Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Contents
1 Board work
1
...

1
...
3 Thursday, 1 October 2015
...
4 Tuesday, 6 October 2015
...
5 Wednesday, 7 October 2015
...
6 Thursday, 8 October 2015
...
7 Tuesday, 13 October 2015
...
8 Wednesday, 14 October 2015
...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...


...
1

...
3

...
7

...
9

...
1

Tuesday, 29 September 2015

1
...

Example
...
Then there must be a point a ∈
A such that a ∈ D1 (0)
...
Next there
must be a point c ∈ A with c ∈ D|b| (0)
...

2
...
Think of a sequence as a collection of a first term, a second
term,
...
, ad infinitum
...
Tail
...
Thus xN , xN +1 , xN +2 ,
...

4
...
Inside vs Outside
...
(In which case the number of terms of (xn )
inside B (x) will automatically be infinity)
...
Note that (i) ⇔ (ii) ⇔ (iii) where

1

(i) (xn ) has finitely many terms outside B (x),
(ii) (xn ) has a tail with zero terms outside B (x), and
(iii) (xn ) has a tail with all terms inside B (x)
...

6
...

7
...
Assume xn → l
...
So every element of the tail is
bounded by |l| + 1
...
Since the maximum
of finitely many finite numbers is also finite, we get a finite bound for
the entire sequence
...
2

Wednesday, 30 September 2015

1
...
an ≤ bn ≤ cn with an → l and cn → l
...

Since an → l, there exists na such that l − < an for all n > na
...
Therefore
bn ∈ B (l) whenever n greater than both na and nb
...
)
2
...

Trivial for c = 0
...
So
can ∈ B (a) for all n > n0
...
)
3
...

Given arbitrary > 0 there exists na such that
|a − an | < /2 for all n > na
...

When n is greater than both na and nb , we get
|an + bn − a − b| ≤ |an − a| + |bn − b| <
...
If an → a then 1/an → 1/a
...

(1) an > a/2,
(2) |an − a| < a2 /2 = β
...
The second when n > n1
...
Once both
conditions are satisfied, we have |1/an − 1/a| <
...

5
...
One way to see
this is to note that −a is a positive number and −an → −a The result
quoted in the slides allows us to say that −1/an → −1/a
...

6
...

(i) b is an upper bound for B
...


1
...
A subsequence of a given sequence is any infinite collection of terms
from the original sequence which retains the ordering of the original
sequence
...
Move forward along a sequence
...
Having picked a term, you cannot go back and pick a term you
had skipped earlier
...
)
3
...

3

4
...
In symbols, if xn → x then xnk → x
...
If a neighbourhood B (x) contains
the tail terms xK , xK+1 , xK+2 ,
...
of the subsequence
from location K onwards
...
Every sequence has a monotone subsequence
...
If infinitely many terms are marked, the marked terms constitute
a monotonically decreasing subsequence
...
Look at terms after
the N th term
...

So we can construct a monotone increasing subsequence
...
A bounded sequence has a convergent subsequence
...
Since the given sequence
is bounded, the monotone subsequence too will be bounded
...
(The limit being the LUB if the subsequence is monotonically
increasing
...
)
7
...

8
...

Recall that if a tail of a sequence lies in a bounded neighbourhood
then the sequence is bounded
...
) If (xn ) is a Cauchy sequence it will have a
tail aN , aN +1 , aN +2 ,
...
In particular,
|xi − xN | < 1 for all i ≥ N
...

lies in the bounded neighbourhood B1 (aN )
...
, |aN −1 |, |an | + 1},
which is a finite number, bounds the entire sequence
...
Convergent sequences are Cauchy
...
For this tail, any two terms are within distance of each
other
...
Cauchy sequences are convergent
...
Since a Cauchy sequence is always
bounded, and a bounded sequence always has a convergent subsequence, (an ) will have a convergent subsequence: ank → l
...

Choose k large enough that
(i) |ank − l| < /2, and
(ii) nk ≥ N so that ank lies in the tail aN , aN +1 , aN +2 ,
...


1
...
Knowing how to add two numbers tells us how to add finitely many
numbers, but not how to add infinitely many numbers
...
+an be the nth partial sum
...
Each term of this sequence makes sense since it is a
just a sum of finitely many numbers
...
If the sequence
(sn ) converges to a limit, say l, define the infinite sum ∞ ai to be the
1
number l
...

2
...
(Harmonic series, telescoping series, geometric series,
etc
...
Recall that a sum of finitely many finite terms is always finite (follows
from induction)
...
So, 1 + 2 + 3 = 1 + 3 + 2 = 2 + 1 + 3 = 3 + 2 + 1
...

n
Now, ∞ ai = limn→∞
1
1 ai if the limit exists
...

Before we take the limit we still have only a finite sum, and so we can
rearrange the (finitely many) terms
...
+ aN ) + α(c1 +
...


4
...

And (sn ) is Cauchy just means that |an+1 + an+2 +
...
In symbols, given
an arbitrary > 0, there exists a constant n0 such that |sm − sn | =
|an+1 + an+2 +
...

5
...
Let an ≤ bn ≤ cn for all n, with
N
∞
1 bn converges as N → ∞
...
Then

∞
1

an = a and

We shall use the Cauchy criterion
...

Similarly, there exists a constant nc such that
|cn+1 + cn+2 + · · · + cm | < for all m ≥ n ≥ nc
...

It follows from the Cauchy criterion that

1
...


Wednesday, 7 October 2015

1
...


This is mostly a matter of recalling the various definitions
...
Similarly, ∞ ai is said to
k
1
converge if and only if the sequence of partial sums ( n ai ) converges
k
to some real number l
...
Since c is always a finite number, the result
follows
...
6

Thursday, 8 October 2015

1
...


∞
1

|an | con-

2
...

Let ∞ an be absolutely convergent
...
Since
−|an | ≤ an ≤ |an |,
the sandwich test for sums tells us that

∞
1

an must also converge
...
If there is a 1 : 1, onto map σ(·) : N → N such that ai = bσ(i) for
all i ∈ N then the sequence (bn ) is said to be a rearrangement of the
sequence (an )
...
Let ∞ an be convergent with all an ≥ 0
...
Then ∞ bn = ∞ an
...
Since the partial sums
1
N
bn are bounded and monotonically increasing, they will converge to
1
the LUB (least upper bound) for the partial sums
...

1

Similarly, the constant ∞ bn is an upper bound for the partial sums
1
N
an
...

1

The result folows by combining the two inequalities
...
The above result assumes that all terms an are nonnegative
...
In other words, if ∞ an is absolutely convergent,
1
and (bn ) is a rearrangement of (an ), then ∞ bn = ∞ an
...
Define An = |an | + an and Bn =
1
|bn | + bn
...

Let us look at the partial sums of An
...
They are also bounded by a finite constant,
N

N

(|an | + an ) ≤

An =
1

N

1

∞

N

(|an | + |an |) = 2
1

|an | ≤ 2
1

|an |
...
The previous result tells us that
1
∞

∞

An =
1

∞

|an | =

Bn , and
1

1

Subtracting gives us the desired result
...

1

6
...
Let An = a1 + · · · an
...

From Abel’s partial sum formula we have
n

n−1

Ai [bi − bi+1 ] + An bn
...
Consider n−1 Ai [bi − bi+1 ] first
...

Note that ∞ −M [bi − bi+1 ] = −M b1 and ∞ M [bi − bi+1 ] = M b1
...

1
Next, note that −M bn ≤ An bn ≤ M bn with −M bn → 0 and M bn → 0
...
Thus
n−1
Ai [bi − bi+1 ] + An bn converges, and hence n ai bi converges too
...
Abel’s test
...
Then an bn converges
...
Let An = a1 + · · · + an
...
By Dirichlet’s test,
ai c i
converges
...

If bn ↓ b work with bn − b = cn ↓ 0 to obtain the convergence of
ai b i = ai b + ai c i
...
7

Tuesday, 13 October 2015

1
...
Let a be a cluster point of A
...

2
...
We say limt→a f (t) = l if:
Informally, f (t) can be brought arbitrarily close to the limit
l by bringing t sufficiently close to a (but not touching it)
...
We make things more precise
...
In symbols, we want |f (t) − l| <
...
In other words, we
say limt→a f (t) = l if:
Given an arbitrary

> 0 there exists a δ > 0 such that
|f (t) − l| <

whenever 0 < |t − a| < δ
...
Meaning of f (Dδ (a)) ⊂ B (l)
...

5
...

Given an arbitrary -neighbourhood of the limit l, there exists
a deleted δ-neighbourhood of a such that f (Dδ (a)) ⊂ B (l)
...
To define limit at a, we need not have a ∈ A
...
What is
limx→0 f (x)? Answer: 1
...
8

Wednesday, 14 October 2015

1
...

Pick a neighbourhood B (l), > 0
...

Further there is a deleted neighbourhood Dδ (c), δ > 0, which maps
into B (l), i
...
,
f (Dδ (c)) ⊂ B (l)
...
And max{|f (c)|, |l| + } will
bound the set f (Bδ (c))
...
Sandwich result
...
If f ≤ h ≤ g
then limx→c h(x) = l too
...
Pick
an arbitrary neighbourhood B (l)
...
By assumption,
there exists a δ1 satisfying
f (Dδ1 (c)) ⊂ B (l)
...
Choose δ =
min{δ1 , δ2 }
...

3
...

Pick an arbitrary sequence (xn ) with xn → c, xn = c
...
Since f (xn ) ≤ h(xn ) ≤ g(xn ), the sandwich theorem
for sequences shows that h(xn ) → l
...
By definition this
means that limx→c h(x) = l
...
limθ→0 cos(θ) = limθ→0

1 − sin2 (θ) = 1
...
limx→a sin(x) = sin(a)
...
We get
lim sin(a + b) = lim sin(a) cos(b) + cos(a) sin(b) = sin(a)
...
limx→a cos(x) = cos(a)
...
We get
lim cos(a + b) = lim cos(a) cos(b) − sin(a) sin(b) = cos(a)
...
For |x| < 1 we have 1 + x ≤ exp (x) ≤ 1 + x + x2
...

2!
3!
4!
11

We calculate lower and upper bounds for the tail series
x3 x4
+
+ ···
...
We have
x3 x4
xn
+
+ ···
3!
4!
n!

|x|3 |x|4
|x|n
+ 3 + · · · + n−1
22
2
2
|x|2 |x|2
≤ |x| 2 + 3 + · · · infinite sum
2
2
2
1
|x|
≤ |x| × 2 ×
2
1 − |x|
≤

2

2

≤
3

|x|
since |x| < 1
...
Since the partial sums are
3!
4!
2
2
trapped between the constants − |x| and |x| , we must have
2
2
−

x3 x4
|x|2
|x|2
≤
+
+ ··· ≤

...

8
...
We say f (·) is continuous at a ∈ A if:
For every sequence (an ) in A which approaches a, the sequence
(f (an )) converges to f (a)
...
Another way of putting it
...

10
...


12

11
...

If f (x) is defined with x in the δ–neighbourhood of a, then f (x) must
belong to the –neighbourhood of f (a)
...
Yet another way to say f (·) is continuous at a
...

13
...

Let (an ) be an arbitrary sequence converging to a
...
In this case, automatically (f (an )) converges
to f (a)
...
We remark, that though f is
continuous at a, limx→a f (a) cannot exist (limits exist only for cluster
points, and a is not one)
...
If a is a cluster point and f is continuous at a then limx→a f (x) = f (a)
...
Pick an arbitrary sequence (an ) that converges to a, without touching it
...
But
that is trivially true since f is continuous at a and (an ) converges to a

---