Numbers and functions Flashcards
cardinality
number of elements in a set
Injective
A function is said to be injective if f(x)=f(y) ⇒ x=y
Surjective
A function f:A->B is surjective if f(A)=B. Each element of the codomain is mapped to by some element of the domain.
Bijective
Both injective and Surjective
Contrapositive
A⇒B, the contrapositive is: not B⇒not A (always true if A⇒B is true)
Converse
A⇒B, the converse is B⇒A (not always true)
How do you prove two sets are equal?
Prove the lhs is a member of the rhs and prove that the rhs is a member of the lhs.
Bounded above
∃M∈ℝ, ∀x∈S, such that x≤M
Bounded Below
∃M∈ℝ, ∀x∈S, such that x≥M
Unbounded above
∀M∈ℝ, ∃x∈S, such that x>M
Maximum
Let S be bounded above and suppose that there exists an upper bound M of S such that M∈S. M is the maximum of S.
infimum
Greatest lower bound
supremum
Least upper bound
minimum
Let S be bounded below and suppose that there exists a lower bound m of s such that m∈S.
Axiom of completeness
Every nonempty set of real numbers which is bounded above has a supremum. Every non-empty set of Real numbers which is bounded below has an infimum.
Bounded
|S(n)|≤M, ∀n∈ℕ
Convergence
∀ℇ>0, ∃n(0)∈ℕ such that ∀n≥n(0), we have |S(n)-𝓁|
Divergence
If the sequence S(n) does not converge to any limit it is said to diverge.
Divergence to ∞
The sequence S(n) is said to diverge to ∞, for which we write S(n)→∞, if for every positive real number H, there exists n(0) such that ∀n≥n(0) we have S(n)>H.
Triangle inequality
|x+y| ≤ |x|+|y|
Sandwich Theorem
Let r(n)→𝓁 and t(n)→𝓁 as n→∞ and suppose that r(n)≤s(n)≤t(n) for all n∈ℕ. Then s(n)→𝓁 as n→∞.
If s(n)→0 as n→∞ and t(n) is a bounded sequence, then s(n)t(n)→? as n→∞
s(n)t(n)→0
If s(n)→𝓁 and t(n)→∞ as n→∞, then s(n)t(n)→? as n→∞
s(n)t(n)→∞
Let s(n)→∞ or s(n)→-∞ as n→∞ and s(n)≠0 for all n. then 1/s(n)→? as n→∞
1/s(n)→0
Put these functions in order of their rate of convergence:
Exponentials, logarithms, factorials, powers
Slowest: Logarithms
Powers
Exponentials
Fastest: Factorials
Bernoulli inequality
For every k∈ℕ and x≥-1, one has (1+x)^k≥1+kx
Little o
Let s(n) and t(n) be two sequences, such that t(n)≠0 for all n. Suppose that as n→∞, s(n)/t(n)→0. Then we can say s(n)=o(t(n)) as n→∞
Big O
Let s(n) and t(n) be sequences. Suppose that there exists C>0 such that for all n∈ℕ, one has |s(n)|≤C|t(n)|. Then one writes s(n)=Ot(n) as n→∞. I.e. s(n)/t(n) is Bounded.
lim(1+1/n)^n
e
increasing sequence
s(n+1)>s(n) for all n∈ℕ
decreasing sequence
s(n+1)
non-decreasing sequence
s(n+1)≥s(n) for all n∈ℕ
non-increasing sequence
s(n+1)≤s(n) for all n∈ℕ
monotone sequence
neither non-increasing nor non-decreasing
True or false:
Every non-increasing sequence which is bounded above is convergent.
FALSE
Every non-decreasing sequence which is bounded above is convergent.
The Bolzano-Weierstrass Theorem
Every bounded sequence has a convergent subsequence, which converges to a limit 𝓵 which is between the bounds of the original sequence.
Cauchy Sequence
s(n) is a cauchy sequence if for all ℇ>0 there exists a n(0) such that |s(m)-s(n)|
Absolute convergence
if the series 𝚺|a(k)| from k=1 to ∞ is convergent then 𝚺a(k) from k=1 to ∞ is absolutely convergent
conditional convergence
If the series 𝚺a(k) from k=1 to ∞ converges but 𝚺|a(k)| from k=1 to ∞ does not then 𝚺a(k) from k=1 to ∞ is conditionally convergent.
The comparison test
Let 𝚺b(k) from k=1 to ∞, be a convergent series of non-negative numbers and suppose that for some constant M>0 we have |a(k)|≤Mb(k) for all k. Then the series 𝚺a(k) from k=1 to ∞ is absolutely convergent.
The alternating series test
Let a(k) be a non-increasing sequence of positive numbers such that a(k)→0 as k→∞. Then the series 𝚺a(k)(-1)^(k+1) from k=1 to ∞ converges.
How would you negate: P⇒Q ?
not-Q & P
How would you negate: (A∩B) ?
not-A OR not-B
How would you negate: A⇔B ?
(not-A & B) OR (A & not-B)
How would you negate: (A∪B) ?
not-A OR not-B
|A|<1 ⇒A^n→? as n→∞
0
True or False: If a(k)→0 as k→∞, then the series 𝚺a(k) from k=1 to ∞ must converge.
FALSE
Although it is true that if 𝚺a(k) from k=1 to ∞ converges, then one must have If a(k)→0 as k→∞.
