Midpoint Polygon Conjecture is false

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Contrary to my expectations, my previous post turned out to be like  Popular-­Lonely primes and Decimal Problem, i.e. I discovered nothing new.

My conjecture is false. Following counterexample is given on pp. 234 of this paper:

Counterexample of the conjecture, taken from: Berlekamp, E. R., E. N. Gilbert, and F. W. Sinden. “A Polygon Problem.” The American Mathematical Monthly 72, no. 3 (1965): 233-41. doi:10.2307/2313689.

As pointed out by uncombed_coconut, the correct theorem is:

Theorem (Berlekamp-Gilbert-Sinden, 1965). For almost all simple polygons there exist a smallest natural number $k$ such that after $k$ iterations of midpoint polygon procedure, we obtain a convex polygon.

The proof of this theorem is very interesting. Till now I thought that proving euclidean geometry theorems using complex numbers was an overkill. But using an N-tuple of complex numbers to represent the vertices of a closed polygon (in given order),  $\mathbf{z} = (z_1,\ldots , z_N)$,  we can restate the problem in terms of eigenvectors (referred to as eigenpolygons) and eigenvalues.  Following are the crucial facts used is the proof:

• An arbitrary N-gon (need not be simple) can be written as a sum of regular N-gons i.e. the eigenvalues are distinct.
• The coefficient of $k^{th}$ eigenvector (when N-gon is written as linear combination of eigenpolygons) is the centroid of the polygon obtained by “winding” $\mathbf{z}$ $k$ times.
• All vertices of the midpoint polygons (obtained by repeating the midpoint polygon procedure infinitely many times) converge to the centroid.
• The sum of two convex components of $\mathbf{z}$ is a polygon. This polygon is the affine image of a regular convex N-gon whose all vertices lie on an ellipse. (as pointed out by Nikhil)
• A necessary and sufficient condition for $\mathbf{z}$ to have a convex midpoint polygon (after some finite iterations of the midpoint polygon procedure) is that the ellipse circumscribing the sum of two convex components of $\mathbf{z}$ is nondegenerate. (The degenerate form of an ellipse is a point. )

For a nice outline of the proof, please refer to the comment by uncombed_coconut on previous post.

Since I didn’t know that this is a well studied problem (and that too by a well known mathematician!) I was trying to prove it on my own. Though I didn’t make much progress, but I discovered some interesting theorems which I will share in my future posts.

Midpoint Polygon Conjecture

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This is going to be yet another ambitious post like New Diagonal Contribution Theorem, and Cross Diagonal Cover Problem.

Let’s begin with following definitions from Wikipedia:

A convex polygon is a simple polygon (not self-intersecting) in which no line segment between two points on the boundary ever goes outside the polygon.

A concave polygon is a simple polygon (not self-intersecting) which has at least one reflex interior angle – that is, an angle with a measure that is between 180 degrees and 360 degrees exclusive.

We know that 3 sided polygons (a.k.a triangles) are always convex.  So, they are not very interesting. Let’s define the following procedure:

Midpoint Polygon Procedure: Given a n-sided simple polygon, join the consecutive midpoints of the sides to generate another n-sided simple polygon.

Here are few examples:

Red polygon is the original polygon, green polygon is the one generated by joining midpoints.

We observe that if the original polygon was a concave polygon then the polygon generated by joining midpoints need not be a convex polygon. So, let’s try to observe what happens when we apply the midpoint polygon procedure iteratively on the 4th case above (i.e. when we didn’t get a convex polygon):

In fourth iteration of the described procedure, we get the convex polygon.

Based on this observation, I want to make following conjecture:

Midpoint Polygon Conjecture: For every simple polygon there exists a smallest natural number $k$ such that after $k$ iterations of midpoint polygon procedure, we obtain a convex polygon.

If the given polygon is convex, then there is nothing to prove since $k=1$, but I have no idea about how to prove it for concave polygons. I tried to Google for the answer, but couldn’t find anything similar. So, if you know the proof or counterexample of this conjecture, please let me know.

Teaching Mathematics

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One of the most challenging and rewarding thing associated with being a math enthusiast (a.k.a. mathematician) is an opportunity to share your knowledge about the not so obvious truths of mathematics. A couple of years ago, I tried to communicate that feeling through an article for high school students.

When I joined college, I tried to teach mathematics to some kids from financially not-so strong family. Since they had no exposure to mathematics, I had to start with  concepts like addition and multiplication of numbers. My experience can be summarized as the following stand-up comedy performance by Naveen Richard:

After trying for about a couple of months to teach elementary mathematics, I gave up and now I discuss mathematics only above the high school level. Last week I delivered a lecture discussing the proof of Poncelet’s Closure Theorem:

Whenever a polygon is inscribed in one conic section and circumscribes another one, the polygon must be part of an infinite family of polygons that are all inscribed in and circumscribe the same two conics.

I had spent sufficient time preparing the lecture, and believed that I was aware of all possible consequences of this theorem. But, almost half way through the lecture one person (Haresh) from the audience of 10 people, pointed out following fascinating consequence of the theorem:

If an n-sided polygon is inscribed in one conic section and circumscribed by the other one, then it must be a convex polygon and no other m-sided polygon (with m≠n) can be inscribed and circumscribed by this pair of conic sections.

This kind of insights by audience motivates me to discuss mathematics with others!

In the praise of norm

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If you have spent some time with undergraduate mathematics, you would have probably heard the word “norm”. This term is encountered in various branches of mathematics, like (as per Wikipedia):

But, it seems to occur only in abstract algebra. Although the definition of this term is always algebraic, it has a topological interpretation when we are working with vector spaces.  It secretly connects a vector space to a topological space where we can study differentiation (metric space), by satisfying the conditions of a metric.  This point of view along with an inner product structure, is explored when we study functional analysis.

Some facts to remember:

1. Every vector space has a norm. [Proof]
2. Every vector space has an inner product (assuming “Axiom of Choice”). [Proof]
3. An inner product naturally induces an associated norm, thus an inner product space is also a normed vector space.  [Proof]
4. All norms are equivalent in finite dimensional vector spaces. [Proof]
5. Every normed vector space is a metric space (and NOT vice versa). [Proof]
6. In general, a vector space is NOT same a metric space. [Proof]

Real vs Complex Plane

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Real plane is denoted by $\mathbb{R}^2$ and is commonly referred to as  Cartesian plane. When we talk about $\mathbb{R}^2$ we mean that $\mathbb{R}^2$ is a vector space over $\mathbb{R}$. But when you view $\mathbb{R}^2$ as Cartesian plane, then it’s not technically a vector space but rather an affine space, on which a vector space acts by translations, i.e. there is no canonical choice of where the origin should go in the space, because it can be translated anywhere.

Cartesian Plane (345Kai at the English language Wikipedia [Public domain, GFDL or CC-BY-SA-3.0], via Wikimedia Commons)

On the other hand, complex plane is denoted by $\mathbb{C}$ and is commonly referred to as Argand plane. But when we talk about $\mathbb{C}$, we mean that $\mathbb{R}^2$ is a field (by exploiting the tuple structure of elements) since there is only way to explicitly define the field structure on the set $\mathbb{R}^2$ and that’s how we view $\mathbb{C}$ as a field (if you allow axiom of choice, there are more possibilities; see this Math.SE discussion).

Argand Plane (Shiva Sitaraman at Quora)

So, when we want to bother about the vector space structure of $\mathbb{R}^2$ we refer to Cartesian plane and when we want to bother about the field structure of $\mathbb{R}^2$ we refer to Argand plane. An immediate consequence of the above difference in real and complex plane is seen when we study multivariable analysis and complex analysis, where we consider vector space structure and field structure, respectively (see this Math.SE discussion for details). Hence the definition of differentiation of a function defined on $\mathbb{C}$ is a special case of definition of differentiation of a function defined on $\mathbb{R}^2$.

Real vs Complex numbers

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I want to talk about the algebraic and analytic differences between real and complex numbers. Firstly, let’s have a look at following beautiful explanation by Richard Feynman (from his QED lectures) about similarities between real and complex numbers:

From Chapter 2 of the book “QED – The Strange Theory of Light and Matter” © Richard P. Feynman, 1985.

Before reading this explanation, I used to believe that the need to establish “Fundamental theorem Algebra” (read this beautiful paper by Daniel J. Velleman to learn about proof of this theorem) was only way to motivate study of complex numbers.

The fundamental difference between real and complex numbers is

Real numbers form an ordered field, but complex numbers can’t form an ordered field. [Proof]

Where we define ordered field as follows:

Let $\mathbf{F}$ be a field. Suppose that there is a set $\mathcal{P} \subset \mathbf{F}$ which satisfies the following properties:

• For each $x \in \mathbf{F}$, exactly one of the following statements holds: $x \in \mathcal{P}$, $-x \in \mathcal{P}$, $x =0$.
• For $x,y \in \mathcal{P}$, $xy \in \mathcal{P}$ and $x+y \in \mathcal{P}$.

If such a $\mathcal{P}$ exists, then $\mathbf{F}$ is an ordered field. Moreover, we define $x \le y \Leftrightarrow y -x \in \mathcal{P} \vee x = y$.

Note that, without retaining the vector space structure of complex numbers we CAN establish the order for complex numbers [Proof], but that is useless. I find this consequence pretty interesting, because though $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as additive groups (and as vector spaces over $\mathbb{Q}$) but not isomorphic as rings (and hence not isomorphic as fields).

Now let’s have a look at the consequence of the difference between the two number systems due to the order structure.

Though both real and complex numbers form a complete field (a property of topological spaces), but only real numbers have least upper bound property.

Where we define least upper bound property as follows:

Let $\mathcal{S}$ be a non-empty set of real numbers.

• A real number $x$ is called an upper bound for $\mathcal{S}$ if $x \geq s$ for all $s\in \mathcal{S}$.
• A real number $x$ is the least upper bound (or supremum) for $\mathcal{S}$ if $x$ is an upper bound for $\mathcal{S}$ and $x \leq y$ for every upper bound $y$ of $\mathcal{S}$ .

The least-upper-bound property states that any non-empty set of real numbers that has an upper bound must have a least upper bound in real numbers.
This least upper bound property is referred to as Dedekind completeness. Therefore, though both $\mathbb{R}$ and $\mathbb{C}$ are complete as a metric space [proof] but only $\mathbb{R}$ is Dedekind complete.

In an arbitrary ordered field one has the notion of Dedekind completeness — every nonempty bounded above subset has a least upper bound — and also the notion of sequential completeness — every Cauchy sequence converges. The main theorem relating these two notions of completeness is as follows [source]:

For an ordered field $\mathbf{F}$, the following are equivalent:
(i) $\mathbf{F}$ is Dedekind complete.
(ii) $\mathbf{F}$ is sequentially complete and Archimedean.

Where we defined an Archimedean field as an ordered field such that for each element there exists a finite expression $1+1+\ldots+1$ whose value is greater than that element, that is, there are no infinite elements.

As remarked earlier, $\mathbb{C}$ is not an ordered field and hence can’t be Archimedean. Therefore, $\mathbb{C}$  can’t have least-upper-bound property, though it’s complete in topological sense. So, the consequence of all this is:

We can’t use complex numbers for counting.

But still, complex numbers are very important part of modern arithmetic (number-theory), because they enable us to view properties of numbers from a geometric point of view [source].

Division algorithm for reals

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You must have seen long-division method to compute decimal representation for fractions. Astonishingly, I never pondered about how one would divide an irrational number to get decimal representation. Firstly, this representation will be approximate. Secondly, we have been doing this in name of “rationalizing the denominator” stating the reason that division by irrationals is not allowed. But, in fact, this is the same problem as faced while analysing division algorithm for Gaussian integers.

Bottom line: Numbers are just symbols. We tend to assign meaning to them as we grow up. Since the set of real numbers, rational numbers and integers  form an Euclidean domain, we can write a division algorithm for them. For example, we don’t have special set of symbols for 3 divided by π, but 3 divided by 2 is denoted by 1.5 in decimals.