## Partial Illumination for the Chords-of-an-Ellipse Problem Tuesday, Aug 31 2010

Back in July, Sam Shah described a beautiful and haunting problem he had encountered at PCMI:

Put n evenly spaced points on a unit circle, with one point at (1,0). Then draw chords from this point to all the other points. Then multiply the chords’ lengths. What do you get?

Mimi Yang drew some great illustrations of the problem, and Mr. Ho made a very slick geogebra applet that illustrates and gets data at the same time.

Sam mentioned an extension problem:

Scale the circle vertically by a factor of $\sqrt{5}$. Scale all the chords too. What is the product of the lengths now?

Mr. Ho outdid himself by creating a geogebra applet for this one too. In fact, he lets you scale the ellipse in all kinds of ways.

It’s really this extension that’s the subject of this post. But first…

Background on the original problem

If you haven’t bumped into this problem yet and it’s not obvious to you how to solve it, you might want to experiment a bit, if nothing else then with Mr. Ho’s applets. I am going to talk about the problem’s results, which are awesome, so I’m warning you now that if the problem is new to you but you read on, I’m going to steal some of your fun.

People posted solutions in the comments to Sam’s original problem statement. The solutions fell into 2 categories:

1) Direct calculation of the product using trigonometry. These methods were able to produce a surprising conjecture about the product of lengths for n points, but were not able to prove it.

2) Recognition that if the points are treated as numbers in the complex plane, they are precisely the nth roots of unity, followed by exploitation of the algebra of the nth roots of unity. These methods were able to prove the conjecture. (For example: Andrew’s comment at Sam’s original post; gasstationwithoutpumps has his/her (?) own post on the subject.)

This is as far as I can get before stating the result; so – SPOILER ALERT.

For n points on the circle, the product of all these chords, most of which have irrational lengths, is a positive integer. Not any positive integer, but n itself.

Here is the roots-of-unity proof, which I’ve done my best to render in a way that’s accessible even if you’ve never worked with roots of unity before. You do need to be familiar with the geometry of complex numbers though.

If you interpret the n equally spaced points on the circle as complex numbers, then they are precisely the nth roots of 1: $\omega, \omega^2, \dots, \omega^{n-1}$ where $\omega$ is the first of the numbers you find if you go around the circle counterclockwise after you leave the positive real axis. (Think about what happens when you multiply this number by itself n times, to convince yourself of this.) Since the point from which all the chords emanate is the number 1, the product of the lengths is the absolute value of the product of the complex numbers $(1-\omega), (1-\omega^2), \dots, (1-\omega^{n-1})$. This product has a startlingly elegant form, which can be seen by noticing that this product is exactly the product $(z-\omega)(z-\omega^2) \dots (z-\omega^{n-1})$ evaluated at $z=1$; and that this product is exactly the monic polynomial that has $\omega, \omega^2, \dots, \omega^{n-1}$ as roots. What are they roots of? Oh right, unity. In other words, they are all zeros of the polynomial $z^n-1$. This polynomial also has 1 itself as a root (because 1 is an nth root of 1 too), so you have to divide it by a factor of $z-1$ in order to get the polynomial that has only the omegas as roots.

$z^n-1=(z-1)(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1})$

$(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1})=\frac{z^n-1}{z-1}=z^{n-1}+\cdots+z+1$

And now, evaluation of this at $z=1$ gives you 1+1+…+1 = n. The product of the chord lengths is the absolute value of this, but this is a positive real number so its absolute value is itself. Thus the product of the chord lengths is n. QED.

The ellipse extension

So everybody who produced a proof to the original problem did it by using the algebra of the nth roots of unity. The first thing that grabbed me about the ellipse extension is that while it is obviously closely related to the original problem, it immediately destroys the linchpin of this method. As soon as the circle gets stretched up to the ellipse, the points are no longer nth roots of unity!

The next thing that grabbed me about it was Tom’s comment on Sam’s original post:

I created my own Geogebra applet to investigate the problem with the ellipse (as you have it written I think). The products turn out to be very interesting:

The lengths of the cords are as follows for certain n:

n length
2 2=2*1
3 6=3*2
4 12=4*3
5 25= 5*5
6 48=6*8
7 91=7*13
8 168= 8*21
9 306=9*34
10 550=10*55

So the conjecture would then be

n n*F_n

where F_n is the n’th Fibonacci number.

WHAT???

(As an aside, the type of reaction I had to Tom’s comment is something we should be cultivating in students. Observation of an unexpected pattern leads naturally to a feverish search for an explanation. This is why it’s so important not to treat a pattern noticed as an established fact: this kills the students’ natural wonderment about the why of the pattern.)

Mr. Ho confirmed Tom’s calculations with his own applet. He had a look at other scale factors besides $\sqrt{5}$ but didn’t find anything. At this point, the problem officially had itself ensconced semi-permanently in the back of my brain. How (the hell) are the frickin Fibonacci numbers arising? If you know something about the Fibonacci numbers, you know that they’re related to the golden ratio, which is related to $\sqrt{5}$, so that kind of makes sense, but c’mon now, only kind of. It’s not like the connection is jumping out at me.

Still, I couldn’t justify spending a lot of time on the problem. I try to stick with math problems that fit into a program of study I’ve given myself, and this one didn’t obviously do that. Nonetheless, it was lodged in my head firmly enough for me to describe it to my excellent colleague Japheth Wood (News from the Math Wizard). The next day, he sent me the following in an email:

Update on the stretched diagonals problem. I wrote a short program in
Python to do some calculations, and I found out the following
patterns, when sqrt{5} is replaced by sqrt{4a+1}:

Conjecture: The product of the diagonals of the stretched n-gon is P_n
= n*T_n, where T_n is an integer sequence defined by:
T_1 = T_2 = 1 and T_{n+2} = T_{n+1} + a*T_n
* This is true in the case of the non-stretched circle: a = 0, and
this recurrence gives P_n = n * 1, which is known.
* This seems true in the PCMI case. 5 = 4*1 + 1, so a = 1, and the
recurrence is Fibonacci.
* The other cases where a is an integer seem true by data I’ve collected.
* The conjecture even seems true when a is not an integer.

# WHAT??!!?

Okay, now the time had obviously come for me to give this problem some serious love. Japheth had just a) exploded onto a completely new level my sense that there’s a lot going on here, and b) given me enough of a direction that I knew some real elbow grease would pay off. I spent quite a few hours this weekend with the problem. What I found is still nowhere near a proof, but it strengthens Japheth’s conjecture and suggests some lines for further investigation. Here’s what I did:

First I found some closed forms for Japheth’s recursive sequence $T_n$:

Closed form #1: Using a standard method involving encoding the recursion $T_{n+2}=T_{n+1}+aT_n$ into a matrix and then diagonalizing the matrix (which I learned – like all the linear algebra I know – from Michael Artin’s Algebra, chapters 3 and 4), I found that

$T_n=\frac{u^n-v^n}{u-v}$

where $u$ and $v$ are the two roots of the polynomial $x^2-x-a$. I didn’t end up using this result but I think it’s very pretty. I’m sure it’s well-known but it’s new to me.

Closed form #2: I got a different formula by directly calculating the first few values of $T_n$ in terms of a, and looking for patterns:

$T_1=1$
$T_2=1$
$T_3=1+a$
$T_4=1+2a$
$T_5=1+3a+a^2$

I put this in a table for ease of calculation and pattern-searching:

$n$ $T_n=$ $1$ $a$ $a^2$ $a^3$ $a^4$
1 1
2 1
3 1 1
4 1 2
5 1 3 1
6 1 4 3
7 1 5 6 1
8 1 6 10 4
9 1 7 15 10 1

This is just a downward-slanted Pascal’s triangle! (Look at what the recursion $T_{n+2}=T_{n+1}+aT_n$ does to a pair of rows to get the next row; this tells you why.) It follows that

$T_n=\displaystyle\sum_{k=0}^{\left \lfloor (n-1)/2 \right \rfloor} \binom{n-k-1}{k} a^k$

The next thing I did was to reason as follows: in the original circle problem, all the power of the roots-of-unity method came from knowing the polynomial $x^n-1$ that has these numbers as roots. Now that the circle has been stretched to an ellipse, the points no longer represent roots of unity; but if I could find the polynomial that had these numbers as roots…

I assumed that $a$ would be such that $\sqrt{4a+1}$ would be a real number, so that using it as the vertical stretch factor would make geometric sense. I put no other restrictions on $a$. Then I just started calculating:

n=2: The points are at 1 and -1; no change when you stretch. The polynomial is $g(z)=(z-1)(z+1) = z^2-1$.

n=3: The points are at 1 and $\frac{-1 \pm \sqrt{3} \mathrm{i}}{2}$. Vertical stretching by a factor of $\sqrt{4a+1}$ makes the latter two into $\frac{-1 \pm \sqrt{4a+1}\sqrt{3} \mathrm{i}}{2}$. So, multiply:

$g(z)=(z-1) \left ( z- \left ( \frac{-1 + \sqrt{4a+1}\sqrt{3} \mathrm{i}}{2} \right ) \right ) \left ( z- \left ( \frac{-1 - \sqrt{4a+1}\sqrt{3} \mathrm{i}}{2} \right ) \right )$

When all is said and done, this simplifies to

$g(z) = z^3 + 3az - (3a+1)$

And so on. I made it to n=8. Let me tell you, n=7 was a dirty job. I looked for ways to make things easier – for example, I wrote the points as $\cos(2\pi k/n)+\mathrm{i}\sqrt{4a+1}\sin(2\pi k/n)$ so that I could take advantage of some symmetry, cross-cancellation and trig identities, and for n=7 I found the polynomial that has $\cos(2\pi k/7)$ as roots, which then gave me some symmetric expressions in the $\cos(2\pi k/7)$s that showed up in the final product, without needing to actually calculate the value of each $\cos(2\pi k/7)$. But still. That case alone had to have taken me at least 2 hours. I messed up several times, thankfully in ways that contradicted each other. But I got my data:

$n$ $g(z)$
1 $z-1$
2 $z^2-1$
3 $z^3+3az-(3a+1)$
4 $z^4+4az^2-(4a+1)$
5 $z^5+5az^3+5a^2z-(5a^2+5a+1)$
6 $z^6+6az^4+9a^2z^2-(9a^2+6a+1)$
7 $z^7+7az^5+14a^2z^3+7a^3z-(7a^3+14a^2+7a+1)$
8 $z^8+8az^6+20a^2z^4+16a^3z^2-(16a^3+20a^2+8a+1)$

Clear patterns are emerging in the structure of these polynomials, what with how
a) the powers of $z$ are falling by 2 and of $a$ are rising by 1
b) the constant term is the negative of the sum of the coefficients of the other terms

Together, these mean we could really regard the whole thing as a sum of terms of the form

$\mathrm{coefficient}*a^k(z^{n-2k}-1)$

The last thing to do is figure out what’s going on with the coefficients. The lead coefficient is always 1, the next (provided it contains a $z$) is counting numbers, and the next (again, provided $z$ is involved) – the sequence 5, 9, 14, 20 – is increasing by counting numbers. This is an echo of the Pascal’s Triangle, so I looked for an expression for these numbers in terms of chooses. I found one: the $k+1$th coefficient of the $n$th polynomial (i.e. the coefficient of the $a^kz^{n-2k}$ term) seems to be

$\dbinom{n-k+1}{k} - \dbinom{n-k-1}{k-2}$

Conjecture:

* If you take the unit circle and place n evenly-spaced points along it, the first at (1,0)
* And you then scale everything vertically by a factor of $\sqrt{4a+1}$, where $a$ is any real number such that $\sqrt{4a+1}$ is real
* And you then regard each of these scaled points as a complex number,
* The monic polynomial $g(z)$ with these numbers as roots will be equal to

$\displaystyle\sum_{k=0}^{\left \lfloor (n-1)/2 \right \rfloor} \left [ \dbinom{n-k+1}{k} - \dbinom{n-k-1}{k-2} \right ] a^k (z^{n-2k} - 1)$

If this conjecture is correct, Japheth’s conjecture follows. (For those of you who like that sort of thing, how does it follow?) So proving this conjecture would settle all the outstanding questions (e.g. it would prove the Fibonacci pattern in the $\sqrt{4a+1}=\sqrt{5}$ case).

But I have no idea how to prove it. Why are the powers of $a$ showing up? How about the chooses? What’s going on here?

Thoughts?

(Darryl Young and Bowen Kerins, the writers of the PCMI problem set, are invited to give the rest of us hints!)

## An Odd Request Wednesday, Aug 25 2010

We interrupt our regularly scheduled programming to ask you an admittedly strange question:

A bunch of brand new young enthusiastic preservice math teachers sit down in front of you. If you could pick 3-5 things about math education that you most want them to understand, what would they be?

Comment with whatever quick and dirty brainstorm you have. I know some of you have given this, or something like it, a stab in the past – in that case I’d appreciate a link. (I remember Dan did this semi-recently – others?)

## The Talent Lie Monday, Aug 9 2010

Back in the fall when I was a baby blogger I wrote a discussion of Carol Dweck’s research about intelligence praise. I did this because I think this research is intensely important. However, I didn’t really let loose on the subject with the full force of what I have to say about it. The truth is I was shy, because a) I’d just had a kind of frustrating conversation on the subject with Unapologetic at Jesse Johnson’s blog, so I was wary of being misunderstood, and b) more embarrassingly, I was excited by the positive response to my previous post about Clever Hans and I didn’t want to alienate any of my new audience.

Now I am a toddler blogger. My godson, with whom I spent the day a few weeks ago, is an actual toddler.

He is profoundly unconcerned with anybody’s opinion of him, and just blazes forth expressing himself (climbing on things; coveting whatever his big sister is playing with; being turned upside down as much as possible) all day long. I am going to take this as inspiration, and commence a series of posts about the idea of “math smarts” and talent and intelligence more broadly. These posts have two central contentions:

1) People constantly interpret mathematical accomplishment through the lens of math talent or giftedness.

2) This is both factually misleading and horrible for everyone.

Tentatively, here is the table of contents for this series. I may edit these titles, add or remove some, and I’ll add links when I’ve got the posts up. But here’s the plan for now:

I. Why the talent lie is a lie; how to understand math accomplishment outside of it
II. How the talent lie is spread (in pop culture, and inside the discipline of mathematics)
III. How the talent lie hurts people who are “good at math”
IV. How the talent lie hurts people who are “bad at math”
V. How to train students to understand math accomplishment outside of the talent lie
VI. Why the talent lie is so entrenched, even though it is stupid and harmful

I should make more precise what I mean by “the talent lie.” It’s really several variants on a fundamental idea. People who are really good at math must have been born with a gift, for example. That they must be extra smart. That being good at math (or not) is something that doesn’t change over time. That being smart (or not) doesn’t change. In short, that your intellectual worth, and the worth of your engagement with the field of mathematics in particular, is an already-determined quantity that’s not up to you. That’s the talent lie.

Some examples of the talent lie at work:
* Any time anyone has ever said, “I’m bad at math.”
* Just about any time anybody makes a big deal about the age by which a young person does something intellectual. (Starts talking, starts reading, starts learning calculus…)

(In that last bullet, the “just about” is there only because of the theoretical possibility that a big deal might get made for a reason other than to prognosticate about the person’s ultimate intellectual worth.)

I give you these examples to show that I am not talking about a fringe, outmoded idea but something very mainstream. I will have much more to say about how the talent lie is manifested in the forthcoming posts.

I expect to spend a long time writing them. This project may take all fall year the next several years. I believe the message I’m communicating is vital for our field and important more broadly as well. It’s also a very personal message. Like all urban educators and all math teachers, I have a lot of first-hand experience with the damage that the labels “not smart” and “not good at math” can inflict. But I am also speaking as someone who spent my early years being seen by others, and regarding myself, as mathematically gifted. This was a heady and thrilling thing when I was in middle school, but I became vaguely aware of the complications by the end of high school, and with hindsight it’s clear that it left me with baggage that took a decade of teaching, learning and introspection to shake. So my own journey is a big part of the story I’m telling here.

I will save the detailed analysis for the forthcoming posts, which means that I am going to defer a lot of clarification and answering-questions-you-might-have for later. But I would like now to articulate in broad terms what I believe needs to change.

According to the Calvinist doctrine of unconditional election, God already decided whether you are going to be damned or saved, and did this way before you were born. Nothing you can do – not a life of good acts, not a wholehearted and humble commitment to acceptance or faith – can have any effect. The most you can do is scan your life for signs of God’s favor, and read the clues like tea-leaves to see if you are chosen or cast away. Modern American culture doesn’t buy this doctrine from a theological point of view, but is 100% bought in when it comes to math. When a person performs mathematically, we obsessively look at the performance, not on its own terms, but as a sign one way or the other on the person’s underlying mathematical worth, a quantity we imagine was fixed long ago.

We need, as a culture, to gut-renovate our understanding of what’s going on when we see people accomplish impressive mathematical feats. Likewise, when people fail at mathematical tasks. We need to stop seeing people’s mathematical performance as nothing more than the surface manifestation of a well-spring of mathematical gifts or talent they may or may not have. Relatedly but even more importantly, we need to stop reading the tea-leaves of this performance to determine these gifts’ presence or absence. This whole game is bunk.

Not only is it bunk but it’s a crippling distraction, for everyone – teachers, students, parents, and our culture as a whole – from the real job of studying, wandering through, becoming intimate with and standing in awe of the magnificent edifice known as the discipline of mathematics.

When you step to the gate and present yourself before it, math doesn’t give a sh*t about the particular profile of cognitive tasks that are easy and hard for you at this moment in time, and you shouldn’t either. There are institutions that are very keen to divine from this profile your worthiness to enter, but this is the curtain they hide behind to make themselves look bigger than they are. It’s time to tear that curtain down.

More on its way. In the meantime here is some related reading:

* I Speak Math recently tackled this same subject. I plan on drawing on some of the research she links.

* Jesse Johnson and I had a conversation about this stuff close to a year ago, and she wrote about it here and here. I’ll go into much more detail on these themes in the coming posts.

* While not as credentialed, the Wizard of Oz nonetheless has a fair amount in common with wolverine wranglers. See if you see what I mean.

## Why They Pay Dan Meyer the Big Bucks Friday, Aug 6 2010

I know it’s kind of redundant to reblog a Dan Meyer post since there are at most three people who may conceivably be reading this who haven’t already read it. (Hi Howard, Steve and mom!)

But there, I did it.

Because that sh*t is so awesome. I couldn’t help myself.

Watch that opening video, and tell me if you’ve ever seen a math problem that compellingly posed.

## There Is No Reason for You To Read This Post (I Mean It) Wednesday, Aug 4 2010

The purpose of this blog has evolved since I started it. I’m just declaring that here for the sake of my own integrity. This is really no use to you.

Originally, the plan was to engage with research on math education from a practitioner point of view. The goal was so I could try to stay vaguely current on the literature by processing it out loud on this blog so when I read something it will be easier to remember.

* First of all, what I engage with has ballooned to include anything I read about math or education, not just the fairly narrow category “math education research,” although I plan on continuing to read math education research and post about it among other things.

* I’ve found myself using this as a platform to articulate thoughts and values about math education that I want to go on record with because I believe they’re important.

* I’ve been using this as a place to hash out and process thoughts about teaching and learning that are inspired not only by reading but by conversations, classes I teach or observe, etc.

* I’ve occasionally indulged the impulse to enthuse about math itself.

* I haven’t done much of this so far, but in the coming months I foresee the need to use this blog from time to time as a place to actively reflect on my own practice.

* Combinations of any and all of the above.

My original intent was for 1 post a week. This has been hard to maintain. I hereby retract this as a declared intention, but retain it as an approximate goal.

While I’m chewing on what this blog has become, it seems appropriate to once again shout out Kate Nowak (my fairy blogmother), and Jesse Johnson (my fairy blog-big-sis).

Okay! Integrity restored! I’ll be back on topic soon…