Saturday, June 29, 2013

The Magic of Algebra

Think of a number from one to ten.
Multiply that number by two.
Add thirteen.
Subtract seven.
Divide by two.
Subtract your original number.
You are thinking of the number three.

This is a classic mind reading trick that many people will learn to do when they are little. The premise is simple; the answer is always three.

Why does this work? As a young child, you probably think back to the arithmetic skills you learned in school and realized that you can't prove this trick because you don't know what number they'll choose.

This is a good time to note that when you learned this trick, the first thing you thought was "why?" Many people see proofs as a tedious process that could be avoided just by accepting things to be true. However, that why spark that went off when you learned the trick is where proofs can be applied. I use proofs so often in my posts primarily for that reason: to foster that why spark.

Anyways, this situation is where you need to define that unknown quantity of the number the spectator is thinking of. How do you define an unknown quantity? Well, this is when you pull out a little tool called algebra. You might not expect to apply algebra to a practical situation, but this is when you can. With algebra, we can stop saying "the number that they are thinking of" and start saying "n."


Now, the first step was to multiply the number by 2. n times 2 is pretty simple; it is 2n.

n • 2

Next, we had to add thirteen. That can't be simplified much.

2n + 13

Now, we subtract seven. 13 - 7 is 6, so we can simplify it to 2n + 6.

2n + 13 - 7
2n + 6

Now, we have to divide by two. 2 and 6 both have a factor of two in it, so we will not need any fractions.

(2n + 6)/2
2(n + 3)/2
n + 3

The last step was to subtract the original number. But, we already defined that number to be n, so we just subtract n.

n + 3 - n

As you see, the n's cancelled, and you ended up with three. Basically, you undid all of the previous operations to end up with one number.

I think that this is a perfect way to show young children magic, but also introduce them to concepts like algebra and proofs by lighting that spark in such a fun and interesting way.

Saturday, June 22, 2013

History of Math: Leonhard Euler

Today, I am giving a talk about Mending Mathematics Education in America at TEDxBushnellPark in Hartford, Connecticut. I did a post on this issue (click here to see it), as well as a huge school project on the issue (click here to see the research paper).

In the talk, I will be bringing up a famous math problem that originated with Leonhard Euler. I won't post about that problem today, but I will give a little story from the life of Euler.

Leonhard Euler was born on April 15, 1707 in Basel, Switzerland. He chose the path of a mathematician (if he didn't, I probably wouldn't be talking about him right now), and ended up as one of the most prolific mathematics writers of all time.

Some of his biggest contributions include standardizing the notation for the number e and π, coming up with Euler paths and circuits, and revamping most of the branches of mathematics that were known in his time. He was said to be able to entertain a child, scratch a cat, and calculate math problems simultaneously.

On this blog, I have mentioned sequences a lot. For instance, the triangular numbers are a sequence. The negative-first powers are a sequence. The powers of 1/2 are a sequence. Let's look at these three sequences:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55...
1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10...
1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128...

We haven't done it before, but a common thing to do is add up all of the numbers in these sequences. In some sequences, this seems odd to be doing, but in other ones, it seems completely understandable. Let's try it with those three.

1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 +...

The individual numbers themselves are going towards infinity, so the sum of them is probably infinity as well. Generally, if the numbers in a sequence are growing, then the sum will be infinite. This type of sequence is called a divergent sequence, and questions like their sum are generally not asked.

What about the next series. This one is a little harder to tell, but if we group some things together, it will be easier to see.

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...

Compare this to the following sequence, which is clearly less than this one.

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...
1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + ...

Since all of the terms in the first sequence seem to be bigger than the second one, then the first one has to sum to a greater total. But, the second sequence simplifies to 1 + 1/2 + 1/2 + 1/2 +..., which will end up as infinity. So, a sequence that has a bigger sum must also be infinite as well. This makes that a diverging sequence.

What about the third one? Well, combining terms together won't do much, but let's try figuring out what the sum is by setting the sequence equal to x.

x = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 +...

Now, what if you were to multiply that by two? You would get:

2x = 2(1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 +...)
2x = 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 +...
2x = 2 + (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 +...)
2x = 2 + x
x = 2

So, the sum of this sequence is not infinity, but rather the natural number two. Since the sequence does not diverge to infinity, it is called a convergent sequence. These are the type that are generally summed to their total, since that number can then be applied.

Anyways, some of Euler's students were analyzing a much more complicated convergent sequence to figure out its sum. They had been adding numbers, and after they got to the seventeenth number in the sequence, they found their answers to be different at the fiftieth decimal place. Since they didn't want to go back through that tedious process, they argued and argued over who got it right.

Eventually, they went to their professor, Euler, to ask him for assistance. After asking for the problem, he mentally calculated out the answer to the fiftieth decimal, and was able to determine who was correct.

I found that story kind of funny, but also really impressive. To be able to calculate that difficult of a problem to that high of a degree shows some real mathematical talent.

Friday, June 21, 2013

Problem of the Week Day 5: Week of 6/17/13 - 6/21/13

Today is the final day of the first 2013 problem of the week! The goal of the problem is to find x, so I have saved the x variable for the last part. In the hard problem, we used x with a subscript, but the plain x will represent the solution to today's problem. We have already covered trigonometry, algebra, number theory, and geometry, so I thought it would be good to devote Friday to probability. I might incorporate some game theory as well on Fridays in later problems.

Take the following set of data:

59, 38, 45, p, 29, g, 27, y + 1.24, q, 41, z + 5.92, 37, b, 21

What is the median of this sequence?

x = _____

First, solve for the following six variables, which will come up in the data set you will be analyzing.

d = a ÷ 1000
g = h ÷ 7
j = u ÷ 2
k = u ÷ 4
r = p ÷ 10
s = q ÷ 3

The following data represents the correlation between the age of test subjects (x) and the average dollar amount that they spend at a nearby shopping mall per visit (y). Some of the data was rounded by the data collectors, and some of it is as accurate as possible, meaning that there will be some longer decimals.

     x                y                   
s q + r + g
t u
q - h q
h - u h - t
r h + t
r - t h - r
d + t p - h
g h - g
d h
2t q + t
k q + t
k u + t
j 4s

Find the function that describes the line of best fit for this data set. Use the traditional y = mx + b to make the equation of the line, using m and b as your variables. Then, determine the average spending of a 65-year-old woman using the equation. The average spending will be the x value that we always end the problem with.

m = ______
b = ______
x = ______

First, use the variables throughout the week to solve for m.

        t – 0.77 – p(100q10 + q12)(q q12)
= –––––––––––––––––––––––––––––––
q0 + 2c – a + b + q6 q10 q11 q14

Now, pretend that there is a room with m people inside of it. Determine what the odds are that in this room, two people share the same birthday (just the month and date, not the year). In other words, solve the birthday paradox for a room of m people. x will represent the odds of this happening.

x = _____%

Thursday, June 20, 2013

Problem of the Week Day 4: Week of 6/17/13 - 6/21/13

Today is day four of the problem of the week! We already covered trigonometry, algebra, and number theory, so I will devote today to geometry. Good luck!

Take the following rectangle:

What is the area and perimeter of this rectangle? Since a and p were already used on the triangle, use b for the area and q for the perimeter (just move up a letter).

b = _____
q = _____

First, determine the values of u and q.

u = n ÷ 1000
q = p ÷ 2

Now, take the following trapezoid, using these values:

What is the area of this trapezoid? Use a for area.

u = _____
q = _____
a = _____

Find the area t of a regular polygon with x3(b - a - c) sides and a perimeter of q. Round to the nearest hundredth.

t = _____

Good luck!

Wednesday, June 19, 2013

Problem of the Week Day 3: Week of 6/17/13 - 6/21/13

Today is day three of the problem of the week! Monday was trigonometry, Tuesday was algebra, so we will make Wednesday be number theory. This branch of math might sound foreign to you, but the concepts within it that I plan to use in the problems have all been taught to you at some point in your life.

Find the Greatest Common Factor of 100y and 100z. We will call this number g.

g = _____

What is the number of dots in a regular t-gon array whose sides are of length h? In other words, look at the figurative families, and find the tth family. Then, find the hth number in that family. Call that number n.

n = _____

Take the following sequence:

x1x2, p + y1, (q + 110x+ 4)/(t), q– q– q10 – q11 – q12 – q13

Now, find the explicit formula for this sequence. For the simplicity of the variables, I will tell you that the formula should be a quadratic function. Use the ax2 + bx + c model, with a, b, and c being the answers to this problem. They are also the coefficients of the equation.

a = _____
b = _____
c = _____

Good luck!

Tuesday, June 18, 2013

Problem of the Week Day 2: Week of 6/17/13 - 6/21/13

Today is day 2 of the problem of the week! Remember to substitute in all of yesterday's answers into today's problems.

Tuesday is the day that I set aside for Algebra. Since the easy level does not use actual material from an Algebra course, I have included things that students need to know for an algebra course (fractions and decimals, order of operations, etc.). I have also tried to make the arithmetic challenging in that section to make the actual task a little harder than normal.

First, calculate the value of m and n using the values of a and p. Reduce the fractions once you have found them.

m = –––––––

= ––––

Now, solve for y and z using the rules for subtracting and dividing fractions.

y = m - n
z = m ÷ n

m = _____
n = _____
y = _____
z = _____

Substitute the values of p and h into the following equation, and then solve for the value of t.

p(ht + 705) = (3t - 56)(6h - 2p)3

t = _____

First, find the value of q0 using the value of p from yesterday.

       9p + 47
q0 = –––––––

Now, qis the zeroth term in a geometric sequence. The recursive formula for this sequence is the following:

q= (½)qn-1

You can find values of this sequence now or at a different time, but you may need up to the 15th number in this sequence. Be prepared to be referring back to this many times throughout the week.

Knowing the values of the geometric sequence, simplify the following rational function (you can leave it factored, but make sure is the only variable):

                   (x - q11)(x - q9)
f(x) = ––––––––––––––––––––––
q11x + p)(x - q10)(x + q11)

Now, determine the vertical and horizontal asymptotes of this function. Use x1x2xand so on for the equations for the vertical asymptotes (if the asymptotes were 2 and 5, but x1= 2 and x2= 5), and y1y2yand so on for the equations for the horizontal asymptotes. Make sure that the asymptotes are arranged from least to greatest in relation to the variable subscripts.

I am not going to put down the variables to find with the blank. Just find as many values of q, x, and y that exist or are necessary for the problem.

Monday, June 17, 2013

Problem of the Week Day 1: Week of 6/17/13 - 6/21/13

Today begins the annual summer Problem of the Week. For more recent viewers, the problem of the week is a math problem I create that has five individual parts of it, each part containing a different branch of mathematics. Rather than posting the whole thing at once, I think it is more fun to spread it out across a whole week. So, there is a problem on Monday, which will find the value of one or more variables. Then, the Tuesday problem will take the values of those variables, and do more math on them to receive the next set of answers. You will have to substitute the answers in, and then solve the new problem. This will continue through the whole week, until Friday, when you finally solve for x and complete the problem. All of the answers get posted one month later along with the regular Saturday posting to go along with it.

For the last two years, I had an easy problem and a hard problem. Since I didn’t have as much math knowledge then, I have added a medium problem. To decide which problem(s) you want to attempt, just self-assess your current math ability. If your ability would compare to a sixth or seventh grader’s, you should do the easy problem. If you think you could handle Algebra I content, try the medium one. If you are confident in Geometry and Algebra II, then the hard one would be best for you. The June ones will also be slightly easier than the August ones, but it shouldn’t be much of a difference.

Because of the diversity of the different parts of the problem, I decided to assign each day of the week a branch of math. Monday will be Trigonometry (or something with triangles), Tuesday will be Algebra (or something that you need to know before taking algebra), Wednesday will be Number Theory, Thursday will be Geometry, and Friday will be Probability/Game Theory. So, all three of today’s problems will involve a triangle.

Many of the concepts have been discussed on the blog before, considering that I won't give a problem if I don't think the topics are somewhat cool and interesting. If I were to ask a question about logarithms (I will not be doing that, so that is the example I chose), search logarithms in the search bar, and you should find a post that discusses that concepts. Also, the Problem of the Weeks in 2011 explained every concept before giving the problem, so that should be helpful.

Without any more stalling, here are the problems:

Look at the triangle below:
Determine the area and perimeter of this triangle. Use a for the area and p for the perimeter. Do not round answers.

a = ____
p = ____

Look at the triangle below:
After finding the value of h, determine the perimeter of this triangle. Use p for the perimeter. Round to the nearest hundredth.

h = ____
p = ____

Look at the triangle below:
First, find the values of s and t. Round them to the nearest whole.

s = ____
t = ____

Next, determine the perimeter of the triangle using the rounded values for s and t. Use p for the perimeter.

p = ____

Remember to save your answers for tomorrow. You will need them!

Saturday, June 15, 2013

Triangular Day: More Figurative Families

Two weeks ago, we talked about the relationships between figurative families. We looked at these in a more basic light, by just analyzing the quick and easy patterns that are noticed. This week, I would like to go a little deeper.

Let's find the explicit formula for the different sequences. We'll start with triangular numbers.

Tn = n(n + 1)/2

We have used this formula in many of the previous posts about triangular numbers, so that one didn't take as much work. What about the square numbers?

Sn = n2

This formula is pretty obvious, considering that the definition of square numbers is a natural number squared.

What about pentagonal numbers? This might take some more work, but it can be found by solving a system of equations. At the end, you would get:

Pn = n(3n - 1)/2

What about hexagonal numbers? Again, this one would take some work. Let's see what it ends up with:

Hn = 2n2 + n

There isn't an obvious pattern right now, but let's rewrite each thing in the terms of n(an - b)/2 with a and b being coefficients and constants in the formula.

Tn = n(1n - (-1))/2
Sn = n(2n - 0)/2
Pn = n(3n - 1)/2
Hn = n(4n - 2)/2

Now the pattern is pretty clear. Each formula is just adding one to a and b. Algebraically, a will always be equal to the number of sides on the figure minus 2, and b will be equal to the number of sides minus 4. I found this pattern to be pretty cool, considering how different the different shapes are.

I just found out that my TEDx talk in India was posted on YouTube. Here is the video of it.

I will also be speaking at TEDxBushnellPark in a week, and I will make sure to post that video when it is available as well.

Saturday, June 8, 2013

Why Has America Dropped to 32nd in Mathematics?

In school for the last four months, we have been doing a project where we find a contemporary issue in America or in the world, and figure out how to fix it through research, conducting an interview, and putting together a research paper, promotional piece, and presentation. My two friends and I decided to do ours on the issue of America having dropped to 32nd in mathematics in comparison to other countries in the world. Watch this video we made to see a general overview of the issue:

I also strongly encourage looking at the following website, which gives a fantastic visual overview of some of the facts and statistics of the issue of our mathematics drop. Click here to view it.

I'm not going to go into detail about the solutions we talked about, but I linked to our research paper at the top of the page. It is pretty long, but it goes through all of the things we could find to fix the problem. I think it is worth skimming.

Our group will be doing our presentation on Monday. I am also giving two TEDx talks over the summer about this issue. I will post links to them once I have done them.

Saturday, June 1, 2013

Triangular Day: Are the Figurative Families that Related?

I don't know if you noticed, but today is a triangular day. It is the first of June, and 1 is indeed a triangular number.

1 is also a special number because it is a square number as well. In fact, if you look at all of the regular polygonal figurative families, one is the first number. It is the first pentagonal number, hexagonal number, and so on. In fact, a dot can represent whatever figure you want it to, which shows the creativity of mathematics.

Because of this specialness to the number one, I thought we should analyze all of the regular polygonal figurative families. Let's look at the first fifteen numbers in each family.


This table just looks like a regular table of numbers. However, there are tons of patterns buried inside of it.

For example, look at each row. If you take the common differences of each one, what do you find?

The very top row has a constant difference of one, as you would expect. The next row, which contains the triangular numbers, has a difference of all natural numbers. The 1 and 3 have a difference of 2, the 3 and 6 have a difference of 3, the 6 and 10 have a difference of 4, and so on. In other words, rows n-1 and n have a difference of n. This is the definition of triangular numbers, so that point is obvious.

The next row, which contains the square numbers, has a difference of all odd numbers. The 1 and 4 have a difference of 3, the 4 and 9 have a difference of 5, the 9 and 16 have a difference of 7, and so on. In other words, rows n-1 and n have a difference of 2n-1. If you want to see a quick proof of it, it isn't too hard. It just requires some algebra. Just do n squared minus n-1 squared, and you should receive 2n-1 to prove it.

n2 - (n - 1)2 = 2n - 1
n - (n2 - 2n + 1) = 2n - 1
n- n2 + 2n - 1 = 2n - 1
2n - 1 = 2n - 1

If you look at the pentagonal numbers, it isn't as obvious what the pattern is. However, the differences are 4, 7, 10, 13, 16, and so on. The differences all differ by three, making the equation 3n-2. This can be proven using similar logic to above.

So, the first row has a difference of 1, which can be rewritten as 0n - (-1). The second row has difference of n, or 1n - 0. The third row has 2n - 1, the fourth row has 3n - 2. Do you see the pattern?

0n - (-1)
1n - 0
2n - 1
3n - 2

The fifth row of hexagonal numbers should have a difference of 4n - 3 if this pattern continues, right? Let's see if it worked.

1 - 0 = 1 = 4(1) - 3
6 - 1 = 5 = 4(2) - 3
15 - 6 = 9 = 4(3) - 3
28 - 15 = 13 = 4(4) - 3
45 - 28 = 17 = 4(5) - 3

The pattern indeed continued. For heptagonal numbers, the difference is 5n - 4, for octagonal numbers, the difference is 6n - 5, and so on.

Instead of looking at the rows this time, let's look at the columns and their differences. The first column has a difference of zero obviously. That is the nature of why one is in every figurative family. The second column has a difference of one. Since this number is the number defining the family itself, that isn't too surprising.

The third column has a difference of three. This is a little strange, but we can keep going. The fourth column has a difference of six. The fifth column has a difference of ten. Now do you see the pattern?

0, 1, 3, 6, 10

These are all triangular numbers! In fact, the triangular numbers have come back in the whole figurative family.

I find this table really cool to analyze. On June 15th (15 is a triangular number), we will revisit this table and look at some other neat identities within it.