑an ( when an+1 ≤an, true for most cases)
→ ∑an≤∞ iff ∫f(x)dx<∞
e.g.1: ∑1/n x∈[2, ∞]≤∫log N dx ≤∑1/n, n∈[1, ∞]
→ logN≤∑1/n≤logN +1
e.g.2: ∑1/np n∈[N+1, ∞]≤∫dx/xp≤∑1/np n∈[N, ∞]
→ 1/(p-1) 1/Np-1≤∑1/np≤1/(p-1) 1/(N-1)p-1
Trick: lose 1 in the index to standardize the sum
E.g. 1: ∑1/np. Here f(x)=1/xp. ∫x-p dx →∞ if p<1 and 1/(p-1) if p>1
Joke: I would like to avoid the chalk I breathe when I erase, so I try not to
change anything on the board.
Also if p=1 → ∫1/xdx=log ℓ = ∞
→ ∑1/np<∞ iff p>1
E.g. 2: ∑1/(n*log n) =∞
f(x) =1/(x log2x) →∑m=2n+1~2n+1 1/(m log2m) < f(x) =1/(x log2x)
< ∑m=2n~2n+1 1/(m log2m)
f(n)=1/(2n n)
∑1/ n(log n)3/2, g(x)=-1/p (log x)-p → g’(x) = 1/ (x (log x)p+1). Hence p>0, ∑
1/(n(logn)p+1)<∞ because ∫2~∞dx/ (x(logx)p+1) <∞. To see this, observe ∫2~∞
dx/ (x(logx)p+1) = -1/p(log ℓ)-p+1/p(log 2)-p
Claim ∑27~∞ 1/ (n log n) =∞. It suffices to show ∫dx/(x log x)=∞. If g(x)
=log(log x). so ∫dx/(x log x) =∫d/dx (log log x) = log log ℓ -log log 27 →∞
as ℓ →∞. We now turn to ∑1/(n(log log n)3/2). If n is sufficiently large, say n
≥n0, then (log log n)3/2< log n →
∑n0~∞1/(n(log log n)3/2) ≥∑n0~∞ 1/ (n log n) =∞.
Remark: for n = 33>ee → log log n>0
∑1/(n(logn (log logn)3/2). Consider g(x) = -1/p (log log x)-p
∑ 1/(n logn (log log n)). Try g(x) = log log log x
Remark 1/ n(∏lognn) can be integrated, others need to use comparison to them
32. Decimal Expansion: Given x∈R, we can write x=R.d1d2…dℓ … for k∈Z, d1,
d2 …dℓ∈{0, 1 … 9}. We really mean x = k+d1/10 + …dℓ/10ℓ + …
The infinite series is always convergent by comparison test for dℓ =9
Remark: If we choose 9 for all digits, then we get 1 as the above calculation
indicates. In other words, we have two possible expansion for 1, namely
1.000… , 0.999… (proof by sum the geometric series)
∀ +ve x, we can find a decimal expansion for x. We give a unique construction
for our expansion if we replace 0 for an infinite string of 9’s.
Construction 34.1:
Step 0: Here is the construction, by the Archimedean Principle, ∃ an integer m
s.t. x≤m (for b=1). now set k = max{n∈{0, 1, …m} : n≤x}
Setp 1: look at 0≤x-k(=.d1d2 …). Put g0=x-k and look at 10g0.
We expect to have 10g0(=d1.d2…). Note that 0≤g0<1. Apply step0 to 10g0.
That is, find nonnegative integer d1 s.t. 0≤10g0-d1<1. Note that since 0≤
10g0<10, so d1∈ {0, … 9} (recursion)
gn=.dn+1dn+2…
Note that only in step 0 the number might exceed 9 (more than 1 digit)
Remark: We just keep approximating the integer part in the SAME scale,
moding it by 1 and left shifting the number (Same scale because we keep left
shifting it by 1)
Remark: Construct the R by Q’s approximation
1/3 = .333… 1/7=.142857 …
Joke: I must do this in my mother language… (can’t do long division). Let me
erase it, so there is no record on what I did.
Theorem 34.2: If X is rational iff the expansion is periodic. More precisely, if x
=a/b, a and b +ve integers iff x = k. d1d2… dm…dn
Joke: If I use left or right, it depends on whether it’s my left or your left and
depends on whether I face you or not. Student, “Use east and west!” Prf: “It
depends on whether it’s your east or my east!” (Stopped, realized sth wrong)
If: (periodic→ rational) will give a finite sum of rational numbers → rational
Iff: (rational→periodic) start from x=a/b. WLOG assume 0<a<b. We want to
prove gn(=dn+1dn+2…)=x=a/b is periodic (refer Construction 34.1)
We argue that rn=bgn is an integer with 0≤rn<b (gn=integer/b).
(by induction a<b → r0 is integer<b and rn+1=10rn –bdn+1<b for max dn+1
satisfying rn+1>0, by contradiction)
Since 0≤gn<a ∧ 0≤rn<b → rn ={0, 1, … b-1}. ∀ continuous b ri’s, there are
at least two ri and rj which coincide. → ... di+1…dj are periodic
Remark: rn is the remainder divided by b so <b