To the dynamics lecture

• About the butterfly effect [XKCD cartoon]
• Sarah pointed out a Scientific american article about roots of polynomials which produce fractal type pictures. See also here. From the article: What Christensen and Derbyshire did was plot the roots of entire families of single-variable polynomials, imposing constraints on the polynomials' degrees and coefficients. For esxample polynomials whose degree 6 polynomials with coefficients taking integer values between -4 and 4. This is an interesting story. It does not seem to be related to dynamical systems (yet?).
• A scene from the Office with billiards in a rectangle (Thanks Andrew)
• Donald about Billiards

To the cryptology lecture

BBC wrote that Alan Turing's papers are released. An exhibit. And an Update from April 23. A couple of pages of Pincocks book are on the exhibit page. There are quite a few Caesar Cypher online tools like Briangle. We mentioned the Kasiski examination in class since the example on the white board just gave an example how repeated patterns allowed to reveal the length of the code word. Here are a few slides from a CS course. We have seen that RSA relies on the difficulty to factor large integers. Here are The RSA Challenges. Can you factor those integers?

To the analysis lecture

• Download page to Xaos demonstrated in class.
• As mentioned in the class, the difficulty of this dimension lecture was the log. Below are the rules of the logarithm we have used to understand the formula

      dim(X) = -log(n)/log(r) 
  

  if n pieces are needed to cover the object X with boxes of length r. Some have wondered how to derive this. The intuition is that one needs

      n = (1/r)^dim(X) 
  

  boxes of size r to cover the space. Now solve this for dim(X) and you get the above formula.
  
  
• Here are the rules of log:

  log(exp(x)) = exp(log(x)) = x  log is inverse of the exp, this defines the log.
  log(x y ) = log(x) + log(y)    Multiplication becomes addition. Secret of the slide rule.
  log(1) = 0                     Multiplicative unit 1 goes into the additive unit 0.
  log(x^y) = y log(x)            Logs allow to define general exponentials xy
  log(1/x) = -log(x)             Reciprocals change sign in log coordinates
  

• Sarah, David and Caroline wondered independently how to get fractals of other dimensions than the ones seen. Fractals of arbitrary dimension between 0 and 1 can be obtained by modifying the Cantor set construction. Start with the unit interval Cut out a piece of side length 1-2a so that two intervals of length a remain. Since we need in step m a number of n=2^m intervals of length r=aⁿ to cover the set, the dimension is

    dim(X) = -log(n)/log(r) = -log(2)/log(a)
  

  For a=1/3, this is the dimension log(2)/log(3) of the usual Cantor set. If a goes to 0, then the dimension goes to zero. If a goes to 1/2, then the dimension goes to 1.
  
  
• It is natural to conjecture that all fractals have irrational dimension, but one can design fractals with rational dimension too.
  
  
• Caroline wondered what is the dimension of the Mandelbrot set. It is interesting that the boundary of the Mandelbrot set has dimension 2. The posterchild of a fractal is technically speaking not a fractal. Caroline also wanted to see examples of higher dimensional fractals. It would be an interesting project to find fractals say in 4 dimension and visualize them in space.
  
  
• Andrew noted during class that the dimension of the Koch snowflake is twice the dimension of the Cantor set. This is because in the Cantor set one interval gets replaced by 2 and in the Koch curve, one interval gets replaced by 4.
  
  
• David tried out what happens if the Koch curve starts with a square etc. If one replaces each side with 4 in each step, the dimension still stays log(4)/log(3). We could adjust the length and cut each side into pieces of length a, 1-2a,a then replace the middle one with a equilateral triangle of length 1-2a.
• David wondered wither one could get curves of ANY dimension between 1 and 2 using other rules? Sarah had asked this also in the break and yes, the answer is that you can get any dimension. For the Cantor set, we just have to scale differently. Also in the Koch curve case, we can make the lines longer and get different fractals with different dimension.
  
  
• David you mentioned that the coastline of a state has been determined to have a specific dimension. I'm not sure how you could possibly determine the actual dimension of a natural object like a coastline, when it is not nearly as regular as our mathematically-defined objects where we know for sure which pieces we are taking out, but let's accept that as a given. But if you can do it for a short distance, aren't there different dimensions in different parts of the coastline? Can you add dimensions (if you have a curve of dimension 0.7 and another half of the curve has dimension 0.9, what is the dimension of the whole curve?). This is an experimental thing. Collect data for different r and measure how long the coast is. The growth rate interpolation is the dimension. It is a statistical measure. The dimension in general depends from point to point. One can imagine parts, where the dimension is close to one, and other, rough parts, where the dimension is close to 2.
  
  
• David also wondered, that the dimension of the Cantor set or Koch curve is one at every step of the construction but in the limit jumps to the fractional dimension. Indeed, it is only in the limit that these fractals appear. However, if we finish after say 20 steps and go about to measure the dimension, we would measure the dimension of the limiting object as long as we do not take boxes smaller than 1/3^20. But thats about the size of an atom.
  
  About the length of the Koch curve: In step k, the length of the curve 3 (4/3)^m. The length goes to infinity. It is surprising that the length of the curve goes to infinity even so the region stays in the same place.
• Our lungs can be approximated by a fractal too. The Lung is located in a relatively small volume of our chest, still, the total surface area is the size of a tennis court and 1500 miles of airways. (Notter, Lung surfactants).
• Karen pointed out connections to Fractal music. Indeed, musicians have used inspiration from Fractal geometry to generate music. here here is a link to a page dealing with this.

  To the topology lecture

  David asks: For basic topological shapes, are we assuming solid shapes or hollow shapes? Assuming solid, I wasn't sure how a hollow sphere works (I'm thinking a hollow sphere with a real thickness -- not just an infinitesimal shell) -- is that zero holes? Or one hole, even though from the outside it looks like a sphere? Is it like a donut? I don't see how to transform a hollow sphere into either a solid sphere, or a donut. I guess the first question is what is its Euler characteristic, but then how do you transform it into something else with the same characteristic? Actually, it seems to me that a hollow sphere WITH a hole in it, is just like a sphere, because then you can stretch it into a single solid shape.... Everything works also with three dimensional shapes. There however one has also to look at three dimensional "spaces". The Euler characteristic is then V-E+F-S. Interestingly this is always zero for three dimensional surfaces without boundary. For three dimensional spaces, the Euler characteristic is not enough to distinguish spaces. What matters also is for example the number of holes as you mention. But this is not all. Even that does not suffice to tell when two three dimensional manifolds are topologically equivalent. The Poincare conjecture, quite recently proven just established that every space in which every closed loop can be contracted to a point must be a sphere. The Klein bottle you showed -- can you fully represent it in 4D space? A Moebius strip is really a 2D shape but you need 3D to fully represent it, right? Yes, one can. One possibility is to use color as the forth dimension and if at the intersection points the colors are different then this shows that there is an embedding in four dimensional space. I read a Lewis Carroll story excerpt (from Sylvie and Bruno Concluded,1893, a lesser-known novel for good reason since it isn't all that good on the whole) about a Klein bottle. He called it "Fortunata's purse" because everything that is outside it is also inside it, so whoever owns it has the whole world in their purse. The text is online here: http://sakredchao.tripod.com/fortunatus.htm A character thinks she understands how to construct one, but ends by saying, "I'll sew it up after tea." Although I've known about the 5 platonic solids, I had never seen the duality between Cube/Octahedron, 12/20, and the self-duality. That was neat, and also neat how it fits into Euler's formula. I started reading Euler's Gem yesterday so perhaps that will answer some of my questions. Thats a very nice book. For the non-convex polyhedra, I wasn't sure WHY you don't count all the edges and vertices but instead consider lines to be self-intersecting. Also, if you DO decide to count all the edges and vertices that you can see on the outside, and not consider it to be self-intersecting, then wouldn't it obey the V+E-F=2, since it has no holes? Regular polytopes are assumed to be made of identical polygons so that every vertex looks the same. These Kepler-Poinsot polytopes can not be realized in space without selfintersections. Good pictures and more information is on the Wikipedia page.

  To the probability lecture

  • The riddle: "Dave has two kids. One is a boy who was born at night. What is the probability that the other is a boy" gave a lot of discussion. Amazingly some of the class (like Charles) have intuition about this. I have a lot of problem to see why the fact that "born at night" makes a difference. Here is the case analysis we have done in class
    
    

    	BB	BG	GB	GG
    NN	X	X	X
    ND	X	X
    DN	X		X
    DD

    You see that there are 7 cases and among these 7 cases, there are 3 cases, where both are boys. The probability is 3/7 which is between 1/3 and 1/2.
    
    Sarah made a detailed analysis and generalized the situation where one knows that the boy is born in some time interval. If the day is divided into n pieces, she found that the probability goes to 1/2. In some sense, by narrowing the time interval in which we know one boy was born, we make the event that this applies to both boys less and less likely and we get to a situation where the chance that the second child is a boy becomes 1/2. The information that one boy is born at 7:45:23 for example makes this boy pretty unique. The chance that the second boy also qualifies is practically zero. So, the second child probability of being a boy is close to 1/2.
  • I had assumed in the combination lock problem that the three numbers are all different. It seems that there are combination locks, where numbers can appear several times. It seems that the masterlock has more structure as Andrew has found.
  • David has an interesting variant of the Martingale strategy: "It seems like with infinite time and money you are guaranteed a win even at terrible odds. You could imagine playing at much worse odds (bet on the number 17 instead of red, and keep doubling your bet if you lose). Eventually you will not only win, but win REALLY BIG -- you'll make MUCH more than you lost , at the 36-to-one payoff for a single number. And that is guaranteed."
  • Here are more comments of David: You mentioned the people who went in with cameras and a computer in their shoe. I read their book -- The Eudaemonic Pie by Thomas Bass -- and it was quite fascinating. As I recall, they couldn't be sure exactly where the ball would land, but they were able to narrow it down to 1/4 of the board, and so by placing bets on those 9 or so numbers, at a payoff on each one of 36-to-1, they could make a killing very quickly. I don't think they were arrested, but I'm also not sure they actually dared play enough to win big -- after all, getting people angry who run gambling joints is not always a good idea :). The book suggests that they COULD have won, but didn't. Then again, maybe they didn't tell us the whole story.
  • The Monty Hall problem: this is one that is SO HARD to explain to people. Here is how I have explained it to people who refuse to accept the answer with 3 doors: suppose you have 1,000 doors -- one car and 999 goats. You pick a door, and Monty Hall then shows you goats behind 998 of the other doors, leaving just the one you picked, and one more. More people (still not all) now believe that it is better to switch. One problem people have is that in real life, the actual game was not quite as simple as the probability puzzle version. The real Monty Hall, for instance, did not always ask if you wanted to switch doors, and we have to wonder if he tended to ask "do you want to switch" more often if you had chosen the car than if you had chosen a goat. He knew where the car was, after all. The other way to convince people, if they really don't believe you, is to offer them to play the game for , as long as they're willing to play at least 100 rounds. When they've lost a lot of money maybe they'll believe you. They think they're making an average of per round, really they're losing .67 (on average).
  • I'd heard the "one child is a boy" problem before from Martin Gardner, but never the "boy born at night" part. That is beautiful -- I followed along on paper and got the same result (3/7 = 42%) as you got. But as you said, it seems truly impossible. Now I wonder: how much more information do we need before we get to 50% ? Is it infinite? Suppose we tell you also that he once had chicken pox, and that he has red hair and dimples. Does that bring us closer to 50%? Further away? I am curious. Will need to do some larger tables. Also Sarah has pondered this question and we convinced ourselves afterwards that we can creep like this more and more to 50 percent. Its fascinating and actually an interesting question how much additional information changes the outcome.
  • I also loved the question asked in class (not sure by whom) that if you know the boy was born during the day, then you also get to 3/7. So this is truly a paradox. Either the kid was born at night or during the day, after all. Either way, the probability of two boys is 3/7, but if you DON'T ask the question, then the odds are 1/3. Not possible! But how can there be paradoxes in simple probability? There must be something we're missing. Now that I see this, I think maybe Martin Gardner did raise this issue but I can't put my hands on it. Conditional probability can be counter intuitive. The events are not completely seperate since one boy can be born at night, the other born during the day. So there are experiments in both. On the line-through-a circle paradox, I don't agree with the 1/3 answer because I think the odds of the angle being any particular angle (the second version of 1/3) are necessarily uniformly distributed. But I'm not sure how to explain the difference between the 1/4 and 1/2 answers. Yes, it is probably the most unintuitive models. Each of the models has its justification and if you set up the experiment according to those rules, you get the probability 1/2 or 1/3 or 1/4. What happens "really"? If you throw the line you place its center, spin and role it at the same time, you probably get some mixture of the three, maybe something close to 1/3. It really depends on how the experiment is setup.
• On June 22, 2013, a reader Jeff wrote: Google led me to your blog at http://www.math.harvard.edu/~knill/teaching/mathe320_2011/blog.html . Unfortunately, your answers to the two-child problem ("Dave has two children. One child is a boy. What is the probability that the other child is a girl?") and Gary Foshee's variant (the boy was born at night) are wrong. The answer is 1/2 to both. And no, I am not confusing the problem with "Dave's oldest child is a boy." The 1/2 answer has a firmly established history that few seem to remember. It goes back to Joseph Bertrand in 1889, and includes when Martin Gardner himself retracted the answer of 2/3 (well, of 1/3; he asked the opposite question). First, let me treat the Monty Hall Problem the same way you treated the Two Child Problem. First, introduce the probability space of all possible events S={CGG, GCG, GGC}, indicating the position of the car and the goats, with P[{CGG}] = P[{GCG}] = P[{GGC}] = 1/3. Let MH={CGG, GCG} be the event where there is a goat behind the door Monty Hall opened, and W={GCG} be the event where you will win by switching. The we get: P[W|MH] = P[W&MH]/P[MH] = (1/3)/(2/3) = 1/2. Yet we know this is wrong. The reason it is wrong, is because it addresses the wrong question. It answers the situation where you asked Monty Hall to open door #3, not the one where he picked it himself. The correct probability space uses S'={CGG1, CGG2, GCG3, GGC2}, MH={CGG3, GCG3}, and W={GCG3}; where the number indicates the door Monty Hall opens. Here, P[{CGG1}] = P[{CGG2}] = 1/6 and P[{GCG3}] = P[{GGC2}] = 1/3. And now we get: P[W|MH] = P[W&MH]/P[MH] = (1/3)/(1/6+1/3) = 2/3. Similarly, in the two child problem, you are just as likely to "know," or "be told,", about either the girl or the boy in a GB family. Or a boy born during the day in the variant. If you do that, you will find the answer is always 1/2. These are all a generalization of Bertrand's Box Paradox, which can be characterized by 1) A set of N random objects that 2) Each possess at least one of two symmetric properties, but might have both; and 3) M is smaller than N/2 have just property #1, or just property #2. The rest have both. When you learn that an object has property #1, the first instinct is to say that the chances are (N-2M)/(N-M) that it has both. But you would say the same thing if you had learned that it had property #2. But if both are true, the law of total probability says the probability it has both is (N-2M)/(N-M) even if you don't learn anything. Yet we know that probability is (N-2M)/N. The paradox is resolved by assigning a probability of 1/2 to learning about either property when the object has both. The Two Child Problem is almost identical to Bertrand's original, only changing N=3 to N=4 and leaving M=1. Monty Hall is identical to the original, but only after realizing that the properties are "there is a goat behind Door #2 or #3". Answer: Gardner has used both variants and got it correct. The second answer leading to 1/2 applies only when changing the problem. Here is a paper about it. Version ii) on page 2 of this article is clearly a different problem. I don't think anybody would interpret the problem like that. In the original problem, "Dave has two kids, one of them is a boy. What is the probability that the other is a boy" the answer is 1/3. If the problem is phrased "Dave has two kids, the first is a boy. What is the probability that the other is a boy", the answer is 1/2. Controversies are always interpretation or language problems which actually should be avoided. Maybe it is a communication problem. For me, the Boy-Girl problem asks for the conditional probability of the event {BB} conditioned to the event {BB,BG,GB} in the probability space Omega = {BB,BG,GB,GG} with uniform probability measure {1/4,1/4,1/4,1/4}. If we agree that this is the mathematical reformulation of the problem then the answer is 1/3. If we do not agree, then we look at different problems and have a language clash. One can always change the problem as Gardner discussed in his second article, but then the problem becomes a different one. His modification is NOT a "retraction". Here is an article which addresses "boy born at night problem".

To the set theory lecture

• Not all slides seem have been seen today: Quicktime of slides (webquality only).
• We have seen the Continuum hypothesis stating that there is no cardinality between the cardinality of the integers and the cardinality of the reals. David asked whether there is a simple proof that the integers have the smallest infinite cardinality. To see this, one has to note that if A is a subset of B then the cardinality of A is smaller or equal than the cardinality of B. Now, given a set B, we can start picking elements and so counting the members. If it stops, then we have a finite set and the cardinality is the number of elements. If it does not stop, then we have found a subset A of B which has by definition the cardinality of the integers. Because A is a subset of B also B must have the cardinality of the integers.
• Here is a Youtube video showing an other proof that the rational numbers have the same cardinality than the natural numbers.
• Some comments and questions of David could be interesting to everybody My first question about the lecture is about plus and times in set theory, or specifically + : why is + defined as the union minus the intersection, rather than just the union? I would have assumed + is OR and x is AND. Wouldn't everything work the same if + is defined as just plain union? Commutativity, association, and distributivity all work as far as I can tell, and the empty set is still the "0". It works only with XOR = + , not OR, which is the union. The union is not good for arithmetic. Aha -- maybe the problem is that with just plain union, there is no inverse? Which of course there isn't, since there's no way to get back to the null set once you have anything at all. So maybe I just answered my own question. yes exactly. It is not a group. Second question: in Cantor's proof of counting the rationals (the triangular version you used, or Cantor's first diagonal proof) -- there are a lot of duplicates, e.g. 1/1 and 2/2 are both listed, and 1/2 and 2/4 are both listed. Why is that not a problem with the proof? We're trying to get an exact one-to-one correspondence, but aren't 1/1 and 2/2 the same fraction? Don't we really want only the ones that are in lowest terms? Good point. Yes, one has to jump over the duplicates. Just cross them out when counting. The Ulam spiral nicely takes care of the fractions larger than one and the negatives, but it still doesn't answer the other issue above. Is it okay to just leave them out as you go, for numbers that are not in lowest terms? How do we know for sure that that still works? Maybe you're leaving out an infinite number that has a different cardinality and therefore there's some problem with the proof. yes, also with the Ulam spiral which is an alternative to the triangular picture, one can just jump over each douplicate Another question: I've always been troubled by the idea that as long as there IS a way of showing one-to-one matching, then the two sets are the same size even if there is also a way of matching that leaves some out. With two finite sets, there is NO way of matching them that leaves some out. Is that perhaps the difference between finite numbers and infinite numbers? Is that the DEFINITION of the difference? For finite sets, the cardinality is equal if they have the same number of elements. Before Cantor, it was not clear how to compare cardinality. The second diagonal proof (non-countability of reals), however, I do not have a problem with -- once you can prove that there is NO WAY to put them in one-to-one correspondence, then clearly they can't be the same size. However, I do have a question that I haven't figured out: why can't you use the same argument for the rationals to show that the rationals aren't the same size as the naturals? Just lay out all the rationals between 0 and 1 in decimal form instead of as fractions, and change one digit for each one.... Why does the proof work for reals but not rationals? very good question. The proof still works but it will just produce you an irrational number, a number not in the original list. We know that already, because sqrt(2) is already irrational. There is actually a small problem with the proof for reals, although it's easy enough to get around. What if the digits that you end up with in the diagonal, after the first digit, are all 8s, so that the new number you get by adding one to each digit gives you something like 0.1999999999 which equals 0.2000000 which MIGHT be on the list. But this we can get around by just saying that if the number is an 8 change it to something other than 9. Yes, well observed. There is always the ambiguity with the decimal expansion. This is not a problem for cardinality purposes. Since there are only countably many of them. On the countability of the algebraic numbers: I'll need to see this better. Perhaps we're looking at a collection of diagonals. I see how you can use a diagonal proof to count all the algebraics of any GIVEN degree, but not that this necessarily follows when you have to count an INFINITE number of degrees, even though all the degrees are finite. For the algebraic numbers, it is a bit more complicated. Yes we first count all solutions to linear equations (the rationals) then all quadratic and then all cubic etc. We can count the linear ones and put them in one line, the quadratic ones in one line etc and then use again the Ulam spiral trick to get through all of them. It is a similar challenge like the rationals. I assume that infinite polynomials are NOT countable, right? That's why you can write PI or E as an infinite series? Yes, the set of infinite polynomials (taylor series) are not countable. Indeed with infinite polynomials we can describe any real number as the geometric series 1/(1-a) = 1+a+a^2+a^3 + .... allows to describe any number between 0 and 1. The continuum hypothesis issue is fascinating. I'd heard that, but I still don't understand how there could possibly be such an infinity in between N and R. Yes, we can not have intuition about it because we can go both ways and have in each case a valid mathematical building. Is R (number of reals) the same size as the number called Aleph-one? And is that the same as the number of sets of naturals? Or is that equality part of the question about the continuum hypothesis? Yes, aleph 0 is the usual name given to the cardinality of the integers, and aleph 1 the cardinality of the reals. The set of all subsets of the reals is even larger and has the cardinality alpha 2 If there were an infinity in between N and R, what would it look like exactly? If we're allowed to decide that it exists, how do we define it? I don't know. I don't think anybody knows. It would be a strange construct because there is a mathematics build on top of ZFC in which this is not possible. Maybe I need to find a book that answers all these questions in easy-to-read form. I learned the 2nd diagonal proof in 9th grade, but I can't remember if we discussed all these ramifications, and then in college we never looked at any of this. If you know a good book please let me know. This is all fascinating. Poor Cantor.... Goedel Escher Bach does a goo job about paradoxa. Also the Book by Eli Maor called "To infinity and Beyond". Your comment about the theorems that make mathematicians go crazy reminded me of a story "The Devil and Simon Flagg" and actually appears in Clifton Fadiman's other collection of mathematical stories, entitled "Fantasia Mathematica". The story is available online at numerous sources, including here. Its interesting that before something is proven there is a remote possibilty that the statement is true but that it is not provable. That could be the case for Goldbach. This is part of the story in the novel of "Uncle Pedros and the Goldbach conjecture". The Epimenides paradox is not in fact a paradox as originally stated. If he'd said "I always lie" that would be a paradox, but by saying "The Cretans are always liars", all we need to solve this logically is to conclude that there is at least one Cretan who is NOT a liar, but that Epimenides IS. I agree this is more precicely done with "I always lie" On the Russell paradox: I understand it, but I've been troubled by this idea of a set that contains itself. Can you give me an example of a set that contains itself? How would you write it down? Here's the empty set: {}. And here's the set of even numbers: {2, 4, 6 ...} and here's the set containing 1, 3, and 12: {1, 3, 12}, and here's the set containing those three sets: { {} {2, 4, 6, ...}, {1, 3, 12} }. And here's the set containing the above set-of-three-sets: { { {} {2, 4, 6, ...}, {1, 3, 12} } } But none of those sets contain themselves. Can you write down a set that contains itself? This is exactly the problem. But the point is that set theory should be able to describe all sets. You could imagine define A={A} without specifying what A is. We can talk about such an object. The problem with the Russell set is that we can not even talk about this because if it exists, then it does not exist and if it does not exist, then it exists. The set A={A} might just not exist. That would be fine for mathematics. On the "no surprise exam" paradox -- isn't it the case that once I've convinced myself, using the induction proof, that there cannot be a surprise exam, you could still give a surprise exam ANYTIME (even on May 3rd, the last day) and it would be a surprise, since I'd already decided it couldn't happen! The fact that there are two reasonings makes it a paradox. One gives a surprise exam, the other does not. Does the paradox mean that we can not do surprise exams ? ...

To the Calculus lecture

• During the lecture, the question came up, when the concept of function was introduced. Caroline reminded me about that and sent the link which tells that it was Johann Bernoulli in a letter to Leibniz in 1694 who gave a first definition and Leonard Euler who clarified. Galileo already had a good understanding of the concept but it needed Descartes to link the geometric and algebraic picture.
  
  
• Several of you have asked about the "Calculus in 20 Minutes" source. Here is a Youtube version
  
• About the 3D printing, see 3d printing project by Elizabeth Slavkovsky.
  
  
• I did forget to bring up the example of the function f(x) = 2^x in class. It has the property that

   D f (x) = f(x+1)-f(x) = 2x+1 - 2x = 2x
  

  It is the exponential function if the Planck constant h is 1. For general h, the exponential function is (1+h)^x/h which has the property that D f = f for

   D f (x) = [f(x+h)-f(x)]/h
  

  For h going to zero, we get to the classical exponential function e^x which has the property that d/dx e^x = e^x.
• What we did in class is calculus with Planck constant 1. This is calculus which could have been developed by the Greeks or even by the Egyptians if somebody would have come up with the concept of function. This calculus is not unfamiliar to us. We all the time work with this, for example, when balancing our checkbooks. In economics, one deals with total cost F and marginal cost f functions which are functions of a quantity q which is of course an integer. The marginal cost f(q) is the derivative of the total cost F(q) and the total cost F(q) is the anti derivative of marginal cost f(q). Nobody cares so much about the fact that one can not buy 4.2323 shoes. The quantity q is always an integer and f(q) = F(q+1)-F(q) as a derivative does very well.
• At the end of the lecture, I mentioned the Poincare sums which run against intuition. David ran these sums in Excel Screenshot. It shows how overflows occur first for base 1000 before the convergence is noticed. With computer algebra systems, this happens also. The point of Poincare is that an experimenter dealing with such data would judge the sum to diverge evenso it does converge and that mathematical statements are therefore often difficult to find experimentally.
  
  
• David also looked at the dominos. How many dominos do you need to get "1 full domino length off the table with 4 dominoes, 2 lengths with 11, and 3 lengths with 31. Is there any sort of pattern to these numbers? We've got 4, 11, 31, 83, and then I found 227 and 616. They do get very big, as you said, very quickly. I couldn't find a pattern other than "vaguely exponential". a search on the Integer sequence encylopedia tells about it.
• Isaac wondered about how one can define 0/0 or infinity/infinity. This is the core question of calculus. I bypassed such questions by looking at calculus without limits. Classically, the derivative for example is defined as (f(x+h)-f(x))/h which is "0/0" for h going to zero. Calculus is all about making sense about this. Isaac also wondered about how to extend algebraic operations to a larger set so that we can for example compute with infinity. This is often done, like in measure theory. One can not define 0*infinity but one can define infinity * infinity = infinity infinity * (-infinity) = - infinity, infinity + infinity = infinity or -infinity - infinity = -infinity. But both infinity/infinity or 0/0 or infinity - infinity can not be defined. Mathematicians can deal with this using the concept of compactification. In complex analysis also one adds a point at infinity to get the Riemann sphere and one can do geometry with this very well. One adds some additional point at infinity. This works topologically but algebraically one has problems to exend the structure to this larger structure. It can be done using nonstandard analysis, where one also computes with infinitely large or infinitely small numbers. But there, one does not add one point infinity but extends things to a larger framework.

To the Algebra lecture

(1,2,3)    (1,3,2)     (2,1,3)    (2,3,1)     (3,1,2)     (3,2,1) 

To the Number Theory lecture

If you look at the p'th row of the triangle and you look at all the entries different from the first and last then they are all divisible by p if p is prime

                          1
                      1       1
                   1      2      1                  <-   Prime
               1      3       3     1               <-   Prime
            1      4      6      4      1
        1      5      10     10     5       1       <-   Prime
    1       6     15      20     15     6      1
1       7     21      35     35      21     7      1  <- Prime

You see for example that 7,21,35 are all divisible by 7. Thats what we prove When doing induction for Fermat's theorem we need exactly this property.

When you are doing prime triplets, how do you prove one of n, n+2, n+4 = is a multiple of 3? How do you do it mathematically in detail? If you look at the reminder of n when dividing by 3, there are three possibilities the reminder can be 0, 1 or 2. In each of the cases we can show that one of the three is divisible by 3: We write n = 3k+1 for example if the reminder is 1:

            n               n+2               n+4
         ---------------------------------------------
         n = 3k + 0     n+2 = 3k + 2      n+4 = 3k + 4          n is divisible by 3
         n = 3k + 1     n+2 = 3k + 3      n+6 = 3k + 5          n+2 is divisible by 3
         n = 3k + 2     n+2 = 3k + 4      n+6 = 3k + 6          n+4 is divisible by 3

Why did people find perfect numbers? What do we do with perfect numbers? Perfect numbers are a way to find large primes. It is a drive for prime records. Every time a new largest prime is found, these are headlines. Like here Originally they were thought to have magic properties. It was more a spiritual quest at first, but once a mathematical problem is established, it becomes a quest, like a holy grail. The quest for odd perfect numbers is especially appealing because it is the oldest theorem. There are so many theorems... to me, people were playing around with the numbers like primes, perfect numbers, etc. Like Wilson's or Fermat's, etc., what do we do with them? Many mathematical quests are first just trying to understand things. It is not that any applications were in mind. If you have a beautiful problem like whether every even number can be written as a sum of two primes, you wonder whether you can prove this. There is no a priory application in mind. Wilson's theorem is beautiful because it allows to define primes in a single formula n divides (n-1)! + 1 if and only if n is prime The original definition was only telling what a prime is not. It is not divisible by any number between 2 and n-1. These are lots of conditions. The Wilson formula is a single formula. When do you use them? It turns out however that many of these questions have applications. Fermat's theorem is essential today for example for secure communications. We will see this. One can use it for example to exchange secrete keys over a public channel so that both parties have a secure key but even if somebody was listening to our conversations, they could not figure out the key. I am wondering if there might be much simpler way to represent all numbers There might be ways to understand numbers better than we do and one can ask whether numbers could be decomposed differently/ One has indeed done that. There are other kind of numbers which also can be decomposed like integers. Examples are Gaussian integers n+i m. With these numbers we can factor 3 for example because 3 = (2+i) (2-i). There are now new prime numbers in the complex plane but they are also not so easy to understand. The structure of primes will occupy mathematicians for a long time to come. It is not impossible that your question will have a surprising answer and that new structures will be implemented which allow to understand primes better. Huge machineries in analysis and geometry were built exactly for that. Many of these theories have applications way beyond where they were originally designed for. They are used to study quantum mechanics or differential equations which help us to predict weather. Prime numbers are used also directly, like for designing music halls in which the sound reaches every seat well.

---

Fermat's theorem and Wilson's theorem which we proved in class are both a bit tough. Take this week and try to dig through the details.

To appreciate it, look at some examples, like (5-1)! + 1 = 25 which has 5 as a factor. The theorem assures that 5 is prime. The numbers get pretty quickly fast. For p=13, already 12! + 1 = 479001601 and this is indeed divisible by 13. For the prime p=101, we have (p-1)! + 1 =

93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000001 and this 155 digit number is divisible by 101.

Wilson needs a bit of modular arithmetic. This means that one can compute with numbers 0,1,...,p-1 for some prime p and keep the numbers always between 0 and p-1. To compute modulo 5 for example, you only use the numbers 0,1,2,3,4 and get 4*3 = 2 because 12 leaves remainder 2 modulo 5. We can of course also add numbers like 4+4 and get 4+4=3 because 8 leaves remainder 3 modulo 5. Wilson's theorem uses the fact that if p is a prime, then any number different from 0 modulo p has a multiplicative inverse. For example, for p=7 then the inverse of 3 is 5 because 3*5 = 15 leaves rest 1 modulo 7.

Sarah has written down a proof of Wilson's theorem which does not need modular arithmetic. The proof essentially introduces modular arithmetic. Sarah also noticed that to show that if n is not prime, then to show that (n-1)! + 1 is not divisible by n requires that one looks also at the case when n=p*p is a square of a prime. In that case (n-1)! is not divisible by n. An example is n=4, where 3!=6 is not divisible by 4. But it is divisible by p and (n-1)! + 1 is then not divisible by p.

Andy H. mailed me an other interesting example: I have a number theory question for you. I have a colleague with a math = tee shirt that says "(pi^4+pi^5)^(1/6)=e. Evaluated this on the calculator and it came out to be true. which shows how calculators can deceive. Calculators are rather poor in handling math. A computer algebra system can compute things with arbitrary precision. Here is the result when computed with Mathematica to 100 digits:

N[(Pi^4 + Pi^5)^(1/6) - E, 100]
1.98471299777560852129322426652 
  20179954506028251710117327842 
  58163490452575964421920645037 
  621054740491*10^-8

The T-shirt identity draws on an other open problem in mathematics. Are the numbers Pi and E independent or is there a relation like Pi^7 E + Pi^6 E^4 + 4 Pi^2 E^9 = 0 which relates Pi and E in a nontrivial manner. Nobody knows.

Is the above almost identity an accident? Probably not. Sometimes there is more behind it. The number E^Sqrt[163] for example is close to an integer and there is a deeper explanation.

David investigated a bit more (with Excel), how frequent the cases are when the product of the first n primes is prime. Such primes are called primordial primes and the product of the first n prime numbers + 1 is called an Euclidean number. The following Mathematica line fills the primordial soup with such primes:

f[n_]:=PrimeQ[Product[Prime[k],{k,n}]+1]; Do[If[f[n],Print[n]],{n,10000}]

Here are all the cases up to 1000. The number is the n such that one plus the product of the first n primes is prime. For n=11 for example, the corresponding primordial prime is 200560490131.

1, 2, 3, 4, 5, 11, 75, 171, 172, 384, 457, 616, 643

For n=643, the prime is already large:

   150348150290083016397381129594893703474294060812823323460778638118649
667830231021456149860472509753329993625194405832165201754015373145785738
484951300769477501812929664685951618603783939796576638369967645283890965
257033525155199587708268045255052279363164920705828657323055979080746273
816028189098637380009424660997874395727065489836186628602623157337049481
248084067637176116235435203878971680598354075902330994265716081713724415
624768925046953663928525767356112150049548223312748032606851993200998042
995693505067462075271476632558683958977265873783047161234475239368099923
730996246872116013695997876976367964374052326197366654103960179797967343
795924610242312443572053469423721404985976621151520064401443640780488933
285556819901161862141199661599612561481679780183115233854648647747676982
148573307336160041499871615231791393641163305888155299303774255749213483
011799066748401744991925888484009804555171687775230013160104461516559395
729428073070275554850824099436184943421086326390677506469173559416771722
937173087275925785425885880671986752656709920594960347151091186595002505
219365482840762556211182732682217051428437598545585616418879292099861682
265732687209536453164270128814350173581304490644992936613633262857757188
562151642977647882764188254772406986097195766887899737873588943640418368
958276958062186010297611776224825402071384479985732823690835127893085115
563494144718629431879182423657254037880163595626079784073060193750365597
290096065848503315291995087803020754851202857569226069432869448816953936
270850796184547260561232901388077711551321356846596017003423587895067316
496768866754655215462534394457087690336325746567977840087941942039366119
079164514432713364850840284009331217726780951939816070716664173582997775
268701460580956420913827907942919206446846789619609363594046660887724613
660905104278049886175614346791112803120362788263346692528508442686573240
645866599716140007783710807291680896182641010865886985391229679367410552
251510264953577227232440637833551790682487055989061498534869384795359845
2500838987057002293989891

About the Pythagorean Triples, Sarah H has found a geometric derivation of the formulas that x=2t m,y=(m²-t², z=(m²+t2) produces all Pythagorean triples. For example 8^2+15^2=17^2 is interpreted as adding (2*15+1) + (2*16+1) pebbles outside as square of 15^2 pebbles to get a square of 17^2 pebbles. This could be the path, the Babylonians took to get the formula.

To the geometry lecture

As several of you have asked for the source, the Mathematica manipulation files are now on the lab page. I will post later a Mathematica Demonstration version which can be used without having Mathematica installed. But note that you can download mathematica and use it as a student of this course.

David writes: Since you used Thales and Pythagoras to prove Hippocrates, I am now thinking of it as the formula "Thales + Pythagoras = Hippocrates" I realize that is a silly idea mathematically, but it might help me remember it all. It would be interesting to know whether Hippocrates statement taken as an assumption would imply both Pythagoras or Thales. David I can see why Morley's result is called a miracle. I have trouble believing it. Why would an equilateral come out of the chaos of ANY possible triangle? Neat. I will look that one up later. I'm also wondering if you get any similarly interesting results by trisecting the angles of a quadrilateral or pentagon or n-gon. No other similar result seems to be known for quadrilaterals. One could wonder whether there is a version for tetrahedra. The theorem is nice and so it is not surprising that it can also fascinate some of the best contemporary mathematicians today. The most elegant proof is by John Conway (the one who invented among other things surreal numbers). Alain Connes, who works on fancy noncommutative geometry and who has the fields medal in mathematics, has given a new proof in 1998. During class, we wondered about Eves proof of the Pythagorean theorem. It had not been quite clear why the parallel. Right after class, Sarah has drawn the crucial picture on the white board Omar has sent me a similar picture. Here is the page of Eves book:

To the arithmetic lecture

Caroline I agree that some of the hardest concepts to teach in math are the simplest ones. I teach an arithmetic class to college students and they still have a hard time understanding negative numbers and fractions. Many of them memorize the rules to perform the operations but lack a deeper understanding of the real number system and its properties. Yes, even the brightest college students and even mathematicians can be at war with fractions. In many of the proofs that sqrt(2) is not rational, we assume that sqrt(2) = p/q and p are q are relatively prime. In class, you did not make this assumption that p and q are relatively prime but the proof still works by looking at the exponent of 2 (odd/even) on both sides of the equation. I think the proof you showed in class is a lot simpler and can be easily understood by students. I did not make that assumption that p,q are relatively prime. It is not needed. We just have to look whether we have an even or odd number of factors 2. Yes Traditionally, the proof is done by first factoring out common factors. I feel the proof is simpler without. Can you give an example of a system where the prime factorization of a number is not unique? I have been thinking of this since the lecture and seems to be an interesting topic. The simplest example are the numbers x+sqrt(5) y where x,y are integers. These numbers can be multiplied and added and have again this form. The number 6 now has two different factorizations 4 = 2 * 2 and 4 = (sqrt(5) -1) (sqrt(5) + 1) The geometric proof of sqrt(2) was great! Yes it is beautiful and does not need the factorization. It has probably been known to the Greeks. Could have been deadly to know this proof. About 1.99999...=2 . I know that many students will hesitate to say this is true. However, when you ask them if .33333...= 1/3, they will immediately say yes! Thats interesting. Because this might be a bridge to convince somebody. The identity 0.33333 = 1/3 can be multiplied by 3 to get 0.9999999 = 1. If the first is accepted, the second should be accepted too. David: How can you tell, in the clay tablet number system, where one digit ends and the next one begins. How would one represent 60? It needs a zero as a place holder. The Egyptian system doesn't have this same problem because they don't really have place value. I'm not sure about that. Spacing must be crucial. Maybe its also context. I also do not see other marks. Sumerian math does not have zero yet. Spacing, context or some other hints must have been enough.

I thought I had read in the Britannica History that Egyptians had, as well as all the unit fractions, also a fraction for 2/3. Unless that was a different civilization. No, you are right. There were other abbreviations used. See also the Wikipedia page My source is Boyer. I have posted some pages from this book here. I didn't know the Mayans had a zero. Neat! Did they use it the same way the Arabic zero was used, as a placeholder anywhere in the place value system? Or was it used in a much more limited fashion? Could it be used as a number all by itself? I know that some systems that used a zero didn't recognize it as an actual number, but ONLY as a placeholder. It think it is only as a place holder in order to be able to represent dates. From This article: "Maya zero is predominantly used in stone inscriptions and screenfold paper manuscripts as a coefficient of Long Count quantities and in counts of the phases, movements, and cycles of celestial bodies including the moon, sun, and Venus, and their various numerical relationships to one another." I used those extensively and still have my set somewhere in the basement. I didn't know anyone else had ever heard of them, though. Yes, Cuisenaire is no more so popular. But I saw it in Montessori schools. I myself went to a public school, had excellent math teachers from primary to high school and am very grateful. I think my first grade teacher introduced the Cuisenaire material on his own initiative. I used Cuisenaire material in high school to explore problems in additive number theory and was even able to discover the pentagonal number theorem with them and was of course disappointed to learn that Euler "did it already".

You stated that 3 + 2, if you add them together, you get 5, and that there is no mystery about 3 + 2 = 2+ 3, the way there is with 3x5=5x3. I'm not sure I agree that the first one is any less mysterious than the second one. It is certainly not obvious to children that the order of addition is immaterial. Indeed, as you talked about how px is not necessarily equal to xp in physics, I was able to think of various real-life situations in which addition would not be the same in both orders either. Transformations like rotations are other examples which do not not commute. In arithmetic, one can also question the commutativity of addition. Its possible to start building up arithmetic on a addition system, where we have no commutativity in addition, but its rather strange and its difficult to bring in multiplication. An important structure with addition and multiplication a "ring" like the integers or polynomials. Even non-associative rings (where a*(b*c) is not equal (a*b)*c like octonions have commutativity in addition. One could for example explore sets X with two noncommutative operations (+,*) such that (X,+,0) and (X,*,1) are noncommutative groups and a*(b+c) = a*b + a*c (b+c)*a = b*a+c*a. Finding such objects is a combinatorial problem if X is a finite set. I don't see any example and do not know whether this combinatorial problem has been studied. We have barely scratched the surface in exploring mathematical structures. I've always loved the arithmetic proof that sqrt(2) is irrational, which I've seen before, but I had never seen the geometrical proof -- thanks! That is also really beautiful. Also I had never seen (or even considered) the proof that the logarithms are irrational, until I saw it in the Britannica book. The proof is so simple! After reading the section in the Britannica I had figured out how to generalize the arithmetic proof to other square-root irrationals. I'm not sure if the geometric proof can be generalized quite as cleanly, though. I'll have to think about that. Also hadn't thought about how to expand it to cube roots but now I see that it really is basically the same, and also explains clearly why the proof of irrationality fails for sqrt(4) or cube-root(27). The geometric proof about the irrationality is from on of Eves book. I do not know a proof for other square roots but it looks reasonable. Would be a nice project on its own. I have never seen it. You say that "little kids" will say that 1.99999... is smaller than 2.0000.... . I wonder if you may be overestimating the average mathematical understanding of older kids and adults. I have asked older kids, adults, my parents.... Many of them are not at all clear on why 1.999... should be equal to 2. Some of them I can convince, but others are never convinced. Would be nice to quantify and make a test like Piaget did and see. The concept of limit is not easy to grasp indeed. You said that algebraic numbers are a "small set" of numbers. Are they countable, equal in size to the rationals? Maybe I should ask you not to tell me yet. I studied sizes of infinity at some point in high school algebra and we looked at Cantor's first and second diagonal proofs to show that rationals are countable and reals are not, but we never looked at the algebraic numbers separately from the reals in general. We will look at this. Yes, algebraic numbers are countable and so a much smaller set than real numbers. This is one of the triumphs of the set theory of Cantor. It is not so easy to show that a particular number like pi is transcendental (not a solution to any polynomial with integer coefficients.)

To the first lecture

Some interesting thoughts were mentioned during class and it turned out that the garden theorem should be clarified a bit more. What is a "flower"? Also interest sparked the "Zeno's paradox", "Euler's formula" V-E+F=2 and "Grandi's series" 1-1+1-1+1-1+..... We will some of it later in the course.

• Sarah: In class last night we briefly discussed one of Zeno's paradoxes of motion (Achilles and the tortoise). When we looked at the paradox you mentioned that the modern concept of the limit has allowed us to solve the paradox. This happens to be a subject that I have read a bit about, and I am under the impression that modern thinkers do not believe that the concept of the limit solves the paradox. As I understand it, the concept of the limit is supposed to solve the paradox by showing (formally) that the sum of an infinite number of terms can be finite, and that the infinite number of distances (or times) in the paradox is just such a sum. If the Zeno's argument were that any sum of infinite terms must itself be infinite, then the formal concept of the limit would indeed solve the two of Zeno's four paradoxes of motion (the dichotomy and the Achilles-tortoise). However, it seems that that was not quite the thrust of Zeno's argument. One can formulate the problem as one of completing an infinite number of tasks. In this interpretation, the issue is that there is no way to complete an infinite number of tasks because there is no final task (i.e. there is no move that Achilles makes that has him reach the tortoise). Thus, on this interpretation, the discussion of limits is beside the point. Reasons for reading Zeno's paradox in this alternative way come from Aristotle's discussion of the paradoxes in his book *Physics *(which has the only surviving account of the paradoxes from ancient Greece). Anyway, I think Zeno's paradoxes are fascinating by themselves, and all the more so because many consider them to remain unsolved. This is a very good analysis. The notion of limit is still difficult today. We see the difficulty in the classroom today and I wanted to make the point that the difficulty of the pioneers in mathematics developing a formalism had a difficult time too. Limits are difficult because it involves the "infinite". And infinity is a concept difficult to grasp. Today we know that if we add up

  1 + 1/2 + 1/4 + 1/8 + 1/16 + ....   , 
  

  we get a limit which is a definite number, namely 2. Similarly, we know that

  1.9999999999999999999....
  

  is the same number like 2. There is no gap between the decimal expansion with infinitely many 9's and 2.00000000000. Piaget type experiments with young kids even up to middle school show however that many would say that 1.999999.... is smaller than 2.00000000... In some sense this is like Archilles and the Tortois. We can argue that 1.999999.... never reaches 2.00000000.... because there are infinitely many steps needed to reach it. Like Zenos paradox with Archilles and the Tortois is a rhetorical" paradox. It appeals to the intuition that we can never can count up to infinity. Zeno would argue that 1.9999999... never reaches 2.0000000... Therefore, there are no numbers larger than 2. But 2.3 for example is larger than 2 and the number 2.3 does not care whether infinitely many steps have been used to get to 2.0000 from approximations 1.9, 1.99, 1.999, 1.9999, ....
  
  We will discuss later the problem of the Pythagoreans with real numbers which are not rational numbers like the square root of 2. Because this number can not be written down completely, (the decimal expansion 1.414213562373095048801688724209698078569671875376948073176679737.... never ends), the Pythagoreans would not accept it and (according to legend) even killed "heretics" who would argue otherwise.
• An other question came up in the break: How do we make sense of the identity 1-1+1-1+1 ... = 1/2? .
  
  The answer is to look at geometric series like 1+1/2+1/4+... and add it up. In this case it is 2. There is a formula for the geometric series 1+a+a²+a³ + ... = 1/(1-a) which can be proven by multiplying both sides with 1-a and noticing that on the left hand side everything except 1 cancels. While the left hand side is only defined for |a| smaller than 1, the right hand side makes sense for all a different from 1. We can now postulate the identity to hold for other values of a also and apply this for a=-1 where the right hand side has the value 1/2. On the left hand side, we have the nonsensical Grandi's series 1 - 1 + 1 - 1 + ..... . What we have seen is quite an important principle: functions defined in one way can be extended and make sense in much larger domains. Mathematicians call this "analytic continuation". Of course this was not yet known to Luigi Grandi who was contemplating probably more about the puzzling fact that different arrangements of the sum lead to different values: (1-1) + (1-1) + (1-1) .... = 0 and 1 + (-1+1) + (-1+1) + (-1+1) .... = 1. I mentioned this example during lecture as one of the "strangest results" I know in Mathematics.
• Isaac: What is the meaning of "base trunc" and "leaves" in trees?
  
  This is not an official terminology in graph theory. There are vertices in a tree which have only one neighbor. To match the picture of a real tree, we single one out and call it the base trunc. The others are then the "leaves". The curvature of such vertices is 1/2. Vertices with two neighbors are "branches" and have zero curvature. Vertices with more than two neighbors are "crotches" and have negative curvature. We generalized the concept of tree and allowed the addition of "flowers", closed loops. The tree becomes then a "plant". Like several trees form a forest, several plants form a "garden. We have seen that in this case the total curvature of a garden is the number of plants minus the number of flowers.
• David: Are we going to go further with the Euler function V-E+F=2 in a future week? Yes, we will look at this again. There is a more general formula which says that for a surface, we have V-E+F=2-2g where g is the "number of holes" in the surface. For a sphere, which has no holes, the answer is V-E+F=2, for a doughnut with one hole we have V-E+F=0 and for a brezel with 2 holes, we have V-E+F=-2. This Euler Poincare formula V-E+F=2-2g is very closely related to our formula "number of plants" minus the "number of flowers". The flowers play the role of the holes. But there is more to it, this number V-F+F is called Euler characteristic. A theorem in differential geometry tells that if we integrate a quantity called "curvature" over the surface, we get this number. We have not discussed this in the first lecture but for a tree, this number can also be expressed as V-E where V is the number of vertices, E is the number of edges.
• Andrew: "I have a question about yesterday's lesson. I understood everything about the forests and how to find curvature. My question is what is the measure of curvature, or rather what do we learn about a plant, tree or forest using it's curvature? Was it just made up by us so that we could talk about the idea of a theorem, or does it have an actual application?"
  
  It actually has applications. It turns out that for large graphs, it can be tedious to compute the number directly and that summing up the curvatures is faster. The number we were computing is called the Euler characteristic. It is defined for any graph. In geometric setups, the Euler characteristic contains topological information. What we have seen in the forest or plant theorems that the sum gives information about the number of components as well as the number of "holes". This number is also important when studying surfaces. The same theorem appears there too if we define curvature of a surface nicely. Last year, we played with the same theorem but in the context of towns. You find in that handout remarks on the connection with surfaces. The theorem we have seen is a prototype of an important class of results in mathematics called Gauss-Bonnet theorems. We can add up local quantitities called curvature to get a global result which does not change if we deform the object. This is a central theme of mathematics. Unfortunately, learning about the classical Gauss Bonnet needs quite a bit of calculus and differential geometry and is usually the culmination of a semester long course. What we did in class with graphs is much easier to grasp. It turns out that it is pretty close to the actual geometric theorem [I work on a proof of the classical result in any dimensions, which uses graph theory and hopefully will be shorter than any known classical proof]. The main ideas in the graph theoretical setup are the same, but we have no technicalities to deal with. We can play and have fun with it in a classroom, where absolutely no calculus, no algebra and no geometry background is required. It just involves adding up numbers. The class was successful in designing a proof of the theorem using induction. I imagine however that the proof can only be found by gifted high school students. It could be a good theme also to introduce the concept of "induction". As we have seen, the induction "seed" is the "seed" of the tree: a graph with only one vertex. We grow a tree and see that adding a branch does not change the total curvature. We can then even allow flowers to grow and still have a result.

In the first lecture, we also saw connections between Math and Art. Here is a movie from an exhibit at Boston