The *difference* between sets and , also known as the *relative complement* of in , is the set defined by

.

If we assume the existence of a universe, , such that all the sets are subsets of , then we can considerably simplify our notation. So, for instance, can simply be written as , which denotes the complement of in . Similarly, , , and so on. A quick look at a few more facts:

- ,
- ,
- if and only if .

The last one is proved as follows. We prove the “if” part first. Suppose . If , then clearly . But, since , we have , which implies . Hence, . This closes the proof of the “if” part. Now, we prove the “only if” part. So, suppose . Now, if , then clearly . But, since , we have , which implies . Hence, . This closes the proof of the “only if” part, and, we are done.

The following are the well-known *DeMorgan’s laws* (about complements):

, and .

Let’s quickly prove the first one. Suppose belongs to the left hand side. Then, , which implies and , which implies and , which implies . This proves that the left hand side is a subset of the right hand side. We can similarly prove the right hand side is a subset of the left hand side, and this closes our proof.

Though it isn’t very apparent, but if we look carefully at the couple of problems whose proofs we did above, we note something called the *principle of duality* for sets. One encounters such dual principles in mathematics quite often. In this case, this dual principle is stated a follows.

Principle of duality (for sets): If in an inclusion or equation involving unions, intersections and complements of subsets of (the universe) we replace each set by its complement, interchange unions and intersections, and reverse all set-inclusions, the result is another theorem.

Using the above principle, it is easy to “derive” one of the DeMorgan’s laws from another and vice versa. In addition, DeMorgan’s laws can be extended to larger collections of sets instead of just pairs.

Here are a few exercises on complementation.

- ,
- if and only if ,
- ,
- ,
- ,
- .

We will prove the last one, leaving the remaining as exercises to the reader. Suppose belongs to the left hand side. Then, and . Now, note that if , then , which implies , which implies . If, on the other hand, , then , which implies , which implies . Hence, in either case, the left hand side is a subset of , and we are done.

We now define the *symmetric difference* (or *Boolean sum*) of two sets and as follows:

.

This is basically the set of all elements in or but not in . In other words, . Again, a few facts (involving symmetric differences) that aren’t hard to prove:

- ,
- ,
- (commutativity),
- (associativity),
- ,
- .

This brings us now to the *axiom of powers*, which basically states if is a set then there exists a set that contains all the possible subsets of as its elements.

Axiom of powers:If is a set, then there exists a set (collection) , such that if , then .

The set , described above, may be too “comprehens ive”, *i.e.*, it may contain sets other than the subsets of . Once again, we “fix” this by applying the axiom of specification to form the new set . The set is called the *power set* of , and the axiom of extension, again, guarantees its uniqueness. We denote by to show the dependence of on . A few illustrative examples: , , , and so on.

Note that if is a finite set, containing elements, then the power set contains elements. The “usual” way to prove this is by either using a simple combinatorial argument or by using some algebra. The combinatorial argument is as follows. An element in either belongs to a subset of or it doesn’t: there are thus two choices; since there are elements in , the number of all possible subsets of is therefore . A more algebraic way of proving the same result is as follows. The number of subsets with elements is . So, the number of subsets of is . But, from the binomial theorem, we have . Putting , we get as our required answer.

A few elementary facts:

- .
- If , then .
- .

EXERCISES:

1.Prove that .

2.Prove that .

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September 11, 2011 at 3:44 am

MattI’m self-studying the book and this is a great reference. Hit a wall on (i) and (ii) of Section 12 and was looking for some online notes. This nailed it. But did you lose steam or is the rest somewhere else? I hope you will complete your mission!

April 15, 2019 at 7:58 pm

prof drd horia orasanuthe question is interesting and have many applications as stated prof dr mircea orasanu and prof drd horia orasanu and further that the complex integral , where f may or may not be analytic, and c may be either a closed curve or an arc. But, c is piecewise smooth and simple. The curve c is called the path of integration or contour.

Suppose that c has initial and final points z = A and z = B, respectively. Then = of any two vectors is defined by a b= c = , where is a unit vector (vector of length 1) pointing perpendicular to the plane of a and b. But as there are two directions perpendicular to any plane, the ambiguity is resolved by the right hand rule: let your fingers point in the direction of the first vector and curl around (via the smaller angle) towards the second; then your thumb indicates the direction of .

A Unit vector is a vector whose magnitude is 1 and point is a particular direction. Without loss of generality, we can assume to be three distinct unit vectors along the x, y, and z-axis relatively.

B

Editor of Duke Mathematical Journal wins Chern Medal Award

Editor of Duke Mathematical Journal wins Chern Medal Award | Duke

bigepsu

July 19, 2018 at 11:29 am

sure here we consider important aspects as say prof dr mircea orasanu and prof horia orasanu as followed with

ON THE THESES OF LAGRANGIAN AND HISTORY

ABSTRACT

Quantities that have direction as well as magnitude are called as vectors. Examples of vectors are velocity, acceleration, force, momentum etc.

Vectors can be added and subtracted. Let a and b be two vectors. To get the sum of the two vectors, place the tail of b onto the head of a and the distance between the tail of a and b is a+b.

Multiplication of a vector by a positive scalar k multiplies the magnitude but leaves the direction unchanged. If k=2 then the magnitude of a doubles but the direction remains the same.

Dot product of two vectors is the product of a vector to the projection of the other vector on the vector. a. b is called the dot product of the two vectors.

a. b = . If the two vectors are parallel, then a. b = and if the two vectors are perpendicular to each other, then a. b = 0

Cross Product

Then,

and

Also,

In cylindrical coordinate systems, a vector , where are the unit

1 INTRODUCTION

m. Imagine that the vector is a force whose units are given in Newtons. Imagine vector is a radius vector through which the force acts in meters. What is the value of the torque , in this case?

n. Now imagine that continues to be a force vector and is a displacement vector whose units are meters. What is the work done in applying force through a displacement ?

o. What is the vector sum of a vector given by 40 m, 30 degrees and a vector given by 12 m, 225 degrees? Use the method of resolving vectors into their components and then adding the components.

3. Consider three vectors:

A. Find .

b. Find .

c. Find .

PROBLEMS OF CALCULUS, Prof. Mircea Orasanu

Here consists that, pentru un sir de descompuneri ale lui de norma tinzand la la zero

Avem limdin norma

Pe de alta parte pentru exista astfel incat sa avem dist de indata, ,iar aceasta implica

Existenta unui morfism intre si aici observatii

.se poate extinde pentru panza cand x= f, y = g, z = h stfel ca matricea || f’u,..| are in

fiecare

On “Theses on Theory and History”

History | Duke University Press News

candozu

July 19, 2018 at 2:14 pm

sure we see these as very interesting for following consequences as any situations as say prof dr mircea orasanu and prof horia orasanu as followed

LAGRANGIAN PROBLEM IN WILDNESS SENSE

ABSTRACT

one of the most highly regarded and expertly designed C++ library projects in the world. — Herb Sutter and Andrei Alexandrescu, C++ Coding Standards

This is the documentation for an old version of boost. Click here for the latest Boost documentation.

Elliptic Integrals of the Third Kind – Legendre Form

Synopsis

#include

namespace boost { namespace math {

template

calculated-result-type ellint_3(T1 k, T2 n, T3 phi);

template

calculated-result-type ellint_3(T1 k, T2 n, T3 phi, const Policy&);

template

calculated-result-type ellint_3(T1 k, T2 n);

template

calculated-result-type ellint_3(T1 k, T2 n, const Policy&);

}} // namespaces

Description

These two functions evaluate the incomplete elliptic integral of the third kind Π(n, φ, k) and its complete counterpart Π(n, k) = E(n, π/2, k).

The return type of these functions is computed using the result type calculation rules when the arguments are of different types: when they are the same type then the result is the same type as the arguments.

template

calculated-result-type ellint_3(T1 k, T2 n, T3 phi);

template

calculated-result-type ellint_3(T1 k, T2 n, T3 phi, const Policy&);

Returns the incomplete elliptic integral of the third kind Π(n, φ, k):

Requires -1 <= k <= 1 and n 1 and φ is not in the range [0, π/2] is currently unsupported and returns the result of domain_error. For this reason it is recomended that you keep φ inside its “natural” range of [0, π/2].

template

calculated-result-type

1 INTRODUCTION

template

calculated-result-type ellint_3(T1 k, T2 n, const Policy&);

Returns the complete elliptic integral of the first kind Π(n, k):

Requires -1 <= k <= 1 and n

For a matrix of size N × N the cost of performing Gaussian elimination is

proprotional to N3. If the matrix is banded with bandwidth M, the matrix

structure can be exploited to reduce the cost to NM2. Gaussian elimination

will be discussed in much greater detail in future lectures. For our matrix A,

N =n2 and M = 2n+ 1 hence a total cost . Iterative methods can also

be pursued － these will depend on the condition number, discussed below.

2 FORMULATION

Poisson Problem 2D

In most practical situations, the solution to the Poisson equation will be required

on general domains, and with more complicated boundary conditions than those

considered here. In this case, some of the solution techniques that will be con-

sidered later on in this course may be more appropriate. However, in those

specific situations where the Poisson equation has to be solved on a rectangle

with sufficiently simple boundary conditions, dicrete Fourier techniques may be

employed to solve the system of equations very efficiently.