The use of Pi in geometry is very common. It is seen in formulas for area and circumference of circles, volume of cones and cylinders and more. Most of us know the value of Pi to be around 3.14, with many people memorizing Pi up to many digits. But where does the value of Pi come from?

The distance around the outside a circle is its circumference. Consider the formula for the circumference of a circle, Circumference = Pi x Diameter. Solving for Pi we get, Pi = Circumference divided by Diameter. Therefore, for a circle of any size, the circumference divided by the diameter is approximately 3.14. But how do we know this?

We can start approximating Pi but considering the perimeter of an regular n-sided figure and dividing that by the length of a diagonal. Take a square with side equal to one. By using the Pythagorean Theorem, or the knowledge of 45-45-90 triangles, the diagonal is approximately 1.41. Therefore the perimeter divided by the diagonal is approximately 2.82.

Now take an octagon, with sides of length equal to one. The length of the diagonal found by taking 2 times the apothem is approximately 2.61. Taking the perimeter of 8 divided by 2.61, we get an approximation for Pi to be 3.06. Notice that this is fairly close to the value of Pi or 3.14.

As the number of sides of a regular polygon increases, the shape more closely approximates that of a circle. When considering a regular polygon with 500 sides, each side of length one, we see calculate the diagonal to be approximately 159.155 and the approximation for Pi to be approximately 3.14157. This is extremely close to the value of Pi at 5 digits, which is 3.14159.

One can continue this process by increasing the number of sides of a regular polygon to get as close of an approximation as possible to the actual value of Pi. I have a book which has the value of Pi calculated out to 1 million digits. Why someone needs to know that, I don't know. For all practical purposes, 3.14 works fine. Also, the fraction 22/7 is often used in calculations involving Pi.

I was working with a student on completing the square and it tends to be a more tricky method of factoring than using the quadratic formula.  There are several steps to remember.

Recall how we factored and solved quadratic equations using the reverse FOIL method. An example of this type of factoring is x^2 + 5x + 4 = 0 factored is (x +4)(x + 1) = 0. Sometimes it's quite difficult to solve quadratic equations using this method, so we can solve by a method known as completing the square.
The idea behind completing the square is to turn a binomial into a perfect square trinomial. For example, consider the binomial x2 + 6x. The perfect square trinomial with the first two terms x2 + 6x is x2 + 6x + 9 because (x + 3)(x + 3) = x^2 + 6x + 9. Notice how we added a 9 to x^2 + 6x. The question we ask ourselves is, “What number squared equals 9?” We know that 3^2 = 9. Also notice that 3 is half of the coefficient of the middle term 6x. We take half of the middle term, square it and
add it to form the perfect square trinomial. Then we factor the trinomial.

Examples: Complete the square and factor the perfect square trinomial.
1. x^2 + 8x

Step one: Take half the coefficient of the middle term. (1/2)(8) = 4.
Step two: Square the result in step one. 4^2 = 16.
Step three: Add the result in step two to the binomial to form the trinomial x^2 + 8x + 16.
Step four: Factor the trinomial. (x + 4)(x + 4) or (x + 4)^2.

 x^2 – 7x
Step one: Take half of the coefficient of the middle term. (1/2)(-7) = -7/2.
Step two: Square the result in step one. (-7/2)^2 = 49/4.
Step three: Add the result in step two to the binomial to form the trinomial x^2 – 7x + 49/4.
Step four: Factor the trinomial. (x – 7/2)(x – 7/2) or (x -7/2)^2


To solve quadratic equations by completing the square we must remember the following:
1. Make sure the coefficient of the squared term is one. If it is not one, we must divide both sides of the equation by the coefficient of that term. For example, if the term is 2x2, we must divide both sides of the equation by 2.
2. Get all variables on one side of the equation and the constants on the other side. This makes sure we have a binomial in the form x2 + bx.
3. Take half of the coefficient of the middle term, square it and add it to both sides of the equation.
4. Factor the perfect square trinomial.
5. Solve the equation using the square root property and check answers by substituting into the original equation.

Note if the coefficient in front of the x^2 term is not 1, must divide the equation by the coefficient before completing the score.

For B to be the multiplicative inverse of A, AB = I and BA = I. So we begin by multiplying the first row of A with the
first column of B to get
(-3)(-5/27) + (4)(1/9) = 15/27 + 4/9

For those having difficulty finding the inverse of a matrix, here's part of a chapter from my book on this topic.

Multiplicative Inverses of Matrices

Recall that the multiplicative inverse of any real number x is the number when multiplied by x equals 1. In this case the multiplicative inverse of x is 1/x. Suppose we have matrix A of the form








 If we multiply A by what is known as the identity matrix, we still get A. We will call the identity matrix I, note the illustration below of AI = A.

 
If A is an n by n matrix, then there exists another matrix A-1, called A inverse, so that AA-1 = I and A-1A = I. So we basically are looking for a matrix that when multiplied by the original matrix, equals the identity matrix and vice versa. The identity matrix is always a matrix with 1's along the diagonal from upper left to lower right and 0's everywhere else.

Example: Show that B is the multiplicative inverse of A where
 







 For B to be the multiplicative inverse of A, AB = I and BA = I. So we begin by multiplying the first row of A with the first column of B to get

(-3)(-5/27) + (4)(1/9) = 15/27 + 4/9
= 15/27 + 12/27
= 1

Next we multiply the first row of A with the second column of B to get
(-3)(4/27) + (4)(1/9) = -12/27 + 4/9
= -12/27 + 12/27
= 0

Now proceed to multiply the second row of A with the first row of B to get
(3)(-5/27) + (5)(1/9) = -15/27 + 5/9
= -15/27 + 15/27
= 0

Finally, multiply the second row of A by the second column of B to get
(3)(4/27) + (5)(1/9) = 12/27 + 5/9
= 12/27 + 15/27
= 1

Therefore B is the multiplicative inverse of A. Notice the illustration of this below.

Here's a little review on graphing and finding equations of perpendicular lines.

Example: Find the equation of the line that passes through the point (2, 3) and is perpendicular to the line
y = 3x - 5.

Solution:
The slope of the line perpendicular to y = 3x – 5 must multiply by 3 equal -1. We think of the slope of the first line as m1 and the slope of the second line as m2, therefore 3(m1) = -1, so m1= -1/3.
Now we know that m = – 1/3, x = 2 and y =3.
Use the formula y = mx + b
3 = (-1/3)(2) + b
3 = -2/3 + b
3 2/3 = b
11/3 =b
The equation of the line perpendicular to y = 3x – 5 that passes through the point (2, 3) is y = (-1/3)x + 11/3.
Notice the graph of both lines below.

Here's a math puzzle


Suppose you go to a hotel and ask for a room and the clerk says that there are no room available.  The clerk then states that if I ask everyone to move to the room next them, then a room will become available. For example, the people in room 1 will move to room 2, from room 2 to room 3 and so on. The people in the last room will move to room 1.  How does a room open up???

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This is a hotel with an infinite number of rooms, therefore noone ever moves into room 1.  It's kind of an absurd concept, which leads to the concept of infinity. It's a very uncertain concept in and of itself.

Suppose we wish to graph the equation y = -3(x - 3)^2 + 5.   We recognize this is a parabola with the vertex at (3, 5) because we can rewrite this as y - 5 = -3(x - 3)^2, which is in the form y - k = a(x - h)^2 (vertex at (h, k)).

The parabola opens up or down, and since a is negative, the parabola opens down. 

Interesting, we can use calculus to determine whether (3, 5) is a maximum or a minimum. If it's a maximum, the parabola opens down, if it's a minimum, the parabola opens up.

y' = -6(x - 3) = -6x + 18

We now find the derivative for a value of x < 3 and one for a value for x > 3

for x = 2, y' = -6(2-3) = -6(-1) = 6

for x = 4, y' = -6(4-3) = -6(1) = -6

Since the slope of the curve at x = 2 is positive and the slope of the curve at x = 4 is negative we have a maximum at x = 3.  Therefore (3, 5) is a maximum and the parabola opens downward.

Just another way to determine how a parabola opens.

I am noticing a disturbing trend in schools today regarding mathematics and teaching of mathematics.  The majority of students I work with tell me how poor their teachers are. Many basically just read things from the book, don't take time to really explain concepts in class or take the time to work with student after school. I even heard of one teacher that simply posts youtube videos on the topic and hands out worksheets with the answer key.  Making matters worse, the school has a tutoring center, but that has been deemed pretty useless as well. When I was in school, teachers seemed more apt and wanting to help students.  Are teachers, in general, getting more lazy? I know there are still plenty of excellent teachers and certainly want to give credit to those fine educators that take their jobs seriously. But I can't overlook what seems to be a higher percentage of poor teachers today than 25 to 30 years ago.

Testing a hypothesis with proportions can be easy.



  For example:  Suppose the manufacturers of a certain brand of candy says that 40% of the pieces in the bag are red and the rest are green.   Suppose a bag of 50 pieces of candy is opened and 15 of them are green.  We can run a test to see if our claim  is true based on the sample size and sample proportion. We will chose to test at 5% significance. 

Step 1:   state your null and alternate hypothesis.  The null hypothesis is what is claimed.

The null hypothesis is that 40% are green. The alternate hypothesis is proportion of green is not 40%


Step 2:  calculate the test statistic

Z = (p^ - p)/standard deviation

p^ is the sample proportion of candy that is green = 15/50 = .30

p is the claimed proportion of candy that is green = .40

square root[(p)(1-p)/n] = standard deviation = 0.0693

test statistic = (p^ - p)/standard deviation = -1.44

Step 3:  compare test statistic with critical value for the test. For 95% and 2 tailed test, this value is 1.96 and -1.96

If the test statistic is greater than 1.96 or less than -1.96 we reject the null hypothesis, otherwise we accept the null hypothesis.

In this case we accept the null hypothesis.

Sometimes there is confusion among students when graphing lines such as x = 5 and y = 4.

The x-axis is the horizontal axis and the y-axis is the vertical axis, yet the graph of x = 5 is a vertical line through 5 on the x-axis. Likewise the graph of y = 4 is a horizontal line through 4 on the y-axis.
This seems to be opposite of what one might expect, but there is a simple reason for this.

For the line x =5, think first of the point (5,0). Now draw the vertical line through this point and pick another point on the line, for instance (5,2), another point would be (5, -1) and so on. Notice that for every point, the only value that changes is the y coordinate, the x coordinate is always 5.  So the vertical line is correct because the x coordinate for every point on that line is 5. The slope of a vertical line is undefined.

The same argument holds true for the equation y = 4.  It's a horizontal line because for every point on that line, the y coordinate is 4.  The point (0,4) is on the line, as are points (1, 4), (-2, 4) and so on.

Consider the concept of  "theoretical probability".


You roll a fair 6 sided die. It is equally likely that any one of the 6 sides, denoted {1, 2, 3, 4, 5, 6} will appear face up.But the actual side that lands face up is not certain. If we roll the die 5 times, the possible outcomes are called the sample space and any occurrence in which the outcome is not known is called an experiment. Rolling a die is an example of an experiment. Therefore, for a single roll of the die, the sample space, denoted by S would be S ={1, 2, 3, 4, 5, 6}.

Suppose the result on a single roll of a die is 2. The next roll of the die is a 4. Each of these results are a subset of the sample space. A subset is called an event and denoted by E. Therefore, the subsets noted above are E ={2} and E = {4}.

The way in which we calculate theoretical probabilities is we divide the number of outcomes in the event divided by the total number of outcomes in the sample space.

What is the probability that a single roll of a fair 6 sided die lands 1 face up?
P(E) = (number of outcomes that result in 1 lands face up)/ (total number of possible outcomes) = 1/6

Example: A coin is flipped 3 times.
a. What is the probability that 2 of the 3 coins land heads up?
First we will figure out the sample space and the number of outcomes in the event.
We'll define the event E = { (H, H, T), (H, T, H), (T, H, H) }
The sample space is S = { (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, T, H), (T, H, T), (T, T, T) }
There are 3 events and 8 outcomes in the sample space, therefore the probability that 2 of the 3 coins land heads up is 3/8.

• Note that an easy way to determine how large the sample space is in this problem is we take the number of
possibilities per coin toss and raise it to the power of the number of tosses (23 = 8).

b. What is the probability that at least 2 coins land tails up?
In this case, we define the event E = { (H, T, T), (T, T, H), (T, H, T), (T, T, T) }
The sample space is the same as in part a, therefore the probability that at least 2 coins land heads up is 4/8 = ½.

Notice the events that 0, 1, 2 and 3 heads appear are listed below with their associated probabilities.
0 heads: E = { (T, T, T) } P(E) = 1/8
1 head: E = { (H, T, T), (T, H, T), (T, T, H) } P(E) = 3/8
2 heads: E = { (H, H, T), (H, T, H), (T, H, H) } P(E) = 3/8
3 heads: E = { (H, H, H) } P(E) = 1/8

Notice if we add the probabilities of all the events, we get 1. The sum of theoretical probabilities of all possible outcomes in a sample space equals 1.

Example: Two fair 6 sided dice as rolled. What is the probability of getting a sum of 6, 7 or 8 ?
Since each die has equally likely outcomes, there are 6 ∙6 = 36 possible outcomes in the sample space.

Notice all of the possible outcomes below.
S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }

The outcomes in S that give us a sum of 6, 7 or 8, which we denote as event E are
E = { (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 4), (4, 3), (1, 6), (6, 1), (2, 5), (5, 2), (2, 6), (6, 2), (3, 5), (5, 3), (4, 4) }

There are 16 outcomes in E and 36 outcomes in S. Therefore, the probability of getting a sum of 6, 7 or 8 is
16/36 =4/9

When graphing a parabola, sometimes it's confusing which way the parabola opens.  Does it open up, down, left or right?

A parabola in the form y = x^2 will open up. For every value of y, there will be 2 values for x.

When x = 3, y = 9.
          x = -3, y = 9
          x = 2, y = 4
          x = -2, y = 4

No matter what value you substitute for x, you will get a positive value for y.

If you plot these points in a graph you will notice the graph is symmetric about the y axis. This means the y axis cuts the parabola in half, mirror images on both sides of the axis.

The graph opens down if in the form -y = x^2.  This is because all the values which were positive before become negative.

You can also think of the parabola opening towards the linear term. In this case, the linear term is y and the parabola opens up or down, and the y axis runs vertically.

The exact opposite argument is true for parabolas in the form x = y^2 or -x = y^2.  In this case the x axis is the axis of symmetry and the parabolas open either right or left (horizontally) and the x axis is the horizontal axis.

Use implicit differentiation to find dy/dx.  We differentiate each x term as usual and when differentiating each y term, you add dy/dx since we are differentiating with respect to x.  Then isolate dy/dx to solve.

x^3 + y^3 - 3xy = 0

3x^2 + 3y^2(dy/dx) - [3x(dy/dx) + 3y] = 0

3y^2(dy/dx) - 3x(dy/dx) = -3x^2 + 3y

(dy/dx)(3y^2 - 3x) = -3x^2 + 3y

dy/dx = (-3x^2 + 3y)/(3y^2 - 3x)

dy/dx = (x^2 - y)/(y^2 - x)

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When multiplying numbers raised to fractional exponents, remember that if the numbers are the same, simply add the exponents. If the exponents are the same and the numbers are different, multiply the numbers and keep the exponent the same.

For example:

6^(2/3) * 6^(3/4)    The bases are the same (6), therefore keep the base and add the exponents

6^(2/3 + 3/4)

2/3 + 3/4 = 8/12 + 9/12 = 17/12

The answer is 6^(17/12)


Another example:

4^(1/3) * 9^(1/3)     The exponents are the same, so keep the exponent as is and multiply the bases

4 * 9 = 36

The answer is 36^(1/3)

The sum of three positive integers is 16. If the first integer is doubled, the sum is 22. If the third integer is tripled, the sum is 20. What are the 3 integers?

Solution:
Let x = First integer
y = Second integer
z = Third integer
x + y + z = 16 (Sum of the integers is 16)
2x + y + z = 22 (Double the first integer and the sum is 22)
x + y + 3z = 20 (Triple the third integer and the sum is 20)

We can eliminate both the y and z variable by subtracting the second equation from the first, therefore
x + y + z = 16
-(2x + y + z = 22)
-x = -6, x = 6

Subtract third equation from the first equation to eliminate the x and y variable, therefore
x + y + z = 16
-(x + y + 3z = 20)
-2z = -4, z = 2

Now substitute 6 for x and 2 for z in the first equation and solve for y.
6 + y + 2 = 16
8 + y = 16
y = 8

The three integers are 6, 8 and 2.