Analysis I Flashcards
(96 cards)
Define the upper bound and the lower bound of a set
Definition Let ∅ != H ⊆ R and K, L ∈ R. We say that
a) K is an upper bound of H if ∀ x ∈ H : x ≤ K ,
Define the following concepts: ”a set is bounded above”, ”a set is bounded below”
. Definition Let ∅ != H ⊆ R. We say that
a) H is bounded above if it has an upper bound, that is ∃ K ∈ R ∀ x ∈ H : x ≤ K ,
b) H is bounded below if it has a lower bound, that is ∃L ∈ R ∀ x ∈ H : x ≥ L,
Define the minimal element of a set
. Definition Let ∅ != H ⊆ R and a ∈ R. We say that
• a is the minimal element (or: least element) of H if a ∈ H and ∀ x ∈ H : x ≥ a.
Notation: a = min H .
Define the maximal element of a set
Definition Let ∅ != H ⊆ R and a ∈ R. We say that
• a is the maximal element (or: greatest element) of H if a ∈ H and ∀ x ∈ H :
x ≤ a. Notation: a = max H .
Define the least upper bound (sup) of a set. What is the least upper bound of a set which
is not bounded above?
Let ∅ != H ⊆ R and suppose that H is bounded above. Then the set of its upper
bounds
B := {K ∈ R | K is upper bound of H}
has minimal element. This minimal element is called the least upper bound of H and
is denoted by sup H or lub H. So
sup H = lub H := min B .
The term sup is from Latin supremum
Define the greatest lower bound (inf) of a set. What is the greatest lower bound of a set
which is not bounded below?
Let ∅ != H ⊆ R and suppose that H is bounded below. Then the set of its lower
bounds
A := {K ∈ R | K is lower bound of H}
has maximal element. This maximal element is called the greatest lower bound of H
and is denoted by inf H or glb H. So
inf H = glb H := max A .
The term inf is from Latin infimum
State the Triangle Inequalities in K, and prove them in R
For any real numbers x, y ∈ R hold
a) |x + y| ≤ |x| + |y| (first triangle inequality)
b) |x − y| ≥ | |x| − |y| | (second triangle inequality)
Proof:
From the definition of the absolute value follows that
−|x| ≤ x ≤ |x| and − |y| ≤ y ≤ |y| .
Adding these inequalities we obtain that
−(|x| + |y|) ≤ x + y ≤ |x| + |y| .
From here follows |x + y| ≤ |x| + |y|.
To prove part b) apply part a) with x − y and y:
|x| = |(x − y) + y| ≤ |x − y| + |y| . From here follows: |x| − |y| ≤ |x − y| .
Similarly (change x with y) we can deduce that:
|y| − |x| ≤ |y − x| = |x − y| .
The last two inequalities imply that| |x| − |y| |≤ |x − y| .
Remark that – applying the first triangle inequality several times – we obtain that
|x1 + x2 + . . . + xn| ≤ |x1| + |x2| + . . . + |xn| (x1, x2, . . . xn ∈ R).
Define the concepts: sequence, n-th term of a sequence
- Definition Let H be a nonempty set.
The functions
a : N → H
are called sequences in H. For an n ∈ N the element a(n) ∈ H is called the n-th term
of the sequence. Its usual notation is an.
Some notations for the sequence a:
a ; (an) ; (an, n ∈ N) ; an ∈ H (n ∈ N)
Define the index-sequence and the subsequence. Give an example for them.
The sequence nk ∈ N (k ∈ N) is called index sequence if it is strictly
monotone increasing, that is
∀ n ∈ N : nk < nk+1 .
3.4. Definition Let a : N → H be a sequence and let (nk) be an index sequence.
Then the sequence
ank ∈ H (k ∈ N)
is called the subsequence of (an) (composed with the index sequence (nk)).
Define the concept of convergence and divergence of a sequence (with inequalities), and
define the limit of a convergent sequence.
The number sequence an ∈ K (n ∈ N) is named convergent if
∃A ∈ K ∀ε > 0 ∃N ∈ N ∀n ≥ N : an ∈ B(A, ε).
The definition can be written using inequalities as follows:
∃A ∈ K ∀ε > 0 ∃N ∈ N ∀n ≥ N : |an − A| < ε .
A number sequence is named divergent if it is not convergent.
n Let an ∈ K (n ∈ N) be a convergent number sequence. The unique
number A in the definition 3.10 is called the limit of the sequence (an), and is denoted
in one of the following ways:
lim a = A , lim an = A , limn→∞
an = A , an → A (n → ∞),
lim(an) = A , (an) → A (n → ∞)
Define the concept: a sequence is bounded
The sequence an ∈ K (n ∈ N) is called bounded if
∃ M > 0 ∀ n ∈ N : |an| ≤ M .
The number M is called a bound of the sequence.
A number sequence is called unbounded if it is not bounded
State and prove the theorem about the connection between the convergent and the
bounded sequences
Every convergent number sequence is bounded.
Proof. Let an ∈ K (n ∈ N) be a convergent sequence and A = limn→∞
an ∈ K. Apply
the definition of convergency with ε = 1:
∃ N ∈ N ∀ n ≥ N : |an − A| < 1 .
Use the second triangle inequality:
|an| − |A| ≤ | |an | − |A | | ≤ |an − A| < 1 ,
from where we have after rearranging
|an| < 1 + |A| (n ≥ N).
Thus obviously
|an| ≤ M (n ∈ N) where M := max{|a1|, |a2|, . . . , |aN−1|, 1 + |A|} .
¤
We remark that the reverse statement is not true. The sequence ((−1)n
) is bounded
but divergent (see example 3.14). Later we will prove that any bounded sequence has
a convergent subsequence (Bolzano-Weierstrass theorem).
Define the zero-sequence
Definition The number sequence an ∈ K (n ∈ N) is called zero sequence if it is
convergent and limn→∞
an = 0.
State and prove the five theorems (Th1-Th5) in connection with zero sequences
m [T1] Let an ∈ K (n ∈ N) and A ∈ K. Then
limn→∞
an = A ⇔ limn→∞
(an − A) = 0 .
Proof. The statement is a simple consequence of the definition of the limit and of the
obvious identity
|an − A| = |(an − A) − 0| .
¤
3.25. Theorem [T2] Let an ∈ K (n ∈ N). Then
limn→∞
an = 0 ⇔ limn→∞
|an| = 0 .
Proof. The statement is a simple consequence of the definition of the limit and of the
obvious identity
|an − 0| = ||an| − 0| .
¤
3.26. Theorem [T3, Majorant Principle] Let an ∈ K (n ∈ N) and bn ∈ R (n ∈ N).
Suppose that (bn) is a zero sequence and that
∃ N0 ∈ N ∀ n ≥ N0 : |an| ≤ bn ,
Then (an) is also a zero sequence.
Proof. Let ε > 0. Since limn→∞
bn = 0, then
∃ N1 ∈ N ∀ n ≥ N1 : bn = |bn − 0| < ε .
Thus for the threshold index N := max{N0, N1} holds:
|an − 0| = |an| ≤ bn < ε .
This means that limn→∞
an = 0. ¤
3.27. Theorem [T4, Sum] Let an, bn ∈ K (n ∈ N) be zero sequences. Then their sum
(an + bn) is also a zero sequence.
36 3. Lesson 3
Proof. Let ε > 0. Since limn→∞
an = 0, then
∃ N1 ∈ N ∀ n ≥ N1 : |an| = |an − 0| <
ε
2
,
and since limn→∞
bn = 0, then
∃ N2 ∈ N ∀ n ≥ N2 : |bn| = |bn − 0| <
ε
2
.
Let N := max{N1, N2}. It will be a good threshold index, because – using the first
triangle inequality – for any n ≥ N holds:
|(an + bn) − 0| = |an + bn| ≤ |an| + |bn| <
ε
2
\+
ε
2
= ε .
This means that limn→∞
(an + bn) = 0. ¤
3.28. Theorem [T5, Product] Let an ∈ K (n ∈ N) be a zero sequence and bn ∈ K (n ∈
N) be a bounded sequence. Then their product (anbn) is a zero sequence.
Proof. Let ε > 0. Since (bn) is bounded, then
∃ M > 0 ∀ n ∈ N : |bn| ≤ M .
Since limn→∞
an = 0, then
∃ N ∈ N ∀ n ≥ N : |an| <
ε
M
.
This N will be a good threshold index, because for any n ≥ N holds:
|(anbn) − 0| = |an| · |bn| ≤ |an| · M < ε
M
· M = ε .
This means that limn→∞
(anbn) = 0.State the theorem about the operations with convergent sequences, and prove the case of
addition.
–
State (without proof) the theorem about the Inequality between the Arithmetic and the
Geometric means
read this from the pdf it is on page 19
[Inequality between the Arithmetic and Geometric Means]
State and prove the Sandwich Theorem
Let an, bn, cn ∈ R (n ∈ N) be real number sequences and suppose that a) ∃ N0 ∈ N ∀ n ≥ N0 : an ≤ bn ≤ cn and that b) (an) and (cn) are convergent and limn→∞ an = limn→∞ cn =: A. Then (bn) is also convergent and limn→∞ bn = A. Proof. Let us start from the inequalities an ≤ bn ≤ cn (n ∈ N, n ≥ N0). After subtracting an we have 0 ≤ bn − an ≤ cn − an (n ∈ N, n ≥ N0). Since limn→∞ (cn − an) = limn→∞ cn − limn→∞ an = A − A = 0 , then (cn − an) is a zero sequence. Using T3 we obtain that (bn − an) is also a zero sequence. Finally limn→∞ bn = limn→∞ ((bn − an) + an) = limn→∞ (bn − an) + limn→∞ an = 0 + A = A
State and prove the theorem about the convergence of the sequences an = c (n ∈ N) and an = 1 n (n ∈ N)
Let a, b ∈ K, a 6= b. Then ∃ r1, r2 > 0 : B(a, r1) ∩ B(b, r2) = ∅ . Proof. Let r1 := |a − b| 2 > 0. Then for every x ∈ B(a, r1) holds (using the second triangle inequality): |x−b| = |x−a+a−b| = |(a−b)−(a−x)| ≥ |a−b|−|a−x| > |a−b|−r1 = |a−b|−|a − b| 2 = |a − b| 2 . thus if r2 := |a − b| 2 > 0, then |x − b| > r2, therefore x /∈ B(b, r2). ¤ After these preliminaries we can formulate the definition of the convergency and of the limit.
Define the geometric sequence. State and prove the theorem about the convergence of
geometric sequences
Let q ∈ K be a fixed number. Then the sequence an := q n (n ∈ N) is called a geometric sequence (with base q or with quotient q). 42 4. Lesson 4 4.16. Theorem The geometric sequence is convergent if and only if |q| < 1 or q = 1. In this case limn→∞ q n = 0 if |q| < 1 1 if q = 1 Proof. The statement of the theorem is trivial if q = 0 or if q = 1. Suppose that 0 < |q| < 1. Then 1 |q| > 1 and – using the Bernoulli inequality (see Theorem 2.2) – 1 |q| n = µ 1 |q| ¶n = µ 1 + 1 |q| − 1 ¶n ≥ 1 + n · µ 1 |q| − 1 ¶ > n · µ 1 |q| − 1 ¶ . After rearranging we have 0 ≤ |q n | = |q| n ≤ 1 1 |q| − 1 · 1 n (n ∈ N). The right side sequence tends to 0. Using the Sandwich Theorem we obtain limn→∞ |q n | = 0. Using Theorem 3.25 we have limn→∞ q n = 0. Suppose that |q| > 1. Once more using the Bernoulli inequality: |q n | = |q| n = (1 + |q| − 1)n ≥ 1 + n · |q| > n · |q| , which implies that the sequence (q n ) is unbounded. Consequently it is divergent. Finally, suppose that |q| = 1 but q 6= 1. Suppose indirectly that (an = q n ) is convergent and denote by A its limit. Then by Theorem 4.1 we have |A| = | limn→∞ q n | = limn→∞ |q n | = limn→∞ |q| n = limn→∞ 1 n = 1 , which implies A 6= 0. On the other hand limn→∞ an+1 = limn→∞ an = A, therefore 0 = A − A = limn→∞ an+1 − limn→∞ an = limn→∞ (an+1 − an) = limn→∞ (q n+1 − q n ) = = limn→∞ q n (q − 1) = (q − 1) · limn→∞ q n = (q − 1) · A . We obtained that 0 = (q − 1) · A , which is a contradiction, because on the right side stands the product of two nonzero numbers. ¤ We have finished the discussion of the geometric sequence. In the following theorems we will discuss some other interesting convergent sequences.
. State and prove the theorem about the convergence of √n a and of √n n
Look for these theorems:
- Theorem
- Theorem
State and prove the theorem about the convergence of n^k· q^n
Look for
4.20. Theorem
State and prove the theorem about the convergence of x^n/n!
4.22. Theorem
Define the monotone sequences (i.e. the different types of monotonicity)
5.1. Definition Let an ∈ R (n ∈ N) be a real number sequence. We say that this
sequence is
• monotonically increasing if ∀n ∈ N : an ≤ an+1
• strictly monotonically increasing if ∀n ∈ N : an < an+1
• monotonically decreasing if ∀n ∈ N : an ≥ an+1
• strictly monotonically decreasing if ∀n ∈ N : an > an+1
• monotone if it is either monotonically increasing or monotonically decreasing
• strictly monotone if it is either strictly monotonically increasing or strictly monotonically decreasing
State and prove the theorem about the convergence of a monotonically increasing sequence
5.4. Theorem
