Odd Sums Revisited

In a previous post, I investigated the sequence of odd numbers:

1, 3, 5, 7, 9, 11, \ldots

In particular, I investigated the ratio between the sum of the first “k terms” and the sum of the next “k terms.” I found that the ratio was always 1/3. I attempted this investigation with the sequence of even numbers and the ratio failed to be constant.

I recently had a brain wave to continue the investigation. Here was my question: “what sequences of numbers have consistent ratios when you sum the first k terms and the next k terms?”

After a bit of investigation, I found another sequence!

2, 6, 10, 14, 18, 22, \ldots

Here is the ratio of the first term and the second term:

\dfrac{2}{6} = \dfrac{1}{3}

The ratio between the first 2 terms and the next 2 terms:

\dfrac{2 + 6 }{10 + 14} = \dfrac{8}{24} = \dfrac{1}{3}

The ratio between the first 3 terms and the next 3 terms:

\dfrac{2 + 6 + 10}{14 + 18 + 22} = \dfrac{18}{54} = \dfrac{1}{3}

We have a constant ratio! What is even more interesting is that this ratio is 1/3, just like before! Below is another sequence that works:

3, 9, 15, 21, 27, 33, \ldots

And another:

4, 12, 20, 28, 36, 44, \ldots

Here is one that uses decimals:

1.1, 3.3, 5.5, 7.7, 9.9, 12.1, 14.3, \ldots

 Or how about negative numbers?

-5, -15, -25, -35, -45, -55, -65, -75, \ldots

In fact, there are an infinite number of different sequences. All of these sequences share the exact same consistency as the original sequence. The ratio will always be 1/3.1 Insights like this are what make mathematics truly beautiful.1



1Below is the method I used to generate these new sequences. First, we need to represent the sequence in general. We will restrict ourselves to arithmetic sequences for this post. Hence, the general sequence can be written as:

a, a+d, a+2d, a+3d, a+4d, \ldots

We want our arithmetic sequence to have a constant ratio. We could rephrase this requirement as “the ratio of the first and second term must be equal to the ratio of the first 2 terms and the next 2 terms.” In symbols, this means:

\dfrac{a}{a + d} = \dfrac{a + (a + d)}{(a + 2d) + (a + 3d)}


\dfrac{a}{a + d} = \dfrac{2a + d}{2a + 5d}
a(2a + 5d) = (a + d)(2a + d)

2a^2+ 5ad = 2a^2 + 2ad + ad + d^2

2a^2 + 5ad = 2a^2 + 3ad + d^2

5ad = 3ad + d^2

0 = -2ad + d^2

0 = d(d - 2a)

Since we want our sequence to be interesting, d ≠ 0 and a ≠ 0. Thus, d = 2a. Hence, any sequence that satisfies the condition d = 2a will exhibit our desired property.2

Now our general sequence looks like this:

a, a+2a, a+4a, a+6a, a+8a, \ldots

a, 3a, 5a, 7a, 9a, \ldots

From this, we see that the ratio will be:

\dfrac{a}{3a} = \dfrac{1}{3}

Since the a’s cancel, our general sequence reduces to our original sequence of odd numbers. Thus, all the properties that we discovered about the sequence of odd numbers will apply to all these new sequences. This also means that the only ratio ever possible is 1/3. Amazing!


2 I have another footnote for two reasons. First, I think it is pretty meta to put a footnote inside a footnote. Second, I actually only showed that the condition d = 2a will force the ratio of the first and second terms equal to the ratio of the sum of the first 2 terms and the next 2 terms.

I neglected to prove that this generalized to the sum of the first k terms and the sum of the next k terms. In other words, I didn’t prove that:

\dfrac{a + (a + d) + (a + 2d) + \ldots + (a + (k - 1)d)}{(a + kd) + (a + (k + 1)d) + \ldots + (a + 2kd)} = \dfrac{1}{3}

It should be obvious to the reader that if you replace d with 2a and factor out an a, the above ratio will reduce to:

\dfrac{1 + 3 + 5 + \dots + (2k - 1)}{(2k + 1) + (2k + 3) + \dots + (4k + 1)}

This is equivalent to the original ratio investigated in the first post.

If anyone is still reading at this point, they may notice that in the original post I didn’t prove that the above ratio was equal to 1/3. You would be correct. The proof is left as an exercise for the reader ;)

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Non-Transitive Dice Follow up

Yesterday I asked you to pick the best die. If you guessed Blue, you would be wrong. Red is all wrong. Likewise, Green is incorrect. The correct answer is that…

There is no best die!

As crazy as this might sound, you have all experience this type of situation before. Consider rock-scissors-paper:

pic1 other

In the above game, there is no best choice. Each symbol beats one symbol and loses to the other. The dice are similar. Consider the following diagram:


This is a chart displaying the possible outcomes between for Blue vs Red. It is clear that he Blue die will win more often, 21 times out of 36. The red die will only win 15 times out of 36. You can construct similar tables for Red vs Green and Green vs Blue. The results are summarized in the following diagram:

pic 3

These dice are called “non-transitive” because the ability to win does not transfer. Red is better than Green, and Green is better than Blue, but Red is not better than Blue.

Warren Buffett is known to be a fan of non-transitive dice. Buffett once attempted to win a game of dice with Bill Gates using non-transitive dice. Buffett suggested that each of them choose one of the dice, then discard the other. They would bet on who would roll the highest number most often. Buffett offered to let Gates pick his die first. This suggestion instantly aroused Gates’s curiosity. He asked to examine the dice, after which he demanded that Buffett choose first.


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Non-Transitive Dice

A normal die has 6 sides with the numbers 1 through 6. However, different dice can have different numbers

Ex: 2, 4, 6, 8, 10, 12

Which die is the best for highest rolling in a head to head match?

Blue: 0, 3, 5, 7, 7, 7

Red: 2, 2, 4, 4, 6, 9

Green: 1, 1, 1, 8, 8, 8

I will post the answer tomorrow.

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Trippy Triples

Have you ever heard of a 3-4-5 triangle? It is a right-angled triangle with sides of length 3 and 4 and 5. It looks like this:


The neat thing about this triangle is that the sides are all whole numbers. This is not always the case for right-angled triangles. For example, a triangle with sides of length 2 and 3 has a hypotenuse (the side across from the right angle) of length 3.6.


A set of whole numbers like (3, 4, 5) that create a right-angled triangle is called a Pythagorean triple. How many triples are there? Well, a triple has to be a right-angled triangle. One way to check if you can construct a right-angled triangle from 3 numbers is to plug the numbers into the Pythagorean Theorem.





Since both sides are equal, (3, 4, 5) is a triple. However, consider (4, 5, 6):




41 \neq 36

Since both sides of the equation are not equal, (4, 5, 6) is not a triple. Or visually, you can see how it is impossible to make a right-triangle with those side lengths:


Ok, so we know that (3, 4, 5) is a triple, but how can we find more? A bit of investigation reveals that you can multiply each number of a triple by two and this new set will also be a triple (6, 8, 10).




What if we multiply by 3? Will (9, 12, 15) be a triple? You bet!


This pattern continues. In general, once you have found a triple, you can generate infinitely many new triples. Pretty cool. But also not cool. Because, although these new triples are new, they are actually just scaled up versions of the original triple. How could we make a different kind of triple? There just so happens to be a formula for generating new triples1. What we find when we use the formula, is that each prime number (3, 5, 7, 11, 13, etc.) generates a unique triple. For example, (5, 12, 13):




Or (7, 24, 25):




Each of these new triples can be scaled up to create an infinite number of triples. However, there are also an infinite number of prime numbers. This means that there are an infinite number of triples that start with a prime, and each of these triples generates an infinite number of scaled up triples. Trippy triples.


1For those of you who want to see how the formula works, here is a behind the scenes look at Pythagorean triples.

We start with the following set of three equations which relates the triples to 2 unknown variables m and n:


B = 2mn

C = m^2 + n^2

Consider A = m^2 - n^2 . We can factor the RHS to get A = (m-n)(m+n). However, A is prime, so the only factors of A are 1 and A. Thus:

m-n = 1

m+n = A

Solving the above system of equations for m and n yields the following:

m = \dfrac{A + 1}{2}

n = \dfrac{A - 1}{2}

In this form, m and n are completely determined by A. Thus, B and C are also completely determined by A.

B = 2 * \dfrac{A+1}{2} * \dfrac{A-1}{2} = \dfrac{A^2 - 1}{ 2}

C = \dfrac{(A+1)^2}{2} + \dfrac{(A-1)^2}{2} = \dfrac{2A^2 + 2}{4} = \dfrac{A^2+1}{2}

This tells us how to find B and C when given any prime A. It also tells us that B and C will be consecutive integers. How interesting.

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Mathematical Toy Part 2

Warning: the following post contains some intense mathematics, caution is advised.

In the previous post, I investigated a simple toy that generated an ellipse. However, how did I know it was an ellipse? I proved it, and here is how.

First, we need to model the two pivot points of the toy. They are both fixed on the x and y-axis. The toy also moves in a circular motion. The pivot on the y-axis is starts at 1 and the pivot on the x-axis is at 0. Hence, we will model the motion of the y pivot by the cosine function and the motion of the x pivot by the sine function. Taken together, this information gives us two points:

(0, \cos(t))

(\sin(t), 0)

I am using t to denote how the position of the pivots change in time. Due to boundary complications, I will only model one half of a complete revolution. However, since the resulting shape is symmetrical, this is model sufficient to determine the curve the toy traces. Thus, 0 ≤ t ≤ π

The handle is attached to a rigid piece of wood. The rigid piece of wood can be modeled by a line segment. Both of the pivot points lie on this line segment. To determine the equation of the line segment, we must use the good old slope formula:

m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}} = \dfrac{\cos(t) - 0}{0-\sin(t)} = \dfrac{\cos(t)}{-\sin(t)}

Then we can use the slope-point formula:

y - y_1 = m(x - x_1)

y - \cos(t) = \dfrac{\cos(t)}{-\sin(t)}(x-0)

y = \dfrac{\cos(t)}{-\sin(t)}x + \cos(t)

Alright! Now we are getting somewhere! The next step is to model the handle. For simplicity, we will model the handle as a point 1 unit away from the x-axis pivot. To determine this point, we can use the distance formula:

d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}

1 = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}

1^2 = (x_2 - \sin(t))^{2} + (y_2 - 0)^{2}

1 = (x_2)^2 - 2x_2\sin(t) + \sin^2(t) + (y_2)^{2}

We can use the previous formula for the equation of the line to determine y2 in terms of x2:

y = \dfrac{\cos(t)}{-\sin(t)}x + \cos(t)


1 = (x_2)^2 - 2x_2\sin(t) + \sin^2(t) + \left( \dfrac{\cos(t)}{-\sin(t)}x_2 + \cos(t) \right)^2

1 = (x_2)^2 - 2x_2\sin(t) + \sin^2(t) + \dfrac{\cos^2(t)}{\sin^2(t)}(x_2)^2 - 2\dfrac{\cos^2(t)}{\sin(t)}x_2 + \cos^2(t)


Rearranging we find that:

1 = \sin^2(t) + \cos^2(t) + (x_2)^2 + \dfrac{\cos^2(t)}{\sin^2(t)}(x_2)^2 - 2x_2\sin(t) - 2\dfrac{\cos^2(t)}{\sin(t)}x_2

1 = \sin^2(t) + \cos^2(t) + (x_2)^2 \left(1 + \dfrac{\cos^2(t)}{\sin^2(t)} \right) - 2x_2 \left( \sin(t) + \dfrac{\cos^2(t)}{\sin(t)} \right)

Using well-known trigonometry identities, the above reduces to:

1 = 1 + (x_2)^2 \left(\dfrac{1}{\sin^2(t)} \right) - 2x_2 \left( \dfrac{1}{\sin(t)} \right)

0 = (x_2)^2 \left( \dfrac{1}{\sin^2(t)} \right) - 2x_2 \left( \dfrac{1}{\sin(t)} \right)

0 = x_2 \left( \dfrac{x_2}{\sin^2(t)} - 2\dfrac{1}{\sin(t)} \right)

Ignoring the trivial solution, we find that x2 is equal to:

x_2 = 2\sin(t)

And solving for y2 yields:

y_2 = \dfrac{\cos(t)}{-\sin(t)}2\sin(t) + \cos(t) = -\cos(t)

Hence, the point in parametric form is:

(2\sin(t), - \cos(t))

We would like to identify what curve these parametric equations will trace. This is equivalent to solving the follow 2 equations in terms of x and y only:

x = 2\sin(t)

y = -\cos(t)

By squaring both sides of both equations, we get the following:

x^2 + 4y^2 = 4\sin^2(t) + 4\cos^2(t) = 4

x^2 + 4y^2 = 4

\dfrac{x^2}{4} + \dfrac{y^2}{1} = 1

Which is the equation of an ellipse. Hence, the handle of the toy traces a path equivalent to an ellipse. Below is an animation I constructed using the parametric equations.


A copious amount of work? Maybe. But what a beautiful proof.

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Mathematical Toy Part 1

A few months ago, I found a toy my grandfather had made.

2014-07-30 14.56.36



The toy was very simple and fun to play with. One thing stuck out to me. The handle seemed to make a circular motion, but the handle stuck out more on one side than the other did. If the handle was tracing a perfectly circular path, then the handle should always stay the same distance away from the block of wood. However, this was clearly not the case.

I took a pencil and (with the help of my uncle) carefully traced out the curve swept by the handle. The picture looked like this.




I instantly recognized the shape, an ellipse!

An ellipse is like a squashed circle. It is still symmetrical and smooth, but one side is longer than the other is. It astonished me that my grandpa could create a toy that would sweep out a perfect ellipse, with only two moving parts.

Below is an animation I created to model the toy.


In the next post, I will prove that the shape doesn’t just resemble an ellipse, but it IS an ellipse.

Endnote: after investigating this interesting toy myself, I discovered its actual name, “the Trammel of Archimedes”

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Odd Sums

Sticking with the theme of investigating quantities that remain constant, let’s investigate the sequence of the odd natural numbers:

1, 3, 5, 7, 9, 11, 13, 15, 17, …

Consider the ratio of the sum of the first 2 terms and the sum of the next 2 terms:

\dfrac{1+3}{5+7} = \dfrac{4}{12} = \dfrac{1}{3}

Consider the ratio of the sum of the first 3 terms and the sum of the next 3 terms:

\dfrac{1+3+5}{7+9+11} = \dfrac{9}{27} = \dfrac{1}{3}

Consider the ratio of the sum of the first 4 terms and the sum of the next 4 terms:

\dfrac{1+3+5+7}{9+11+13+15} = \dfrac{16}{48} = \dfrac{1}{3}

If you try a few more examples, you will see that this pattern continues. You always get 1/3. In math words, we could say that the ratio of the sum of the first “k terms” and the sum of the next “k terms” is always 1/3. I think that is quite interesting. But what if we had more terms on the bottom? Maybe twice as many on the bottom as on the top?

Consider the ratio of the sum of the first 2 terms and the sum of the next 4 terms:

\dfrac{1+3}{5+7+9+11} = \dfrac{4}{32} = \dfrac{1}{8}

Consider the ratio of the sum of the first 3 terms and the sum of the next 6 terms:

\dfrac{1+3+5}{7+9+11+13+15+17} = \dfrac{9}{72} = \dfrac{1}{8}

Again, if you try a few more examples, you will see that this pattern continues. You always get 1/8. The ratio of the sum of the first “k terms” and the sum of the next “2k terms” is always 1/8. Therefore, our ratio changed (from 1/3 to 1/8) but the consistency remained. What if we had 3 times as many numbers on the bottom, or 4 times as many? Would the ratio still be constant? If you work it out, you will find that yes, the ratio is always constant, but it is also different each time.

  • 3 times as many numbers on the bottom: the ratio is 1/15
  • 4 times as many: 1/24
  • 5 times as many: 1/35
  • 6 times as many: 1/48

You get the idea. One more point of interest. We can form a sequence from the denominators of these constant ratios. It would go like this:

3, 8, 15, 24, 35, 48, …

Do you notice the pattern? Each new term is one less than a perfect square (24 is one less than 5*5 = 25). What an interesting and beautiful structure to emerge from just the ratio of odd numbers!

I attempted this same trick with even numbers, and let me tell you, that was a huge disappointment. No consistency to be found.

\dfrac{2+4}{6+8} = \dfrac{6}{14} = \dfrac{3}{7}

\dfrac{2+4+6}{8+10+12} = \dfrac{12}{30} = \dfrac{2}{5}

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