Chapter 1
Preliminaries
1.1
Quantiers
1
Sentential logic/propositional calculus:
A, B are absolute statements about the state of aairs. Any expression like A is (globally)
true or false.
How to express more delicate ideas: relations between specic objects
0.2 Logic and inference
0.2.1
13
Set operations and logical connectives.
.
.
intersection: A B = cfw_x . x A and x B
.
.
union: A B = cfw_x . x A or x B
.
.
complement: Ac = cfw_x . x A
/
.
.
dierence: A \ B = cfw_x . x A and x B = A B c
/
.
.
product:
0.2 Logic and inference
11
Indirect proof: Proof by contrapositive.
(A = B )
(B = A),
so show B = A directly.
Example 0.2.3 (contrapositive). 3n + 2 odd = n odd.
The contrapositive is: n even = 3n + 2 even.
1. Assume n is an even integer.
2. Then n = 2k ,
0.2 Logic and inference
9
A = B means that whenever A is true, B must also be true, i.e., it CANNOT be
the case that A is true B is false: (A = B ) (A and B ). This means that the truth
table for = can be found:
A
B
B
A and (B )
(A and B )
A
B
A = B
T
T
F
Chapter 0
Pre-preliminaries
0.1
Course Overview
The study of the real numbers, R, and functions of a real var, f (x) = y where x, y real.
Given f : R R which describes some system, how to study f ?
Need rigourous vocab for properties of f (denitions)
Ne
7.5 Approximation by polynomials
is a multi-index, denoting (1,2) =
2
x1 x2 ,
2
129
etc.)
Denition 7.5.19. Call E a fundamental solution of P i P E = . (What is E ?)
Then a solution of P u = f can be found by convolving with E :
P (E f ) = P E f
by the m
7.5 Approximation by polynomials
127
Step (3) If f C (K ), x K , and > 0, then there is a function gx B such that
gx (x) = f (x) and gx (t) > f (t) , for t K .
Proof. Note that x and are xed. Since A B and A satises the hypothesis of
the previous theorem,
7.5 Approximation by polynomials
125
Proof. By Weierstrass Thm, there is a sequence cfw_gn of real polynomials converging
uniformly to |x|. Then let
fn (x) := gn (x) gn (0).
7.5.2
The Stone-Weierstrass Theorem
We will isolate the properties of the polyno
7.5 Approximation by polynomials
(iii) For every > 0, limn
|x|
123
gn (x) = 0.
If f gn is the weighted average of f (y x), with weights given by gn , then (iii) shows
that gn concentrates all weight at the origin as n . In fact, (iii) is equivalent to
(ii
7.5 Approximation by polynomials
121
Convolution is a smoothing operation.
Theorem 7.5.5. For f, g R(D), f g is continuous, D compact.
Proof. First, prove it for the case when f, g are continuous. Fix > 0 and c D. Compact
support gives a bound for the con
7.4 Power series
119
Theorem 7.4.10. If f is analytic and f (x) = 0, x (a, b), then f 0.
Theorem 7.4.11. If f, g are analytic, then so are f g , f g , f /g (for g = 0), and f g
(for dom f Im g )
7.4 Exercise: #
1.
Recommended: #
120
Math 413 Sequences and
7.4 Power series
7.4
117
Power series
7.4.1
Radius of convergence
an xn , where x is a variable.
Denition 7.4.1. A power series is a series of the form
The nth term of a power series is fn (x) = an xn (rather than just an ).
an xn is a family of series, o
7.3 Uniform convergence
115
Theorem 7.3.15 (Baby) dominated convergence thm). Suppose fn R[a, b] for every
unif
0 < a < b < , and suppose fn f on every compact subset of (0, ). If g
R[0, ), then
|fn | g = lim
fn (x) dx =
n
0
f (x) dx.
0
Proof. HW.
Theore
7.3 Uniform convergence
7.3.4
113
Spaces of functions
The previous result may be rephrased as: C [a, b] is dense in R[a, b] in the
C [a, b] dense in R[a, b] in the
-norm?
H (x) =
For any f C (R), sup |H (x) f (x)|
1 -norm.
Is
No: consider the Heaviside
7.3 Uniform convergence
7.3.2
111
Criteria for uniform convergence
Theorem 7.3.5 (Cauchy Criterion). cfw_fn converges uniformly on I i
> 0, N
m, n N = |fn (x) fm (x)| < , x.
Proof. HW. (), use ineq. (), use pointwise Cauchy Crit to obtain the limit f .
7.3 Uniform convergence
109
Example 7.3.2. Let fn (x) = xn on I = [0, 1]. Then cfw_fn converges pointwise and
f (x) =
0,
0 x < 1,
1, x = 1.
So a sequence of continuous functions can converge pointwise to something which is not
continuous! In fact, fn C (
7.2 Numerical Series and Sequences
Fix > 0. Since
107
bn converges, choose N such that n N = |n | < , so that
|en | |a0 n + . . . anN 1 N +1 | + |N anN + + an 0 |
+ |N anN + + an 0 |.
Now since an 0, for N xed and n > 1, we can make |N anN + + an 0 | < .
7.2 Numerical Series and Sequences
105
a+ and
n
negative terms, so separate into
a , and reindex so that each is decreasing.
n
For x 0, form rearrangement as follows:
1. Add positive terms until sum exceeds x. Stop as soon as
J
j =1
a+ x.
j
2. Add negativ
7.2 Numerical Series and Sequences
103
Shifting one unit to the right, the region under the graph is contained in the rectangle, so
0 An f (n), for n N.
This gives
N +n
N +n1
N +n
f (x) dx
f (k )
N +1
k=N +1
N +n1
f (x) dx
f (k )
N
k=N
and so the two s
7.2 Numerical Series and Sequences
101
Theorem 7.2.13. Suppose we have a vector space (X, ). Then X is complete i every
absolutely convergent series in X converges.
Proof. () Suppose that every Cauchy sequence in X converges and that
k=1
converges. Must s
7.2 Numerical Series and Sequences
99
NOTE: series are not multiplicative like sequences are:
an bn =
an
bn ,
because
1 + a1 b2 + a2 b2 + . . . an bn = (1 + a1 + + an )(1 + b2 + + bn ).
(More cross terms on right.)
Theorem 7.2.7 (Increasing & bounded). If
7.2 Numerical Series and Sequences
7.2
7.2.1
97
Numerical Series and Sequences
Convergence and absolute convergence
Denition 7.2.1. An (innite) series is a sum of a sequence cfw_ak :
ak = a0 + a1 + a2 + . . . .
k=0
To make it clear that the terms of the s
7.1 Complex Numbers
95
Denition 7.1.5. We can extend |x| to |z |:
|z | := (z z )1/2
or
|x + iy | :=
x2 + y 2 .
Note: | | : C R+ is a continuous function.
Theorem 7.1.6.
1. |z | = |z |.
2. |zw| = |z |w|.
Proof. HW: show |zw|2 = |z |2 |w|2 and take
.
3. | R
Chapter 7
Sequences and Series of Functions
7.1
Complex Numbers
1
7.1.1
Basic properties of C
Consider the plane
.
.
R2 := cfw_(x, y ) . x, y R.
This space already has a vector space structure:
(x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 )
a(x1 , y1 ) =
6.3 Improper Integrals
6.3
91
Improper Integrals
Denition 6.3.1. Roughly, an improper integral is an integral which can only be dened
on a domain by taking a limit of its values on subdomains.
Example 6.3.1. f (x) = x has an integrable singularity at x =
6.2 Properties of the Riemann Integral
89
Putting this minor result together with the previous two shows that
D(I (f ) = f,
but I (D(f ) = f + c,
so integration and dierentiation are almost inverse operations.
Theorem 6.2.6 (Integration by parts). If f, g
6.2 Properties of the Riemann Integral
Since was arbitrary, this gives
inequality. For cf , use
87
f dx
f1 dx +
cMi (xi xi1 ) = c
f2 dx. Similarly for the other
Mi (xi xi1 ), etc.
(ii) maxcfw_f, g , mincfw_f, g R[a, b].
Proof. Let h(x) := maxcfw_f (x),
6.1 Integrals of continuous functions
85
Theorem 6.1.12. Let f R[a, b], m f M . If is continuous on [m, M ] and
h(x) := (f (x) for x [a, b], then h R[a, b].
Proof. Fix > 0. Since is uniformly continuous on [m, M ], nd 0 < < such that
|s t| < = |(s) (t)| <
6.1 Integrals of continuous functions
83
Denition 6.1.8. A norm on a vector space is a function from X to R that satises
(i)
x 0, x = 0 x = 0.
(ii) ax = |a| x , a R.
(iii)
x z x y + y z , x, y, z X .
Example 6.1.1. On R, absolute value is a norm.
For x =
Chapter 6
Integration
6.1
Integrals of continuous functions
1
6.1.1
Existence of the integral
Let f (x) be a function dened on [a, b]. We want to dene its integral
b
a
f (x) dx.
Denition 6.1.1. A partition P of the interval [a, b] is a nite set of points