Wednesday, February 26, 2014

Well ordering principle

This week we talked about the well ordering principle present in geometric and arithmetic sequences.

The statement is: 

For every set X, there exists a well-ordering with domain X.

How do we prove it then?

Take the set A of all well-orderings of subsets of X: an element of A is an ordered pair (a,b) where a is a subset of X and b is a well-ordering of a. A can be partially ordered by continuation. That means, define E  F if Eis an initial segment of F and the ordering of the members in E is the same as their ordering in F. If E is a chain in A, then the union of the sets in E can be ordered in a way that makes it a continuation of any set in E; this ordering is a well-ordering, and therefore, an upper bound of E in A. We may therefore apply Zorn's Lemma to conclude that A has a maximal element, say (M,R). The set M must be equal to X, for if X has an element x not in M, then the set M∪{x} has a well-ordering that restricts to R on M, and for which x is larger than all elements of M. This well ordered set is a continuation of (M,R), contradicting its maximality, therefore M = X. Now R is a well-ordering ofX.


In 9.1 about sequences and summation notation, we can see that every set of numbers can be organized because there is a relationship existing in between. 

 Example:

Find the relationship between 1,3,5,7... and 1,2,3,4

Fathering putting the two sequences side by side, we can see the relationship 

An = 2n-1



Fun Post- Pascal's Triangle!

Today lets talk about pascal's triangle!
Consider the triangle in Figure 1, called Pascal's triangle. It consists of numbers where each entry is the sum of the two entries above it.
Do you recognize the numbers in each row? This a quick way to generate the coefficients of (x+y)nfrom algebra!
And, better yet, you can use them as a quick way to calculate the powers of 11, since 11=10+1. Notice that 11, 121, 1331, and 14641 are all powers of 11...
Presentation Suggestions:
Draw several rows. Ask what 114 and 115 are! As a challenge, 116 is harder; you have to carry...
The Math Behind the Fact:
The generation of Pascal's triangle works because in long multiplication of a polynomial by (x+y), you end up adding adjacent coefficients of the polynomial together. The Fun Fact Multiplication By 11 is based on this idea.
Figure 1

Tuesday, February 25, 2014

9.4 mathematical formula

In this section we will study a mathematical proof called mathematical induction. 
The principles of mathematical induction: 
Let Pn be a statement involving the positive interger n. If
1. P1 is true 
2. Pk is assumed to be true
3. P(k+1) is proved to be true 
Then Pn must be true for all positive integers n
 Ex 1:
Find P(k+1) for Pk:Pk= [k^2 (k+1)^2]/4
P(k+1):P(k+1)= (k+1)^2(k+2)^2/4
 Ex 2: 1+3+5+...+(2n-1) = n^2
 Step 1 prove the statement is true for n= 1
            S1= 1
 Step 2 assume true for n= k
           1+3+5+...+(2k-1) = k^2
Step 3 prove true for n= k+1
           1+3+5+...+(2(k+1)-1) = (k+1)^2
           1+3+5+...+(2k-1) +(2(k+1)-1) =( k+1)^2
           Substitute the assumption for n= k
           Get k^2 +(2(k+1)-1) = ( k+1)^2
                  K^2+ 2k + 1= K^2+ 2k + 1
          Therefore the statement is true for all n as in positive integers. 

Thursday, February 20, 2014

9.2 arithmetic sequence

Today lets go into arithmetic sequences!
Definition: the difference between consecutive terms are the same! 
      A2-A1=A3-A2= A4-A3=d
How to find the nth term
      An= dn+c
       (c= a1-d)
Ex: find a formula for the nth term of the sequence with d= 3 and a1= 2
An= 3n+c 
C= 2-3= -1
An= 3n-1
The sum of a finite arithmetic sequence:
An= n/2(a1+an)
Ex: find the sum of intergers from 1 to 100
Sn= 100/2* (1+100)= 5050

Recursive sequence


A recursive sequence {f(n)}_n, also known as a recurrence sequence, is a sequence of numbers f(n) indexed by an integer n and generated by solving a recurrence equation. The terms of a recursive sequences can be denoted symbolically in a number of different notations, such as f_nf(n), or f[n], where f is a symbol representing thesequence.

The idea of sequences in which later terms are deduced from earlier ones, which is implicit in the principle of mathematical induction, dates to antiquity.

In the case of linear recurrence equations such as the recurrence

 F_n=F_(n-1)+F_(n-2)

(with F_1=F_2=1) generating the Fibonacci numbers, it is possible to solve for an explicit analytic form of the nth term of the sequence. Some special classes of recurrence equations have analytic solutions for specific parameters, but solutions for a general parameter is not known. An example of this type is the logistic equation

 x_n=rx_(n-1)(1-x_(n-1)),

which has known exact solutions only for r=-2, 2, and 4. It is not known how to solve a general recurrence equation to produce an explicit form for the terms of the recursive sequence, although computers can often be used to calculate large numbers of terms through brute force (combined with more sophisticated techniques such as caching, etc.).

RecursiveSequences

Historically speaking, the Fibonacci numbers F_n (left top figure), which are one of the most well-known such sequences, predate Leonardo Fibonacci's 1202 discovery by more than a millennium, having arisen around 200 BC in work by Pingala (Wolfram 2002, pp. 890-891). In the late 1800s and early 1900s, investigations into the foundations of mathematics led to the formal definition of so-called recursive functions. However, a systematic investigation of the types of possible behavior was apparently not undertaken until the work of Wolfram (2002), with the exception of a few isolated sequences such as Hofstadter's Q-sequence Q(n) in 1979 (right top figure) and the Hofstadter-Conway $10,000 sequence a(n) in 1988 (bottom figure).

Wednesday, February 19, 2014

9.1 sequence and summation notation

Sequence:
 Domain infinite sequence: positive intergers
              Finite sequence: the first n positive intergers only
 Ex : find the two terms of a sequence of an= 3n-2
        A1= 3(1)-2= 1
        A2= 3(2)-2= 4
 Ex: write an expression for the apparent pattern 
       N= 1 2 3 4 ...n
       Terms: 1 3 5 7 ... An
       Pattern: 2n-1
Factorial:
 N!= 1*2 *3 *4...*(n-1)*n 
 0!=1 
Summation notation
                                n
Ai= a1+a2+a3+a4+...+an
                     i=1

I= index of summation
N= upper limit
1= lower limit


Friday, February 14, 2014

Google site

So this is the site our group has made for chapter 8 review!
Good luck on your tests!
https://sites.google.com/a/maranathastudents.org/chapter-8-review/

Saturday, February 8, 2014

Crypotography Breakthrough!

Hey Guys today let us talk about the mysterious cryptography!!!
As a graduate student at the Massachusetts Institute of Technology in 1996, Amit Sahai was fascinated by the strange notion of a “zero-knowledge” proof, a type of mathematical protocol for convincing someone that something is true without revealing any details of why it is true. As Sahai mulled over this counterintuitive concept, it led him to consider an even more daring notion: What if it were possible to mask the inner workings not just of a proof, but of a computer program, so that people could use the program without being able to figure out how it worked?
The idea of “obfuscating” a program had been around for decades, but no one had ever developed a rigorous mathematical framework for the concept, let alone created an unassailable obfuscation scheme. Over the years, commercial software companies have engineered various techniques for garbling a computer program so that it will be harder to understand while still performing the same function. But hackers have defeated every attempt. At best, these commercial obfuscators offer a “speed bump,” said Sahai, now a computer science professor at the University of California, Los Angeles. “An attacker might need a few days to unlock the secrets hidden in your software, instead of a few minutes.”
Secure program obfuscation would be useful for many applications, such as protecting software patches, obscuring the workings of the chips that read encrypted DVDs, or encrypting the software controlling military drones. More futuristically, it would allow people to create autonomous virtual agents that they could send out into the computing “cloud” to act on their behalf. If, for example, you were heading to a remote cabin in the woods for a vacation, you could create and then obfuscate a computer program that would inform your boss about emails you received from an important client, or alert your sister if your bank balance dropped too low. Your passwords and other secrets inside the program would be safe.
“You could send that agent into the computing wild, including onto untrusted computers,” Sahai said. “It could be captured by the enemy, interrogated, and disassembled, but it couldn’t be forced to reveal your secrets.”
As Sahai pondered program obfuscation, however, he and several colleagues quickly realized that its potential far surpassed any specific applications. If a program obfuscator could be created, it could solve many of the problems that have driven cryptography for the past 40 years — problems about how to conduct secure interactions with people at, say, the other end of an Internet connection, whom you may not know or trust.
“A program obfuscator would be a powerful tool for finding plausible constructions for just about any cryptographic task you could conceive of,” said Yuval Ishai, of the Technion in Haifa, Israel.
Precisely because of obfuscation’s power, many computer scientists, including Sahai and his colleagues, thought it was impossible. “We were convinced it was too powerful to exist,” he said. Their earliest research findings seemed to confirm this, showing that the most natural form of obfuscation is indeed impossible to achieve for all programs.
Then, on July 20, 2013, Sahai and five co-authors posted a paper on the Cryptology ePrint Archive demonstrating a candidate protocol for a kind of obfuscation known as “indistinguishability obfuscation.” Two days later, Sahai and one of his co-authors, Brent Waters, of the University of Texas, Austin, posted a second paper that suggested, together with the first paper, that this somewhat arcane form of obfuscation may possess much of the power cryptographers have dreamed of.
“This is the first serious positive result” when it comes to trying to find a universal obfuscator, said Boaz Barak, of Microsoft Research in Cambridge, Mass. “The cryptography community is very excited.” In the six months since the original paper was posted, more papers have appeared on the ePrint archive with “obfuscation” in the title than in the previous 17 years.
However, the new obfuscation scheme is far from ready for commercial applications. The technique turns short, simple programs into giant, unwieldy albatrosses. And the scheme’s security rests on a new mathematical approach that has not yet been thoroughly vetted by the cryptography community. It has, however, already withstood the first attempts to break it.
Researchers are hailing the new work as a watershed moment for cryptography. For many cryptographers, the conversation has shifted from whether obfuscation is possible to how to achieve it.
“Six or seven years ago, you could have looked at this question and wondered if we’ll ever know the answer,” said Leonard Schulman, of the California Institute of Technology in Pasadena. “The fact that there’s now a plausible construction is huge.”
Too Powerful to Exist
When Sahai started thinking about obfuscation 17 years ago, the first task was simply to define it. After all, users can always learn something about a garbled version of a program simply by feeding it inputs and seeing what comes out.
The most natural, and also the strongest, definition was the idea of a “black box” obfuscator, which would jumble a program so thoroughly that a person with the best available computational resources could figure out nothing at all about it, except for what might be gleaned from inputs and outputs. You could not figure out the value of a password hidden inside the software, unless that password was one of the program’s outputs, nor could you reassemble parts of the program to compute anything meaningful other than what the program was originally designed to compute.
A black box obfuscator, if it existed, would be immensely powerful, providing instant solutions to many cryptography problems that took decades to figure out or in some cases remain unsolved. Take, for example, public key encryption, whose development in the 1970s paved the way for Internet commerce. Prior to its creation, two people who wanted to communicate secretly had to meet in advance to choose an encryption scheme and share a secret key for encoding and decoding messages. Public key encryption allows you to announce a key to the entire world that permits people you’ve never met to send you messages that only you can decrypt. The innovation so revolutionized cryptography that its early developers have been recognized with one award after another.

8. 3 finding inverse of a matrix


Hey today we learnt to find the A^-1 of A, commonly called the inverse of A.
Definition:
 If there exists A^-1 x A= I(n) = A^-1xA,
 A^-1 is called the inverse of A
How do you find an inverse matrix:
1. Write the nX2n matrixm( matrix A on the left and Identity matrix on the right with a dotted line separating the two)
2. Reduce A to I using elementary row operations on the entire matrix set. The result will be [I: A^-1] 
    Note : if this is not possible, A is not imversible
3. Check your work 

Here are two examples 
In example one, we used the procedures and found out the inverse
 Example two does not have an inverse!!!!!
 Note how the last matrice set is not the identity, so there is no inverse for this matrix. 
 Have you learnt something? 

Wednesday, February 5, 2014

How to find the determinant of a matrix?

For the firs part of our class today we learnt how to find the determinant of special size matrices.
1. For 2X2 square matrix, [ a a1 
                                          a2 a3] : det (A) = aa3-a1a2
2. For 3X3 matrix: 1. Write out the first two columns on the right side of the matrix
                             2. Subtract the left three diagonals from the left three diagonals
3. Triangular matrix: the upper one has zeroes below the main entry, and the lower has zeroes above
                                multiply each number in the main entry of the matrix 
See the example problem