Tau vs. Pi

I recently watched a video explaining the difference between pi and a new constant called tau, which is the equivalent of two times pi. The argument for tau being used is that many equations include two times pi, and that using tau would eliminate this problem and make the equation look simpler. The video itself is somewhat humorous, but it is also very interesting. Enjoy!

Pascal's Triangle

In class this week, we learned about Pascal's Triangle. This is a very useful method of expanding binomials, as I will demonstrate with a quick example.
Imagine that you have the following binomial:

(a + b)^5

To find the coefficients for this expansion, you would have to consult Row 5 of Pascal's triangle:

The coefficients for Row 5 are:

1  5  10  10  5  1

Now, you must place these coefficients before each term. The first term (a) will decrease in power from left to right, while the second term (b) will increase in power from left to right.

(1*a^5*b^0) + (5*a^4*b^1) + (10*a^3*b^2) + (10*a^2*b^3) + (5*a^1*b^4) + (1*a^0*b^5)

When simplified, this is equal to:

a^5 + 5(a^4)(b) + 10(a^3)(b^2) + 10(a^2)(b^3) + 5(a)(b^4) + b^5

Finally, we have expanded this binomial using Pascal's triangle. Expanding the binomial by using the F.O.I.L. method would have also worked, although it would have taken significantly longer and required a much greater amount of work also.

Well Ordering Principle

The well ordering principle states that "every non-empty subset of N has a smallest (or first) element" (http://www.proofwiki.org/wiki/Well-Ordering_Principle).

This principle is what we use during mathematical induction. The process for mathematical induction is essentially the following:
Prove that the statement is true for n = 1.
Assume that statement to be true for n + 1.
Prove that the statement is true for n + 1.
By doing this, you will prove that the statement is true for all natural numbers n.
I have found two great videos that explain the process for mathematical induction below. The first one is a bit over 9 minutes long, so I have also included a shorter video if you would not like to spend as much time watching the video.

Longer video:

Shorter video:

Why is 0 Factorial Equal to 1?

The factorial of k! is defined as the product of all positive integers from 1 to k. This means that k! can also be written as the following:

k! = k*(k - 1)!

Here is an example:

k = 2

2*(2-1)! = 2

By using this rule, if we set k = 1, then the formula results in the following:

k = 1

1*(1-1)! = 1

1*(0)! = 1

0! = 1

Therefore, 0! is equal to 1.

(Source: http://www.zero-factorial.com/whatis.html)

Finding the n-th Term and Sum of Sequences

Today in class, we learned how to find the nth term of an arithmetic sequence and also how to find the sum of a sequence.
I will start of by explaining the process for finding the nth term of an arithmetic sequence. To provide an example, I will work with the following sequence: 2, 4, 6, 8, ... n. To find the nth term of this sequence, we must use the following equation:

In this equation, a represents the first time of the sequence, and d represents the common difference. The common difference is "the constant added to each element of an arithmetic progression to obtain the next" (definition from: http://en.wiktionary.org/wiki/common_difference).

In this case, a is equal to 2, since 2 is the first term in our sequence. Also, d is equal to 2, because we must add 2 to each term in order to obtain the next term.

Therefore, our equation looks like this:

2 + (n - 1)2

(or, when simplified:)

2n

Now, if we would like to find the 10th term of this sequence, we simply substitute n with 10 in the equation, yielding:

2(10) = 20

Next, I will explain the process of finding the sum of n terms of a sequence (I will use the same sequence as above for this example also).

Taking a closer look at this equation, we can see that this is the same as the following:

(n/2)(a₁ + a_n) 

To obtain a_n, or the nth term, we can simply use the first equation.

Again, let us set n = 10. This means that a_n = 20.

All we have to do now is substitute these values into the sum equation:

((10)/2)(2 + 20) = 5(22) = 110

Now, we have determined the sum of this arithmetic sequence as well.

Sequences and Sigma Notation

In class on Wednesday, we learned about sequences and summation notation. By using sequences, one can determine the nth term when given a sequence of numbers.
Also, one can determine the sum of a set of numbers by using summation (sigma) notation. Here is a simple example:

The number at the top, 3, is called the upper bound or limit. The number at the bottom, 0, is called the lower bound or limit. In this case, n is the variable that will be used; as you can see, n is set to 0 to begin with.
The process of finding the sum requires you to substitute every value of n (in increments of 1) from the lower limit to the upper limit into the expression to the right of the sigma symbol.
For this example, we would start by substituting 0 (n's starting value) into the expression (2+3n). This yields:

2 + 3(0) = 2

We would repeat this process for every value of n until n was equal to 3:

2 + 3(1) = 5
2 + 3(2) = 8
2 + 3(3) = 11

Now, we take all of the numbers that we have produced and add them together:

2 + 5 + 8 + 11 = 26

Finally, we have determined that the sum of every number from 0-3 when substituted into the expression (2+3n) is equal to 26.

VoiceThread

The following is a link to Group 3's VoiceThread:
https://voicethread.com/?#q.b5433149.i27621334

Chapter 8 Review Group 3

Follow this link to visit Group 3's review website!

No Solution or Infinitely Many Solutions in Matrices

When solving a system of equations using matrices, there can be a unique solution, infinitely many solutions, or no solution.
In row-reduced form, the matrix (if it has a unique solution) will consist of a diagonal of 1's surrounded entirely by zeros, and a column of constants to the far right; here is an example:

| 1  0  0 | 4 |
| 0  1  0 | 2 |
| 0  0  1 | 7 |

A matrix with infinitely many solutions has more variables than non-zero rows; here is an example:

| 1  0  3 | 4 |
| 0  1 -2 | 3 |
| 0  0  0 | 0 |

Finally, a matrix with no solution will have a row of zeros on the left side, while the constant on the right side is not 0. A matrix with no solution is called inconsistent; here is an example:

| 1  0  3 | 4 |
| 0  1 -2 | 3 |
| 0  0  0 | 2 |

When graphed, the three types of solutions will look similar to the following image, where consistent and independent means unique solution, consistent and dependent means infinitely many solutions, and inconsistent means no solution.

Inverses of Matrices

This week, we learned how to find the inverse of a matrix.
The inverse of a matrix A is written as A^-1. When a matrix is multiplied by its inverse, or when the inverse is multiplied by the original matrix, the resultant matrix should have 1's in its main diagonal, while every other entry is 0.
An easy method of finding the inverse of a matrix is to juxtapose a matrix in the desired format to it, and then to perform the basic row operations until the matrix portion on the left looks like the desired matrix. The matrix portion on the right will be the inverse. This might sound a bit confusing, so I will explain the process with a simple 2 x 2 matrix I found on the Internet (by the way, not all matrices are invertible, so some might not have an inverse at all).

Here are the steps to solving this matrix:

1. Add the first row multiplied by -3 to the second row.
2. Divide the second row by -2.
3. Add the second row multiplied by -2 to the first row.

Now, we have determined that the matrix on the right side of the vertical line is the inverse of the original matrix.
You can always verify that you have correctly determined the inverse matrix by either multiplying the original matrix by its inverse or vice-versa; if the resultant matrix consists entirely of 0's except for 1's in the main diagonal, then the inverse is correct!

Cryptography Worksheet

In class, we went through several aspects of cryptography:

1. We found a decryption matrix from an encryption matrix by finding the inverse of the encryption matrix.
2. We split the encrypted message into groups of 3, and then multiplied each group of 3 by the decryption matrix in order to find their respective values.
3. We used the code 1=A, 2=B, 3=C, and so on, to determine the alphabetical counterpart of each number and to subsequently determine the final message.
4. We selected our own invertible matrix and multiplied the resultant number groups from step 2 by this matrix to encrypt the message ourselves.

Here is a picture of this worksheet:

Determinants of Matrices

Today we learned how to find the determinant of any square matrix.
It is fairly easy to find the determinant of a 2 x 2 matrix, where the determinant in the matrix
| w  x |
| y   z |
is wz - xy.
For a 3 x 3 matrix, several methods of finding the determinant can be used:

• You can use a shortcut in which the first two columns of the matrix are written to the right side of the last column of the matrix, and then diagonals are formed in order to individually subtract the products of the upward diagonals from those of the downward ones.
• You can use minors and cofactors to solve any square matrix n x n, and this method can be applied to a 3 x 3 matrix, where n = 3.
• If the matrix is triangular, you can find the product of the middle diagonal of the matrix, and this will be the determinant.

Below is an image of the practice problems from today's lesson, which I have completed in UPAD lite:

The following is a link to an Educreations video made in our group:

http://www.educreations.com/lesson/view/determinant-of-a-matrix/16850106/?s=ciem9G&ref=app