Homology: counting holes in doughnuts and why balls and disks are radically different.

November 24, 2014 by . 0 comments

There are some questions that are really easily posed, have an obvious answer, but are in fact really, really hard to answer in a mathematically satisfactory way. Two examples are:

  1. How many holes does a doughnut have?
  2. Are a ball and a disk “the same thing”? Meaning: can I deform the first to make it the second in a way that locally preserves its structure, i.e. without tearing, and without “squishing” things too much?

The intuitive (and actually correct) answers are: one, and no. However, in order to prove them true, we’ll have first to formalize the questions in mathematical language, and then to develop a theory that will allow us to work on them.

Let’s start with the first question. “A doughnut” can be interpreted in two ways: it could be a filled doughnut, that is the space \(S^1\times D^2\), where \(S^1\) denotes the circle, and \(D^2\) the \(2\)-dimensional disc , or it could be a hollow doughnut, corresponding to the torus \(T^2=S^1\times S^1\). Both of them are examples of topological spaces, so we will generalize the question to: given a topological space \(X\), how many holes does \(X\) have? If you don’t know what a topological space is, just take \(X\) to be one of the examples of doughnuts given above, unless otherwise specified.

The next question we have to answer is: what is a hole? We have many different examples of things we would like to consider holes. One is our hole in the doughnut, but we could also take the space, \(\mathbb{R}^3\), and take away a ball from it, or a infinitely long filled cylinder. Notice that the last two examples are fundamentally different, in the following way: if we take away an infinite cylinder from the space and draw a closed curve around it, we will never be able to deform it to a path not containing the cylinder, or to a single point, without “breaking” it, while we deform every closed curve to a single point in the very last example. However, if we were to put a sphere around the ball we removed in the last example, then we could not deform it to a point, while we could do it for every sphere in our space without a cylinder.

Noticing that a closed path looks very much like a circle, we can use this to distinguish between various kinds of holes. We will informally call an \(n\)-dimensional hole an \(n\)-sphere \(S^n\) that cannot be deformed to a single point without tearing it, where we define: $$S^n=\left\{x\in\mathbb{R}^{n+1}:\lvert x\rvert^2=1\right\}$$ Notice that \(S^1\) is the circle, and \(S^2\) is the sphere. Now, spheres look like really simple spaces, but are in fact quite difficult to work with. However, there’s something we can do to avoid working directly with spheres while keeping intact the essence of what we have said until now: we can cut up spheres in smaller “triangular” pieces. We make the following definition:

Definition: Let \(v_0,\ldots,v_n\in\mathbb{R}^{m}\), where \(m\ge n\). We define the affine \(n\)-dimensional singular simplex \([v_0,\ldots,v_n]\) as the closed convex hull of the points \(v_0,\ldots,v_n\), that is: $$[v_0,\ldots,v_n]=\left\{\sum_{k=0}^nt_kv_k:t_k\in[0,1]\forall k,\ \sum_{k=0}^nt_k=1\right\}$$ We also define the standard \(n\)-simplex as \(\Delta^n=[e_0,\ldots,e_n]\subset\mathbb{R}^{n+1}\), where \(e_i\in\mathbb{R}^{n+1}\) are the standard basis elements.

Notice that the standard \(2\)-simplex is in fact a triangle, and the standard \(3\)-simplex is a tetrahedron. So this definition gives a sensible generalization of what a triangle of dimension \(n\) should be. Let’s now cover the circle \(S^1\) with \(1\)-simplices, and the sphere \(S^2\) with \(2\)-simplices. We immediately notice that they have something special with respect to a random collection of simplices: they have no boundary, that is, the triangles composing them have the sides glued together in such a way that they cancel each other. This leads us to make the next definition:

Definition: The \(i\)-th face of an \(n\)-simplex \(a=[v_0,\ldots,v_n]\) is the \((n-1)\)-simplex \(a^{(i)}[v_0,\ldots,v_{i-1},v_{i+1},\ldots,v_n]\). The boundary of the simplex is the (formal) sum of \((n-1)\)-simplices: $$ \partial a=\sum_{k=0}^n(-1)^ka^{(k)}$$

Notice that the boundary of a simplex is a sum of simplices of lower dimension, so we would like to define some set of simplices where we are allowed to take sums in a sensible way. Also, we would like our simplices to live in our topological space \(X\), and not only in \(\mathbb{R}^m\). So we define the group of \(n\)-simplices in \(X\), denoted by \(S_n(X)\), as the free abelian group generated by continuous maps \(\sigma:\Delta^n\to X\). This simply means that the object (called chains) of \(S_n(x)\) are finite sums of simplices of dimension \(n\) in \(X\) (this is why we take maps from the standard simplex to \(X\)), and that we can sum two such objects in the obvious way, for example if we have two chains \(\sigma_1\) and \(\sigma_2\), then we have: $$(2\sigma_1+\sigma_2)+3\sigma_2=2\sigma_1+4\sigma_3$$ The boundary defines then a map (in fact, a group homomorphism) from \(S_n(x)\) to \(S_{n-1}(X)\), defined on the generators \(\sigma:\Delta^n\mapsto X\) as: $$\partial \sigma=\sum_{k=0}^n(-1)^k\sigma^{(k)}$$ where \(\sigma^{(k)}\) is simply the map \(\sigma\) restricted to the \(i\)-th face of \(\Delta^n\). The sequence of groups \(S_n(X)\) together with the boundary maps defines a sequence: $$\ldots\stackrel{\partial_{n+1}}{\longrightarrow} S_n(X)\stackrel{\partial_{n}}{\longrightarrow} S_{n-1}(x)\stackrel{\partial_{n-1}}{\longrightarrow}\ldots\stackrel{\partial_{2}}{\longrightarrow} S_1(X)\stackrel{\partial_{1}}{\longrightarrow} S_0(X)\to 0$$ where the composition of two consecutive arrows gives the zero map (this can be easily checked by writing down what happens to a generator and rearranging a couple of sums). This kind of sequence is really important in some areas of mathematics, and they have a special name: they are called chain complexes. As we have seen, of all the elements in the chain complex we want to consider those with no boundary, that is, elements in: $$Z_n=\ker(\partial_n)=\{c\in S_n(X)|\partial c=0\}\subseteq S_n(X)$$ We call these elements cycles. They represent in some sense “closed things”, like circles, spheres, but also for example tori (i.e. our hollow donuts) and similar stuff. Moreover, we would like to identify two cycles whenever they represent, for example, two closed paths that can be deformed in such a way that the first becomes equal to the second. Notice that if this is the case, then during the deformation the first curve will draw some kind of annulus, which we can cover with \(2\)-simplices and (as a chain) will have as boundary the first path minus the second one. This leads us to the idea of identifying two cycles whenever their difference is a boundary. Thus we define the homology groups of \(X\) by: $$H_n(X)=Z_n/B_n$$ where \(B_n=\partial_{n+1}(S_{n+1}(X))\) is the set of boundaries.

These groups have a great deal of nice properties. First of all, they are topological invariants. This means that if two spaces are “essentially the same” (the technical term is homeomorphic, meaning that there exist a continuous bijection with continuous inverse between the two), then their homology groups are equal. Another similar thing is that if \(Y\) is a subspace of \(X\), and we can deform \(X\) to \(Y\) without deforming \(Y\), then \(X\) and \(Y\) have the same homology groups. The other useful properties are mostly too technical to be stated here without making this post excessively long. Also, unfortunately, we don’t have enough tools to compute the homology groups of spaces more complicated than, say, finite unions of points, or balls, or \(\mathbb{R}^n\). So when needed I will just state the results, and if you are interested in the computations, you can try to consult one of the bibliographical references I will give at the end.

Whew! We’ve come a long way from the original question of counting the number of holes in a doughnut! But finally we can answer the question. From our definitions, the \(n\)-th homology group should more or less count the number of \(n\)-dimensional holes in the space. For example, for the filled doughnut we have: $$H_n(S^1\times D^2)=\begin{cases} \mathbb{Z}&\text{for $ n=0,2$} \\ 0 &\text{else } \end{cases}$$ The \(0\)-th group doesn’t really matters to us, in fact all it does is to count the number of “pieces” of which our space is made of. The \(1\)-st homology group, however, gives us the information we needed: up to equivalence, there is exactly one \(1\)-chain which is not a boundary (we get \(\mathbb{Z}\) because we can also count twice the same chain, or three times, etc.). This means that there is one circle (or something analogous) that cannot be deformed to be a point, and thus that there is exactly one \(1\)-dimensional hole. Similarly, for the hollow doughnut we have: $$H_n(S^1\times S^1)=\begin{cases} \mathbb{Z} &\text{for $n=0,2$}\\ \mathbb{Z}^2 &\text{for $n=1$}\\ 0 &\text{else}\end{cases}$$

Thus we have one \(2\)-dimensional hole (the whole hollow part of the doughnut), and two \(1\)-dimensional holes (the circle going around the central hole of the doughnut, and the circle going around the hole formed by the hollow part).

As a bonus, we can also use homology to answer our second question. As I have stated, two homeomorphic spaces (“essentially the same”, remember?) have the same homology groups. Now if a ball \(B^3\) in space were homeomorphic to a disk \(B^2\), then removing exactly one interior point in each of those spaces in a sensible way we would again get homeomorphic spaces. However a ball without a point deforms nicely to a sphere, and a disk without a point to a circle, and we have: \begin{gather}H_n(S^1)=\begin{cases}\mathbb{Z} &\text{if $n=0,1$}\\ 0 &\text{else}\end{cases},\\ H_n(S^2)=\begin{cases} \mathbb{Z} &\text{if $n=0,2$}\\ 0 &\text{else} \end{cases}\end{gather}

which is exactly what we would expect from our intuition about \(n\)-dimensional holes. Since these are not equal, the two spaces cannot be homeomorphic, and thus we are done.

Those two problems we just solved are two of the many applications of homology theory, and indeed of the larger framework, which is called algebraic topology. If you want to know more on the subject, here are three books you can try to read:

  • Allen Hatcher, Algebraic Topology
  • Glen E. Bredon, Topology and Geometry
  • Edwin H. Spanier, Algebraic Topology

More than Infinitesimal: What is “dx”?

November 3, 2014 by . 12 comments

Problem

Many people have asked this question, and many will continue to do so. It is the natural question of someone first learning the subject of calculus: what is “\(\mathrm{d}x\)”, and why is it everywhere in calculus?

Frankly, it’s mostly Leibniz’s fault. Leibniz, a brilliant philosopher and mathematician who may or may not have invented calculus depending on whom you ask, introduced the notation. His view of derivatives was as the ratio of related infinitesimals. In slightly more modern terms, $$\lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}=\frac{dy}{dx}.$$ Unfortunately, people have carried this view for an unhealthily long amount of time. A branch of analysis known as “nonstandard analysis” found a clever and eponymously nonstandard way to make the idea of the infinitesimal rigorous. Still others have questioned the validity of the law of the excluded middle, which says that a proposition must be either true or false. By not accepting this law, some finagling and rather nonstandard logic can bring about the idea of an infinitesimal. The list goes on. In short, there is a myriad of “nonstandard” ways to realize \(\mathrm{d}x\).

This brings us to a much more refined question: is there a standard way to define \(\mathrm{d}x\)?

We will defend the claim that the answer is a resounding “yes.” What’s more, we will attempt to demonstrate that the concept is intuitive and natural, even to those relatively new to the subject of analysis. more »

Matching Theory

October 1, 2014 by . 1 comments

Matching theory is an active field in mathematics, economics, and computer science. It ensured a [Nobel memorial prize][1] for [Alvin E. Roth][2] and [Lloyd S. Shapley][3] in 2012. The theory is applied in the real world to match students and colleges, doctors and hospitals, and to organize the allocation of donor organs. It all started with a beautiful paper by [David Gale][4] and Lloyd Shapley in 1962: [College Admissions and the Stability of Marriage][5] (read it!). In this post, we study the simplest version of what is known as the stable marriage problem and take a look at inherent conflicts. The exposition follows chapter 22 of the wonderful book [Game Theory][6] by Michael Maschler, Eilon Solan, and Shmuel Zamir.

What are stable matchings?

There is a set \( G \) of \( n \) girls and a set \( B \) of \( n \) boys. All girls and boys are heterosexual and desperate enough to prefer every member of the opposite sex to staying single. Each girl \( g \) has preferences over the boys, represented by a [total order][7] \( \succeq_g \) on \( B \). Similarly, each boy \( b \) has preferences over the girls, represented by a total order \(\succeq_b\) on \( G \).

We want to pair up the girls and the boys. Formally, a matching is simply a [bijection][8] \( M \) from \(G\) to \(B\) and if \( f(g)=b \), we say that \( g \) and \( b \) are matched (under \( M\)). Now, nobody can force the girls and boys to be together, so we have to look at matchings in which the boys and girls are relatively content with whom they get. To be precise, a matching is stable if we cannot find a girl and a boy who prefer each other under their preference ordering to whomever they are matched with.

Are there any?

Before we go into a deep study of stable matchings, we ought to make sure that there are stable matchings. We do so by giving an explicit algorithm for finding a stable matching, the boy courtship algorithm.

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Playing with Partitions: Euler’s Pentagonal Theorem

August 25, 2014 by . 8 comments

Introduction

Today we will play a small game which is really really simple. We will add up numbers to make numbers. And to keep the matters simple we will only deal with counting numbers \(1, 2, 3, \ldots\) As an example \(2 + 3 = 5\), but to make the game challenging we would reverse the problem. Let’s ask how we can split \(5\) into sum of other numbers. For completeness we take \(5\) also as one of the ways to express \(5\) as sum of numbers. Clearly we have the following other ways $$ 5 = 4 + 1 = 3 + 1 + 1 = 3 + 2 = 2 + 2 + 1 = 2 + 1 + 1 + 1 = 1 + 1 + 1 + 1 + 1$$ so that there are \(7\) different ways of adding numbers to make \(5\). We don’t take into account the order of summands. Also one number can be repeated if needed.

Partitions of a Number

It turns out that a whole generation of distinguished mathematicians (Euler, Jacobi, Ramanujan, Hardy, Rademacher etc) were also interested in playing the above game. And needless to say, they made the whole thing very systematic by adding some definitions (its their silly habit so to speak). Following their footprints we say that a tuple \((n_{1}, n_{2}, \ldots, n_{k})\) of positive integers is a partition of a positive integer \(n\) if $$n_{1} \geq n_{2} \geq \cdots \geq n_{k}\text{ and }n_{1} + n_{2} + \cdots + n_{k} = n$$ Thus \((3, 2), (3, 1, 1)\) etc are partitions of \(5\). If \((n_{1}, n_{2}, \ldots, n_{k})\) is a partition of \(n\) then we say that each of the \(n_{i}\) is a part of this partition, \(k\) is the number of parts, \(n_{1}\) is the greatest part and \(n_{k}\) the least part of this partition.

The mathematicians were really not so interested in finding individual partitions of a number \(n\), but were rather interested in finding out the total number of partitions of \(n\). The example above deals with \(n = 5\) and clearly there are \(7\) partitions of \(5\). You should convince yourself by taking a slightly larger number, say \(n = 8\) or \(n = 10\), that finding the number of partitions of any given number \(n\) is not that easy. Especially writing out each partition of \(n\) and then counting them all is very difficult. You may always feel that probably you have missed one of the partitions. Mathematicians however found out a smart way to count partitions. We explore this technique further.

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Binary quadratic forms over the rational integers and class numbers of quadratic fields.

August 23, 2014 by . 3 comments

I wrote an article with Irving Kaplansky on indefinite binary quadratic forms, integral coefficients. At the time, I believe I used high-precision continued fractions or similar. It took me years to realize that the right way to solve Pell’s equation, or find out the “minimum” of an indefinite form (and other small primitively represented values), or the period of its continued fraction, was the method of “reduced” forms in cycles/chains, due to Lagrange, Legendre, Gauss. It is also the cheapest way to find the class number and group multiplication for ideals in real quadratic fields, this probably due to Dirichlet. For imaginary quadratic fields, we have easier “reduced” positive forms.

A binary quadratic form, with integer coefficients, is some $$ f(x,y) = A x^2 + B xy + C y^2. $$ The discriminant is $$ \Delta = B^2 – 4 A C. $$ We will abbreviate this by $$ \langle A,B,C \rangle. $$ It is primitive if \({\gcd(A,B,C)=1. }\) Standard fact, hard to discover but easy to check: $$ (A x^2 + B x y + C D y^2 ) (C z^2 + B z w + A D w^2 ) = A C X^2 + B X Y + D Y^2,$$ where \({ X = x z – D yw, \; Y = A xw + C yz + B yw. }\) This gives us Dirichlet’s definition of “composition” of quadratic forms of the same discriminant, $$ \langle A,B,CD \rangle \circ \langle C,B,AD \rangle = \langle AC,B,D \rangle. $$ In particular, if this \({D=1,}\) the result represents \({1}\) and is (\({SL_2 \mathbb Z}\)) equivalent to the “principal” form for this discriminant. Oh, duplication or squaring in the group; if \({\gcd(A,B)=1,}\) $$ \langle A,B,AD \rangle^2 = \langle A^2,B,D \rangle. $$ This comes up with positive forms: \({ \langle A,B,C \rangle \circ \langle A,-B,C \rangle = \langle 1,B,AC \rangle }\) is principal, the group identity. Probably should display some \({SL_2 \mathbb Z}\) equivalence rules, these are how we calculate when things are not quite right for Dirichlet’s rule: $$ \langle A,B,C \rangle \cong \langle C,-B,A \rangle, $$ $$ \langle A,B,C \rangle \cong \langle A, B + 2 A, A + B +C \rangle, $$ $$ \langle A,B,C \rangle \cong \langle A, B – 2 A, A – B +C \rangle. $$

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Two points determine a line, three a quadratic — what has that got to do with CDs?

July 24, 2014 by . 5 comments

In this post I describe how simple facts about polynomials are applied in correcting errors, for example scratches on compact disks.  The same technique is used in many other places, e.g. in the 2-dimensional QuickResponse  bar codes.

Elements

The two facts from algebra that we need are:

Theorem 1. A polynomial of degree \(n\) has at most \(n\) zeros.

Theorem 2. If \((x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)\) are \(n\) points on the \(xy\)-plane such that \(x_i\neq x_j\) whenever \(i\neq j\), then there is a unique polynomial \(f(x)\) of degree \(<n\) such that \(f(x_i)=y_i\) for all \(i\).

So, if \(n=2\), we want the polynomial \(f(x)\) to be linear. In that case, the graph \(y=f(x)\) will be the line passing through the points \((x_1,y_1)\) and \((x_2,y_2)\). Similarly, when \(n=3\) we want the polynomial to be (at most) quadratic, and we want its graph to pass through the given three points. Finding the coefficients of such a quadratic is not too arduous an exercise in linear systems of equations. For general \(n\) there is a known formula for the polynomial \(f(x)\) called Lagrange’s interpolation polynomial . The uniqueness of such a polynomial follows from Theorem 1. If \(f_1(x)\) and \(f_2(x)\) were two different polynomials of degree \(<n\) passing through all these \(n\) points, then their difference \(f_1(x)-f_2(x)\) is also of degree \(<n\) and vanishes at all the points \(x_i, i=1,2,\ldots,n\), which is impossible by Theorem 1.

Extending a message using a polynomial

The applications I discuss are about communication. We have two parties, a transmitter and a receiver. The transmitter wants to send a message to the receiver. We assume that they have in advance agreed upon a method of coding the messages to sequences of numbers \(y_1,y_2,\ldots,y_k\) for some natural number \(k\). The simplest way of communicating would be for the transmitter to simply write this list of numbers to a channel that the receiver can later read. The channel could be something like a note that you pass to a classmate or it could be compact disk, where the transmitter just writes the numbers. It could be something fancier like a radio frequency band, or an optical fiber, but we ignore the physical nature of the channel here.

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Zabreiko’s lemma and four fundamental theorems of functional analysis

June 25, 2014 by . 0 comments

There are no doubts that open mapping theorem, closed graph theorem, bounded inverse theorem, uniform boundedness principle are the fundamental theorems of functional analysis. All of them are similar in the sense that they explicitly or implicitly use the Baire category theorem, but in the details they are different. However, all these theorems can be formulated as continuity of certain seminorms defined on a normed spaces.

The proof of the lemma is very similar to the usual proof by Banach-Schauder of the open mapping theorem and is actually stronger than the usual consequences of the Baire category theorem in basic functional analysis. The lemma can be used to easily prove the aforementioned theorems.

I would like to thank the user t.b. for drawing our attention to this lemma and adaptation of the original proof from the general case of linear metric spaces, which he gave as an answer to the question Direct Approach to the Closed Graph Theorem on the main site.

Lemma. (Zabreiko, [1], [2]) Let \(X\) be a Banach space and let \(p: X \to [0,\infty)\) be a seminorm. If for all absolutely convergent series \(\sum_{n=1}^\infty x_n\) in \(X\) we have \[ p\left(\sum_{n=1}^\infty x_n\right) \leq \sum_{n=1}^\infty p(x_n) \in [0,\infty] \] then \(p\) is continuous. That is to say, there exists a constant \(C \geq 0\) such that \(p(x) \leq C\lvert x\rvert _{X}\) for all \(x \in X\).

Proof. Let \(A_n = p^{-1}([0,n])\) and \(F_n = \overline{A_n}\). Note that \(A_n\) and \(F_n\) are symmetric and convex because \(p\) is a seminorm. We have \(X = \bigcup_{n=1}^\infty F_n\) and Baire’s theorem implies that there is \(N\) such that the interior of \(F_N\) is nonempty.

Therefore, there are \(x_0 \in X\) and \(R \gt 0\) such that \(B_R(x_0) \subset F_N\). By symmetry of \(F_N\) we have \(B_{R}(-x_0) = -B_{R}(x_0) \subset F_n\), too. If \(\lvert x\rvert \lt R\) then \(x+x_0 \in B_{R}(x_0)\) and \(x-x_0 \in B_{R}(-x_0)\), so \(x \pm x_0 \in F_{N}\). By convexity of \(F_N\) it follows that \[ x = \frac{1}{2}(x-x_0) + \frac{1}{2}(x+x_0) \in F_N, \] so \(B_R(0) \subset F_N\). But, in fact, we can prove much more: \[ B_{R}(0) \subset A_N \] Suppose \(\lvert x\rvert \lt R\) and choose \(r\) such that \(\lvert x\rvert \lt r \lt R\). Fix \(0 \lt q \lt 1-\frac{r}{R}\), so \(\frac{1}{1-q} \frac{r}{R} \lt 1\). Then \(y = \frac{R}{r}x \in B_{R}(0) \subset F_N = \overline{A_N}\), so there is \(y_{0} \in A_N\) such that \(\lvert y-y_0\rvert \lt qR\), so \(q^{-1}(y-y_0) \in B_R\). Now choose \(y_1 \in A_N\) with \(\lvert q^{-1}(y-y_0) – y_1\rvert \lt q R\), so \(\lvert (y-y_0 – qy_1)\rvert \lt q^2 R\). By induction we obtain a sequence \( (y_k)\subset A_N\) such that \[ \left\lvert y - \sum_{k=0}^n q^k y_k\right\rvert \lt q^n R \quad \text{for all }n \geq 0, \] hence \(y = \sum_{k=0}^\infty q^k y_k\). Observe that by construction \(|y_k| \leq R + qR\) for all \(k\), so the series \(\sum_{k=0}^\infty q^k y_k\) is absolutely convergent. But then the countable subadditivity hypothesis in \(p\) implies that \[ p(y) = p\left(\sum_{k=0}^\infty q^k y_k\right) \leq \sum_{k=0}^\infty q^k p(y_k) \leq \frac{1}{1-q} N \] and thus \(p(x) \leq \frac{r}{R} \frac{1}{1-q} N \lt N\) which means \(x \in A_N\), as we wanted.

Finally, for any \(x \neq 0\), we have with \(\lambda = \frac{R}{\lvert x\rvert (1+\varepsilon)}\) that \(\lambda x \in B_{R}(0) \subset A_N\), so \(p(\lambda x) \leq N\) and thus \(p(x) \leq \frac{N(1+\varepsilon)}{R} \lvert x\rvert \), as desired. \(\blacksquare\)

Now the proof of the fundamental theorems becomes an easy exercise.

  1. The open mapping theorem. Hint: set \(p(y) = \inf_{x\in T^{-1}(y)}\lvert x\rvert \).
  2. The bounded inverse theorem. Hint: set \(p(x) = \lvert T^{-1}(x)\rvert \).
  3. The uniform boundedness principle. Hint: set \(p(x) = \sup_{i\in I}\lvert T_i (x)\rvert \).
  4. The closed graph theorem. Hint: set \(p(x) = \lvert T(x)\rvert \).

Detailed solutions can be found in theorems \(1.6.5\), \(1.6.6\), \(1.6.9\) and \(1.6.11\) in An Introduction to Banach Space Theory Graduate Texts in Mathematics. Robert E. Megginson.


[1] П. П. Забрейко, Об одной теореме для полуаддитивных функционалов, Функциональный анализ и его приложения, 3:1 (1969), 86–88

[2] P. P. Zabreiko, A theorem for semiadditive functionals, Functional analysis and its applications 3 (1), 1969, 70-72)

When do n and 2n have the same decimal digits?

June 11, 2014 by . 3 comments

A recent question on math.stackexchange asks for the smallest positive number \( A \) for which the number \( 2A \) has the same decimal digits in some other order.

Math geeks may immediately realize that \( 142857 \) has this property, because it is the first 6 digits of the decimal expansion of \( \frac 17 \), and the cyclic behavior of the decimal expansion of \( \frac n7 \) is well-known. But is this the minimal solution? It is not. Brute-force enumeration of the solutions quickly reveals that there are 12 solutions of 6 digits each, all permutations of \( 142857 \), and that larger solutions, such as \( 1025874 \) and \( 1257489 \), seem to follow a similar pattern. What is happening here?

Stuck in Dallas-Fort Worth airport last month, I did some work on the problem, and although I wasn’t able to solve it completely, I made significant progress. I found a method that allows one to hand-calculate that there is no solution with fewer than six digits, and to enumerate all the solutions with 6 digits, including the minimal one. I found an explanation for the surprising behavior that solutions tend to be permutations of one another. The short form of the explanation is that there are fairly strict conditions on which sets of digits can appear in a solution of the problem. But once the set of digits is chosen, the conditions on that order of the digits in the solution are fairly lax.

So one typically sees, not only in base 10 but in other bases, that the solutions to this problem fall into a few classes that are all permutations of one another; this is exactly what happens in base 10 where all the 6-digit solutions are permutations of \( 124578 \). As the number of digits is allowed to increase, the strict first set of conditions relaxes a little, and other digit groups can (and do) appear as solutions.

Notation

The property of interest, \( P_R(A) \), is that the numbers \( A \) and \( B=2A \) have exactly the same base-\( R \) digits. We would like to find numbers \( A \) having property \( P_R \) for various \( R \), and we are most interested in \( R=10 \). Suppose \( A \) is an \( n \)-digit numeral having property \( P_R \); let the (base-\( R \)) digits of \( A \) be \( a_{n-1}\ldots a_1a_0 \) and similarly the digits of \( B = 2A \) are \( b_{n-1}\ldots b_1b_0 \). The reader is encouraged to keep in mind the simple example of \( R=8, n=4, A=\mathtt{1042}, B=\mathtt{2104} \) which we will bring up from time to time.

Since the digits of \( B \) and \( A \) are the same, in a different order, we may say that \( b_i = a_{P(i)} \) for some permutation \( P \). In general \( P \) might have more than one cycle, but we will suppose that \( P \) is a single cycle. All the following discussion of \( P \) will apply to the individual cycles of \( P \) in the case that \( P \) is a product of two or more cycles. For our example of \( A=\mathtt{1042}, B=\mathtt{2104} \), we have \( P = (0\,1\,2\,3) \) in cycle notation. We won’t need to worry about the details of \( P \), except to note that \( i, P(i), P(P(i)), \ldots, P^{n-1}(i) \) completely exhaust the indices \( 0. \ldots n-1 \), and that \( P^n(i) = i \) because \( P \) is an \( n \)-cycle.

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Green’s Theorem and Area of Polygons

June 4, 2014 by . 10 comments

A common method used to find the area of a polygon is to break the polygon into smaller shapes of known area. For example, one can separate the polygon below into two triangles and a rectangle:

Figure 1

By breaking this composite shape into smaller ones, the area is at hand: $$\begin{align}A_1 &= bh = 5\cdot 2 = 10 \\ A_2 = A_3 &= \frac{bh}{2} = \frac{2\cdot 1}{2} = 1 \\ A_{total} &= A_1+A_2+A_3 = 12\end{align}$$

Unfortunately, this approach can be difficult for a person to use when they cannot physically (or mentally) see the polygon, such as when a polygon is given as a list of many vertices.

 

Formula

Happily, there is a formula for the area of any simple polygon that only requires knowledge of the coordinates of each vertex. It is as follows: $$A = \sum_{k=0}^{n} \frac{(x_{k+1} + x_k)(y_{k+1}-y_{k})}{2} \tag{1}$$ (Where \({n}\) is the number of vertices, \({(x_k, y_k)}\) is the \({k}\)-th point when labelled in a counter-clockwise manner, and \({(x_{n+1}, y_{n+1}) = (x_0, y_0)}\); that is, the starting vertex is found both at the start and end of the list of vertices.)

It should be noted that the formula is not “symmetric” with respect to the signs of the \({x}\) and \({y}\) coordinates. This can be explained by considering the “negative areas” incurred when adding the signed areas of the triangles with vertices \({(0,0)-(x_k, y_k)-(x_{k+1}, y_{k+1})}\).

In the next sections, I derive this formula using Green’s Theorem, show an example of its use, and provide some applications.

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Welcome to the Math.SE Blog!

June 3, 2014 by . 1 comments

Welcome to the Math StackExchange community blog!

Four years ago today, Dan Dumitru proposed the creation of Math.StackExchange. The site has evolved into a community with hundreds of thousands of questions and answers, and over 30 thousand questions and answers appearing each month.

Now, we have a blog.

This blog provides a way of going beyond the Q&A format to allow exposition and discussion. There might be posts about mathematics, or Math.SE, or a book review, or whatever seems appropriate. Anyone can contribute to the blog. This blog is written and edited by community members, and we are actively soliticing both one-time and regular contributors! So if you have an idea of something you’d like to hear about or read about, or if you would like to contribute, check out the blog chat room and the Blog FAQ thread on meta.

I’m looking forward to see what we make here.

 

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