Lecture 3
2
2.1
Dierentiation
Dierentiation in n dimensions
We are setting out to generalize to n dimensions the notion of dierentiation in one dimensional calculus. We begin with a review of elementary onedimensional calculus. Let I R be an open interval
Lecture 18
We begin with a quick review of permutations (from last lecture). A permutation of order k is a bijective map : cfw_1, . . . , k cfw_1, . . . , k . We denote by Sk the set of permutations of order k . The set Sk has some nice properties. If Sk
Lecture 17
Today we begin studying the material that is also found in the Multilinear Algebra Notes. We begin with the theory of tensors.
4.3
Tensors
Let V be a ndimensional vector space. We use the following notation. Notation. V k = V V .
k times
(4.14
Lecture 16
4
4.1
Multilinear Algebra
Review of Linear Algebra and Topology
In todays lecture we review chapters 1 and 2 of Munkres. Our ultimate goal (not today) is to develop vector calculus in n dimensions (for example, the generalizations of grad, div,
Lecture 15
We restate the partition of unity theorem from last time. Let cfw_U : I be a collection of open subsets of Rn such that
U= U . (3.169)
I Theorem 3.30. There exist functions fi C0 (U ) such that
1. f1 0, 2. supp fi U , for some , 3. For ever
Lecture 14
As before, let f : R R be the map dened by 0 if x 0, f (x) = e1/x if x > 0. (3.162)
This is a C inf (R) function. Take the interval [a, b] R and dene the function fa,b : R R by fa,b (x) = f (x a)f (b x). Note that fa,b > 0 on (a, b), and fa,b =
Lecture 13
Let A be an open set in Rn , and let f : A R be a continuous function. For the moment, we assumethat f 0. Let D A be a compact and rectiable set. Then f |D is bounded, so D f is welldened. Consider the set of all such integrals: # = cfw_ f : D
Lecture 12
So far, we have been studying only the Riemann integral. However, there is also the Lebesgue integral. These are the two basic integral theories. The Riemann inte gral is very intuitive and is usually adequate for problems that usually come up.
Lecture 11
We review some basic properties of the Riemann integral. Let Q Rn be a rectangle, and let f, g : q R be bounded functions. Assume that f, g are R. integrable. We have the following properties of R. integrals: Linearity: a, b R = af + bg is R.
Lecture 10
We begin todays lecture with a simple claim. Claim. Let Q Rn be a rectangle and f , g : Q R be bounded functions such that f g . Then f g. (3.49)
Q Q
Proof. Let P be a partition of Q, and let R be a rectangle belonging to P . Clearly, mR (f ) m
Lecture 9
We quickly review the denition of measure zero.
A set A Rn is of measure zero if for every > 0, there exists a covering of A by
rectangles Q1 , Q2 , Q3 , . . . such that the total volume v (Qi ) < . Remark. In this denition we can replace rect
Lecture 8
3.2 Riemann Integral of Several Variables
Last time we dened the Riemann integral for one variable, and today we generalize to many variables. Denition 3.3. A rectangle is a subset Q of Rn of the form Q = [a1 , b1 ] [an , bn ], where ai , bi R.
Lecture 7
We continue our proof of the Inverse Function Theorem.
As before, we let U be an open set in Rn , and we assume that 0 U . We let
f : U Rn be a C 1 map, and we assume f (0) = 0 and that Df (0) = I . We summarize what we have proved so far in t
Lecture 6
We begin with a review of some earlier denitions. Let > 0 and a Rn . Euclidean ball: B (a) = cfw_x Rn : x a < Supremum ball: R (a) = cfw_x Rn : |x a| < = I1 In , Ij = (aj , aj + ). (2.66) (2.67)
Note that the supremum ball is actually a rectan
Lecture 5
2.3 Chain Rule
Let U and v be open sets in Rn . Consider maps f : U V and g : V Rk . Choose a U , and let b = f (a). The composition g f : U Rk is dened by (g f )(x) = g (f (x). Theorem 2.9. If f is dierentiable at a and g is dierentiable at b,
Lecture 4
2.2 Conditions for Dierentiability
In this lecture we will discuss conditions that guarantee dierentiability. First, we begin with a review of important results from last lecture. Let U be an open subset of Rn , let f : U Rn be a map, and let a
Lecture 2
1.6 Compactness
As usual, throughout this section we let (X, d) be a metric space. We also remind you from last lecture we dened the open set U (xo , ) = cfw_x X : d(x, xo ) < . Remark. If U (xo , ) U (x1 , 1 ), then 1 > d(xo , x1 ). Remark. If
Lecture 1
1
1.1
Review of Topology
Metric Spaces
Denition 1.1. Let X be a set. Dene the Cartesian product X X = cfw_(x, y ) :
x, y X .
Denition 1.2. Let d : X X R be a mapping. The mapping d is a metric on
X if the following four conditions hold for al
Lecture 19
We begin with a review of tensors and alternating tensors.
We dened Lk (V ) to be the set of k linear maps T : V k R. We dened
e1 , . . . , en to be a basis of V and e , . . . , e to be a basis of V . We also dened 1 n cfw_e = e1 ek to be a
Lecture 20
We begin with a review of last lecture.
Consider a vector space V . A tensor T Lk is decomposable if T = 1
1 k , i L = V . A decomposable tensor T is redundant of i = i+1 for some i. We dene I k = I k (V ) = Span cfw_ redundant k tensors . (
Lecture 21
Let V , W be vector spaces, and let A : V W be a linear map. We dened the pullback operation A : W V . Last time we dened another pullback operator having the form A : k (W ) k (V ). This new pullback operator has the following properties: 1. A
Lecture 38
We begin with a review from last time. Let X be an oriented manifold, and let D X be a smooth domain. Then Bd (D) = Y is an oriented (n 1)dimensional manifold. We dened integration over D as follows. For n (X ) we want to make sense c of the in
Lecture 37
6.10 Integration on Smooth Domains
Let X be an oriented ndimensional manifold, and let n (X ). We dened the c integral , (6.136)
X
but we can generalize the integral ,
D
(6.137)
for some subsets D X . We generalize, but only to very simple subs
Lecture 36
The rst problem on todays homework will be to prove the inverse function theorem for manifolds. Here we state the theorem and provide a sketch of the proof. Let X, Y be ndimensional manifolds, and let f : X Y be a C map with f (p) = p1 . Theore
Lecture 35
Before moving on to integration, we make a few more remarks about orientations. Let X, Y be oriented manifolds. A dieomorphism f : X Y is orientation preserving if for every p X , the map dfp : Tp X Tq Y (6.85)
is orientation preserving, where
Lecture 34
6.6 Orientation of Manifolds
Let X be an ndimensional manifold in RN . Assume that X is a closed subset of RN . Let f : X R be a C map. Denition 6.21. We remind you that the support of f is dened to be supp f = cfw_x X : f (x) = 0. (6.69)
Since
Lecture 33
6.5 Dierential Forms on Manifolds
Let U Rn be open. By denition, a k form on U is a function which assigns to each point p U an element p k (Tp Rn ). We now dene the notion of a k form on a manifold. Let X RN be an n dimensional manifold. Then,
Lecture 32
6.4 Tangent Spaces of Manifolds
We generalize our earlier discussion of tangent spaces to tangent spaces of manifolds. First we review our earlier treatment of tangent spaces. Let p Rn . We dene Tp Rn = cfw_(p, v ) : v Rn . (6.31)
Of course, we
Lecture 31
6.3 Examples of Manifolds
We begin with a review of the denition of a manifold. Let X be a subset of Rn , let Y be a subset of Rm , and let f : X Y be a continuous map. Denition 6.6. The map f is C if for every p X , there exists a neighborhood
Lecture 30
6 Manifolds
6.1 Canonical Submersion and Canonical Immersion Theo rems
As part of todays homework, you are to prove the canonical submersion and im mersion theorems for linear maps. We begin todays lecture by stating these two theorems. Let A
beverages
Article
Evaluation of Fermentation Products of Palm Wine
Yeasts and Role of Sacoglottis gabonensis Supplement
on Products Abundance
Ogueri Nwaiwu 1, *, Vincent I. Ibekwe 2 , Ekperechi S. Amadi 2 , Angela C. Udebuani 3 ,
Ferdinand C. Nwanebu 2 ,