The problem is to prove:

*The intersections of two perpendicular tangents to an ellipse form a circle.*

Of course, this can be computed by Analytic Geometry and I carry this out below. It is no fun, however. E.g., you can find the radius of the circle if the claim is correct, take a point on the circle, compute the two tangents to the ellipse and check that they are perpendicular.

Let us find a more geometrical solution! At first, there is a well known „construction“ of ellipses folding a circular paper. You can do it with actual paper. Cut a circular paper. Mark a point A inside the circle. Then fold the paper, so that the boundary lies on A. I.e., the point P is reflected along the folding line to A. The line is the middle perpendicular of AP. If you do that often enough the folds will outline the green ellipse. There are even videos on Youtube showing this.

For the proof, you need that ellipses are the set of points where the sum of distances to A and M is constant. In our case, it is „obviously“ constantly equal to the radius of the circle. We have the following.

*The set of all middle perpendiculars to a fixed point A and points P on a circle is the set of tangents to an ellipse.*

Now you know how to get two perpendicular tangents. In the following construction, I did just that. The four blue lines are four middle perpendiculars. They are constructed by selecting a point P on the black circle, the line PA and a rectangular to that line. Then the blue lines are the four middle perpendiculars on A and the four intersections of our green lines with the circle.

We have to prove that

- the midpoint Z on AM is the center of the rectangle
- and that the rectangle has the same diameter, independent on the choice of P on the black circle.

Then the corners of the blue rectangle will always be on the same circle.

Now, we stretch the blue rectangle by a factor 2 from the center at A. The point Z will be stretched to the point M then. The blue rectangle will become a rectangle through the intersections as in the following construction.

It is now „quite obvious“ that M is the center of the blue rectangle. Thus Z was the center of the 1/2 times smaller rectangle. This proves our first claim.

For the second claim, we need to show that the length of the diagonal of the large blue rectangle does not depend on the choice of the red point. The diagonal has the length

\(\sqrt{c^2+d^2}\)

by Pythagoras. Now we have another claim.

*The sum of the squares of the lengths of each two perpendicular secants of a fixed circle that meet in a fixed point inside the circle is constant.*

To prove that have a look at the dashed rectangle with diagonal AM. Using this rectangle and Pythagoras you can „easily“ express the diagonal of the blue rectangle by the length of AM and the radius of the circle. This proves the second claim.

The images in this posting have been done with C.a.R. (Compass and Ruler), a Java program developed by the author. It allows beautiful exports of images and the automatic creation of polar sets for sets of lines. That feature was used to compute the ellipse in the second image. The first ellipse was done using the two focal points. C.a.R. has also ellipses defined by 5 points, or even by an equation or a parameterization.

I promised to compute the problem using Analytic Geometry. I am using the Computer Algebra system Maxima via my Euler Math Toolbox for this.

The first method computes two perpendicular tangents to the ellipse with the equation

\(x^2 + c^2 y^2 = 1\)

To find a tangent perpendicular to a vector v, we can maximize the expression

\(v_1 x + v_2 y\)

on the ellipse using the method of Lagrange. If the vector v has norm 1 the value of this maximum will be the distance of the tangent from the origin. The Lagrange equations fot his are

\(v_1 = 2 \lambda x, \quad v_2 = 2 \lambda c^2 y, \quad x^2+c^2 y^2 = 1\)

After solving this, we get

\(v_1 x + v_2 y = 2 \lambda (x^2+c^2y^2) = 2 \lambda\)

Then we do the same with the orthogonal vector, i.e., we maximize

\(-v_2 x + v_1 y\)

We then show that the sum of squares of these two values is constant.

In EMT and Maxima, this is the following code.

>&solve([v1=2*la*x,v2=2*la*c^2*y,x^2+c^2*y^2=1],[x,y,la]); ... > la1 &= la with %[1] 2 2 2 sqrt(v2 + c v1 ) ------------------ 2 c >&solve([-v2=2*la*x,v1=2*la*c^2*y,x^2+c^2*y^2=1],[x,y,la]); ... > la2 &= la with %[1] 2 2 2 sqrt(c v2 + v1 ) ------------------ 2 c >&factor(la1^2+la2^2) 2 2 2 (c + 1) (v2 + v1 ) -------------------- 2 4 c

Thus the circle of intersections has the radius

\(r = \dfrac{\sqrt{1+c^2}}{c} = \sqrt{1 + \dfrac{1}{c^2}}\)

This is confirmed by the special case of tangents parallel to the axes.

There are several other methods. One is to construct the tangents using tangents to a circle. For this, the ellipse needs to be stretched by 1/c into y-direction. It will then become a circle. We need to compute points on the image of our circle with radius r under this mapping, then take the tangents to the mapped ellipse and map back. Below is the plan of such a construction.

The computations are very involved, however.

Another method to find both tangents is the following: We look at all lines through a given point (e.g. determined by their slope) and find the ones that intersect the ellipse only once. The product of the two slopes that solve this problem should be 1. Again, this is a complicated computation.

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The trick is to use a number of states S0, S1, …, Sn, and probabilities P(i,j) to get from state Si to state Sj. S0 is the starting state, and Sn is the state we want to reach, in our case the Yahtzee. For a first attempt, we use the number 6s we have on out 5 dice. Then we have 6 states, S0 to S5. With a bit of combinatorics, we can compute the probabilities P(i,j) as

\(p_{i,j} = p^{j-i} (1-p)^{n-i-(j-i)} \binom{n-i}{j-i}\)

If we compute that with Euler Math Toolbox we get the following matrix P.

>p=1/6; >i=(0:5)'; j=i'; % This is the matrix of probabilities to get from i sixes to j sixes. 0.4018776 0.4018776 0.1607510 0.0321502 0.0032150 0.0001286 0.0000000 0.4822531 0.3858025 0.1157407 0.0154321 0.0007716 0.0000000 0.0000000 0.5787037 0.3472222 0.0694444 0.0046296 0.0000000 0.0000000 0.0000000 0.6944444 0.2777778 0.0277778 0.0000000 0.0000000 0.0000000 0.0000000 0.8333333 0.1666667 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000 1.0000000

This matrix allows us to simulate or compute results about probabilities for 6-Yahtzees. E.g., we can start with no 6 in S0. After one dice throw, the first row of P yields the distribution of the expected states. We can do dice throws by applying P to the right of our distribution vector.

>v=[1,0,0,0,0,0].P 0.4018776 0.4018776 0.1607510 0.0321502 0.0032150 0.0001286 % After two more throws, we get the following. >v.P.P 0.0649055 0.2362559 0.3439886 0.2504237 0.0911542 0.0132721

This tells us that the chance to be in state S5 after three throws is just 1.3%.

How about the average time that we have to wait if we keep throwing the dice until we get a 6-Yahtzee? This can be solved by denoting the waiting time from state Si to S5 by w(i) and observe

\(w_i = p_{i,n} + \sum_{j \ne n} p_{i,j} (1+w_j)\)

where n=5 is the final state. Obviously, w(5)=0. Moreover, the sum of all probabilities in one row is 1. Thus we get

\(w_i – \sum_{j=1}^{n-1} w_j = 1\)

Let us solve this system in Euler Math Toolbox.

>B=P[1:5,1:5]; >w=(id(5)-B)\ones(5,1) 13.0236615 11.9266963 10.5554446 8.7272727 6.0000000

The average waiting time for a 6-Yahtzee is approximately 13 throws. If you already got 4 sixes, the average waiting time for the fifth is indeed 6.

We can interpret this in another way. Observe

\(w = (I-B)^{-1} \cdot 1 = (I+B+B^2+\ldots) \cdot 1\)

The sum converges because the norm of B is clearly less than 1. So we interpret this as a process

\(w_{n+1} = B \cdot (w_n + 1)\)

Suppose a waiting room. We have w(n,i) persons in state Si at time n. Now, one more person arrives at each state and we process the states according to our probabilities. On the long run, the number of persons in state S0 will become the waiting time to get out of the system into the final state.

Unfortunately, the probabilities are not that easy to compute when we ask for a Yahtzee of any kind. E.g., if we have 66432 we will keep the two sixes as 66. But we may throw 66555 and switch to keep 555. There is a recursion for the probability to get from i same dice to j same dice. I computed the following matrix which replaces the matrix P above.

>P 0.0925926 0.6944444 0.1929012 0.0192901 0.0007716 0.0000000 0.5555556 0.3703704 0.0694444 0.0046296 0.0000000 0.0000000 0.6944444 0.2777778 0.0277778 0.0000000 0.0000000 0.0000000 0.8333333 0.1666667 0.0000000 0.0000000 0.0000000 0.0000000 1.0000000

Note that I have only the states S1 to S5 here. With the same method, I get the waiting times.

>B=PM[1:4,1:4]; >w=(id(4)-B)\ones(4,1) 11.0901554 10.4602273 8.7272727 6.0000000

Thus we need only about 11 throws to get any Yahtzee. The probability to get one in three throws is now about 4.6%.

>[1,0,0,0,0].PM.PM.PM 0.0007938 0.2560109 0.4524017 0.2447649 0.0460286

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Ist Mathematik wie das Erkennen von Mustern in scheinbarer Unordnung? Der Gedanke wirkt verlockend. Es ist nicht von der Hand zu weisen, dass beides Beharrlichkeit und Geduld (engl. persistence) erfordert. Er scheint auch zu erklären, warum manche in Mathematik talentierter sind als andere. Sie haben einfach das bessere Auge. Was mir daran gefällt ist, dass es somit möglich erscheint, durch Wiederholung und mit der Zeit Erfolge in Mathematik zu erzielen. Das ist ohne Zweifel ein guter und positiver Gedanke.

Dennoch möchte ich widersprechen. Beharrlichkeit und Geduld sind nämlich für alles Lernen notwendig. Das gilt für das Tennisspielen oder den Spracherwerb ebenso wie für das Trompete Blasen. Damit wird es als Leitfaden speziell für die Mathematik untauglich. Eigentlich ist es insgesamt zu sehr eine Binsenweisheit als dass es nützlich wäre. Die Weisheit von Hollywood-Filmen „You can reach everything you really want!“ ist schon zweifelhaft. Als didaktische Richtschnur für Mathematik hilft sie nicht viel.

In Wahrheit ist Mathematik eine Sammlung von Techniken, mit denen wir versuchen, die Welt zu erklären. Dieses Grundmuster beginnt beim 1,2,3-Zählen und zieht sich bis zur Quantenmechanik. Wir bauen dabei immer auf dem auf, was wir von den Alten übernommen haben. Beharrlichkeit und Geduld, gewürzt mit ein wenig Kreativität, helfen uns, neue Möglichkeiten zu entdecken. Dadurch entwickelt sich die Mathematik fort. Die besten Mathematiker sind die, die das umfassenste Wissen auf ihrem Spezialgebiet mitbringen, gepaart mit dem Drang, dieses Wissen auf Neues anzuwenden und anzupassen. Bloses Warten auf einen Einfall hilft meist überhaupt nichts.

Für den Lehrer bedeutet das, das er Mathematik als Technik darstellen soll, mit der man etwas anfangen kann. Wenn das nicht gelingt, verlieren die Schüler das Interesse. Oder, besser gesagt, das Interesse beschränkt sich auf das in den Tests Benötigte.

Was aber kann man mit der Mathematik anfangen? Hier sind wir bei der entscheidenden Frage für einen guten Unterricht. In einem Mathe-Wettbewerb zu brillieren oder später selber Lehrer zu werden, kann nur eine Teilantwort auf diese Frage sein. Ich habe an anderer Stelle so viel über Mathematik versus Welt geschrieben, dass ich hier nicht darauf eingehe. Aber ein Lehrer, der darauf keine überzeugte Antwort zu geben weiß, wird niemals ein guter Lehrer.

]]>Auf der anderen Seite stehen die altbekannten Bremser. Viele Lehrer und Eltern scheinen sich Sorgen zu machen, dass „Kulturtechniken“ verloren gehen. Als Beispiele werden oft das Rechnen und die Rechtschreibung genannt, gerne auch das handschriftliche Schreiben und die schöne Schrift. In den Chor mischen sich die Anhänger der sprachlichen Ausbildung in Latein und Griechisch, den Grundlagen des christlich-römischen Abendlandes und des logisch, analytischen Denkens. Das Internet mit der vorherrschenden Sprache Englisch und dem Video als Hauptmedium ist das grasse Gegenteil dieses Erziehungsentwurfs.

Auf dieser Basis ist es sehr schwer, über eine Einbindung des „Digitalen“ in die Schule vernünftig zu diskutieren. Ich würde raten, erst einmal verbal und emotional abzurüsten. Weder ist ein Tablett im Unterricht der Untergang des Abendlandes, noch kann ein Unterricht rein über vernetzte Medien gelingen. Wie immer liegt die Wahrheit doch in der Mitte.

Wer Jugendliche im Netz beobachtet, sieht doch, dass sie die Möglichkeiten der neuen Geräte im Wesentlichen zur sozialen Interaktion nutzen. Das Surfen ist eben nicht gleichzusetzen mit dem Fernsehkonsum. Smartphones sind deswegen so beliebt, weil man Nachrichten senden und empfangen kann. Die überwiegende Zeit verbringen Jugendliche in sozialen Netzen wie WhatsApp oder Facebook.

Also ist es offensichtlich, dass ein Smartphone im Unterricht stört. Es ist einfach gleichzusetzen mit dem Schwätzen, einer unvermeidlichen Unsitte, mit der Lehrer seit Urzeiten kämpfen. Die Schüler nutzen ihre Smartphones auf genau diese störende Weise im Unterricht, wenn man es nicht ausdrücklich untersagt. Sie nutzen sie auch als Ablenkung von den Hausaufgaben. Diese Trivialität des guten Lehrens und Lernens wird gerne hergenommen, um Smartphones in Bausch und Bogen zu verdammen. Wir sollten statt dessen dahingehend wirken, dass die dauerende soziale Interaktion über ein Netz da unterbleibt, wo der Kontakt mit dem Gegenüber oder die Konzentration auf eine Sache wichtiger sind.

Die Anhänger des händischen Rechnens und der Beherrschung eines korrekten Deutsch sollten über diese Lernziele durchaus einmal nachdenken. Rechner und Schreiber waren in den ganz alten Zeiten zwar angesehene, aber dennoch rangniedrige Spezialisten. Heute ist der allgemeine Erwerb dieser Fähigkeiten allerdings eine Grundlage für unser modernes Arbeits- und Gemeinwesen, also die Voraussetzung für ein funktionierendes Bürgertum. Ohne Lesen und Schreiben zu beherrschen und mit Zahlen umgehen zu können, ist eine höhrere Ausbildung nicht möglich. Die Ausbildung aller Bürger in diesen Techniken ist also absolut notwendig für die Gesellschaft, in der wir leben.

Nun schreitet aber die Technik voran und auch die Welt als Ganzes. Ich habe selbst als Berufsmathematiker seit 40 Jahren keine schriftliche Division oder Multiplikation mehr durchgeführt. Mein Latein, das ich in neun langen Jahren erworben und verfeinert habe, brauche ich sehr selten und nur für Hobbies, und mein fünfjähriges Altgriechisch habe ich komplett vergessen – bis auf die Buchstaben, weil sie in der Mathematik verwendet werden. Meine Rechtschreibung habe ich im Wesentlichen nach der Schule verfeinert und ich schaue auch gerne mal im Online-Duden nach. Ein wesentlicher Teil der Kommunikation findet in Englisch statt, einer Sprache, die nolens volens (die Eliteerziehung schimmert immer durch) die allgemeine Sprache einer zusammenwachsenden Welt geworden ist. Meine altsprachliche Ausbildung behindert mich im Englischen nur. Übrigens ging sie auch auf Kosten einer naturwissenschaftlichen Bildung. Auch schreibe ich fast nie mit der Hand, und wenn, dann unleserlich. Tafelvorlesungen versuche ich mit Druchbuchstaben zu schreiben. Das geht genauso schnell und ist leserlicher.

Man schüttet natürlich nicht das Kind mit dem Bade aus. Selbstverständlich ist die Grundschule dazu da, die Grundlagen dieser Kulturtechniken zu erwerben. Es geht nicht ohne. Es ist lediglich so, dass wir uns überlegen müssen, was uns wichtiger ist. Wollen wir weiterhin stupide Techniken einüben, die in der Welt „draußen“ später nie benötigt werden? Oder wäre es da nicht besser, die nutzbringende Anwendung der zur Verfügung stehenden Techniken zu lehren? Damit ich nicht falsch verstanden werde, füge ich hinzu, dass für Kunst, Musik und zum Beispiel so etwas Exotisches wie Kaligraphie in der Schule Platz sein muss. Wir müssen auch das wertschätzen lernen. Aber gerade deswegen sollten wir Inhalte aus dem Unterricht entfernen, die nur Zeit verschwenden.

Und sobald man diesen Schritt vom Arbeits-, Schreib- und Rechenknecht hin zu einem selbständig denkenden, suveränen Menschen gemacht hat, wird klar, dass die digitale Umwelt Teil der Schule werden muss.

]]>Assume you are living as a small animal on a flat surface moving just a few meters in each direction. You will, e.g., discover that walking 10 steps into one direction, turning right at a rectangular angle (that is the angle that splits the straight angle into two equal parts), then walking another 10 steps and continuing like this brings you back to were you have been. If you are a really clever animal you will discover the Pythagoras and all of the beauty of plane geometry. Why would you want to do this? The simple answer is that a mathematical model of your world is the only way to describe what you see and predict what you expect to happen. It will be enormously useful. It does not „explain“ your world, nor „is it“ your world. It is just your model of the world, accurate enough to do science that you cannot do in any other way.

As soon as we as men look up and a bit further to the horizon, we discover how ships vanish behind the horizon. So the surface we live on does not stretch straight in all directions when we define „straight“ as the way a ray of light propagates. The surface bends somehow. Doing more accurate observations will yield a more accurate model where we live on a ball instead of a flat surface. With just a little effort we can even come up with a rather precise measurement of the diameter of this ball. And we can now predict that walking along a square as above will not exactly bring us back to our starting point. To test this will take some effort, however, as well as to test the prediction that the angles in a triangle no longer add to 180 degrees if the triangle sides follow the surface. We might even predict that we are not capable of reaching India from Spain with the resources of our sailing ships unless we are lucky to find land in between. It is an aesthetically pleasing and useful „explanation“ of our world, but as we now know it is not accurate. Moreover, we cannot „see“ the ball that we live on and have only indirect proof of that fact, such as the shadow of the earth on the moon among a lot of other phenomena. So it is just another model which helps us describe and predict.

The same happens to the planets, the moon and the sun. The first model with the earth in the center had to be made too complicated to explain what we really see in the sky. It turns out that there is a far more simple way to describe and predict the skies. We only have to put the sun into the center and the planets in circles around them. But wait! More accurate measurements show that this is again not a precise way to describe what we see. Indeed, the planets seem to be moving on ellipses with the sun in one focal point. How can this be? The genius that found a simple „explanation“ valid for centuries was Newton. He gave us a model with an interaction called gravity between bodies that decreases with the inverse square of the distance. Using mathematical tools, he showed that the result was just what we see.

Again, this model had flaws. E.g., the movements of Mercury are not exactly explained. This fact was known for a long time. But it took another update of the model by another genius, Einstein, to explain this effect. We always encounter new phenomena that do not fit with our model. Currently, we have a problem with invisible mass and energy in the large universe, and also with awkward behavior in the tiny things.

The history of this goes on and on. E.g., the special theory of relativity is just a model to explain why we do not measure a different speed of light when we move relative to the light source. The combination of time and space into a four-dimensional mathematical model yields very pleasing formulas to describe and predict this. Without these formulas, modern technology would not work. But I would strongly argue against stating that our world „is“ a four-dimensional space-time. This is just a model that helps us to describe the phenomena we see and predict the outcome of new experiments and observations. One of the most striking predictions was the curvature of light around stars by Einstein. He just found a more accurate way to model the straight lines that light rays follow, and to model our measurement of time.

In conclusion, our mathematical models are useful, but they are not identical to the world. Why then are they so elegant? An explanation for this may be that they smoothen statistical facts that we are incapable of seeing. E.g., we describe air by pressure and flow, or even chaotic turbulence, with simple formulas while in fact air is formed by movements of zillions of individual particles. On a more elementary level, we „neglect“ the resistance of air. But even if we find the one world formula like in the „standard“ model I bet it will only yield an approximation of the world, and we will soon discover that it does not represent the complete truth.

However, I could be wrong with this bet. We might be able to model the world as far as is forever possible to us humans. As useful as such a model might be, it will fail to „explain“ the world on a grand scale. We will have to be contempt with our approximate math models which served us so well over the centuries.

]]>Let me first tell you the finite version. It is about a very nice trick which first seems impossible. Assume you are in a group of friends (i.e. finitely many, of course) with blue and red hats. Each of your friends can see all the other hats, but not his own. Your problem is that you want each member of the group to announce the correct color of his or her hat, one after the other. This is clearly not possible, but you can make it so that each but one announces the correct color. If you want to find a solution to this problem, go ahead. I’ll be waiting.

You certainly found the trick: The first person announces „red“ if the number of red hats he sees is even, and „blue“ if it is odd. You take this first round. At this point, all your friends will know the color of their hat. Neat, isn’t it? Moreover, you get a 50% chance to get your color right.

The video I am discussing generalizes this to an infinite number of friends, countable many. Now you can play the even-odd-trick only if you are in non-standard analysis and have a non-standard infinite natural number at your disposal. All standard friends will then get their hats correctly. But this is using a theory which is even more fantastic than the solution I show you below. If you up for a thought adventure go ahead and find a solution. Hint: You have to use the axiom of choice.

The proposed solution uses an equivalence relation on all sequences in the set {red,blue}. Two such sequences are considered equivalent if they are equal from some point on, i.e., they differ only in a finite number of places. This yields equivalence classes and you agree with your friends on a representative element of each and every equivalence class. Using this, each of your friends can come up with the same sequence of colors, and this sequence will agree with the correct sequence of hat colors from some point on, i.e., differ only at finitely many places. Note that all of you have the same sequence! So you, on number one position, only need to announce if you see an even or an odd number of discrepancies to this sequence. Then everyone will know if his hat color agrees with the color in the sequence or not.

A trained set theorist will immediately spot that the axiom of choice is needed to define a function Phi that maps any sequence to one representative. If this function exists it will compute the same representative even if we change a finite number of colors in the sequence. That is why the trick above works.

However, this is all too absurd for me to be considered of any value. Other mathematicians may disagree and find this to constitute a nice application of set theory. To me, it’s useless garbage. The problem is that there is absolutely no constructive way to compute the above wonder function Phi. To compute it you need to inspect the complete sequence at once. And this is not possible. You cannot get close to Phi(S) by inspecting the first N elements of the sequence S like you can get closer and closer to the square root of 2. After N elements of S, you know nothing at all about the Phi(S) you can choose.

In the video, there is another algorithm where you and your infinite number of friends announce their color at the same time. By announcing the color in the representative sequence they will get „at most a finite number“ of colors wrong. In the finite case, this means they might get all wrong. This is again a useless generalization to the infinite case.

There is another trick if friend N can only see friends with a higher number as if you are standing in line. Again, if the first in the row announces the even or odd number of discrepancies, every following announcement (one after the other) can be made correctly. This does also work in the finite case as you will observe easily.

Finally, here are some interesting videos about infinity.

]]>Teaching at the university level is completely different from teaching in schools. We assume that adults are capable of autonomous, self-guided learning. Thus we teach by „reading“ to them. In math, that is writing on a blackboard. Students are expected to make notes to conserve our wisdom for their studies at home. In math, this means copying every word, formula, and figure from the blackboard, carefully avoiding transmission errors and possibly correcting the errors we make on the blackboard.

*Simple copying is a boring, time-consuming, and inefficient way to learn. *

The old-timers often disagree. They argue that it has a learning effect, especially if it is followed by preparing a careful summary at home, something that only very few students do and only in their favorite subjects. Nowadays, students have to be present about 20 hours a week and cannot be expected to spend the same time to prepare summaries, and then the same amount of time again for solving exercises.

The truth is that most university teachers are too lazy to prepare each lesson separately. So they reproduce the same script year after year. A minority does indeed prepare the lesson mentally with the result that they talk and write at the same time, but with different content. This is a most common mistake with slides. Note:

*When using slides you must read aloud everything that is written on the slide!*

Otherwise, your audience will be reading and not listen to you.

Blackboards do not foster this mistake in the same way as slides, because you have to turn your back to the audience while writing, and it feels strange to talk without eye contact. Some do it nevertheless. But yet you are easily tempted to talk and explain while the students still copy your content from the blackboard. You should always remember that you have just uncovered your writing by stepping aside from the blackboard. The following is very obvious.

*When using the blackboard be silent while the students copy from the blackboard!*

Since we are talking about the current state and the mistakes that are done in teaching we should become aware that the times are changing. Be aware that after the Bologna process the universities changed drastically. When I studied math I took two subjects, such as topology and probability theory. In these subjects, I had to pass a final examination with no grading. The other subjects in my secondary subject I did not take too seriously. They had to wait until I prepared for the final examination, sometimes all at once in one written test.

Nowadays, it is not uncommon to have six subjects at the same time, like complex functions, algebra, numerical math, an introduction to statistics, and two classes in the side subject. All these modules have tests at the end of the term, and all must be passed and are graded. The grades usually go into the final grade of your degree.

*Students have less time for your class than you think.*

Most modules in our university have 5CP with 4 hours of presence. The semester has 30CP, so that adds to 24 hours of presence. You need a bit of time to learn or do exercises. I would say you need at least another 18 hours just to keep up with the subject. So you should not be surprised if students do not spend as much time on your subject as you expect.

We also have classes with 10CP in the first semesters and 6-7 hours of presence. This is much better. But even then this is only 1/3 of the total amount of studying. So students can only spend two full days on your subject, almost one day completely taken by copying your content from the blackboard.

**Towards better Teaching**

The first change would be to have fewer subjects at the same time. There is simply no way a student can learn a lot of different math subjects at the same time in any sustainable way. The student will concentrate on a few of them, maybe one or two. It might be yours that is neglected!

*The number of graded exams should be limited to two or three in each semester.*

There might be one or two more subjects but they should not be graded and the exams should be very easy. And in no way should they count for the final grade.

This concentration of fewer exams can actually be achieved by collecting modules from several semesters into one bigger module in one semester. The problem is often to organize this, i.e., to make room for larger modules. The staff that is responsible for organization finds it easier to squeeze small modules into the available space. We should not let them do this.

*Math takes time and concentration on one subject without too much distraction.*

The next step is not to waste the time of your students. If you write your script onto the blackboard and they have to copy it from there in the class this process is simply a waste of time. Students do not think while they write. If you want to get the feeling for this process go into a research talk and write down everything that appears on the blackboard, or even worse, on the slides. Do that while the speaker is explaining what he just did and try to split your attention to reading, writing and listening. Research talks are not meant to be written down verbally. Some take notes, but nobody writes down all presented content. So why are you expecting your students to do just that?

*Have a script available for the students and explain it in the class!*

By explaining I mean any communication that is not written down by the students. It is about looking, thinking, imagining and if possible discussing. It may include small sketches or mind maps. The important thing is that it is activating the process of reflection on the subject. One of the best ways of doing this is by connecting it to previous knowledge or experience. In any case, these explanations are not meant to be written down and read later. For this, we use the script.

I have done classes without a script and actually am doing one right now. I am completely aware and ashamed of doing so. As a lame excuse, this is not one of the classes I do on a regular basis. Moreover, I try to cope with this defect by carefully waiting until the blackboard content has been copied. Only then I start to explain what I did and encourage the students to think and reflect on what they just saw. Moreover, I give them time to recover while I carefully clear the blackboard. The good side effect is that I do not cover an excessive amount of details. It is still enough since this class is one of the rare modules with 10CP.

*You and the students should know what you expect your students to accomplish.*

Do you expect your students to study every detail of every proof you give? That is exactly what many teachers do. Of course, it is not possible to achieve this completely. Consequently, the process of learning is a frustrating thrive for perfection. The argument behind this ambitious goal is to keep the standards as high as possible. Asking for less sounds like giving in to the „mediocre students of today“. But this attitude does not work and hinders the students from becoming true mathematicians.

*Use the exercises to make your goals as clear as possible!*

You should set clear goals for your students. These should be goals they can reach. So, device exercises that actually are possible to do. Do not underestimate the positive motivation of a goal that has successfully be mastered and the negative impact of a thrive for perfection which is impossible to reach. In fact, it is not possible to completely master a subject in math. Luckily, there will always be some open question in our subject, even on an „elementary“ level.

*Get the students interested *in* the interesting math that is not covered in your class.*

The part of the subject that you have covered in your class should be a good basis, but it is never complete. Make that clear! And make clear that you expect the basics to be mastered, and that those basics are covered in the exercises that you gave. You can add exercises that you do not expect to be solved but by one or two students in your class. You expect the students to understand and try these problems to practice their persistence. These exercises should be clearly marked, however.

In the same way, you might extend your class to subjects which are not covered in the exam. Make clear that you do this because these subjects are interesting. Make sure that they really are. Be aware, that the students will only accept this if they get the feeling that they can actually accomplish the subjects that are part of the exam. Otherwise, they will claim that you are wasting their time with unnecessary stuff which just confuses them.

**The Future**

The future has already started. We now have the internet. It has its good and bad sides. Usually, we tend to completely ignore the net. We argue that its content is not reliable enough. In view of our own background, it is not surprising that we want the students to use books and research papers. But these limits are no longer appropriate. Restricting students to printed material is wishful thinking.

*Be aware of the Internet!*

In the net, you will find academies with thousands of videos about all topics in math. Many of them are actually very well done and enjoyable ways to learn a topic. The idea of the academies or the virtual universities is „self-guided“ studying. You can stop a video or a presentation and repeat it at any time. No doubt do we meet problems. Some videos, even by high-class universities, are actually very bad. But so are the classes, to begin with, and they do not get better by recording them. The purpose of a video in the net can only be to design teaching on a very high level and make that public. It takes effort and time to do this.

There is also material like scripts and exercises publicly available in abundance. Clearly, this material might not always meet the requirements or follow the presentation of the attended class. But students tell me that with a bit of search they always find something that helps them better understand my teaching. I have no quarrels with that.

*Use the net for social interaction!*

This is what young people today are used to. Social media is by far the biggest part of the net, and the most attractive. You can view that as an extension of learning groups or student meetings. Sadly, is it often also a replacement for the actual being-together of people. But for us teachers, it is a chance that we just are beginning to explore. Currently, I use our platform only to present material like exercises, scripts and code snippets, and occasionally links. It is a one-way communication. The students have their own meeting points.

This ends my „manifest“ towards a better teaching of math. This blog does not get many comments. But it may be worth to discuss a little bit about the teaching of math in the comments. You are welcome to do so!

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Assume we have a set of Bridge cards (52 cards with 4 colors from 2 to 10, J, Q, K, Ace). This is dealt to four players, each player getting 13 cards. Answer the following two questions.

- Assume one player gets an ace of any color. What is the probability to get a second ace?
- Assume one player gets the ace of spades. What is the probability to get another ace?

My take on this is the following guideline:

*Device a repeating experiment that mimics your problem!*

In this case, we deal the 52 cards „randomly“ over and over again. We assume that every deal has the same probability. And I assume that you know that there are

\(\displaystyle \binom{52}{13} = \frac{52\cdot\ldots\cdot40}{2\cdot\ldots\cdot13} = 635’013’559’600\)

possible ways to select 13 cards from 52 cards. We have assumed that all of these deals come with the same probability. The frequentistic approach is to assume that we deal each distribution with the same frequency on the long (very long) run. So we can just assume we deal each one exactly once, and get our probability (the expected ratio of „good cases“) by

probability = „number of good cases“ / „number of all cases“

For problem 1, „all cases“ are the cases where there is one ace, and „good cases“ are the cases where there are two or more aces. For problem 2, „all cases“ are the cases with the ace of spaces and „good cases“ the cases with the ace of spades and another ace.

We are then faced with counting, a problem of combinatorics. In this case, we can count the result. In other cases, there is only the option of a Monte-Carlo simulation. Let us do one in EMT.

>function simulate1 (N) ... $ c=1:52; n1=0; n2=0; $ loop 1 to N $ c=shuffle(c); $ k=sum(c[1:13]<=4); $ if k>=1 then $ n1=n1+1; $ if k>=2 then n2=n2+1; endif; $ endif; $ end; $ return n2/n1; $ endfunction >simulate1(1000000) 0.370341025387

EMT may be an easy language to do the programming, but it is not the fastest one. And it is easy only if you know the basic syntax of a matrix language. E.g., the command c[1:13]<=4 returns an array b of 13 zeros and ones (where we simulate the four aces by the numbers 1 to 4). Its element b[i] will be one if and only if c[i]<=4. Summing up we get the number of elements in the first 13 shuffled cards that are smaller or equal to 4. In the code, n1 is then the number of times where one ace showed up and n2 the number times where two showed up.

It is important to note that we discarded all shuffles with no ace in the first 13 cards. It is like a condition in Bayesian reasoning. We are asking for the probability that there are two or more aces under the condition that there is one ace.

The change for the second problem is only in one line. We have to check if a specific ace (say the number 1) is in the array.

>function simulate2 (N) ... $ c=1:52; n1=0; n2=0; $ loop 1 to N $ c=shuffle(c); $ k=sum(c[1:13]<=4); $ if any(c[1:13]==1) then $ n1=n1+1; $ if k>=2 then n2=n2+1; endif; $ endif; $ end; $ return n2/n1; $ endfunction >simulate2(1000000) 0.561442110359

The function any() returns 1 if any element of the argument is non-zero. Since c[1:13]==1 tests for the elements to be equal to 1 (the ace of spaces) we get the right answer.

As you see, the outcome of the experiment is quite different. This is why the problem is interesting.

Now we want to do the combinatorial job of counting and computing the probabilities. We expect our simulations to get „close“ to the „correct“ probabilities. The mathematics behind this is very basic combinatorics. My feeling is that this kind of reasoning should be in the mathematical class of any higher education. If you do not know the result you should learn it.

The number N(k) of deals for player number 1 which contain exactly k aces is as follows.

\(\displaystyle N_k = \binom{4}{k}\binom{48}{13-k}\)

The reasoning behind this is as follows: We have to select k from the 4 aces, and 13-k cards from the 48 cards that are no aces. If we do this we get all possible selections with 13 cards that contain exactly k aces.

Finally, the probability in problem 1 is („good cases / all couting cases“)

\(\displaystyle p_1 = \frac{N_2+N_3+N_4}{N_1+N_2+N_3+N_4} \approx 0.3696\)

as computed by EMT with

>function N(k) := bin(4,k)*bin(48,13-k) >sum(N(2:4))/sum(N(1:4)) 0.369637191337

This is close enough to our Monte-Carlo simulation.

Note that N(2:4) returns the array [N(2), N(3), N(4)] automatically in a matrix language. This works since the function bin() „maps“ to elements of array arguments. The sum of N(0) to N(4) is the number N of all deals. This allows a simplification with a new insight.

\(\displaystyle p_1 = 1 – \frac{N_1}{N-N_0} = 1 – \frac{a_1}{1-a_0}\)

where a(k)=N(k)/N is the probability to get exactly k aces. Of course, 1-a(0) is the probability got the one or more aces. So the right-hand side is what we expect as the probability for our Problem 1. If you are not as frequentistic as I am you may have come up with this formula immediately.

Let us tackle the Problem 2 now. This is even easier. We are now only dealing with 51 cards (all but the ace of spades), and we deal only 12 cards. It is easy to count the number of deals with none of the remaining aces. The answer is

\(\displaystyle p_2 = 1 – \frac{\binom{48}{12}}{\binom{51}{12}} \approx 0.56115\)

This should now be obvious (after some thought), and our simulation was close enough to this.

So the probabilities are indeed different. This seems counter-intuitive at first thought. But let us try to find a good argument.

Let us look at the complementary probabilities 1-p(1) and 1-p(2). In both problems, we then have to divide the number of times we have exactly one ace („bad cases“) over the number of times we have at least one ace („all counting cases“). But both numbers change between Problem 1 and 2. The bad cases in Problem 1 are simply 4 times the bad cases in Problem 2. But the counting cases in Problem 1 are less than 4 times the counting cases of Problem 2. It is not 4 times as likely to have any 3 aces than to have the ace of spades plus any two aces. Since the numerator is smaller, 1-p(1) is larger, and p(1) is consequently smaller.

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Three prisoners A, B and C are told that two of them will come free on the next day. You are prisoner A. You cannot stand the waiting time and ask a guard to tell you the name of one of the prisoners who will not come free, but not your name, of course. The guard says: B will come free. Now, what is the chance that you will come free too?

The impulsive answer is 1/2. After some thought, it is 2/3 again. Or maybe it is still 1/2? There are good arguments for both. And does „probability“ make any sense at all?

The decision about the two lucky prisoners has been made before A asked, hasn’t it? Asking does not change the chance. So it is still 2/3, assuming that the two prisoners are chosen at random with equal probabilities among the three. Right?

If you have decided in favor of 2/3, I change the question a bit to confuse you. Assume, it was 10 prisoners and 9 become free. After you ask the guard, he sets 8 of them free. You are alone with the last one and claim that the chance for both of you to get free is 9/10. Would you think this is a sensible claim?

To confuse a bit more assume that the guard is free to say any name. He says A. Now what is the meaning of you saying that your probability to become free is 2/3?

What makes this problem hard to treat is the notion of „probability“. For me, probabilities make sense only if there is an experiment going on. I am what they call a frequentist. Now, what could be the experiment in this case? Clearly, it is the selection of the two prisoners by an external force. Without further knowledge, A is among the selection with probability 2/3, i.e., in 2/3 of the cases on average.

The important question is if the knowledge that B is among the selected prisoners changes our experiment. We can argue that it does. The options that A/C are selected is no longer possible. We have only two options left, A/B and C/B. So in 1/2 of the possible outcomes of our experiment, A will come free. What this tells us is that the probability for A to come free is 1/2 provided we know that B will come free.

But does this change of our experiment (discarding A/C) really reflect what is happening in the problem? This can only be clarified by studying the exact question in the problem.

In the online journal, the question was formulated carefully: Does it make sense for A to ask? Will he know more after he gets an answer?

And this is not a question we can answer without assumptions. The reason is that it depends on the preference of the guard in the case B/C. If he does not select to say B or C with equal probability A does indeed know more after asking. But if he does A cannot gain any further knowledge. Our chance is still 2/3.

Let me elaborate that. We start our experiment by selecting 2 of the 3 prisoners. A is among the selected with probability 2/3. Now we ask the guard. In the B/C case, we assume the guard says B with probability p. Now a bit of thinking shows that he will say B in 1/3+p/3 of all outcomes of our experiment, and C in 1/3+(1-p)/3.

- Assuming he says B, A will be set free in (1/3)/(1/3+p/3) = 1/(1+p) of these cases. For p=1/2 this is 2/3 as expected. For p=1 it is 1/2.
- Assuming he says C, A will be set free in (1/3)/(1/3+(1-p)/3)=1/(2-p) of these cases. For p=1/2 this is 2/3 again. For p=1 it is 1. Indeed, if the guard always says B in the case B/C, we know for sure that we get free if he says C.

As always, the problem turns out to be more complicated than it appeared at first sight.

For those of you who still think this is rubbish and the probability must be 2/3 to get free because the selection has been made beforehand, I have good news. You are also right!

Let us compute. A comes free in 1/(1+p) of the cases where the guard says A, and he says that in (1+p)/3 of the cases. A does also come free in 1/(2-p) of the cases where the guard says B, and he does so in (2-p)/3 of all cases. If you add all the cases where A gets free you end up with 2/3.

We have just fine tuned our knowledge about the chances if the guard says B or C. In the case p=1/2, the result is quite easy. The guard will say B or C with probability 1/2, and A gets free in 2/3 of the cases, no matter if the guard says B or C.

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I friend gave me a geometric problem which turns out to boil down to the figure above. We have three arbitrary circles C1, C2, C3. Now we construct the three green lines through the two intersection points of each pair of these circles. We get lines g11, g12, g13. These lines intersect in one point. Why?

As you know, the argument for the middle perpendicular lines on the sides of a triangle goes like this: Each middle perpendicular is the set of points which have the same distance to two of the corners. So if we intersect two of them in P then d(P,A)=d(P,B) and d(P,B)=d(P,C), which implies d(P,A)=d(P,C). As usual, d(P,A) denotes the distance from P to A. Thus P is also on the third middle perpendicular. Note that we need that P is on the middle perpendicular on AB *if and only if* d(P,A)=d(P,B).

A similar argument is possible for the angle bisectors. These rays are the set of points with equal distance to two sides. For the heights, such an argument is not available. The standard proof goes by constructing a bigger triangle where the heights are middle perpendiculars. By the way, this proof stops working in Non-Euclidean hyperbolic geometry, where the fact still holds.

Can we make up a proof similar to these proofs for our problem? It turns out that this is indeed possible. The correct value is the following:

\(f(P,C) = d(P,M_C)^2-r_C^2\)

where r(C) is the radius of the circle C, and M(C) is its center. To complete the proof, we only need to show that the line through the intersection of two circles C1 and C2 is the set of all points P such that f(P,C1)=f(P,C2). Then the proof is as easy as the proofs above.

There are several ways to see this.

We could use a result that I call chord-secant-tangent theorem which deals with products of distances of a point on a secant or chord to the circle. But it is possible to get a proof using the Pythagoras only. In the image above we have

\(d(P,Q)^2+d(M,Q)^2 = d(P,M)^2, \quad d(S,Q)^2 + d(M,Q)^2 = r^2\)

Thus

\(f(P,C) = d(P,M)^2 – r^2 = d(P,Q)^2-d(S,Q)^2\)

where C is the circle. Now, if we have two intersection circles, the right-hand side of the equation is the same for both circles, and thus also the left-hand side.

We have seen that f(P,C1)=f(P,C2) for all points on the green line.

But we have to prove the converse too. For this, we observe that D(P,C1)=D(P,C2) implies that P is on a circle around M(C1) and on another circle around M(C2). The two circles meet only in points on the green line.

There is also another way to see that f(P,C1)=f(P,C2) defines the green line. If you work out this equation analytically, you see that it is equivalent to an equation for a line. I leave that to you to check.

Note that there is a second situation where the result does hold too.

In this case, we need f(P,C) for P inside C. It will be negative, but f(P,C1)=f(P,C2) still holds for all points on the line through the intersection, and even if P is on the circles.

There is the following special situation.

It can be seen as a limit case of the previous situation. But it can also be proved by observing that all the tangents have the same length between the intersecting point and the tangent point.

Here is another situation.

The green lines are the sets of points such that f(P,C1)=f(P,C2) for two of the circles. It is quite interesting to construct these lines. I leave that to the reader.

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