Calc Week 1 Flashcards
To denote natural numbers
fancy capital N with a line with the first one
Sequence and how we denote
a sequence a(n) is a function whose domain is all natural numbers
- an infinite list of number with definite ordering
{a sub n} or (a sub n)
a sequence’s terms and how we denote them
a(1), a(2), a(3)….
a sub 1, a sub 2, a sub 3, ….
1 - Oscillation Sequence
2- exponential decay sequence
3-Fibonacci sequence
4- harmonic sequence
5 - alternating harmonic sequence
1 - alternates from 1 value to another
{ 4 + (-1) ^ n }
2- gets close to 0 fast
{ 1 / 10^ n}
3- the 2 numbers before it add up to the next
4- {1 / n}
5- alternates from negative to positive
{ (-1)^n / n }
null sequence and how to identify one
a sequence {a sub n} converges to zero if
for every epsilon > 0, there exists a positive number m such that
| a sub n| < epsilon
whenever n >= m
for any positive distance epsilon, if you go far enough along the sequence, you will eventually reach a cutoff (m) where the terms of sequence are within epsilon distance of zero
identify one by :
finding its limit
- “plug” in the limit (what has the biggest weight, its reciprocal is multiplied to the top and bottom)
-convert to a function (usually when we would have to use L’H with the sequence because we cannot)
- multiplying by conjugate
(subtracting or adding infinity, multiply time the opposite sign of the same thing on top and bottom)
converge
a sub n - L| < epsilon whenever n >= m
{a sub n } converges to L if for every epsilon > 0 there exists a positive number m such that
diverges
{ a sub n } diverges to infinity if for every natural number c there is a cutoff m such that a sub n > c when every n >= m
limit
if {a sub n} converges to L
L is the limit of {a sub n} as n approaches infinity
Finding the limit
- calculation (theorem, just plug in the limit)
-converting to a function(usually to use L’H rule with it) - multiplication by conjugate(subtracting or adding infinity, multiply times the same thing (on top and bottom but with opposite sign)
algebraic limit theorem
Let {a sub n} and {b sub n} be sequences such that a ub n -> a and b sub n -> b as n -> infinity
then:
lim n -> inf ( a sub n + b sub n) = a + b
lim n -> inf (a sub n - b sub n) = a - b
lim n->inf (a sub n * b sub n) = ab
lim n->inf ( c * a sub n) = ca
lim n->inf ( a sub n/ b sub n) = a/b (as long as we do not get indefinite/undetermined)
bounded
there exists real numbers m and M such that
m <= a sub n <= M for all n in natural numbers
unbounded
if {a sub n} is not bounded
bounded above
if {a sub n} <= M for all n
bounded below
if m <= a sub n for all n
theorem of convergence then…
let {a sub n} be a convergent sequence.
then {a sub n} is bounded