Consider \(f\) and \(A\) and \(f(A)\). What is the identity? Can there always be an inverse?
No, consider \(f(x)=x^2\) and \( A = \{ 0 < x < 1 \}\). We see that \(f^{-1}(f(A)) = \{ -1 < x < 1 \}\).
What condition can guarantee that you can always find an inverse?
From a new idea, how many more ideas can be formed in combination with existing ideas?
Set of existing ideas = [ identity, inverse, transitive, union, intersection, equality, less than or equal, subset ]
Language
- Use of language in math
- https://assets.press.princeton.edu/chapters/gowers/gowersI2.pdf
Boolean Logic
- Statements that are True or False
- When we write
- \(A\), means \(A\) is true
- \(\lnot A\) means \(A\) is false
- \(A \land B\) means \(A\) and \(B\) is true
- \(A \lor B\) means \(A\) or \(B\) is true
- \(A \rightarrow B\) means \(\lnot A\) or \(B\) is true
Negation
- A and !A
- When introduced to a new definition, do consider the negation as part of the elements you must permute to see the connections
What is proof by contradiction
- Suppose \(A\) is the statement we are trying to prove and we know B and A^B are true
- Suppose A is false ^(A,B) = F, then we have
- A B A^B
- ? T T
- F T F
- TODO: Continue and finish
Implication and Contrapositive
- A -A B -B A->B C
- F T F T T T
- F T T F T T
- T F F T F F
T F T F T T
A Truth table is a table of the domain and range of statement values
Two statements are equivalent when they are the same function
The implication is interesting because it correlates A and B in a way
- When we write \(f(A,B),\) we mean \(f(A,B)=true\)
To prove \(f(A,B)=true\), we need to plug in all variables to see if it is constant and true in all cases
To prove \(A \rightarrow B = true\)
- 1 if \(A\) is false \(A \rightarrow B = true\)
- This means we only need to be concerned with the case where \(A\) is true.
- 2 if \(A\) is true \(A \rightarrow B = true\) only when \(B\) is true, requiring us to show \(B\) is true (This links A and B/that B can never be false when A is true)
- This also mean when \(B\) is false, \(A\) must also be false because otherwise \(A \rightarrow B = False\)
From this relationship can we draw another implication (which satistfies 1&2)?
- To show an implication is true, we only need to be concerned with when hypothesis is true
- \(!B \rightarrow !A\)
1.1 Sets and Functions
DEF Set
- \(A\) a set is a collection of objects
- \(a \in A\) if the element \(a\) is in A
- \(A \subset B\) if B contains all the elements of A
- \(A = B\) if they contain the same elements
- \( \cup \cap \setminus \) are binary operations \(S \times S \rightarrow S\)
- \( \varnothing \)
PROPERTIES Set
- \( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)
- \( A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\)
- \( ( A \cup B )^c = (A^c \cap B^c) \)
- \( ( A \cap B )^c = (A^c \cup B^c) \)
- \( A \setminus (B \cup C) = (A \setminus B) \cap (A \setminus C)\)
- \( A \setminus (B \cap C) = (A \setminus B) \cup (A \setminus C)\)
DEF Function
- Mapping from a set \(A\) to \(B\)
- Such that an element of \(A\) maps to exactly one element in \(B\)
1.2 Mathematical Induction
1.3 Finite and Infinite Sets
DEF 1.3.1 Finite Infinite
- A set is finite iff there is a bijection between the set and \(\{1...n\} n \in \mathbb{N}\)
THM 1.3.2 Cardinality of a finite set is unique
- Given \(A \subset B\) it is not possible to form a bijection from B to A
THM 1.3.3 \(\mathbb{N}\) is infinite set
THM 1.3.4 Cardinality and Set operations
- Cardinality of finite members are preserved under disjoint union and set difference by a subset
- Cardinality of infinite finite sets
- ? inf
- Lets see how to contrapositive flips within the argument
THM 1.3.5 \(\subset\) and conditions for Finite/Infinite
- \(INF \subset S \rightarrow S\) is infinite
- \(S \subset FIN \rightarrow S\) is finite
- Conclusion (finite/infinite) drawn about one set from another set based on ordering
DEF Denumerable, countable, uncountable
- A set \(S\) is denumerable - bijection from \(S\) to \(\mathbb{N}\)
- countable - finite or denum
- uncountable - not countable
THM 1.3.8 \(N \times N\) is denumerable
- You have comeup with an interesting function. How to show
\(N(a,b) = b + sum_prev_diags\)
\(diag(a,b) = a+b-1 = d\)
we know \(b \leq diag(a,b) \)
if diag is same, then it is clear N is injective
suppose diag is different, let diag2 bigger than diag1 so \(b1 < diag1 < diag2\)
\(b1+diag0..1 = b2 + diag0..1+diag1..2\)
\(b1 = b2 + diag1..2 > diag2\) contradiction
b1 cannot make up for the distance btw two diff diagonals
THM 1.3.9 \(T \subset S\)
- Conclusion (countable/uncountable) drawn about one set from another set based on ordering
THM 1.3.10
- Illustration on using the various properties of sets and bijections and subsets and it's effect on countable uncountable
THM 1.3.11 Rational numbers are countable
2.1 Algebraic and order properties of R
2.1.1-2.1.3 Algebraic properties of field R
- Uniqueness of 1,0, inverse
- 0*a = 0
- a*b = 0 means one must be 0
2.1.4 \(\sqrt{2}\) is irrational
Order of \(\mathbb{R}\)
- Defined in terms of positive number
- Relation
2.1.7 Properties of order on R
- if \(a > b\) and \(b > c\) then \(a > b > c\) Transitive
- if \(a > b\) then \(a+c > b+c\)
- if \(a > b\) and \(c > 0\) then \(ca > cb\)
- if \(a > b\) and \(c < 0\) then \(ca < cb\)
No smallest positive real number can exist
2.1.9 Thm
If \(0 \geq a \gt e \forall e \gt 0\) then \(a=0\)
2.1.10 Thm
If \(ab > 0\) then \(a > 0\) \(b > 0\) or \(a < 0\) \(b < 0\)
Remark: Arithemetic-Geometric mean
- \(\sqrt{ab} \leq \dfrac{1}{2}(a+b)\)
- \((a_1 a_2 ... a_n)^{1/n} \leq \dfrac{1}{n}(a_1+a_2+..+a_n)\)
Remark: Bernoulli's Inequality
- If \(x > -1\) then \((1+x)^n \geq 1+nx\) for all \(n \in \mathbb{N}\)
2.2 Absolute Value and Real Line
Definition of ABS
- \(|a|\) is a function from \(\mathbb{R} \rightarrow \mathbb{R}\) such that
- \(a > 0 |a| = a\), \(a < 0 |a| = -a\), \(a=0 |a| = 0\)
2.2.3 Triangle Inequality
2.2.4 Corallaries of Triangle Inequality
- \(||a|-|b|| < |a-b|\)
- \(|a-b| < |a|+|b|\)
2.2.5 Triangle Inequality for N sums
- \(|a_1 + a_2 + ... + a_n| \leq |a_1|+|a_2|+ ... +|a_n|\)
2.2.7 E neighborhood
- e neighborhood of \(a \in \mathbb{R}\) is the set \(V_e(a) := \{ x \mid |x-a| < e \} \)
2.2.8 if x is in every e neighborhood of a then x is a
2.3 Completeness Property
Def 2.3.1 Bounded above, below, bounded sets
- Given an subset, \(S\), of \(\mathbb{R}\) it is bounded if ...
Def 2.3.2 Supremum Infimum
- Supremum least upperbound
- if u is upper bound of \(S\), then \(sup S \leq u\)
- Infimum greatest lowerbound
Remark: Uniqueness of suprememum infimum
Remark: Consider the contrapositive of least upper bound condition of definition of suprememum
Lemma 2.3.3
- What is smaller than supremum is not a upper bound
- Contrapositive of definition of Supremum
Lemma 2.3.4
- Upperbound u is supremum iff \(\forall e > 0\) there exists \(s_e \in S\) sutch that \(u-e < s_e\)
2.3.6 Completeness Property of R
- Every nonempty set of real numbers that has upper bound has a supremum in \(\mathbb{R}\)
Lets break down definition supremum
- \(sup S\) is a upper bound of \(S\)
- if u is upper bound of \(S\), then \(sup S \leq u\)
- What are the inputs that define supremum? (S set)
- What are the output of the definition? (real value)
- Can anything other than input output influence or server some connection?
2.4 Application of Supremum
2.4.1
- \(a + supS = sup\{ a+s | \forall \ in S \}\)
- Suppose \(a \leq b \forall a b \in A B \). Then \(sup A \leq inf B\)
2.4.2
- \(f(x) \leq g(x) \forall x \in D\) then \(sup f(D) \leq sup g(D)\)
- This does not guarantee \(sup f(D) \leq inf g(D)\)
2.4.3 Archemdean Property
- For any \(r \in \mathbb{R}\) there is \(n_r \in \mathbb{N}\) such that \(r \leq n_r\)
- There is no real number larger than all natural numbers
2.4.4 Corollary
- \(S = \{ \dfrac{1}{n} | n \in \mathbb{N} \}\) then \(inf S = 0\)
2.4.7 Sqrt of 2 is real number
- \( \ sqrt{2}\) is a real number
- This is a consequence of the Supremum property of \(\mathbb{R}\)
- Can show
2.5 Intervals
DEF Interval
- (a,b) := {x | a < x < b}
- (a,b), (a,b], [a,b), [a,b]
- (-inf,b), (-inf,b], (a,inf), [a,inf), (-inf,inf)
2.5.1 THM Characterization of Interval
- if \(S\) has the property, then \(S\) is an interval
- property: if \(a,b \in S\) and \(a < b\) then \([ a, b ] \subset S\)
DEF Nested Intervals
- A sequence of intervals \(I_n, n \in \mathbb{N}\) is nested if \( ... I_3 \subset I_2 \subset I_1\)
QUESTION
- Why must these intervals be index by N?
- Supremum does not require
- Does any other part require us?
2.5.2 THM Nested Interval Property
- If \(I_n = [a_n,b_n], n \in \mathbb{N}\) is a sequence of closed bounded intervals, there is \( c \in I_n \forall n \in \mathbb{N}\)
2.5.3 THM Uniqueness of Common Nested Interval Point
- If \(I_n = [a_n,b_n], n \in \mathbb{N}\) is a sequenc of closed bounded intervals, and \(inf \{b_n - a_n | n \in \mathbb{N} \} = 0\)
- then \( c \in I_n \forall n \in \mathbb{N}\) is unique
2.5.4 Set of \(\mathbb{R}\) is not countable
- Assume is countable
- \( \{x_1, x_2, ... \} = S\)
- want to find a number that is not in the set that includes all numbers
- \( c \in \forall I_n \) then make each \(s \in S\) be the negation of \(\in \forall I_n\)
DEF Binary representation
- A bijection?
- homorphism under addition/sub?
2.5.5 THM \(\mathbb{R}\) is uncountable
3.1 Sequences and Their Limits
3.1.1 DEF Sequence
- Sequence in \(S\) is a function from \(\mathbb{N} \rightarrow \S\)
- \(( s_n | n \in \mathbb{N} )\)
- inductively/recursively defined sequences \(x_{n+1} = x_n + 5\)
3.1.3 DEF Limit of Sequence
- \(L\) is limit of sequence \((s_n)\) if \( \forall e \: \exists m \) s.t. \(m < i \rightarrow |L-s_i| < e\)
- If \((s_n)\) has a limit, it is convergent
- If \((s_n)\) doesn't have limit, it is divergent
- \(Lim(s_n+x_n) = s+x\)
- Effect of limit on s_x operated on by a constant
- Effect on limit on s_x operated by another sequence
- \(s_n < x_n \rightarrow s < x\)
- \(s_i \in V_e(L)\)
- \((s_n) is a ordered set, how about ordered subset?\)
- Uniqueness of Limit
3.1.4 Uniqueness of Limit
I started off with a diagram and realized, hey maybe the distance between the 2 different limits,(A,B), will show that the sequence will have to exist in 2 impossible places
It came more down to if the sequence is within 1 limit Ve(A), it cannot be in the other Ve(B)
- Let \(m\) be the point where \(|A-s_m| < e\) and \(|B-s_m| < e\) for \(e = \dfrac{|B-A|}{2}\)
- \( |B-A| = |B-s_m+s_m-A| \leq |B-s_m| + |A-s_m| < \dfrac{|B-A|}{2} + \dfrac{|B-A|}{2} = |B-A| \)
3.1.5 Explanation of Convergence of Limits in terms of Neighborhoods
3.1.8 DEF Tail of a Sequence
- Given a sequence, \(X = (x_1, x_2,...)\), the m-tail of X is a sequence, \((x_m, x_{m+1}, ...)\)
- We denote this as \(X_m\)
3.1.9 THM Tail of a Sequence has same limit as sequence
- for all epsilon, the tail or the original sequence will give an value that will guarantee \(|L-x_i| < e\)
3.1.10 THM
- Let \(x_n\) be a sequence. If \((a_n)\) converges to \(0\) and \(|x_n -x| \leq a_n\) then \((x_n)\) converges to \(x\)
- Not a big fan of theorem. (using a another sequence to show a sequences converges)
- I think one can build a more constructive approach with smaller theorems (see below)
- With application of Squeeze Theorem
- \(Lim(|x_n-x|) = 0 \rightarrow Lim(x_n-x)=0 \rightarrow Lim(x_n)=x\)
- This shows \(Lim(x_n)=x\)
3.2 Limit Theorems
3.2.0 Introduction
- Develop techniques to expand the collection of convergent sequences
- Find more examples
3.2.1 DEF Bounded sequence
3.2.2 THM Convergent sequence implies bounded
- Suppose \(Lim(s_n) = s\)
- \((s_n)\) is bound in \(V_e(s)\) for \(i > K(e)\)
- Clearly we can find a bound for elements \(\{ s_1, s_2, ... , s_{K(e)} \}\)
- Merge these two bounds for a bound for all of \((s_n)\).
- Let \(B1\) be bound for set \(A\). Let \(B2\) be bound for set \(B\).
- Merge of \(B1\) and \(B2\) gives bound for \(A \cup B\)
3.2.3 THM Real number operation on sequences and limit
- TODO: sum, diff, constant mult
- mult: if \((x_n)\) and \((y_n)\) are convergent then \((x_n * y_n)\) converges to \(x*y\)
- \(|xy-(x_n)(y_n)| = |xy -(x_n)y + (x_n)y -(x_n)(y_n)| \leq |xy -(x_n)y| + |(x_n)y -(x_n)(y_n)| \leq |y||x - (x_n)| + |x_n||y - (y_n)|\)
- \(|y||x - (x_n)| \leq \dfrac{e |y|}{2}\)
- \(|x_n||y - (y_n)| \leq \dfrac{e (e+|x|)}{2}\)
- \(|x_n| - |x| < |x_n - x| < e\)
- \(|x_n| < e + |x|\)
- \(|y||x - (x_n)| + |x_n||y - (y_n)| \leq e/2+e/2 = e\)
- div
3.2.4 THM order on sequence corresponds to order on limit
- if \((x_n) \geq 0\) then \(Lim(x_n) \geq 0\)
3.2.5 THM Inequality is preserved for sequences and their limits
- \((x_n) \leq (y_n) \rightarrow Lim(x_n) \leq Lim(y_n)\)
- Can show \((x_n) > (y_n)\) is not true if \(Lim(x_n) > Lim(y_n)\) because we can select neighborhoods of \(x\) and \(y\) s.t all elements of the neighborhood will \(V_e(x) > V_e(y)\)
3.2.6 LEMMA
- \( (z_n) \leq (x_n) \leq (y_n) \rightarrow Lim(x_n) \leq Lim(x_n) \leq Lim(y_n)\)
3.2.7 THM Squeeze Theorem
- If \((z_n) \leq (x_n) \leq (y_n)\) and \(L = Lim(z_n) = Lim(y_n)\) then \(Lim(x_n)\) exists and is equal to \(L\)
- If \(z_n\) and \(y_n\) are found within \(V_e(L)\) then \(x_n\) is also in this neighborhood
- This means that for \(i > min( Ky(e), Kz(e) )\), \(x_i\) will be in the \(V_e(L)\)
3.2.8 Examples of non convergent sequences
3.2.9 Convergence and abs
- If \(Lim(x_n) = x \rightarrow Lim(|x_n|) = |x|\)
- \( | x_i - x| < e \)
- \( | |x_i| - |x| | < e \)
- \((\dag)\) Can we show \( | |x_i| - |x| | < | x_i - x|\)?
- I got quite dissatistfied after looking at this proof and demotivated
- I think the reason is it doesn't really explain why or arrive at the solution in a natural/discoverable way
- I think when I am not satisfied with a proof or an explanation, I should really dig deeper and analayze it closer
- But at the same time not let it get the best of me.
- Don't be results driven
- Below is the aforementioned neat proof from a book
- \(|b| = | b - a + a | < |b-a| + |a|\)
- \(|b| - |a| < |b-a|\)
- \(|a| = | a - b + b | < |b-a| + |b|\)
- \(|a| - |b| < |b-a|\)
- \(|b| - |a| > -|b-a|\)
- Useful Lemmas
- Keep in mind these two properties of absolute value. \(|-x| = |x|\) and if \(b < 0 -b = |b|\)
- How to show \(a > |b|\) iif \(a > b\) and \(a > -b\)
- Continuing from \((\dag)\)
- 1 Consider \(|a-b|\) if \(a\), \(b\) have same sign. Then it is clear \(|a-b| = | |a|-|b| |\)
- 2 Suppose they do not have the same sign and let \(b < 0\) then \(|a-b| = | a+|b| | = |a|+|b|\)
- \(|a|+|b| > = |b|-|a|\) and \(|a|+|b| > = |a|-|b|\) so \(|a-b| = |a|+|b| > ||a|-|b||\)
3.2.10 Convergence and sqrt
- If \((x_n)\) converges to \(x\) then \((\sqrt{x_n})\) converges to \(\sqrt{x}\)
- \(x_n\) is a positive sequence
- \(e \geq x-x_n = (\sqrt(x)-\sqrt(x_n))(\sqrt(x)+\sqrt(x_n)) \geq (\sqrt(x)-\sqrt(x_n))\)
3.2.11 Convergence and ratio
- if \(Lim(\dfrac{x_{n+1}}{x_n}) = r < 1\) and \(x_n > 0\) then \(Lim(x_n)=0\)
- \(\dfrac{x_{n+1}}{x_n}-r < + e\)
- choose \(e,\) st \(r+e < 1\)
- \(\dfrac{x_{n+1}}{x_n} < r < r + e < K < 1\)
- \(y_1 = x_{n+1} < K * x_n\)
- \(y_i = x_{n+i} < K^i * x_n\)
- if we can show Lim(K^n) = 0
- if a=b x
- Show for K<1 Lim(K^n)=0
- We note that if \(K < 1\) then there is \(a,b\) s.t \(K < \dfrac{a}{b} < 1\)
- Let use try to show \(Lim( (a/b)^n ) = 0\)
- -
- pick \(0 < 1/r < t/r < e\)
- \(1/r < a/b\)
- lets forget about a
- why is proving something about the smallest element often involve inspecting the largest element
- its basically showing b^x is unbounded for any integer b>0
- consider the set of all multiples of b less than M
- produce the largest or smallest element, the boundary case right before
- -
- Lemma: \((\dfrac{1}{b}^{n})\) for \( n \in \mathb{N}\) converges to 0
- We will show for all \(E > 0\) there is \(m\) s.t. \(\dfrac{1}{b}^{m} < \dfrac{1}{r} < E\)
- 1 Suppose this is not the case and there is a lower bound for all elements of \(\{\dfrac{1}{b}^{n}\}\) this is equivalent to the following 2
- 2 Suppose there is an upper bound for \(\{b^n\}\). Then, it has supremum call it \(S\)
- choose e such that \(S-e = S/\sqrt{b} < b^x < =S\)
- This means \(b^{x+1} > b*S/\sqrt{b} = \sqrt{b}*S > S\) which is a contradiction
EX 3.2.21
- if \(|x_n - y_n| < e\)
- transitivity property
- \(a-b < e\), \(b-c < e\), \(a-c < 2e\)
- \(R(a,b) = |a-b| < e\)
- What properties does this relation have?
- reflexive, transitive, identity,
- \(R(a,b) = R(a-b,0)\)
- \(R(a,b) + R(0,c) = R(a,b+c) = R(a+c,b)\)
3.3 Monotone Sequences
3.3.0 Introduction
- Develop techniques to show sequence is convergent even if the value is not known
- Evaluate the limit by other methods once it is known
3.3.1 DEF monotone sequence
- Let \((x_n)\) be a sequence
- If \(i > j\) then \(x_i \geq x_j\), we say \((x_n)\) is increasing
- If \(i > j\) then \(x_i \leq x_j\), we say \((x_n)\) is decreasing
- \(x_n\) is monotone if it is increasing or decreasing
3.3.2 THM Monotone sequence is convergent
- increasing sequence is convergent
- Converges to the supremum
- decreasing sequence is convergent
3.3.5 Sequence that converges to sqrt 2
- \(x_1=1\) \(x_{n+1} = 2/x_{n}\)
- Then limit of \(x_n\) is equal to limit of \(x_{n+1}\) (lim of tail of sequence is equal to lim of sequence)
- \(x = 2/x \rightarrow x^2 = 2\)
- The issue is that this sequence is not monotone nor we know it converges
- Can you find a monotone bounded sequence, that when limit is taken between the tail and
3.3.6
- Euler's number
- \(e_n = (1+1/n)^n\)
- TODO: Look into more detail
3.4 Subsequences and Boltzano Weistrauss
3.4.1 DEF Subsequence
- Let \((x_n)\) be a sequence
- Given \(n_1 < n_2 < n_3\) ..., \((x_{n_1} , x_{n_2}, x_{n_3},...)\) is a subsequence of \((x_n)\)
3.4.2 THM Sequence convergence and subsequence convergence
- If a sequence \((x_n)\) converges to \(x\), then any subsequence converges to \(x\)
3.4.4 THM TFAE
- \((x_n)\) does not converge to \(x\)
- There is a \(e\) where there always is unending \(x_n \not\in V_e(x)\)
- For all \(M\), there is \(n > M\) and \(P\) is false
- There is \(e\) and a subsequence that lies outside of \(V_e(x)\)
3.4.5 Divergence Criteria
- \((x_n)\) has two subsequences whose limits are not equal. Then \((x_n)\) is divergent
- All subsequences of convergent sequence must converge to same limit
- \((x_n)\) is not bounded
3.4.6 THM Monotone Subsequence
- A \((x_n)\) sequence has a monotone subsequence
- Let \(x_a\) be the first element where for all \(x_a > x_i\) for \(a < i\)
- And let them be \(\{ x_a, x_b, ... \} = S\)
- If there are no such elements of \(S\), we can construct an increasing sequence since for any xa there is some xa
- If there are finite elements of \(S\), we can work with subsequence which starts with last element of S
- If there are infinite elements of \(S\), the elements of \(S\) a form a decreasing sequence
3.4.7 THM Bounded sequences has convergent subsequence
- Every sequence has a monotone subsequence
- PF1: This monotone subsequence is bounded because the sequence is bounded
- Bounded monotone sub sequence is convergent
- PF2: Nested Intervals based Theorem? TODO
- let sequence be bound from 0 .. M (Since x_n is bounded can easily add a constant to sequence to adjust)
3.4.9 THM Boltazno Weistrauss
- Let \((x_n)\) be a bounded sequence. If every convergent subsequence of \((x_n)\) converges to \(x\), then \((x_n)\) converges to \(x\)
- Suppose \((x_n)\) doesn't converge to \(x\) then there is \(e\) where for all \(N\) there is some \(x_i \not\in V_e(x)\)
- These elements has a convergent subsequence and it must converge to x. But all the elements lie outside \(V_e(x)\)
3.5 Cauchy Criterion
3.5.0 Introduction
- Way to show convergence of a limit without knowing the limit
- Our goal is to show a Cauchy sequence is equivalent Convergent
3.5.1 DEF Cauchy Sequence
- Sequence \((x_n)\) is a Cauchy sequence if for all \(e\), there is \(H\) such that if \(i,j > H\) then \(|x_i-x_j| < e\)
3.5.2 THM Convergent Sequence is Cauchy
- \(|x_i-L| < e/2\), \(|L-x_j| < e/2\)
- \(|x_i-L+L-x_j| < |x_i-L|+|L-x_j| < e/2+e/2\)
3.5.3 THM Cauchy sequence is Bounded
- Let \(x_n\) be our sequence, consider e. Then for \(i,H+1 > H\), \(x_i \in V_e(x_{H+1})\)
- So the sequence's bounds can be determined by {x1,...xH, x{H+1}+e, x{H+1}-e }
3.5.4 THM Cauchy sequence is Convergent
- We have shown cauchy sequence is bounded so it has a convergent subsequence \((x_{s_n})\)
- Suppose this subsequence converges to \(L\)
- Then for any \(e/2\) we can find \(K\) such that \(i > K |L-x_{s_i}| < e/2\)
- Also since the sequence is cauchy we can find H such that for \(i,k > H\) \(|x_{i}-x_{k}| < e/2\)
- Let X be max(K,H) and both these conditions hold true
- \(|L-x_{k}| < |L-x_{s_i}|+|x_{s_i}-x_{k}| < e\)
3.5.7 DEF Contractive Sequence
- A sequence \((x_n)\) is contractive if for \(0 < C < 1\) \(|x_{n+2}-x_{n+1}| < C|x_{n+1}-x_{n}|\)
3.5.8 THM Contractive Sequence is Cauchy
- Consider series \(1/2+1/4+1/8\)
- The distances are \(c+c^2+c^3\)
- This will show that our contractive sequence is bounded
- Why? We know distances converge to zero, so find H where from then on the distance is less than 1
- Then, from H, the sequence must be found in a S neighborhood around X_H, where S is the infinte series of c^n
- How about the sequence \(c^n+c^{n+1}+c^{n+2}...\)?
- Does this converge to zero? Yes.
- Let \(S = c^0+c^1+c^{2}+c^{3}...\)?
- Then \(c^n+c^{n+1}+c^{n+2}... = c^n * S\)?
- Lim(c^n)=0 Lim(S)=S so Lim( c^n+c^{n+1}+c^{n+2}...) = 0*S = 0
- Then for any \(e\) there is \(H\) such that for n>H \( c^n+c^{n+1}+c^{n+2}... < e\)
- for any \(i,j > H\) \(|x_i-x_j| < |x_i-x_{i+1}| + ... + |x_{j-1}+x_{j}| < < c^n+c^{n+1}+c^{n+2}... < e\)
- For \( < < \)
- Since the distances also converge to zero, we can find i>H2 s.t. d_i<1=e
- Then \(d_2 < c d_1\), \(d_3 < c*d_2 < c^2*d_1 < c^2\)
- \(d_n < c*d_{n-1} < c^{n-1}*d_1 < c^{n-1}\)
- So our sequence is Cauchy
3.5.9 Estimation of limit based on Contractive
\(x_n\) will converge to \(L\) so if we take the sum of the distances \(|x_1-x_2|+|x_2-x_3|+..\), they should eventually go to L
- \(|L-x_n| < e\) for \(n > H\)
3.5 EXERCISES
3.6 Properly Divergent Sequence
3.6.1 DEF Divergent Sequence
- \((x_n)\) is divergent to \(\infty\) if for all \(e > 0\), there is \(H\) such that when \(n > H\), \(x_n > e\)
\((x_n)\) is divergent to \(-\infty\) if for all \(e < 0\), there is \(H\) such that when \(n > H\), \(x_n < e\)
What about sum of divergent and convergent sequence (xn) + (sn)
- \(M+L-e < M+s_n < x_n+s_n\)
- \(|L-s_n| < e\)
- \(L-e < s_n < L+e\)
- Since we can freely chose, \(M = N-L+e\), the sum is divergent
3.6.3 THM
A monotone sequence is divergent iff it is unbounded
3.6.4 THM
- \(x_n < y_n and x_n\) divergent to \(\infty\) the \(y_n\) divergent
- \(y_n < x_n and x_n\) divergent to \(\infty\) the \(y_n\) divergent
3.6.5 THM
- \( Lim(x_n/y_n) = L\) And suppose x_n is divergent. Then \(y_n\) must also be divergent
/ and divergence
- Interesting view at relationship between
- Pick any M we want to show that at some point x_n is larger/less than M
- Choose H so that y_n meets both conditions
- \(M < y\)
- \(|x/y-L| < e \rightarrow L-e < x/y < L+e\)
- If M is positive
- \((L-e)*M < (L-e)*y < x < (L+e)*y\)
- Choose M = M/(L-e)
- \(y < M < 0\)
- \(|x/y-L| < e \rightarrow L-e < x/y < L+e\)
- \( (L-e)*M > (L-e)*y > x > (L+e)*y\)
3.6 Exercises
3.7 Introduction to Series
MAP
- \((s_n)\), series, is a sequence of \(s_i = x_1+...+x_i\). with \((x_n)\) a sequence.
3.7.1 DEF Series
- If \((x_n)\) is a sequence, let \(s_n = x_1 +...+ x_n \)
- If \((s_n)\) converges it is called sum of the series, otherwise its called divergent
3.7.2 EX Geometric Series
- Series on sequence \(r^n\).
- Is it contractive is r<1?
EX Natural series
- TODO \(\sum{\dfrac{1}{n}}\) is unbounded
- TODO \(\sum{(\dfrac{1}{n})^2}\) is bounded
- p series \(\sum{(\dfrac{1}{n})^p}\)
- p series is convergent p>1
- p series is divergent 0
- alternative harmonic series is convergent
PRP \(\sum{x_n}\) convergent \(\rightarrow\) \(Lim(x_n)=0\)
- It is easy to see this if \(x_n\) were all positive because the sum wil be infinite otherwise
- If a sequence is convergent, the distance between the sequences will converge to zero
- This is definition of Cauchy
The book proves it by using \(s_{n+1} - s_{n} = x_{n+1}\) = \(Lim(s_{n+1} - s_{n}) = Lim(x_{n+1}) = Lim(x_n)\)
- I think this is a valuable perspective as well
PRP Series is convergent \(\iff\) series is bounded (if \((x_n)\) is positive)
Series is convergent -> series is bounded
Series is bounded. Series is monotone -> Series is convergent
PRP Series, order and convergence
Suppose \(0 = < x_n < y_n\). Then the following is true
- \(\sum{y_n}\) convergent \(- > \) \(\sum{x_n}\) convergent
- bounded, monotone
- Book proof: use cauchy
- \(\sum{x_n}\) divergent \(- > \) \(\sum{y_n}\) divergent
- \(\sum{x_n}\) divergent -> not monotone or not bounded -> not bounded -> \(\sum{y_n}\) is also not bounded -> not convergent
- Another proof: Is also contrapositive of prior statement
PRP Series, ratio and convergence