Analysis 1 Flashcards
Axioms for the real numbers:
What are the addition axioms?
A1 a + b = b + a [+ is commutative]
A2 a + (b + c) = (a + b) + c [+ is associative]
A3 a + 0 = a [zero and addition]
A4 a + (−a) = 0 [negatives and addition]
Axioms for the real numbers:
What are the multiplication axioms?
M1 a · b = b · a [· is commutative]
M2 a · (b · c) = (a · b) · c [· is associative]
M3 a · 1 = a [the unit element and multiplication]
M4 If a ≠ 0 then a · 1/a = 1 [reciprocals and multiplication]
Axioms for the real numbers: What are the non-addition/multiplication axioms? (2 of them)
D a · (b + c) = a · b + a · c [· distributes over +]
Z 0 ≠ 1 [to avoid total collapse]
Properties of real numbers:
Prove using the axioms
(1) If a + x = a for all a then x = 0 (uniqueness of zero element)
(1) We have
x = x + 0 by A3
= 0 + x by A1
= 0 by the hypothesis, with a = 0.
Properties of real numbers:
Prove using the axioms
(2) If a + x = a + y then x = y (cancellation law for addition, which implies uniqueness of
additive inverse of a).
(2) We have y = y + 0 by A3 = y + (a + (−a)) by A4 = (y + a) + (−a) by A2 = (a + y) + (−a) by A1 = (a + x) + (−a) by hypothesis = (x + a) + (−a) by A1 = x + (a + (−a)) by A2 = x + 0 by A4 = x by A3.
Properties of real numbers:
Prove using the axioms
(3) −0 = 0.
(3) 0 + 0 = 0 by A3 with a = 0. Hence −0 = 0 by A4 and (2).
Properties of real numbers:
Prove using the axioms
(4) − (−a) = a.
(4) We have (−a) + a = a + (−a) by A1 = 0 by A4, and (−a) + − (−a) = 0 by A4. Now appeal to (2) (cancellation law for addition).
Properties of real numbers:
Prove using the axioms
(5) − (a + b) = (−a) + (−b)
proof
Properties of real numbers:
Prove using the axioms
(6) If a · x = a for all a 6= 0 then x = 1 (uniqueness of multiplicative identity)
Use M1-4
Properties of real numbers:
Prove using the axioms
(7) If a 6= 0 and a · x = a · y then x = y (cancellation law for multiplication, which implies
uniqueness of multiplicative inverse of a)
Use M1-4
Properties of real numbers:
Prove using the axioms
(8) If a 6= 0 then 1/ (1/a) = a
Use M1-4
Properties of real numbers:
Prove using the axioms
(a + b) · c = a · c + b · c.
(9) Note that the statement is like Axiom D but with multiplication on the other side. We have (a + b) · c = c · (a + b) by M1 = c · a + c · b by D = a · c + b · c by M1 twice.
Properties of real numbers:
Prove using the axioms
(10) a · 0 = 0
(10) We have a · 0 + 0 = a · 0 by A3 = a · (0 + 0) by A3 = a · 0 + a · 0 by D. Now appeal to (2) (cancellation law for addition)
Properties of real numbers:
Prove using the axioms
(11) a · (−b) = − (a · b). In particular (−1) · a = −a
(11) We have (a · b) + (a · (−b)) = a · (b + (−b)) by D = a · 0 by A4 = 0 by (10). Also (a · b) + (−(a · b)) = 0 by A4. Now appeal to (2).
Properties of real numbers:
Prove using the axioms
· b). In particular (−1) · a = −a.
(12) (−1) · (−1) = 1
(12) We have
(−1) · (−1) = −(−1) by (11)
= 1 by (4)
Properties of real numbers:
Prove using the axioms
(13) If a · b = 0 then either a = 0 or b = 0 (or both). Moreover, if a 6= 0 and b 6= 0 then
1/(a · b) = (1/a) · (1/b)
(13) Assume for a contradiction that a, b ≠ 0 but a · b = 0. Then 0 = (1/a · 1/b) · 0 by (10) = 0 · (1/a · 1/b) by M1 = (a · b) · (1/a · 1/b) by hypothesis = ((b · a) · 1/a) · 1/b by M2 = (b · (a · 1/a)) · 1/b by M2 = (b · 1) · 1/b by M4 = b · (1/b) by M3 = 1 by M4, This contradicts Axiom Z. The second assertion has been proved along the way
There is a subset P (the (strictly) positive numbers) of R such that, for a, b ∈ R,
P1
P2
P3 ???
P1 a, b ∈ P =⇒ a + b ∈ P;
P2 a, b ∈ P =⇒ a · b ∈ P;
P3 exactly one of a ∈ P, a = 0 and −a ∈ P holds
Describe what we mean by > and ≥ using the (strictly) positive numbers set and the non-negative numbers set
There is a subset P (the (strictly) positive numbers) of R such that, for a, b ∈ R
We write a < b (or b > a) iff b − a ∈ P and a ≤ b (or b ≥ a) iff b − a ∈ P ∪ {0}
(the non-negative numbers)
Explain each property of the order on R (reals)
Reflexivity
Reflexivity: a ≤ a.
Proof. a − a = 0 ∈ P ∪ {0}, by A4
Explain each property of the order on R (reals)
Antisymmetry
Antisymmetry: a ≤ b and b ≤ a together imply a = b.
Proof. If a − b = 0 or b − a = 0, then a = b by properties of addition. Otherwise
P 3 a − b and P 3 b − a = −(a − b) (by properties of addition). Now apply P3.
Explain each property of the order on R (reals)
Transitivity:
Transitivity: Assume a ≤ b and b ≤ c. Then a ≤ c, and likewise with < in place of ≤.
Proof. We have c−a = c+ (−a) = c+ 0 + (−a) = c+ (−b) +b+ (−a) = (c−b) + (b−a)
by properties of addition. So the result for < follows from the definition of < and P3.
The proof for ≤ is the same except for the need to allow also for the trivial cases
a = b and/or b = c.
Explain each property of the order on R (reals)
Trichotomy
Trichotomy: Exactly one of a < b, a = b and b < a holds.
Proof. This follows from P3 and the definition of
The following statements hold: (1) 0 < 1 (equivalently, 1 ∈ P), (2) a < b if and only if −b [ ]. In particular a > 0 iff [ ]. (3) a < b and c ∈ R implies a + c [ ]. (4) a < b and 0 < c implies ac < [ ]. (5) a² > 0, with equality iff [ ] (6) a > 0 iff 1/a [ ]. (7) If a, b > 0 and a < b then 1/b [ ] Claims (2)–(4) also hold with ≤ replacing <
(1) 0 < 1 (equivalently, 1 ∈ P),
(2) a < b if and only if −b < −a. In particular a > 0 iff −a < 0.
(3) a < b and c ∈ R implies a + c < b + c.
(4) a < b and 0 < c implies ac < bc.
(5) a² > 0, with equality iff a = 0.
(6) a > 0 iff 1/a > 0.
(7) If a, b > 0 and a < b then 1/b < 1/a.
Claims (2)–(4) also hold with ≤ replacing
Prove the following:
(1) 0 < 1 (equivalently, 1 ∈ P),
(1) By trichotomy, exactly one of (i) 1 < 0, (ii) 1 = 0, (iii) 0 < 1 holds. Axiom Z rules
out (ii). Suppose for a contradiction that 1 < 0. Then −1 = 0 + (−1) ∈ P. We deduce
that (−1) · (−1) ∈ P by P2. But 1 = (−1) · (−1) (by 1.2(12)), so 0 < 1 and we have a contradiction to trichotomy.
Prove the following:
(2) a < b if and only if −b < −a. In particular a > 0 iff −a < 0
(2) By properties of addition,
a < b ⇐⇒ b − a ∈ P
⇐⇒ (−a) − (−b) ∈ P
⇐⇒ (−a) > (−b).
Prove the following:
(3) a < b and c ∈ R implies a + c < b + c
(3) a + c < b + c iff 0 < (b + c) − (a + c) = b − a iff a < b
Prove the following:
(4) a < b and 0 < c implies ac < bc
(4) a < b and c > 0 implies bc − ac = (b − a)c > 0, by P2.
Prove the following:
(5) a² > 0, with equality iff a = 0
(5) Note a² = 0 iff a = 0, by {If a · b = 0 then either a = 0 or b = 0 (or both). Moreover, if a ≠ 0 and b ≠0 then
1/(a · b) = (1/a) · (1/b).}.
Assume a ≠ 0. Then we have a² = a·a = (−a)·(−a).
Since either a > 0 or −a > 0 it follows that a² > 0 by P2.
Prove the following:
(6) a > 0 iff 1/a > 0
Assume for contradiction a > 0 but 1/a < 0. Then −1 = −(a · (1/a)) = a · (−1/a) > 0
by P2 which is impossible by (1). Likewise we obtain a contradiction if a < 0 and
1/a > 0.
Prove the following:
(7) If a, b > 0 and a < b then 1/b < 1/a
Use
a < b and 0 < c implies ac < bc
and
a > 0 iff 1/a > 0
What is Bernoulli’s Inequality?
Let x be a real number with x > −1 and let n be a positive integer. Then
(1 + x)ⁿ ≥ 1 + nx.
Prove Bernoulli’s Inequality
We shall prove the inequality by induction—note that the inequality is trivially true
when n = 1.
Suppose that, for k ∈ N,
(1 + x)ᵏ ≥ 1 + kx
holds for all real x > −1. Then 1 + x > 0 by {a < b and c ∈ R implies a + c < b + c} and kx² ≥ 0 as k > 0 and x² ≥ 0 by
{a² > 0, with equality iff a = 0.}
(1 + x)ᵏ⁺¹ = (1 + x) (1 + x)ᵏ by definition
≥ (1 + x) (1 + kx) by hypothesis and {a < b and 0 < c implies ac < bc} (≤ version)
= 1 + (k + 1) x + kx²
by A1–A4
≥ 1 + (k + 1) x by {Inequality rules}.
Hence the result follows by induction.
Define the modulus
real numbers
The modulus |a| of a ∈ R is defined by
{ a if a > 0
|a| = {0 if a = 0
{ -a if a < 0
Prove the following facts about the modulus:
(1) | − a| = |a|;
(2) |a| ≥ 0
(1) and (2) are immediate from the definition and the fact that a > 0 iff −a < 0
Prove the following facts about the modulus:
(3) |a|² = a²;
We can prove (3) and also (4), by using P3 to enumerate cases; recall too that (−a)(−a) = a² for any a.
Prove the following facts about the modulus:
(4) |ab| = |a||b|;
We can prove (3) and also (4), by using P3 to enumerate cases; recall too that (−a)(−a) = a² for any a.
Prove the following facts about the modulus:
(5) −|a| ≤ a ≤ |a|
For (5), note that
{ −|a| ≤ 0 ≤ a = |a| if a ≥ 0,
{ −|a| = a < 0 ≤ |a| if a < 0
Prove the following facts about the modulus:
(6) if c ≥ 0, then |a| ≤ c iff −c ≤ a ≤ c. if c > 0, then |a| < c iff −c < a < c
Assume first that |a| 6 c, Then, by (5{−|a| ≤ a ≤ |a|} and transitivity of 6, we get
−c ≤ a ≤ c. Conversely, assume −c ≤ a ≤ c. Then −a ≤ c and a ≤ c. Since |a| equals
either a or −a, we obtain |a| ≤ c.
The case with < in place of ≤ is handled similarly.
What is the triangle law/inequality?
Real
(1) Let a, b ∈ R. Then
|a + b| ≤ |a| + |b|.
Prove the triangle inequality
Real
Proof. (1): We have
−|a| ≤ a ≤ |a| and − |b| ≤ b ≤ |b|.
Adding (if a < b, then ac > bc if and only if c < 0) and using properties of addition we get
−(|a| + |b|) ≤ a + b ≤ (|a| + |b|).
Now use 2.5(6){(6) if c ≥ 0, then |a| ≤ c iff −c ≤ a ≤ c. if c > 0, then |a| < c iff −c < a < c}
What is the reverse triangle law/inequality?
Real
(2) Let a, b ∈ R. Then
|a + b| ≥ ||a| − |b||
Prove the reverse triangle inequality
Real
By the Triangle Law,
|a| = |a + b + (−b)| ≤ |a + b| + |(−b)| = |a + b| + |b|,
so |a|−|b| ≤ |a+b|, and likewise, reversing the roles of a and b, we get |b|−|a| ≤ |b+a| = |a+b|.
Now use the fact that |c| is either c or −c always, and apply this with c = |a| − |b|
What are the triangle inequality and reverse triangle inequality for complex numbers?
z + w| ≤ |z| + |w| and |z + w| ≥ ||z| − |w||
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Upper bound
b is an upper bound of S if s ≤ b for all s ∈ S;
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Lower bound
b is a lower bound of S if b ≤ s for all s ∈ S;
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Bounded above
S is bounded above if it has an upper bound;
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Bounded below
S is bounded below if it has a lower bound;
Given:
Let S ⊆ R and b ∈ R.
Define the following:
Bounded
S is bounded if it is bounded above and below.
Let S ⊆ R and b ∈ R.
What is the Supremum?
Let S ⊆ R. Then α is the supremum of S, denoted sup S, if
(sup1) s ≤ α for all s ∈ S [α is an upper bound of S]
(sup2) s ≤ b for all s ∈ S implies α ≤ b [α is the least upper bound of S]
Note: sup S is unique if it exists.
What is the Completeness Axiom for the Real Numbers?
Let S be a non-empty subset of R which is bounded above. Then sup S
exists
Note the necessity for the exclusions here. The empty set has no supremum because it has no least upper bound ((sup2) fails). A set which is not bounded above cannot have a supremum because it has no upper bound ((sup1) must fail).
Let S ⊆ R and b ∈ R.
What is the Infimum?
Infimum (= greatest lower bound).
Analogous definitions apply here, by replacing ≤ by > in the definitions above. Let S be a subset of R and α ∈ R. Then α is the infimum of S, written inf S, if
(inf1) α ≤ s for all s ∈ S [α is a lower bound for S]
(inf2) if b ≤ s for all s ∈ S then b 6 α [α is the greatest lower bound of S]
Assume ∅ ≠ S ⊆ R and let s₀ ∈ R
Define the maximum of S
We say s₀ is the maximum of S, and write
s₀ = max S, if
(max1) s₀ ∈ S [s₀ belongs to S]
(max2) s ≤ s₀ for all s ∈ S [s₀ is an upper bound of S]
If a set S is empty or is not bounded above then max S cannot exist
Assume ∅ ≠ S ⊆ R and let s₀ ∈ R
Define the minimum of S
Similarly we say a non-empty set S which is bounded below has a minimum, min S, if there exists s₀ ∈ S such that s₀ ≤ s for all s ∈ S.
Given a, x ∈ R we may interpret |x − a| as ….
Given a, x ∈ R we may interpret |x − a| as the distance from x to a.
For b > 0, and a, x ∈ R,
|x − a| < b ⇐⇒ a − b < x [ ]
For b > 0, and a, x ∈ R,
|x − a| < b ⇐⇒ a − b < x < a + b
For b > 0, and a, x ∈ R,
|x − a| < b ⇐⇒ a − b < x < a + b
prove it
{ if c ≥ 0, then |a| ≤ c iff −c ≤ a ≤ c. if c > 0, then |a| < c iff −c < a < c}
gives |x − a| < b iff −b < (x − a) < b and this holds iff a − b < x < a + b.
So by considering whether this holds for different values of b we can assess how good an
approximation x is to a.
Let S be non-empty and bounded above (so sup S exists). Then, given ε > 0, there exists
sε (in general depending on ε) in S such that
sup S − ε < [ ]
sup S − ε < sε ≤ sup S.
Prove that there exists a unique positive real number α such that α² = 2
End of pg15
Strategy:
Step 1: S ≠ ∅ and S is bounded above, by 2.
Step 2: Assume for contradiction that α² < 2. Note also that α ≥ 1 > 0 since 1 ∈ S. Let
h > 0.
Step 3: Assume for contradiction that α² > 2. Let h > 0.
Explain the incompleteness of Q
The set Q of rationals does not satisfy the Completeness
Axiom with respect to the order it inherits from R. If it did,
T := { q ∈ Q | q > 0 and q² < 2 }
would have a supremum in Q. The proof in 4.10{proof of existence of root 2} works just as well for T as it does for S.
But we know there is no rational square root of 2.
Describe the theorem of the existence of nth roots of real numbers
Any positive real number has a real nth root,
for any n = 2, 3, 4, . . .
Theorem (the Archimedean Property of the Natural Numbers).
(i) N is not bounded above.
(ii) Let ε > 0. Then there exists n ∈ N such that 0 < 1 / n < ε
Prove
Theorem (the Archimedean Property of the Natural Numbers).
(i) Assume for a contradiction that N is bounded above. Then sup N exists by the
Completeness Axiom. By the Approximation Property with ε = 1/2, there exists k ∈ N
with
sup N − 1 / 2 < k ≤ sup N.
But then k + 1 ∈ N and k + 1 > sup N + 1 / 2, a contradiction.
For (ii), exploit the fact that 1/ε cannot be an upper bound for N.
Theorem (compare with the well-ordered property of N)
Minima and maxima of subsets of Z
(i) Every nonempty subset S of Z which is bounded below has a minimum (least element).
(ii) Every nonempty subset S of Z which is bounded above has a maximum (greatest
element) .
Prove:
(i) Every nonempty subset S of Z which is bounded below has a minimum (least element).
(ii) Every nonempty subset S of Z which is bounded above has a maximum (greatest
element) .
Proof. i) We know that inf S exists (by applying the completeness axiom to {−s : s ∈ S}. So by the approximation property with ε = 1 there is some n ∈ S such that
inf S ≤ n < inf S + 1.
It is enough to show that inf S = n, since then inf S ∈ S and so inf S = min S. Assume for
a contradiction that n ≠ inf S, so that n > inf S and hence n = inf S + ε where 0 < ε < 1.
By the approximation property again, there exists some m ∈ S such that
inf S ≤ m < inf S + ε = n.
Since n > m we have n − m > 0 and so n − m ≥ 1 because n − m is an integer, so
n ≥ inf S + 1, which contradicts our first inequality for n.
The proof of (ii) is similar.
What are the density properties?
Density properties.
(i) Given a, b ∈ R with a < b there exists x ∈ Q such that a < x < b.
(ii) Given a, b ∈ R with a < b there exists y ∈ R \ Q such that a < y < b
When are two sets equinumerous?
Let A and B be sets. We say A and B are equinumerous (notation A ≈ B) if there is
a bijection f : A → B. Note ≈ has the properties of an equivalence relation.
What does the symbol A ≤ B? If A and B are two sets
Note:
Not less than or equal to, concave version of the symbol
Given sets A and B we shall write A ≤ B if there is an injection f : A → B. Intuitively
this says that B is at least as big as A.
What is a finite set?
Let A be a set. We call A finite if either A = ∅ or there exists n ∈ N such that
A ≈ {0, . . . , n − 1} (or equivalently if A ≈ {1, . . . , n}).
What is an infinite set?
A set which is not finite is said to be infinite
What is another way to define finiteness of a set, using maps?
Another way to capture the notion of finiteness is to say that A is finite iff every
injective map from A to A is surjective: the Pigeonhole Principle holds for A
Any subset of a finite set is [ ]
finite.
any non-empty finite subset of R must be [ ] (in fact, it contains a [ ])
Hence any subset of R which is not bounded above must be [ ]
bounded above
largest element
infinite
By the Archimedean property, N is not bounded above, and hence is [ ]
infinite
We call a set A
• countably infinite if [ ];
• countable if [ ]
• uncountable if A is [ ]
- countably infinite if A ≈ N;
- countable if A ≤ N; {curvy ≤}
- uncountable if A is not countable
(1) A is countable (that is, [ ]) iff A is finite or [ ].
(2) If A ≤ B and B ≤ A then A ≈ B.
A ≤ N {curvy ≤}
countably infinite
State 5 countably infinite sets
(a) N;
(b) N>⁰:= N \ {0};
(c) { 2k + 1 | k ∈ N } (the odd natural numbers);
(d) Z;
(e) N × N
Prove that the following sets are countably infinite:
N>⁰:= N \ {0};
the successor function, n 7→ n + 1, is a bijection from N
to N \ {0}
Prove that the following sets are countably infinite:
{ 2k + 1 | k ∈ N } (the odd natural numbers);
note that the map 2k + 1 7→ k is injective and maps the given set onto N
Prove that the following sets are countably infinite:
Z
We can define a bijection f from Z to N by
f(k) = { −2k if k ≤ 0,
{ 2k − 1 if k > 0
Prove that the following sets are countably infinite:
N x N
by setting up a bijection f : N × N → N. Define f by
f((m, n)=2ᵐ(2n + 1) − 1.
Injectivity of f: 2ᵐ¹ (2n₁ + 1) = 2m2 (2n2 + 1) implies (by uniqueness of factorisation in N (assumed)) that 2ᵐ¹ = 2ᵐ² so that m₁ = m₂ (make use of laws of indices for the last step) and 2n₁ + 1 = 2n₂ + 1, whence n1 = n2. So (m₁, n₁) = (m₂, n₂).
Surjectivity of f: take k ∈ N. Assume first that 2 -| (k + 1). Then k is even and so
k = 2⁰(2n + 1) − 1, for some n ∈ N. Now assume 2|(k + 1). Then there exists m such that
2ᵐ | (k + 1) and 2m+1 -| (k + 1) (use the fact that { 2ᵐ | m ∈ N } is not bounded above. Then k + 1 = 2ᵐ(2n + 1), for some n ∈ N.
Let A and B be countable sets. Let f : A → N and g : B → N be injective.
Claim 1: Assume A and B are disjoint. Then A∪B is [ ]. [Disjointness not essential, but it simplifies the proof.]
countable
Prove that:
Let A and B be countable sets. Let f : A → N and g : B → N be injective.
Claim 1: Assume A and B are disjoint. Then A∪B is countable. [Disjointness not essential,
but it simplifies the proof.]
Define
h(x) = { 2f(x) if x ∈ A,
{ 2g(x) + 1 if x ∈ B.
Then h is an injection from A ∪ B to N.
Let A and B be countable sets. Let f : A → N and g : B → N be injective.
Claim 2: A × B is [ ].
countable
Prove that
Let A and B be countable sets. Let f : A → N and g : B → N be injective.
Claim 2: A × B is countable
Define h: A × B → N by
h((a, b)) = 2ᶠ⁽ᵃ⁾ ⁺ ¹3ᵍ⁽ᵇ⁾ ⁺ ¹ 1 for a ∈ A, b ∈ B.
Then uniqueness of factorisation in N>0
implies h is injective.
Theorem: Q is countable
Prove it
Proof. We write Q as the disjoint union
Q>0 ∪ {0} ∪ Q<0
(>0 is superscript)
By {Let A and B be countable sets. Let f : A → N and g : B → N be injective.
Claim 1: Assume A and B are disjoint. Then A∪B is countable.}, it will be enough to prove that Q>0 (and so, likewise, Q<0) is countable.
We can write each element of Q>0 as p/q where p, q ∈ N, p > 0, q > 0 and p/q is expressed
in its lowest terms. Then p/q 7→ 2ᵖ3ᵠ is an injection into N. So Q>0 ≤(curvy}) N as claimed.
Is the set of the reals countable?
No
What is a real sequence?
A sequence of real numbers is an assignment
n |→ α(n) of a real number α(n) to each n = 1, 2, . . .. Thus a sequence is a function α: N ≥1 → R (where N≥1 = N \ {0} = {1, 2, 3, . . .}) and we call α(n) the nth term of the sequence. (Sometimes later we shall work with sequences with terms labelled by n = 0, 1, 2, . . . instead.)
We shall usually write an in place of α(n) and then say that α defines the sequence
(aₙ) = (a₁, a₂, a₃, . . .)
also written (aₙ)ₙ>₁ — that is, we specify the sequence by its terms. Note that the terms in order determine the sequence.
What is a complex sequence?
A complex sequence is defined in the same way as the reals however the terms aₙ are taken from C
What is the tail of a sequence?
Given a sequence (aₙ) and any k ∈ N we can form a new sequence (bₙ) by chopping off the first k terms a₁, . . . , aₖ of (aₙ) and relabelling. That is, bₙ = an+k for all n. We call (bₙ)
a tail of (aₙ).
Define the convergence of a real sequence
Convergence of a real sequence. Let (aₙ) be a sequence of real numbers and let L ∈ R. Then we say that (aₙ) converges to L if
∀ε > 0 ∃N ∈ N ∀n > N |aₙ − L| < ε.
Here N can, and almost always will, depend on ε. Note that we can replace ‘n > N’ by n ≥ N and/or ‘|an −L| < ε’ by |an −L| ≤ ε in this definition without changing the meaning. However it is crucial that ε should be strictly greater than 0.
What is the limit of a sequence?
When an → L we say that L is the limit of (aₙ) and we write
L = lim n→∞ aₙ or just L = lim aₙ.
We say (aₙ) converges if there exists L ∈ R such that aₙ → L as n → ∞. We say (aₙ)
diverges if it does not converge.
The limit is unique
What is the tails lemma?
Tails Lemma. Let (an) be a sequence.
(i) If (aₙ) converges to a limit L then each tail of (aₙ) converges, to the same limit L.
(ii) Assume some tail (bₙ) = (aₙ₊ₖ) converges. Then (aₙ) converges.
Prove the tails lemma
pg22
What is the simple form of the sandwiching lemma?
Let (bₙ) and (cₙ) be real sequences. Assume
cₙ → 0 and that 0 ≤ bₙ ≤ cₙ for all n. Then bₙ → 0.
Prove the simple form of the sandwiching lemma
Proof. Let ε > 0 and pick N so that |cₙ − 0| < ε for all n > N. Then, for n ≥ N,
−ε < 0 ≤ bₙ ≤ cₙ = |cₙ| < ε.
and so |bₙ − 0| < ε for all n > N.
[What this says is that, given ε, an N that works for (cₙ) also works for (bₙ).]
What is the . Theorem (uniqueness of limit)?
Let (aₙ) be a sequence and suppose that aₙ → L₁
and aₙ → L₂ as n → ∞. Then L₁ = L
Prove the uniqueness of limits
end pg 24
pg 25
Assume that (aₙ) is a sequence which converges to L. Then (|aₙ|) converges too, to [ ]
|L|.
Prove that
Assume that (aₙ) is a sequence which converges
to L. Then (|aₙ|) converges too, to |L|.
By the Reverse Triangle Law,
||aₙ− |L|| ≤ |aₙ − L|.
Now apply the Sandwiching Lemma, or argue directly from the convergence definition
Limits;
What is the Preservation of weak inequalities?
Assume that an → L and bₙ→ M and that an ≤ bₙ
for all n. Then L ≤ M.
Prove the Preservation of weak inequalities
pg 25
What is the general sandwiching lemma?
Assume that (xₙ), (yₙ) and (aₙ) are real sequences
such that xₙ ≤ aₙ ≤ yₙ for all n. Assume that lim xₙ = lim yₙ = L. Then (aₙ) converges
to L.
Prove the general sandwiching lemma
Outline Given ε > 0 we can find N such that for all n ≥ N we have |xₙ − L| < ε and |yₙ − L| < ε. Then, for n > N, L − ε < xₙ ≤ aₙ ≤ yₙ < L + ε, so |aₙ − L| < ε
Describe
Proposition (a convergent sequence is bounded)
Assume (aₙ) converges. Then
∃M ∈ R ∀n |aₙ| ≤ M.
(This says that (aₙ) is bounded, meaning that its set { aₙ | n = 1, 2, . . . } of terms is bounded.)
If (aₙ) is not bounded then (aₙ) diverges
Prove
Proposition (a convergent sequence is bounded)
pg26
Let aₙ be a sequence of real numbers
We say ‘aₙ tends to infinity’ and write
aₙ → ∞ as n → ∞ if ….
∀M ∈ R ∃N ∈ N ∀n ≥ N aₙ > M.
Here we tend to think of M as being a very large positive/negative number.
Let bₙ be a sequence of real numbers
we write bₙ → −∞ if
∀M ∈ R ∃N ∈ N ∀n ≥ N bₙ < M.
Here we tend to think of M as being a very large positive/negative number.
For α a real number we define n^α to be…
exp(α log n)
Describe how n^α behaves as n → ∞
We have
(i) if α < 0, then n^α → 0 as n → ∞;
(ii) if α > 0, then n^α → ∞ as n → ∞.
Prove:
We have
(i) if α < 0, then n^α → 0 as n → ∞;
(ii) if α > 0, then n^α → ∞ as n → ∞.
pg27
Powers with exponent n. Let c be a positive constant.
(i) If c < 1 then (cⁿ) [ ]
(ii) If c = 1 then (cⁿ) [ ]
(iii) If c > 1 then cⁿ→ [ ]
(i) If c < 1 then (cⁿ) converges to 0.
(ii) If c = 1 then (cⁿ) converges to 1.
(iii) If c > 1 then cⁿ→ ∞.
Theorem (convergence of complex sequences)
Let (zₙ) be a sequence of complex numbers and write zₙ = xₙ + iyₙ, so (xₙ) and (yₙ) are real sequences. Then (zₙ)
converges if and only if (xₙ) and (yₙ) both converge.
Prove
Theorem (convergence of complex sequences)
End pg27
Informally, what is the definition of a subsequence?
Let (aₙ)ₙ≥₁ be a (real or complex) sequence. Informally, a subsequence
of (aₙ)ₙ≥₁ is a sequence (bᵣ)ᵣ≥₁ whose terms are obtained by taking infinitely many terms
from (an)ₙ≥₁, in order.
What is the formal definition of a subsequence?
Formally, a subsequence (bᵣ)ᵣ≥₁ of the sequence (aₙ)ₙ≥₁ is defined by a map f : N → N
such that f is strictly increasing (meaning that r < s implies f(r) < f(s)), so that
bᵣ := aₙᵣ, where nᵣ = f(r).
Expressing this another way, we have a infinite sequence of natural numbers
n₁ < n₂ < n₃ < .. . and the sequence (bᵣ) = (aₙᵣ) has terms
aₙ1, aₙ2, aₙ3, . . . .
Describe
Proposition (subsequences of a convergent sequence)
Let (aₙ) be a sequence.
(i) Assume (aₙ) converges to L. Then every subsequence (aₙᵣ) of (aₙ) converges, to the
same limit L.
(ii) Assume (aₙ) has subsequences which converge to limits L and M where L ≠ M. Then (aₙ) does not converge
Prove
Proposition (subsequences of a convergent sequence)
pg29
Algebra of limits: Assume that (aₙ) and (bₙ) are real or complex sequences and assume that (aₙ) converges to L and (bₙ) converges to M. Then the following hold as n → ∞
Constant rule
(i) (constant) If aₙ = a (constant) for all n then aₙ → a.
Algebra of limits: Assume that (aₙ) and (bₙ) are real or complex sequences and assume that (aₙ) converges to L and (bₙ) converges to M. Then the following hold as n → ∞
Addition rule
(ii) (addition) aₙ + bₙ → L + M.
Algebra of limits: Assume that (aₙ) and (bₙ) are real or complex sequences and assume that (aₙ) converges to L and (bₙ) converges to M. Then the following hold as n → ∞
Scalar multiplication rule
(iii) (scalar multiplication) caₙ → cL for any constant c.
Algebra of limits: Assume that (aₙ) and (bₙ) are real or complex sequences and assume that (aₙ) converges to L and (bₙ) converges to M. Then the following hold as n → ∞
Subtraction rule
(iv) (subtraction) aₙ − bₙ → L − M.
Prove the constant rule in the algebra of limits
(i) Immediate from convergence definition
Prove the Addition rule in the algebra of limits
End pg 30
Prove the Scalar multiplication rule in the algebra of limits
End pg 30
Prove the Subtraction rule in the algebra of limits
Use Scalar multiplication and addition (with c = -1)
(1) Assume aₙ → 0 and (bₙ) is bounded. Then aₙbₙ → [ ] as n → ∞
aₙbₙ → 0
(2) Assume aₙ → L and L ≠ 0. Then there exists N such that n > N implies |aₙ| > |L|/2 (and in particular an ≠ 0) and hence ∃N ∈ N ∀n ≥ N 1 / |aₙ| < [ ] (so (1/|an|) has a [ ] tail).
2/ |L|
bounded
Assume that (aₙ) and (bₙ) are real or complex sequences and assume that (aₙ) converges to L and (bₙ) converges to M. Then the following hold as n → ∞
Product rule
(vi) (product) aₙbₙ → LM.
Assume that (aₙ) and (bₙ) are real or complex sequences and assume that (aₙ) converges to L and (bₙ) converges to M. Then the following hold as n → ∞
Reciprocal rule
(vii) (reciprocal) If M ≠ 0, then 1/bₙ → 1/M.
Assume that (aₙ) and (bₙ) are real or complex sequences and assume that (aₙ) converges to L and (bₙ) converges to M. Then the following hold as n → ∞
Quotient rule
(viii) (quotient) aₙ/bₙ → L/M if M ≠ 0.
Let (aₙ) be a sequence of
positive real numbers. The following are equivalent:
(a) aₙ → ∞ as n → ∞;
(b) 1/aₙ [ ]
1/aₙ → 0 as n → ∞.
For any rational number a > 0, log n/nᵃ → [ ] as n → ∞
0
As a rule of thumb, when it comes to the behaviour of functions f(x) for large x: Polynomials Trig Hyperbolics Logs Exponentials
trig functions & constants < logarithms
< polynomials < positive exponentials & hyperbolic functions.
What is O notation?
Let (an) and (bn) be real or complex sequences. We write
aₙ = O(bₙ) as n → ∞ if there exists c such that for some N
n ≥ N ⇒ |aₙ| ≤ c|bₙ|.
Note that, if aₙ = O(bₙ) and bₙ → 0 as n → ∞, then aₙ → 0 as n → ∞ too.
What is o notation?
We write
aₙ = o(bₙ) as n → ∞ if aₙ/bₙ is defined and
aₙ / bₙ → 0
as n → ∞.
Let (aₙ) be a real sequence
Define monotonic increasing
(aₙ) is monotonic increasing if aₙ ≤ aₙ₊₁ for all n;
Let (aₙ) be a real sequence
Define monotonic decreasing
(aₙ) is monotonic decreasing if aₙ ≥ aₙ₊₁ for all n;
Let (aₙ) be a real sequence
Define monotonic
(aₙ) is monotonic if it is either monotonic decreasing or monotonic increasing
What is the Monotonic Sequence Theorem?
Let (aₙ) be a real sequence.
(i) Assume (aₙ) is monotonic increasing. Then (aₙ) converges if and only if it is bounded above (that is, there exists a finite constant M such that an ≤ M for all n).
(ii) Assume (aₙ) is monotonic decreasing. Then (aₙ) converges if and only if it is bounded below.
Prove the monotonic Sequence Theorem
end pg 34
A real sequence which has a tail which is monotonic increasing converges iff [ ]
it is bounded above.
A real sequence which is monotonic increasing and not bounded above tends to [ ]
∞
What is the scenic view point theorem?
Let (aₙ) be a real sequence. Then (aₙ) has a
monotonic subsequence.
Prove the scenic view point theorem
pg37
What is the Bolzano–Weierstrass Theorem (for real sequences)?
Let (aₙ) be a bounded real
sequence. Then (aₙ) has a convergent subsequence.
What is the Bolzano–Weierstrass Theorem (for complex sequences)?
Let (zₙ) be a bounded
sequence in C. Then (zₙ) has a convergent subsequence.
What is a Cauchy sequence?
Let (aₙ) be a real or complex sequence. Then (aₙ) is a Cauchy sequence if
it satisfies the Cauchy condition:
∀ε > 0 ∃N ∈ N ∀m, n ≥ N |aₙ − aₘ| < ε.
It is not sufficient in the Cauchy condition just to consider adjacent terms, that is, only
to consider m = n + 1
Facts about Cauchy sequences:
A Cauchy sequence (aₙ) is [ ]
bounded.
Facts about Cauchy sequences:
A [ ] sequence is a Cauchy sequence
convergent
Facts about Cauchy sequences:
Let (aₙ) be a Cauchy sequence and assume that (aₙ) has a subsequence (aₙᵣ) which converges, to L say. Then (aₙ) converges to [ ].
L
What is the Theorem (Cauchy Convergence Criterion)?
Let (aₙ) be a (real or complex) sequence. Then
(aₙ) is convergent ⇐⇒ (aₙ) is a Cauchy sequence
What is a convergent series?
Let (aₖ) be a real or complex sequence. Let
sₙ = a₁ + a₂ + · · · + aₙ = ⁿΣₖ₌₁ aₖ (n ≥ 1).
Then we say that the series Σₖ≥₁ aₖ converges (or as shorthand that the series Σaₖ
converges) if the sequence (sₙ) of partial sums converges. If sₙ → s, then we write
∞Σₖ₌₁ aₖ = s. If (sₙ) fails to converge, then we say Σₖ≥₁ aₖ diverges.
Theorem (terms of a convergent series).
(i) Assume Σₖ≥₁ aₖ converges. Then aₖ → 0 as k → [ ]
(ii) A sufficient condition forΣₖ≥₁ aₖ to diverge is that [ ]
∞
aₖ does not go to 0
Does Σ1/k converge?
Harmonic series - diverges
Assume aₖ is real. Then
(sₙ) is monotonic increasing if and only if [ ]
aₖ ≥ 0 (for k ≥ 2)
if aₖ is non-negative, then Σₖ≥₁ aₖ converges if and only if its partial sum sequence
(sₙ) is [ ]
bounded above.
What is the comparison test. simple form?
Assume 0 ≤ aₖ ≤ Cbₖ, where C is a positive constant. Then Σₖ≥₁ bₖ convergent implies Σₖ≥₁ aₖ convergent. Also Σₖ≥₁ aₖ divergent implies Σₖ≥₁ bₖ divergent
What is Theorem (Cauchy criterion for convergence of a series)?
Let (aₖ) be a real or
complex sequence with partial sum sequence (sₙ). Then Σₖ≥₁ aₖ converges if and only if
∀ε > 0 ∃N ∈ N ∀n > m ≥ N |aₘ₊₁ + · · · + aₙ| = |sₙ − sₘ| < ε
What is absolute convergence?
Let (aₖ) be a sequence of real or complex numbers. We say Σₖ≥₁ aₖ converges absolutely if Σₖ≥₁ |aₖ| converges.
What is the Theorem (absolute convergence implies convergence)?
Let (aₖ) be a real or
complex sequence. Then
Σₖ≥₁ |aₖ| converges ⇒ Σₖ≥₁ aₖ converges.
When does Σk⁻ᵖ diverge or converge?
diverges if p ≤ 1 and converges if p > 1
What is Leibniz’ Alternating Series Test?
The series Σ (−1)ᵏ⁻¹uₖ converges if
(i) uₖ ≥ 0,
(ii) uₖ₊₁ ≤ uₖ,
(iii) uₖ → 0 as k → ∞
What is the comparison test, limit form?
Let aₖ and bₖ be strictly positive and assume that
aₖ / bₖ → L, where 0 < L < ∞.
Then
Σaₖ converges ⇐⇒ Σbₖ converges
Prove the comparison test, limit form
top pg 44
What is D’Alembert’s Ratio Test, for series of strictly positive term?
Let aₖ > 0. Assume that lim(k→∞) aₖ₊₁ / aₖ exists and equals L. Then 0 ≤ L < 1 =⇒ Σaₖ converges; L > 1 =⇒ Σaₖ diverges; L = 1 =⇒ the test gives no result
What is the ratio test using absolute convergence?
Let aₖ be non-zero real or complex numbers and assume that
lim(k→∞) |aₖ₊₁ / aₖ|
exists and equals L.
(Here we allow L = ∞.) Then L < 1 =⇒ Σaₖ converges absolutely, and hence converges; L > 1 =⇒ Σaₖ diverges; L = 1 =⇒ the test gives no result.
What is the integral test theorem?
Assume that f is a real-valued function defined on [1,∞)
with the following properties:
(i) f is non-negative and decreasing;
(ii) ᵏ⁺¹∫ₖ f(x) dx exists for each k ≥ 1 [for future reference: this holds if f is continuous].
Let
sₙ = ⁿΣₖ₌₁ f(k) and Iₙ = ⁿ∫₁ f(x) dx.
Let σₙ = sₙ − Iₙ. Then (σₙ) converges to a limit σ, where 0 ≤ σ ≤ f(1).
What is Corollary: the Integral Test?
Assume that, f : [1, ∞) → [0, ∞) is monotonic decreasing and such that ᵏ⁺¹∫ₖ f(x) dx exists for each k. Then Σf(k) converges
if and only if (Iₙ) converges.
What is Euler’s constant,γ?
Apply the Integral Test Theorem in the special case that f(x) = 1/x; certainly f is non-negative and monotonic decreasing, and its integral exists over
any interval [k, k + 1]. We have that
γₙ := 1 + 1/2 + 1/3 + · · · + 1/n − log n → γ, where γ is a constant between 0 and 1.
Define a rearrangement of a series
Take any series Σaₖ and let g : N>0 → N>0 be a
bijection. Let bₖ = a𝓰₍ₖ₎. Then Σbₖ is said to be a rearrangement of Σaₖ
If Σaₖ is absolutely convergent then all rearrangements of Σaₖ [ ], and the value of the sum is [ ] by rearrangement
converge
not changed
Define the exponential function
eᶻ = ∞Σₖ₌₀ zᵏ/k!
Define sine using series
sinz = ∞Σₖ₌₀ (-1)ᵏ z²ᵏ⁺¹/(2k+1)!
Define cosine using series
cos z = ∞Σₖ₌₀ (-1)ᵏ z²ᵏ/(2k)!
Define sinh z using series
cosh z = ∞Σₖ₌₀ z²ᵏ/(2k)!
Define cosh z using series
sinh z = ∞Σₖ₌₀ z²ᵏ⁺¹/(2k+1)!
If Σaₖ converges to s and Σbₖ converges to t, then Σ(aₖ + bₖ) converges, to [ ]. Similar arguments apply to Σ(aₖ − bₖ) and to Σcaₖ, where c is a constant.
s+t
Define cosine and sine of z in terms of e
cos z = 1/2(eᶦᶻ + e⁻ᶦᶻ)
sin z = 1/2i (eᶦᶻ - e⁻ᶦᶻ)
eᶦᶻ = cosz + isinz
Define cosh and sinh of z in terms of e
cos z = 1/2(eᶻ + e⁻ᶻ)
sin z = 1/2i (eᶻ - e⁻ᶻ)
Define The radius of convergence of a power series
R = { sup{ |z| ∈ R | Σ|cₖzᵏ| converges } if the sup exists,
{ ∞ otherwise
Let Σ cₖzᵏ be a power series with radius of convergence R (> 0). Then
Σ|cₖzᵏ| converges for [ ] and hence [ ] converges for |z| < R
|z| < R
Σ cₖzᵏ
Σ cₖzᵏ
Σ cₖzᵏ diverges if …
|z| > R
What does disc of convergence mean? Interval of convergence?
We refer to { z ∈ C | |z| < R } as the disc of convergence. For real power series we
have an interval of convergence
What is the Differentiation Theorem for (real) power series?
Let Σ cₖxᵏ be a real power
series and assume that the series has radius of convergence R where 0 < R ≤ ∞. Let
f(x) := ∞Σₖ₌₀ cₖxᵏ
Then f(x) is well defined for each x with |x| < R and moreover the derivative f’(x) exists for each such x and is given by
f’(x) = ∞Σₖ₌₀ kcₖxᵏ⁻¹
In words, we say that f’(x) is obtained by term-by-term differentiation
e^[ ]i = 1
2π
