Math E-320: Spring 2012
Teaching Math with a Historical Perspective
Mathematics E-320:
Instructor: Oliver Knill
Office: SciCtr 432

To the project

The projects tickle in. The program being pretty free, students have traditionally been able to chose from a variety of formats, from prose to picture book, worksheet, presentation. This year, one student (Luke Hatfield) wrote a website where some of the concepts are illustrated. More about the projects to come here.

To the computing lecture

  • At the end of the lecture, I mentioned Povray, the ray tracing program. The program can be obtained here. Here is a simple example.

    To the dynamics lecture

    To the cryptology lecture

    To the analysis lecture

    To the set theory lecture

    To the Calculus lecture

    To the Algebra lecture

    The main goal of the algebra lecture is to learn about the problem of solving polynomial equations like the quadratic, cubic or quartic which will lead us to groups. Groups are algebraic objects which can come to live as symmetry groups or in puzzles like the Rubic puzzle or the 15 puzzle. Here is something about the floppy

    David suggested: to make a triangle with six different colors to help with visualizing the rotations and reflections: draw the three angle bisectors to get six regions and color each a different color. Some students (not all) might find that easier to visualize than just having the letters at the corner. And of course I wouldn't start out with the terminology "group" -- I'd just let them experiment with combining operations and seeing the great patterns in the product table. David also mentioned: Sam Loyd published a great book of puzzles in 1928 called "Sam Loyd and His Puzzles: An Autobiographical Review", which I have in my library. The 15 puzzle (which he calls the 14-15 puzzle after the challenge of reversing those two squares) is the first chapter, and he has a great essay on a variety of people's attempted solutions that, of course, don't work -- ending with the reason why he stopped agreeing to listen to people's solutions (someone threatened to punch him in the nose if Loyd didn't believe that the fellow had solved it) Yes, the "god number" had had been in the news several times. Each time people would bring down the upper bound or bring up the lower bound. I have a link to the article on the exhibits page.

    An important class of groups are permutations. Lets take the group of permutations of three elements. We write the result of the permutation starting with (1,2,3) as
    (1,2,3)    (1,3,2)     (2,1,3)    (2,3,1)     (3,1,2)     (3,2,1) 
    
    The permutation (1,2,3) is the identity. If we (1,3,2) and compose with (2,1,3), then 1 -> 1 -> 2 and 2-> 3 -> 3 and 3 -> 2 -> 1 resulting in (2,3,1).

    David notices " that this looks like another way of writing the symmetry group for equilateral triangles, with the corners labeled 1,2,3 instead of A,B,C. We have the same operations, so (1,3,2) is a reflection around the 2-3 line, for instance, and (2,3,1) is a rotation of 120 or 240 degrees depending on our definition of direction..

    Sarah mentioned a puzzle which involves the group of permutations. Lets call it the "swapper problem". I formulate it in my own words: 20 boxes are alligned in a closed room. Each box has a number 1 to 20 inside visible only when opening the box. The boxes are randomly permuted and arranged on a line so that we can say box number 1, box number two etc. Now, 20 people, numbered 1 to 20 and a "swapper" have the task to find a strategy to solve the following problem: after arranging a common strategy, the swapper goes first into the room, looks at all the numbers and is allowed to swaps two boxes of choice, then leaves, not be seen and heared of again. After that, each of the 20 people enters the room, each to look into 10 boxes of his choice. This has to happen in such a way that every person finds its number among the 10 opened boxes. This happens so that each person will close the 10 opened boxes again and not leave any traces before exiting and allowing the next person to enter. Like this every of the 20 people has so the same information available when entering the room. No information exchange between any of the parties is allowed to happen from the moment on that the swapper starts going into the room to look into all boxes to possibly swap two boxes.

    Mathematically, the problem asks whether one can change a permutation with 2m elements with one transposition so that 2m people numbered 1,...,2m can independently pick each m elements from the 2m and be sure that person k choses the number k.

    It looks impossible but can be solved. The solution is here.

    Caroline noticed that the number of solutions 3 was not addressed: - When you solve the cubic equation using Tartaglia and Cardano's formula, you'll get two solutions for the quadratic in u^3. Substituting back, you can then get two solutions for the cubic. Is there a formula to get exact value of the third solution? The formula should get all the solutions. The substitution for u gives you also a choice. So there are two places, where roots appear. There are 3 solutions in general. One could think that the solution formula gives 4 since twice a solution of a quadratic are involved. Can you explain how you get the 192 = 4!*8 for the Floppy puzzle? and for the Rubik cube? You can permute all the 4 corners arbitrarily. this is 4!. Then you can turn the edges in two way. This would give 2^4 possibilities. But there is a parity thing happening. The signature of the permutation plus the number of flips is even. This stays true during every move. Therefore, we get only half of 2^4*4! The Rubik cube is similar. Look at all the possible positions of the corners and edges. There are parity issues also here which shows that not all possible permutation/orientations are possible. You can not turn a single corner by 120 degrees for example (this is a "quark" state). You need 3 quarks (a Baryon) to be seen. Similarly, one can not turn a single edge by 180 degrees. This is also a quark state. One needs to quarks (a "meson") to get a state which can be realized.

    To the Number Theory lecture

    [ Added Mai 12, 2012 following Slashdot story: About the Goldbach conjecture: Blog entry of Tao, and arxiv paper: Every odd number is a sum of 5 primes.

    A few interesting questions by Karen: In Pascal's triangle. I didn't understand when you made = a smaller triangle inside of all the 1's. What does that do? 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
  • Please send questions and comments to knill@math.harvard.edu
    Math E320| Oliver Knill | Spring 2012 | Extension School | Harvard University