Happy End Problem

As this is the last post of the year, I thought it would be appropriate to end with a post on the "Happy End Problem." I will explain the problem, which is pretty cool in itself, and then talk a little bit about the history behind it.

The initial question was if any five points were placed on a plane with no three of them in a straight line, will four of those points always form a convex quadrilateral? For example, in the following image:

The following convex quadrilateral can be formed:

Will this always work? As usual, I encourage you to grab a scrap piece of paper and try out a few examples. Have fun with it. Get creative! You will end up finding that no matter how you position the five points, you cannot get a combination without a convex quadrilateral.

Why is this true? In fact, there is a very easy way to prove it. Let's analyze three cases.

The first case is the top left one in the red, where the five points form a convex pentagon. In this instance, connecting any four of the points will form a convex quadrilateral by nature.

The second case is the top right one in the blue, where one point is located in between the four outside points. The illustration shows the inside point being included in the quadrilateral, but it could have just as easily been made as just the four outside points. This will continue to work for any combination of this nature by logic.

The third case is the bottom one in the yellow, where two points are enclosed in a triangle. When you draw a line between the two center points, two of the outside points will end up on one side and one will be on the other. Using the two outside points as your third and fourth vertices will form a quadrilateral without flaw.

Now, you might be wondering if one can prove a similar case with a convex pentagon. Could it be done with six? Seven? Eight? Turns out, nine points are required for it to work every time. As you can see below, eight points is just one too few.

What about convex hexagons? Or heptagons? Or octagons? Or chiliagons (1000-sided polygons)? Well, what many mathematicians will do from here is look for a formula to figure out how many points are required for a given n-gon. We know that for a triangle (n = 3), just 3 points are needed (all triangles are convex). For a quadrilateral (n = 4), we proved that 5 points are needed. For a pentagon (n = 5), I mentioned that 9 points are needed. Do you see the pattern?

3, 5, 9, ...

It is not easy to spot at first, but what if I subtract one from each of those terms:

2, 4, 8, ...

They are all now powers of two! This pattern seems to fit the formula A_n = 2^n-2 + 1. Plugging six in for n would give:

A₆ = 2^6-2 + 1
A₆ = 16 + 1
A₆ = 17

This formula predicts that seventeen points would be required for a hexagon. Mathematicians would then work to try to prove that this is the case. Further, they would try to prove that for any value of n, the A_n formula holds true.

George Szekeres (1911-2005; a Hungarian-Austrailian mathematician and analytical chemist) and Esther Klein (1910-2005, another Hungarian-Austrailian mathematician) worked together to prove that all values of n will have a finite A_n output (there will be a number of points that creates the ability for a convex n-gon to be formed), but they could not get this bound down to the formula above. Soon after this proof was published, Szekeres and Klein married each other, which inspired the name "Happy End Problem."

Paul Erdös (1913-1996; a Hungarian mathematician), possibly one of the most influential of the twentieth century), was able to prove successfully that with 71 points, a hexagon can always be drawn.

Sixty years later, Ronald Graham (born 1935; a Californian mathematician) and his wife, Fan Chung, decided to take a swing at the problem. While on a plane ride to a math conference in New Zealand, they were able to lower Erdös's bound to 70 points, which doesn't sound like much, but it brought the problem back into the minds of mathematicians. It was also ironic that another achievement pertaining to the Happy End Problem was from a couple.

Daniel Kleitman (born 1934; an applied mathematician at MIT) and Lior Pachter (born 1972; an Israeli mathematician and molecular biologist at Berkeley College) worked together to lower the upper bound to 65 points. The number was then lowered to 37 points, and has yet to be lowered further.

Although lots of progress has been made on this problem, the overarching proof still has not yet been found. There has been no counterexample to the A_n formula, and there has certainly been no guaranteed formula to generate the future values. People often wonder what a mathematician actually does for his/her job. A big part of it is trying to figure out the answers to these unsolved problems, which can often be understood by the average person. Try playing around with it and you might make a discovery too.

Carl Friedrich Gauss: The Child Prodigy

One of the most famous mathematicians of all time is the German mathematician Carl Friedrich Gauss (1777 - 1855). Gauss was one of the leading number theorists of all time, as well as a contributor to algebra, statistics, geometry, analysis, and applied mathematics.

There is a piece of mathematical folklore (which may or may not be 100% accurate) that involved a child Gauss. It is a great story, highlights a great point, and shows the intelligence of a great mathematician.

A fifth grade teacher is teaching a class, and started to get frustrated with the students. So, in an attempt to punish them, she demanded that they add up all of the numbers from 1 to 100. This is a daunting task for the average person. She expected to have the students start working on the problem, and she could leave and take a break.

As she was about to walk out the door, the young Gauss raised his hand and declared that the answer is 5050. The teacher was stunned. After checking his work, they found that 5050 was the correct answer.

How did he do it? Well, he visualized a horizontal line with all 100 numbers:

1   2   3   4   5   ...   96   97   98   99   100

And then he took the second half of that line (51 - 100) and flipped it around underneath to look like so:

    1     2     3     4     5   ...   46   47   48   49   50

100   99   98   97   96   ...   55   54   53   52   51

Each of these vertical columns is its own addition problem. And in all fifty columns, the sum is 101. So, the sum of the numbers from one to one-hundred is the same as fifty 101's, or 50 x 101. Since 50 ends in a zero, it is a pretty quick computation: 50 x 101 = 5050. And there is the answer.

I think this is a great story when it comes to historical mathematicians, regardless of how true it is. Gauss did go on to study triangular numbers, which are the sums of consecutive integers up to a point. And since triangular numbers are absolutely fascinating, this story is a great way to begin an endeavor in that topic.

Law of Sines

A few months ago, I posted the proof of the Law of Cosines, which is an extremely important aspect of trigonometry. If you do not know what sines and cosines are, it is a very easy concept. Click here to view the post discussing that (which is also the post with the proof of the Law of Cosines).

But this law only works if you are given all three sides of the triangle or two sides and the enclosed angle. What if you are given two sides and a non-enclosed angle, or two angles and one side (three angles isn't enough information to generate side lengths)? How could you approach this problem?

This is where you use the Law of Sines. This law goes as follows:

This was taught to me last year in school, and I immediately wondered what the proof was. Though the law of cosines one was a bit clunky, I found that this proof was quite simple and elegant. So, I thought that it would be great to share.

Since it would require many diagrams, I thought it would be easier to just watch a video of it. It is pretty short, and explains the proof well.

A big part of the reason why most of the cool stuff I post isn't taught in school is that it is not mandated in the curriculum or test standards. Of course, I do believe there are changes that need to be made to these (click here for my Capstone research paper explaining those). However, the Law of Sines is something already taught in school. Same with the Law of Cosines.

These proofs, especially the sine one, fit right into the curriculum. The Law of Sines is already being taught, so why not take an extra 5 minutes to explain the proof? Or even better, explain the basic thought process behind the proof and have the students generate the formula (which works really well for the Quadratic Formula as well). This increases the students' ability to understand and apply the concept, as well as making it fun and interesting. On top of that, the common core standards do want students able to "construct viable arguments," which is the whole purpose of proofs. I think that this proof is not only interesting, but shows that cool math stuff can be integrated into the classroom while keeping it relevant and obedient to the standards.

Time to Double Your Money!

Last year, I did a post about a number called e. This number is around 2.71828, and has many applications to calculus and finance. In the other post, I discussed its application to finding compound interest. Click here to see that post.

You might remember that the formula for finding the amount of money you have when your interest gets compounded continuously is:

Pe^rt

P = amount of money originally deposited
r = interest rate
t = time (years)

This is a prettier formula, but a more practical formula, which also has variable n for the number of times the money was compounded in the year is:

P(1 + r/n)^nt

For instance, if you deposited 1000 dollars in the bank, and your money got compounded every year with a 3% interest rate, after 30 years, you would have:

P = 1000
r = 3% = .03
t = 30
n = 1

P(1 + r/n)^nt
1000(1 + .03/1)¹⁽³⁰⁾
1000(1 + .03)³⁰
1000(1.03)³⁰
1000(2.42726)
2427.26

So, after thirty years, you would have about $2427.26, which means you were able to more than double your money! This might be a little surprising that it is possible for your money to double if you leave it alone for long enough.

You might be wondering how long it will take to double. How many years does it have to sit there? In other words, what value of t makes that equation equal to 2P? We deposited P, so to make it double, we must get 2P.

2P = P(1 + r/n)^nt

In our scenario, we deposited $1000 with an annually compounding interest rate of 3%. So, plug all of this in and we get:

2P = P(1 + r/n)^nt
2(1000) = 1000(1 + .03/1)^1t
2000 = 1000(1.03)^t
2000/1000 = (1000(1.03)^t)/1000
2 = (1.03)^t
ln(2) = ln((1.03)^t)
ln(2) = t • ln(1.03)
ln(2)/ln(1.03) = t
24 ≈ t

Note: this computation required something called logarithms. They look weird, but are very easy to understand. I explained them in my post on Benford's Law. Click here to read it.

So, it will take 24 years to double. So, let's set a rule to this - figure out a simple formula where you can figure out how long it will take for your interest to double, assuming it is compounded annually.

2P = P(1 + r/1)^1t
2P/P = (P(1 + r)^t)/P
2 = (1 + r)^t
ln(2) = ln((1 + r)^t)
ln(2) = t • ln(1 + r)
ln(2)/ln(1 + r) = t

This is as simplified as the equation will get without using calculus. And this doesn't look very simple anyways. But, let's look at a graph of it. It is easier to see as a picture than a messy jumble of logarithms.

This is a logarithmic function. However, it looks very much like a rational function, or a function that is the quotient of two polynomials (for example, 1/x is a rational function). So, let's try to find a rational function that fits this blue curve.

You can play around with it on your graphing calculator if you want, but I will just tell you that the function that fits it best is 72/(100r). Here are the two graphs:

As you can see, the two graphs are practically touching. In fact, all the way up through 0.5, they are very close together, as you can see here:

Since no bank on this planet offers 50% interest rates (if anyone has heard otherwise, please contact me), the 72/(100r) should be a good approximation for any of our purposes. Even to see the difference in the ones and tens, I had to set the graph below two. So, they are very close together.

This 72/(100r) equation can look even better. You might remember that r is currently in decimal form. To turn it into a percentage, you must move the decimal over twice, or multiply it by 100. So, dividing the interest percentage into 72 will give the same approximation.

Let's try it out on the original example. We said that it was a 3% interest rate. 72 ÷ 3 = 24, and we did conclude that it would take about 24 years. For a 6% interest rate, it would take about 72 ÷ 6 = 12 years. For a 4.5% interest rate, it would take about 72 ÷ 4.5 = 16 years. Couldn't be easier!

This rule is normally called the Law of 72. I find it very cool because you don't even need to know the amount of money you deposited to figure out this time. It doesn't matter. You don't even need to convert the interest rate into a decimal. I think this is a really intriguing and practical formula.

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.

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.

T_n = 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?

S_n = 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:

P_n = n(3n - 1)/2

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

H_n = 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.

T_n = n(1n - (-1))/2
S_n = n(2n - 0)/2
P_n = n(3n - 1)/2
H_n = 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.

Bonus:
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.

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.

Name	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
Triangular	1	3	6	10	15	21	28	36	45	55	66	78	91	105	120
Square	1	4	9	16	25	36	49	64	81	100	121	144	169	196	225
Pentagonal	1	5	12	22	35	51	70	92	117	145	176	210	247	287	330
Hexagonal	1	6	15	28	45	66	91	120	153	190	231	276	325	378	435
Heptagonal	1	7	18	34	55	81	112	148	189	235	286	342	403	469	540
Octagonal	1	8	21	40	65	96	133	176	225	280	341	408	481	560	645
Nonagonal	1	9	24	46	75	111	154	204	261	325	396	474	559	651	750
Decagonal	1	10	27	52	85	126	175	232	297	370	451	540	637	742	855

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

n²  - (n² - 2n + 1) = 2n - 1
n² - 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.

Triangular Day: The Oblong Numbers

I don't know if you noticed, but today is a triangular day. It is April 6th, and six is the third triangular number.

You might remember from December when I brought up figurative families. The triangular numbers are a figurative family because each number can be represented by a triangular array of dots. Each array forms a geometric figure.

I also talked about square numbers as being a figurative family. Each number in that sequence can be represented by a square array, which is another geometric figure.

What about a rectangle? Considering that any number can be turned into a rectangular array, it would not be as interesting of a family. However, what about a rectangle whose array has just one more row than it does columns. For instance:

•  •  •  •
•  •  •  •
•  •  •  •

This has four rows and three columns, making it fit into this family. These numbers are called the "oblong numbers."

The oblong numbers go as follows:

2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132...

Here is a little property of oblong numbers. Divide each number in that sequence by two. You get:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66...

You end up with the triangular numbers. Why is this? Remember the formula used to find any triangular number?

T_n = n(n + 1)/2

We said that a number is oblong if its array had one more row than it had columns. So, an oblong number must be factorable into two numbers that are just one apart. We can denote these two factors as n and n+1. This makes the oblong number equal to:

O_n = n(n + 1)

After dividing by two, we ended up with the triangular numbers.

I enjoy trying to find ways to combine these different families together. For instance, take the first triangular number, first square number, and first oblong number. This would be 1, 1, and 2 respectively. We will add the first two together, and then subtract the last one.

1 + 1 - 2 = 0

Let's try that with some more:

T₁ + S₁ - O₁ =     1 + 1 - 2         = 0
T₂ + S₂ - O₂ =     3 + 4 - 6         = 1
T₃ + S₃ - O₃ =     6 + 9 - 12       = 3
T₄ + S₄ - O₄ =     10 + 16 - 20   = 6
T₅ + S₅ - O₅ =     15 + 25 - 30   = 10
T₆ + S₆ - O₆ =     21 + 36 - 42   = 15

Do you see the pattern? Each of the answers is a triangular number. In fact, it is one less than the triangular number used in the example.

Why is this? When I saw it, I first thought that we could check by using algebra. Take the explicit formula for each sequence:

T_n = n(n + 1)/2
_Sn = n²
_On = n(n + 1)

Our goal is to end with T_n-1, so we will use n(n - 1)/2 to denote that.

[n(n + 1)/2] + [n²] - [n(n + 1)] = [n(n - 1)/2]

Let's simplify all of the brackets so we don't have parentheses to work with.

[n(n + 1)/2] + [n²] - [n(n + 1)] = [n(n - 1)/2]
[1/2n² + 1/2n] + [n²] - [n² + n] = [1/2n² - 1/2n]

Now, we will combine like terms.

[1/2n² + n² - n²] + [1/2n - n] = 1/2n² - 1/2n
1/2n² - 1/2n = 1/2n² - 1/2n

And there is our proof. Though oblong numbers are not the family of a regular polygon, it is still an interesting sequence to look at.

Benford's Law

Take any random quantity. Maybe a population of a city, a mass of a planet, a distance from a star, or an amount of twitter followers. Now, take the first digit of this number. What are the odds that it will start with a 1, or a 2, and so on?

One would think that it would be a one in nine chance for each digit. If you are randomly selecting this number, the first digit is just a random number selection between one and nine.

However, this reasoning does not work. Try finding these quantities yourself. Or, just go to http://testingbenfordslaw.com. If you check all of the different areas that are possible, you will see that the smaller the digit, the more commonly it appears.

Here are the approximate odds for each digit:

1: 30.1%
2: 17.6%
3: 12.5%
4: 9.7%
5: 7.9%
6: 6.7%
7: 5.8%
8: 5.1%
9: 4.6%

Since this is a math blog, something we would want to do is find a pattern between these numbers. This could probably be done on a graphing calculator, using similar techniques to the post on data analysis.

It is clear that a line of best fit would not be the solution. If we connected the points with a curved graph, it would look like this:

You can see that this graph gets really close to the y-axis, but it does not seem to touch it. Similarly, it gets really close to the x-axis, but doesn't touch it. Thus, the x and y-axes would be called asymptotes of this graph.

But, an asymptote could be found in a rational function, radical function, exponential function, logarithmic function, hyperbolic function, trigonometric function... So, it is hard to define this graph solely based on the presence of asymptotes.

The function that does work with this graph is a logarithmic one. A logarithm is basically the opposite of an exponent. For instance:

5² = 25
log₅(25) = 2

Just knowing that the equation is logarithmic doesn't seem to narrow it down a lot, because there are so many different types of logarithms. However, there are three types which are seen the most frequently. In fact, most scientific calculators contain just these three types.

Logarithm Type	Base	Simple Notation	Standardized Notation	Applications
Natural Logarithm	e (the irrational number ≈ 2.71828)	loge(x)	ln(x)	Calculus, Statistics, Physics, Chemistry, Economics
Common Logarithm	10	log10(x)	lg(x)	Algebra, Engineering, Geology, Spectroscopy, Music
Binary Logarithm	2	log2(x)	lb(x)	Discrete Mathematics, Computer Science, Information Theory

If we confine the possible equations to one of these, it narrows it down a lot. And this assumption is correct. The equation that the graph fits is:

y = lg(1 + 1/x)

(Remember that x is the starting digit and y is the percentage for that digit)

I was pretty surprised about this equation. But, all of the percentages listed above are the outcome of this equation.

Benford's Law can be used in biology, accounting, law, economics, etc. However, a more fun way to use it is to turn it into a game.

Tell someone that you will get the numbers 1, 2, and 3 and they will get the numbers 5, 6, 7, 8, and 9 (and nobody gets 4). You then have them come up with random quantities that they wouldn't know, and you look up the number (you can use Google or WolframAlpha for this). Every time it is a 1, 2, or 3, you win a point and every time it is a 5, 6, 7, 8, 9, they get a point. They think they have around a 2:1 advantage over you, but you are really the one with the 2:1 advantage.

Click here to see this game played on the show Scam School.

Handshake Problem

Let me start by giving a little math problem. n people are in a room, and each one has to shake every other person's hand. How many handshakes will it take for all n people to have shaken every other hand?

Let's first try a few numbers. For two people, it would obviously just take one handshake. For three, it would take three handshakes. For four, it would take six handshakes. For five, it would take ten handshakes.

Do you see the pattern? 1, 3, 6, and 10 are the triangular numbers. In fact, this pattern always continues. 

If you look at it logically, you will see why. Person 1 has a number of hands to shake. Person 2 would have to shake all of the hands except for person 1's (it was already counted). Each person has to shake one less than the one before until there is just one left.

Adding up these handshakes will be a sum of the first n natural numbers, which is the definition of triangular numbers. 

I found it cool that a famous number sequence could be applied to a practical problem like this. 

History of Mathematics: Charles Dodgson

Mainly, I talk about proofs, patterns, and ways to apply mathematics to the real world. However, there are other ways that mathematics is cool too. Not only is the actual content interesting, but the stories are as well.

Today, I will tell the story of a mathematician named Charles Lutwidge Dodgson. Dodgson was born in 1832 in Cheshire, England. He graduated from Christ Church College at Oxford, and then began his career as a mathematician, lecturing and tutoring at Oxford.

Dodgson was not the type of mathematician who often made breakthroughs and discoveries. However, he did find some interesting things in mathematics and logic. One of them, I actually talked about in a previous post, but never mentioned that Dodgson was the mind behind it. Click here to see it.

He is also known for a method of election he developed. In America, we just vote for the candidate of our choosing, which is called simple plurality. Yet, we run into problems in elections such as the Bush-Gore-Nader election of 2000. In this election, most Nader supporters preferred Gore to Bush and most Bush supporters preferred Gore to Nader, even if it was just by a little bit. So, Gore was never the least favorite of anybody, while Bush was the least favorite of most Nader supporters and probably many Gore supporters. Using a ranking method rather than a plurality method, we can find a winner that the most people are satisfied with.

Yet, this method can fail in elections such as the Obama-Romney election of 2012. The leading third party candidate was Gary Johnson, but was supported by only a small percentage of the population. But, Obama fans would rank Johnson as their second favorite in order to give as little support as possible to Romney, and vice versa. This may have led to a win for Gary Johnson, which very few people would be satisfied with.

So, mathematicians in the field of game theory are always struggling to find a perfect election method. There are dozens of ideas out there, one of which is created by Charles Dodgson. His method makes use of finding a Condorcet winner. If on every ballot, a certain candidate is ranked higher than another one, then this beaten candidate is eliminated. For example, if all Americans preferred Barack Obama to Jill Stein, the Green Party candidate, then Stein would be eliminated. If you can eliminate every candidate in this way, the candidate still standing is the Condorcet winner.

Since it is near impossible to have a Condorcet winner, this method is impractical. However, Dodgson extended it by saying that after eliminating everyone possible, you begin swapping rankings on people’s ballots until you are able to have a Condorcet winner. The candidate that requires the least swaps wins. This method will eliminate weak third party candidates, but still take them into strong consideration.

On a different note, Dodgson was also known for his ability to write. He wrote dozens of famous math textbooks, and compiled texts for undergraduate students. But he also enjoyed implementing his mathematical knowledge into fantasy writing. As a man who loved children, he would tell stories that had some mathematics and logic infused in them.

One of his favorite children was named Alice Liddell, whose father was the Dean of the school where Dodgson taught. He began telling her stories about a girl named Alice, which were always continued every time they saw each other. Years later, he gave her a written manuscript of the story for Christmas.

In the story, there were many mentions of mathematics. For instance, there is one point in the story where Alice is three inches tall. She finds out that she must eat from a mushroom to grow back to normal, but half of the mushroom stretches her neck and half shrinks her torso. She must find the correct proportions to grow properly.

This is the foundation of Algebra. In fact, the word “Algebra” comes from an Arabic algebra book whose title translates to “Restoration and Reduction.” And she must use the concepts in this book, by finding how much of each side of the mushroom is necessary to make her neck proportions equivalent to her torso proportions. This resembles an algebraic equation.

Dodgson even used his ability to play with words in the story, by describing the branches of arithmetic as ambition, distraction, uglification, and derision.

These manuscripts ended up getting published, and eventually found their way to the queen of England: Queen Victoria. After reading them, she demanded that she receive every book written by this man. To her surprise, she ended up with a huge stack of mathematics textbooks.

This book that the queen loved, that has many mathematical references, that is written by a mathematician from Oxford, is titled Alice and Wonderland. And you might know Charles Dodgson better by his pen name, Lewis Carroll.

Bonus: Lots of these historical mathematicians have a few funny stories in their successful careers as well. Here is one about Charles Dodgson.

Because of his friendliness to children, Dodgson was a popular guest at parties. One time when he was invited to a party in London, he decided to crawl into the room as a surprise to the kids. However, he crawled into the wrong household, where a group of adults were also having a party.

A Dumbing Down of the Riemann Hypothesis

Today is my first post on math in the news. I recently came across this article on the Riemann Hypothesis, which I had planned to talk about in India, but didn't get a chance to. Let me give a brief background and then I will share the article.

Back in 2000, Clay Mathematics Institute of Providence, Rhode Island announced the Millennium Prizes, which consisted of seven problems that had been stumping mathematicians for a long time. They set aside a million US dollars for any person who solved one of the problems.

I find it interesting just on its own that you can become wealthy as a mathematician. Other than the Nobel Economics Prize, math has its own way of getting a million dollars.

One of the more popular of these problems is the Riemann Hypothesis, which I wanted to talk about today. I will try to explain here what the Riemann Hypothesis is (it is a difficult concept, but online sources complicate it drastically), and then show the article.

First off, you might remember the number i, which is the square root of -1. This is not a normal variable that can just replace anything you want; it is always the square root of -1. You may have heard in math class the term "real number." A number that has just an i in it are imaginary numbers, like 2i or -5i.

A little over a year ago, I did a post about the complex plane. This takes our horizontal number line from first grade and makes it our x-axis. It then takes these imaginary numbers and makes those the intervals of the y-axis.

A point on the x-axis is a real number, represented with the letter a. A point on the y-axis is an imaginary number, represented with the term bi. A point that is just floating around somewhere not on one of these lines is a complex number. You can write it with the expression a + bi, with a being the number it lines up with on the x-axis and bi being the number it lines up with on the y-axis.

Say you had to take the equation y = x^3 - 2 and start plugging in values for x (replacing the x with a number and then figuring out what it equals). Most people would start plugging in real numbers like 0, 1, 2, 3, -1, -2, -3, and so on. However, this Riemann Hypothesis requires us to open up our minds a little bit. Rather than just plugging real numbers into equations, we have to plug complex numbers into equations.

The Riemann Zeta Function is the equation we are plugging these numbers into (a little side-note: the Riemann Zeta Function has a 2π in it for any tauists). For the Riemann Hypothesis, it is concerned to find when this equation equals zero. This is called the zero of the equation.

The real numbers that are zeros of this equation are called the trivial zeros. They are equal to -2, -4, -6, -8, and so on. The complex numbers that are zeros of this equation are called the non-trivial zeros. As far as we know, these non-trivial zeros all have different b values in our a + bi, but the a value always seems to be 1/2.

The Riemann Hypothesis is simply asking the question is there a non-trivial zero of the Riemann Zeta Function whose a value is not equal to 1/2. Imagine winning a million dollars after submitting a hundred page paper that answers just a yes or no question.

It seems like proving either side would be extremely difficult. This article gives a nice explanation of how they are going about proving the yes side of it.

http://www.rdmag.com/news/2012/11/supercomputing-solve-superproblem-mathematics

The article brings up a very good way to do it. If you find just one time where the a value is not 1/2 and it is complex, then the statement is proven. So, Yuri Matiyasevich decided to turn the supercomputers on and start cranking out values. They have not found any values without the 1/2, but they also cannot mathematically prove that it always is the 1/2, and that is where the dispute lies.

Let me finish by saying what I find cool about the Riemann Hypothesis. So what if some weird function might have a consistency with its x-intercepts? But there is a very practical and interesting connection.

You may notice how with the prime numbers, there really isn't any relation between them. I mean, the Fibonacci numbers are the sum of the two before it, the powers of two are one more than the sum of all the ones before it, the triangulars are the sum of the natural numbers, the squares are the sum of the odd natural numbers, but the primes have no relation like that. I have always wondered why that is, or if there was one.

If the Riemann Hypothesis gets solved, it will shine a light on the distribution of prime numbers. We will be able to see if they do have a pattern or if there is no pattern. Some might find the technological, algebraic, or financial parts of this problem interesting, but I think the practical aspect is really cool.