21-127 Concepts
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Created by:
smartin014 on April 26, 2011
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Evil course.
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75 terms
English | Math / Symbols |
|---|---|
 Bounded Set | For all elements s in the set, there exists a real value M such that |s| < M |
| Increasing Function | f(x) < f(x') whenever x < x' |
| Decreasing Function | f(x) > f(x') when x < x' |
| Nonincreasing Function | f(x) ≥ f(x') whenever x < x' |
| Nondecreasing function | f(x) ≤ f(x') whenever x < x' |
| Set Equality | Prove S⊆T, then prove T⊆S. Do this by considering the first set, then deriving the second. |
| AGM Inequality | If x,y ∈ R then 2xy ≤ x^2 + y^2 and xy ≤ ((x+y)/2)^2 If they are both non-negative, (xy)^.5 ≤ (x+y)/2 |
| Triangle Inequality | |x + y| ≤ |x| + |y| |
| g composed with f | f: A->B, g: B->C g o f => h: A->C, h(x) = g(f(x)) for x ∈ A. |
| Composition of two surjections | surjection |
| Composition of two injections | injection |
| Composition of two bijections | bijection |
| Surjective function | f: A-> B, ∀y ∈ B, ∃x∈A such that f(x) = y |
| Injective Function | f: A-> B. ∀x,y ∈ A, f(x) = f(y) => x = y |
| Well-ordered Set | Has a least element |
| Direct Proof | Assume P, derive Q |
| Proof by Contrapositive | Assume not Q, derive not P |
| Proof by Contradiction | Show that P and not Q can never be true |
| Proof by Induction | Base case: P(1) is true Inductive Hypothesis: Assume true for some n. Show P(n) => P(n+1) |
| (g o f)^(-1) | f^(-1) o g^(-1) |
| f has an inverse | is logically equivalent to f is a bijection, and implies that the inverse is unique. |
| Countability | A is countable iff A is finite or there exists a bijection between A and a set known to be countable. |
| Cardinality | If there exists a bijection between A and B, |A| = |B|. |
| Known countable sets | The natural numbers, integers, rational numbers are countable. The real numbers and the irrational numbers are uncountable. |
| PowerSet(A) | PowerSet(A) has the cardinality 2ⁿ if A has the cardinality n. |
| Diagonalization | Proof technique to disprove surjectivity. Ensure every ƒ from A to B will miss an element of B. |
| Snaking | Proof technique to show countability. Show that every element will be hit in a finite number of steps. |
| Monotone | An increasing or decreasing function. (A strictly monotone function is strictly increasing or decreasing.) |
| Canter-Bernstein-Schroder Theorem | If ∃ injections from A to B and from B to A, then A and B are bijections. |
| Injectivity and Cardinality | ƒ: A -> B injective => |A|≤|B| |
| Surjectivity and Cardinality | ƒ: A -> B surjective => |A|≥|B| |
| Partition | 1. The intersection of any two distinct sets is the empty set. 2. The union of all sets is the universe. |
| Permutation | A permutation of S is the bijection from S to S. An n element set has n! permutations. |
| Permutations with repetition | For n element, k of which are distinct: n!/(n-k)! |
| n choose k | n!/(k!(n-k)!) Used for cases where order/distinct elements don't matter. |
| Chairperson Identity | k(n choose k) = n( (n-1) choose (k-1)) |
| Summation Identity | sum from{i = 0} to{n} (i choose k) = (n+1) choose (k+1) |
| (n choose k) is also | n choose (n-k) |
| Pascal's Identity | (n choose k) + (n choose (k-1)) = ((n+1) choose k) |
| Pascal's Triangle | 2ⁿ = sum of elements in each row, from the 0th row. Symmetric. The nth row has elements from (n choose 0) to (n choose n). |
| Stars and Bars | (n + k - 1) choose (k - 1) given n objects being distributed in k groups with empty groups possible. |
| Binomial Theorem | (x+y)ⁿ = sum from{k = 0} to{n} (n choose k) x^(n-k) y^k |
| Notations for permutations | Two line, cycle. |
| Do permutations commute? | Only disjoint cycles can commute. |
| Identity permutations | (1), usually not written |
| Transposition | Two element permutation |
| Permutation as combination of transpositions | (x₁ x₂ ... x₋) = (x₁ x₋) ... (x₁ x₃) (x₁ x₂) |
| If a|b and b|c, | then a|c. |
| If c|a and c|b, | then c| (ma+nb) |
| PI(x) | Number of primes ≤ x |
| Euclidean Algorithm | 1. (a,b) = (b,r) if r = a - bq 2. (a,0) = a |
| Fundamental Theorem of Arithmetic | All natural numbers bigger than 1 can be written uniquely as products of powers of primes. |
| Linear Diophantine Equation: ax + by = c, (a,b) = d | Only has solution if d|c. Solutions: if x₀, y₀ are solutions, then all x = x₀ + (b/d)n and y = y₀ + (a/d)n are solutions too. |
| R is a binary relation on A and B iff | R ⊆ A x B |
| Notation: (a,b) ∈ R | R(a,b) = aRb |
| R is an equivalence relation [R(a,b) is (a~b)] iff | it fulfills reflexivity (xRx), symmetry (xRy => yRx), and transitivity (xRy, yRz => xRz) |
| a ≡ b (mod m) iff | m | (a-b) |
| Z_m | Set of congruence classes mod m |
| a ≡ b (mod m) => | a + c ≡ b + c (mod m), and ac ≡ bc (mod m) |
| Division in modular arithmetic | (c,m) = d, ac ≡ bc (mod m) => a ≡ b (mod (m/d)) |
| a ≡ b and c ≡ d (mod m) => | a + c ≡ b + d (mod m), and ac ≡ bd (mod m) |
| Linear Congruences | ax ≡ b (mod m), (a,m) = d. d|b <==> d incongruent solutions x = x₀ + (m/d) t, 0≤t≤(d-1) |
| Chinese remainder theorem | Given equivalences mod pairwise prime numbers, there exist numbers fulfilling all conditions mod the product of all the pairwise prime numbers. |
| Modular Inverses | ax ≡ 1 (mod m) given (a,m) = 1 |
| Wilson's Theorem | If p prime then (p-1)! ≡ -1 (mod p) |
| Fermat's Little Theorem | p prime, a integer, a not divisible by p => a^(p-1) ≡ 1 (mod p) |
| FLT corollaries | 1. a^(p-2) is the inverse of a (mod p) 2. ax ≡ b (mod p) => x ≡ ba^(p-2) (mod p) |
| P(AUB) | P(A) + P(B) - P(A∩B) |
| Bertrand's Ballot Problem: 'a' gets more votes than 'b', what is the probability 'a' never trails during the election? | (a-b+1) choose (a+1) |
| P(A|B) | P(A∩B) / P(B) |
| Mutually exclusive events | P(A∩B) = 0 |
| Bayes' Theorem | P(A_i|B) = P(A_i∩B) / P(B), P(B) = P(B|A₁) P(A₁) + P(B|A₂) P(A₂) + ... |
| Independence Proofs | 1. P(A|B) = P(A) 2. P(B|A) = P(B) 3. P(A∩B) = P(A)P(B) |
| Expected Value | Sum over all possible events of P(event)*Value(event) |
| Multinomial Coefficient | (n choose k₁, k₂, k₃ ...) = n! / (k₁! k₂! k₃!...) |
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