Analysis 1 period 1.1 Flashcards
Definition of subsets
Suppose that A and B are sets. A is a subset of B if for all x ∈ A it follows that x ∈ B.
A proper subset of B, denoted by A proper subset B if for all x is an element of A it follows that x ∈ B and there exists x ∈ B s.t x is not an element of A.
What kind of operations do you have on sets?
- Intersection
- Union
- Complement
- Minus
- Disjoint
Guideline for proofs with sets
- If you want to proff a = B, proof that a is a proper subset of B and otherwise
- If you want to proof a statement with for all, then take an arbitrary situation and proof the statement for that situation
- If you want to proof a statement with there exists, try to find a specific situation that fits the statement
- If you want to proof an ‘If then’ statement you need to assume the if statement and proof the then statement
- If you want to proof an if and only if statement you need to proof both ways.
What is a relation?
A relation h is a set of ordered pairs(x,y) where x ∈ A and y ∈ B. This relations links two sets to each other. h: A -> B
The visualization is called the graph of a relation, also called a curve. Two relations are equal if they have the same graph.
What is a function?
A relation f is a function if it doesn’t contain two different ordered pairs with the same first coordinate. In math, f is a function if and only if: for all (x1,y1) (x2,y2) ∈ f with x1 = x2 it follows that y1 = y2
A relation f is a function when you can say that every x value has an unique y value
Domain
An other word for domain is preimage
The domain/preimage of a function f, denoted by Df is defined by Df = {x|(x,y) ∈ f}
Range
Other word for range is image
Denoted by Rf is defined by: Rf = {y|(x,y)∈ f and x ∈ Df}
How do you find the maximum domain of f?
There are 3 functions you have to take into account
- For n^√g(x) with n is even, it must hold that g(x) >= 0
- For log a (g(x)) is must hold that g(x) > 0
- For f(x) / g(x) it must hold that g(x) =/ 0
=/ -> not equal , >= -> equal or bigger than
Injective
If ∀x1,x2 ∈A; f(x1)=f(x2) it follows that x1 = x2
Surjective
f is surjective If ∀ y∈B ∃x∈A: f(x)=y. So in case of surjection Rf = B
Bijection
When f is both injective and surjective
you have to proof both the ways in order to conclude that f is bijective
How do you proof injection?
- x1,x2 ∈A; f(x1)=f(x2) be given
- Use algebraic rules to work to the conclusion x1=x2
- Conclude that f is injective
If you can’t reach this conclusion, you must give an example where two different x values give the same y value. This example concludes that f is not injective
How do you prove surjection?
- Let y∈B be given
- Use algebraic rules isolate x and conclude x∈A
- Conclude that f is surjective
If you can’t reach this conclusion, you must give a example where a number y∈B cannot be reached with f with domain Df. This example concludes that f is not surjective
When is a function even and when is and function odd?
- f is an even function if and only if f(-x) = f(x) for all x ∈ A.
- f is an odd function if and only if f(-x) = -f(x) for all x ∈ A
Proof even
- Let x ∈ A be given
- Define f(-x) and use algebra to prove that f(-x) = f(x)
If f is neither, then give an example of a number x ∈ A which gives f(-x) =/ f(x) and f(-x) =/ -f(x)
Proof odd
- Let x ∈ A be given
- Define f(-x) and use algebra to prove that f(-x) = -f(x)
If f is neither, then give an example of a number x ∈ A which gives f(-x) =/ f(x) and f(-x) =/ -f(x)
Monotone functions
- f is increasing<=> f(x1) <= f(x2) for all x1,x2 ∈ A; x1<x2
- f is strictly increasing <=> f(x1) < f(x2) for all x1,x2 ∈A; x1<x2
- f is decreasing <=> f(x1) >= f(x2) for all x1,x2 ∈A; x1<x2
- f is strictly decreasing <=> f(x1) > f(x2) for all x1,x2 ∈ A; x1<x2
- f is constant <=> f(x1) = f(x2) for all x1,x2 ∈A
If a function is increasing or decreasing then it is monotone. Same counts for strictly monotone
Proving monotone functions
There are three steps
- use a draft: start with f(x1) < f(x2)
- Use algebra to get a statement about x1 and x2
* x1 < x2 f is strictly increasing
* x1 > x2 f is strictly decreasing
* if you can’t conclude any of these options, f is none of them - Now we can start with the real proof:
1.1 In case of non constant: Start with the expression you found about x1 and x2 and work in the opposite direction until you reached f(x1) < f(x2) and draw your conclusion
1.2 In case of a constant function: Choose three coordinates (x1,y1), (x2,y2) and (x3,y3) where the function goes up, down and up again or down up and down again
Absolute value of any real number
Denote by |x| and is defined by
|x| = { x if x >= 0
-x if x < 0
The absolute value function of f(x) = |x| with x ∈ R
Bounded functions
Give the options
consider the function f: A -> B
* f is bounded below <=> there exists M ∈ R: f(x) >= M for all x ∈ A
* f is bounded above <=> there exists M ∈ R; f(x) <= M for all x ∈ A
* f is bounded <=> there exists M ∈ R+; |f(x)| <= M for all x ∈ A
f is bounded <=> there exists M∈R+; -M <= f(x) <= M for all x ∈ A
So f is bounded <=> f is bounded below and bounded above
supremum and infimum versus maximum and minimum
Consider a bounded function f : a-> B
* If the global maximum of f exists, then sup f = max f
* If the global minimum of f exists, then inf f = min f
Composition function
Suppose functions f : A->R and g : B = f(A) -> B. The composition function of g on f, g ○ f
g○f, maps set A into R.
In math: g ○ f : A->R
The function g ○ f is defined by (g○f)(x) = g(f(x))
Mathematical induction
if p(n) is a statement for all n∈N s.t
a. p(1) is true, and
b. for each m∈N, if p(m) is true, then p(m+1) is true
then p(n) is true for all n∈N
General mathematical induction
if p(n) is a statement for all n>=k for k∈N s.t.
a. p(k) is true, and
b. or each natural number m >=k, if p(m) is true, then p(m+1) is true
then p(n) is true for all n >= k
Procedure mathematical induction:
- Basis of the induction: Show that p(k) is true
- Induction hypothesis (IH): Assume that for a certain m>= k that p(m) is true
- Inductive step: Prove that p(m+1) is true
Binomial coefficient
(n k) n! / (k! * (n-k)!)
where k and n are integers s.t
0 <= k <= n
Strong induction
If p(n) is a statement for all n ∈N and m ∈ N s.t.
a. p(1), p(2),….,p(m) are true
b. For all k>=m if p(i) is true for i ∈ {1,…..,k} then p(k+1) is true
then p(n) is true for all n ∈ N
Procedure strong induction
- Basis of the induction: Show that p(n) is true for all n<=m
- Induction hypothesis (IH): Assume for a certain k>=m s.t. p(i) is true for i∈{1,…..,k}
- Inductive step: Prove that p(k+1) is true
Inverse function
A function f is invertible if and only if its inverse relation is a function. f^(-1) denotes the inverse of f.
A function f is invertible if and only if f is one-to-one. That is, if f is injective
Inverse relations
The inverse of a relation S is the relation T, where T = {(y,x)|(x,y)∈S}. The graph of T is obtained from the graph of S by reflecting the graph of S along the diagonal line y=x
Procedure finding the inverse function
If we have an injective function f and we wish to find its inverse, we perform three steps:
1. replace f(x) by y
2. Interchange the variables x and y
3. Solve for y and label it f^(-1)(x)
Definition: Image
Suppose the function f : A -> B
The image of the function f is defined by
f(C) = {f(x)∈B : x∈C}
Procedure: Image
- Find the inverse relation f^(-1) and write it in terms of y so f^(-1)(y)
- Solve the equation or inequalities in terms of y
- Put your answer in set notation for the image
Definition: preimage
Suppose the function f : A -> B
The preimage of the function f is defined by
f^(-1)(C) = { x∈A : f(x) ∈ C}
Procedure: preimage
- Solve the equation or inequalities in the terms of y
- Put your answer in set notation for the preimage
Inverse of bijections
- Suppose that f : A -> B is a bijection; then
1.1 (f-1 ○ f )(x) = x for all x ∈ A
1.2 (f ○ f-1)(x) = y for all y ∈ B - The function f : A -> B is a bijection if and only if f-1 : B -> A is a bijection
What are the triangle equalities
- |a + b| <= |a| + |b|
- |a - b| >= |a| - |b|
- ||a| - |b|| <= |a - b|
