Theorem about a circle, three chords and a midpoint

Posted by Mathoman, Friday 29 January 2010 at 13:36 - Riddles And Exercises - Tags

Here is a nice exercise in plane geometry. As often in mathematics the statement is quite simple — but the proof is not!

Let $\scr{C}$ be a circle, A,B two distinct points on $\scr{C}$ and M be the midpoint of the chord [AB]. Take two other chords, [PQ] and [SR], that pass through M. Let C (resp. D) be the intersection of [AB] with [PS] (resp. [RQ]).
Prove that M is the midpoint of the chord [CD].

Chords of a circle, midpoint of segments, plane geometry, a theorem in euclidean geometry, butterfly theorem

Suprising! If M is the midpoint of [AB] then also of [CD].

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Christmas Riddle

Posted by Mathoman, Friday 25 December 2009 at 13:57 - Riddles And Exercises - Tags

X-mas is the time to crack nuts. At least that's what normal people do. But mathematicians enjoy cracking very peculiar nuts, like the following one. Show that the equation below is true for every positive integer.

$\sum_{k=0}^n\left$\begin{array}{c}2n+1\\2k+1\end{array}\right$(2k+1)\:=\:2^{2n-1}(2n+1).$

As always in maths, the slogan is short is beautiful, i.e., you should find a solution that is as short and elegant as possible, without using much computations. For this exercise ou may get inspiration from Santa Claus who has to distribute gifts into ChristmasStockings...

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An exercise about groups

Posted by Mathoman, Sunday 20 December 2009 at 12:45 - Riddles And Exercises - Tags

A topological group is a set G with a group structure and a topology such that the binary law

$G \times G \rightarrow G ,\;\; (x,y) \rightarrow xy,$

and the map of the inverse

$G \rightarrow G ,\;\; x \rightarrow x^{-1},$

are continuous. In other words both structures, the algebraic and the topological one, are linked in a natural way. So the following question is of interest.

Question

Are there topological groups which are isomorphic as groups and homeomorphic as topological spaces but not isomorphic as topological groups?

Answer

Yes. The proof goes in three stages:

Let G and H be dense subsets of $\mathbb{R}$ and $f :\, G \rightarrow H$ a monotonic bijection. Then f is a homeomorphism.

Without restriction of generality we can suppose that f is increasing. We prove that it is continuous. Let $x_0\in G$ and $\epsilon>0$ . Since H is dense in $\mathbb{R}$ we have $H\cap\,]f(x_0)-\epsilon,f(x_0)[\,\neq\emptyset$ . Thus there exists $y_1\in H\cap\,]f(x_0)-\epsilon,f(x_0)[\,$ . Similarly there exists $y_2\in H\cap\,]f(x_0),f(x_0)+\epsilon[\,$ . Since f is onto we can write $y_k=f(x_k)$ with $x_k\in G, k=1,2$ . Let $\delta=\min(x_0-x_1,x_2-x_0)$ . Then for all x in G
$\begin{align*}x_0-\delta<x<x_0+\delta \;\;\;\Longrightarrow\;\;\;& f(x_0-\delta)<f(x)<f(x_0+\delta)\\ \Longrightarrow\;\;\;&y_1=f(x_1)\leq f(x)\leq f(x_2)=y_2\\ \Longrightarrow\;\;\;&f(x_0)-\epsilon<f(x)<f(x_0)+\epsilon\,, \end{align*}$
which proves that f is continuous in $x_0$ . The proof of the continuity of the inverse function $f^{-1}$ is the same.
Let G and H be countable and dense subsets of $\mathbb{R}$ . Then they are homeomorphic.

First we write
$\begin{align*} G&=\{x_0,x_1,x_2,\ldots\}\;\;\;\;\;(*)\,,&H&=\{y_0,y_1,y_2,\ldots\}\;\;\;\;(**)\,. \end{align*}$
We will now list G and H in a new way, and The aim is to make a monotonic bijection (then it is automatically a homeomorphism). For this we proceed as follows.
- k=0. We take $x'_0=x_0,\;y'_0=y_0$
- k=1. We take $x'_1=x_1$ . To choose $y'_1$ we look at the order of $x'_0$ and $x_1'$ .
  If $x'_1<x'_0$ then we take as $y'_1$ any element in H which is smaller than $y'_0$ .
  If $x'_1>x'_0$ then we take as $y'_1$ any element in H which is greater than $y'_0$ .
- k=2. We take as $y'_2$ the first element of $H\setminus\{y'_0,y'_1\}$ listed in (**). To choose $x'_2$ we look at the order of $y'_0,y'_1,y'_2$ .
  If $y'_2$ is smaller than $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is smaller than $x'_0$ and $x'_1$ .
  If $y'_2$ is greater than $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is greater than $x'_0$ and $x'_1$ .
  If $y'_2$ is between $y'_0$ and $y'_1$ we take as $x'_2$ any element in G which is between $x'_0$ and $x'_1$ .
- k=3. We take as $x'_3$ the first element of $G\setminus\{x'_0,x'_1,x'_2\}$ listed in (*). To choose $y'_3$ we look at the order of $x'_0,x'_1,x'_2,x'_3$ . There are 24 possible orders for these four numbers.
  If $x'_3<x'_0<x'_1<x'_2$ we take as $y'_3$ any element in H which is smaller than $y'_0,y'_1,y'_2$ .
  If $x'_2<x'_3<x'_0<x'_1$ we take as $y'_3$ any element in H which is bewteen $y'_2$ and $y'_0$ .
  And so forth.
The topological groups $G=\mathbb{Q}+\mathbb{Q}\sqrt2$ and $H=\mathbb{Q}+\mathbb{Q}\sqrt3$ fulfill our wishes.

After what has been said before, G and H are homeomorphic as topological spaces. Clearly they are isomorphic groups. But they are not isomorphic topological groups. In fact, suppose that there exists an isomorphism of topological groups $f :\, G \to H$ . One proves easily by induction that f(n)=nf(1) for every integer n, and then that f(r)=rf(1) for every rational r. Therefore by continuity
$f(\sqrt2)=\sqrt2f(1),\;\;\;\;\;\lightning$
a contradiction in $H=\mathbb{Q}+\mathbb{Q}\sqrt3$ .

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Cargo-cult science training

Posted by Mathoman, Friday 4 December 2009 at 14:59 - Maths And Society - Tags

Here is very interesting article about "modern ideas" of teaching, written by João Magueijo a physics professor at Imperial College London. Unfortunately, his observations are so true. It seems that the same problem, i.e., the rise of bureaucrats and educationalists, occurs not only in the UK but also in Germany and France.

As one commentator puts it so rightly: The whole development is part of a broader crisis of our societies becoming increasingly unable to discern the values on which they rest so comfortably and to protect their vulnerable creators.

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An exercise about the temperature distribution on the earth

Posted by Mathoman, Friday 23 October 2009 at 14:55 - Riddles And Exercises - Tags

Heute mal ein bisschen climatology! The following question, funny and useless, is dedicated to my friend A. Wirth, who has left pure mathematics in order to use his mind for applied sciences like oceanography and meteorology ;-)

Exercise: Suppose that the earth is a perfect ball and that the temperature on the earth's surface is a continuous function. Prove that there are infinitely many pairwise disjoint sets {A,B} with A and B points on the earth surface such that the temperature in A and B is the same and such that the distance between A and B is 1000 km.

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Mixing wine and beer

Posted by Mathoman, Saturday 22 August 2009 at 11:33 - No Maths - Tags

Before summer I wrote a blog entry about mixing wine and beer. french friends told me that such a mixture can't exist, putting beer into wine is a sacrilege... It is okay to drink one after the other, but never, never simultaneously! The order though, wine first or beer first, depends on the contry:

Germany: Wein auf Bier das rat ich dir, Bier auf Wein das lasse sein.
France: Bière sur vin est venin, vin sur bière est belle manière.
England: Beer after wine, and you’ll feel fine, wine after beer and you’ll feel queer.

So I was very suprised when during summer brazilian friends introduced me to espanhola, a drink that they like to mix on the beach and which contains exactly wine and beer! The taste depends on the proportions (mostly more beer than wine), the color looks rather dirty and flakes swim around in the glass...


Those brazilians mix...	...wine with beer!

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Comatrix and adjugate matrix

Posted by Mathoman, Tuesday 14 July 2009 at 13:38 - Riddles And Exercises - Tags

The comatrix com(M) of a n x n-matrix M is the n x n-matrix whose entry in (l,k) is $\small{(-1)^{l+k}}$ times the determinant of the matrix which you get by deleting in M the line l and the row k.
It is the transpose of the comatrix which is of interest to us; it is called adjugate matrix, and it is shown in every lecture on linear algebra that it has the following fundamental property:

$^t\text{com}(M)\:M\;=\;M\:^t\text{com}(M)\;=\;\det(M)\:I\:.$

If you work with matrix coefficients in a ring R, it follows that the matrix M is invertible in the matrix ring precisely if the scalar det(M) is invertible in the ring R. For example over $\small\mathbb{Z}$ the invertible matrices are those whose determinant is 1 oder -1 ist.

Exercise: Prove that com preserves matrix multiplication, i.e.,

com(I) = I et com(MN) = com(M) com(N).

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Dimension of the Commutator Space

Posted by Mathoman, Thursday 9 July 2009 at 14:02 - Riddles And Exercises - Tags

Here is a beautiful exercise from matrix theory.

Let A be a n x n matrix. Show that the dimension of the commutator space of A (i.e., the set of all matrices that commute with A) is at least n.

There is a proof that works on any field. But in the special cases of real or complex numbers there are alternative proofs.

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A sunday child

Posted by Mathoman, Sunday 5 July 2009 at 20:08 - Riddles And Exercises - Tags

Here is a personalised birthday gift for my father, in form of a mathematical riddle.

Today, sunday 5th of july 2009, is my father's birthday. He was born on a sunday of a leap year. What is his age today?

In order to solve that exercise you can use the fact that I am older than twenty-three, that my father was older than twenty-three when he took upon him the responsibility of becoming my father and finally that he isn't hundred years old.

Anyway, happy birthday, Daddy!

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Deserve your dessert!

Posted by Mathoman, Wednesday 1 July 2009 at 11:30 - Riddles And Exercises - Tags

We all know those little riddles that entertain mathematicians at the end of lunch at the cafeteria. Here is one of them:

Connect the following nine points by four straight lines, without lifting your pencil.

°             °             °

°             °             °

°             °             °

Easier said than done. The solution in the video below shows that you should not get contained in your thinking habits...

MathOMan connects 9 points with 4 lines

Smiling Bin is even more witty and takes her revenge by challenging me with the following task, which must be impossible for any specialist on connectivity questions in topology.

Without lifting your pencil draw a circle and its center (but not more).

Here is the video where she shows us her tricky solution:

Bin draws a circle and its midpoint

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An extension to Picard's theorem?

Posted by Mathoman, Monday 15 June 2009 at 17:38 - Conjecture - Tags

At the end of my article Hyperelliptic action integral, Annales de l'institut Fourier 49(1), p.303–331, I state the following conjecture:

Conjecture about a singularity
Let D be the open unit disk in the complex plane and $U_1,U_2,\,\dots\,,U_n$ be an open cover of the puntured disk D*= D\{0}. Suppose on each open set $U_j$ there is an injective holomorphic function $f_j$ such that $df_j=df_k$ on every intersection $U_j\cap U_k$ . Then those differentials glue together to a meromorphic 1-form on D.

It is evident that the 1-form is holomorphic on D*. In the case that its residue at the origin vanishes, the conjecture follows because the singularity can be at most a pole, as can easily be derived from Picard' big theorem (cited below). But I am unable to prove the conjecture for the case of a non-zero residue.
I would be grateful for any proof or counter-example — well, trule spoken, a counter-example would be less welcome since I hope that the conjecture is true. I am guided by my simple geometric intuition on Riemann surfaces...

In 1880 Charles Emile Picard (1856-1941) proved the following theorem.

Picard's Big Theorem
A holomorphic function takes in any neighborhood of an essential singularity every complex value, with at most one exception, infinitely often.

A typical example for Picard's theorem

The function defined by

$\:f(z)=e^{1/z}=\sum_{k=0}^{\infty}\:\frac1{k!z^k}\;$

is holomorphic on $\mathbb{C}\backslash0$ and has an essential singularity in $0$ . Is there a value that is does not take (Picard says "at most one exception")? Yes, since $f(z)\neq0$ for all $z\in\mathbb{C}\backslash0$ , there is an exceptional value and it must be zero; according to Picard's theorem there are, for any complex value $w\neq0$ and any $\epsilon>0$ , infinitely many complex numbers $z$ such that $0<|z|<\epsilon$ and $f(z)=w$ .

Direct compuation of this example

In fact, for the example above the theorem of Picard is not really necessary, because we can also see what happens by direct computation: let $w$ be a non-zero complex value and let $\epsilon>0.$ There existe two real numbers $r>0$ and $\varphi$ such that

$w=re^{i\varphi}.$

For $n \in \mathbb{N}$ let $u_n=\ln r+i(\varphi+2\pi n)$ and $z_n=1/{u_n}.$ Thus we have $\lim_{n\to\infty}z_n=0.$
Hence

$f(z_n)=e^{u_n}=e^{\ln r+i(\varphi+2\pi n)}=re^{i \varphi}=w.$

By taking $n$ big enough we see that $w$ has infinitely many pre-images in the punctured disk $0<\,|z|\,<\epsilon\,.$

A less simple example

Denote by P the set of all prime numbers and consider the following function

$g(z)=\sum_{p \in P}^{}\frac{1}{p!z^p}\;.$

Since there is an essential singularity we can apply Picard's Theorem.
But a direct computation seems impossible...

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Sophisticated rhythms perfectly performed

Posted by Mathoman, Sunday 14 June 2009 at 18:50 - No Maths - Tags

Short musical interlude — or rather a machine interlude since these guys look like tin soldiers and play the most complex rhyhmical patterns with the precision of a swiss watch. The drum group Top Secret Drum Corps from Basel is extremely precise, on a local and on a global level.

Just for comparison a video of the brazilian Batucada or Bateria Batala, where I played the Caixa (snare drum) for some years. We don't have the same precision as the swiss but nevertheless a lot of fun ;-) Viva la Samba in Paris!

At the beginning of the video clip there ere only the repiniques and the caixas, the girls in the first row start join in later with another rhythm on their big drums called surdos.

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Jokes About Mathematics

Posted by Mathoman, Wednesday 10 June 2009 at 09:16 - Fun And Jokes - Tags

Classification

There are three kind of people: those who know how to count and those who don't.
There are two classes of people: those who think that mankind can be divided in two classes and those who doubt it.
There are 10 sorts of people: those who understand the binary notation and those who don't.

How many mathematicians do you need to change a light bulp?
None. Mathematicians can't change light bulps but they can prove that it is feasable.

How many researchers in analysis do you need to change a light bulp?
Three. One for the existence, one for the uniqueness and one to determine the initial conditions.

How many researchers in numerical analysis are necessary to change a light bulp?
3.9967 (after six itérations).

How many constructivists are necessary for the change a light bulp?
Impossible. They don't belive in infinitesimal rotations.

How does a bourbakist change a light bulp?
Light bulp changing is a special case of the more general problem concerning the maintenance of an electric system. In order to establish a lower and an upper bound of the number of persons necessary for such a task, we must check whether the conditions of lemma 2.1 (disponibility of the staff) and the conditions of corollarys 2.3.55 (staff's motivation) are fulfilled. Only in the case where all those conditions are fulfilled we can obtain the result by applying the theorem from section 3.11.23. We recall that the upper bound is to be taken in a measure space equipped with the weak-*-topology.

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Let's multiply!

Posted by Mathoman, Wednesday 3 June 2009 at 15:39 - Maths for everyone - Tags

One should think that elementary school taught us all that is to be known about basic operations on numbers: adding, substracting, multiplying and dviding. But there are still some surpises left! Here are three distinct ways to compute the product of two integers.

The classical multiplication of the perfect pupil

written multiplication of two numbers

The lazy pupil's method (or Ancient egyptian multiplication)

How to use the peasant multiplication method

Instruction:

Method of Karatsuba (published in 1962)

Multiplication with Algorithm of Anatolii Karatsuba, Karatsuba Algorhythmus

Remark
The up-shot of all this is the reduction of our computation to the products of one-digit numbers. On a computer the choice of an efficient algorithm can reduce considerably the time necessary to perform the program — when numbers with several milliards of digits are involved the saved time can be of the order of some days! Computing with such big numbers is not only interesting from a theoretical viewpoint, but has practical applications such as in cryptologie.

Exercices

Why does the lazy man's method work? Both factors play different roles; when you are given two numbers what casting do you propose?
With the help of the Karatsuba's multiplication compute the product 3116 x 1014. Explain in general, why the Karatsuba method works.
When you use the classical multiplication of the perfect pupil to compute the product of two n-digit integers, how many multiplications of one-digit numbers do you need to do?
By re-iteration of Karatsuba's method (separating the factors into, two, fours, eight, etc. parts) one constructs an slgorithm, the socalled Karatsuba-Algorithm. >When you apply it to the product of two n-digit intergers, how many single-digit products do you need? Compare the classical algorithm with the one of Karatsuba and show in particular, that the one of Karatsuba is much faster, being of the order $O\left(n^{1,58}\right)$ .

Solutions
Here are the answers to this exercice in pdf-format.

Last but not least, a video about another multiplication method which produces a beautiful calligraphy — therefore we call it chinese multiplication!

The basic idea of the so-called chinese multiplication is this: a set of n parallel lines intersects another set of m parallel lines in nxm points. So there is nothing mysterious about that so-called chinese multiplication. Note that in the examples they always choose numbers with small digits like 112 or 2131 because with numbers like 89 or 897 there would be too many lines to draw!

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Another bet you can win

Posted by Mathoman, Sunday 24 May 2009 at 12:05 - Riddles And Exercises - Tags

The photos to my recent post on the circumference remind me of another bet about a beer glass. Just ask your fellow drinkers the following question:

There are two glasses, one filled with beer and the other one with the same quantity of wine. One takes a spoonful of the beer and pours it into the wine; then one makes the the same in the other direction, i.e., one transfers a spoonful of that new wine-beer mix back into the glass filled with beer. Now in the first glass the beer is spoiled by a small quantity of wine and in the second glass the wine is spoiled by a small quantity of beer. Which glass is more polluted?

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Categories

Theorem about a circle, three chords and a midpoint

Christmas Riddle

An exercise about groups

Cargo-cult science training

An exercise about the temperature distribution on the earth

Mixing wine and beer

Comatrix and adjugate matrix

Dimension of the Commutator Space

A sunday child

Deserve your dessert!

An extension to Picard's theorem?

Sophisticated rhythms perfectly performed

Jokes About Mathematics

Let's multiply!

Another bet you can win

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