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Recreational MathematicsThe CasasAlvero conjectureThe CasasAlvero conjecture states that a univariate, complex polynomial f of positive degree n that has a nontrivial gcd (i.e., a root in common) with each of its derivatives f', f'', ..., f^{(n1)} must be of the form a(xb)^{n}. The conjecture is known to hold if n=q p^{e} with q in {1,2,3,4} and p a prime larger than q, except possibly for q=4 and p=7. For q=1,2 this is proved here, and for q=3,4 the statement can be proved in a similar fashion. Below is an applet, written by my Bachelor student Johan P. de Jong, with which you can get a feeling for why the conjecture might be trueif you try to make roots of f coincide with roots of its derivatives, then the roots seem to have the tendency to "collapse" into a single point. Leftclicking adds a root or increases the multiplicity of an existing root, rightclicking deletes a root or decreases the multiplicity of an existing root, and the roots of derivatives of f are depicted in ever lighter shades of grey. The applet uses Michael Thomas Flanagan's Java library for complex polynomials.Tree chompChomp is a game that can be played on any partially ordered set with a a smallest element. Two players alternatingly remove an element of the set, together with all larger elements. The player who is forced to remove the minimum loses the game. Classical chomp is played on a chocolate bar, and very hard to analyse. See this page for mathematics and links and an applet. Another special of chomp case is graph chomp. A yet more special case, which should be analysable by good high school students, is tree chomp, for which I wrote an applet once. He who takes the last vertex wins the game; can you find a strategy?A poem about mathematicsArrowsA mathematician's most common mistake takes products to sums, and subthings to quotients; semisimplicity can only fake the lurking, neglected emotions. But don't despair! A oneway road back is there, though narrow: immediate rereversal of the arrow. A threegap proof??Every once in a while you read a mathematical statement that you just have to try and prove yourself. This happened to me when browsing through the latest issue of Nieuw archief voor de wiskunde. In this article, in which Krzysztof Apt passionately explains why everyone should post their papers on the arxiv, he also recalls the "ThreeGap Theorem", which states the following: let b be a positive real number, let a be a nonnegative real number smaller than b, and let n be a nonnegative integer. Remove from the circle R/bZ (R=real numbers, Z=integers) the points (cosets of) ka for k=0,...,n. Then the lengths of the connected components of what remains attain at most three distinct values. Put more operationally: stepping around a circle of length b with constant step size a, and deleting each point where you step, you cut the circle into smaller and smaller pieces. The theorem says that, after every step, the pieces take only three different lengths.Very unscientifically, I haven't done any research as to whether this is a famous theoremI just had to think it over myself. Would you say that what follows is a proof? Let L(n,a,b) be the set of lengths. So for instance L(0,a,b)=L(n,0,b)={b} and if a is in the open interval (0,b) we have L(1,a,b)={a,ba}. For fixed positive n, let P(n) be the statement that the union of L(n,a,b) and L(n1,a,b) has cardinality at most 3 for all a,b. Yes, we are going to prove something stronger! We will prove P(n) by induction on n; suppose that P(m) holds for all m smaller than n, and fix a,b as above. If a=0 then L(n,a,b)=0 and we are done. So we may assume that a is strictly between 0 and b. Write b=q*a+r with q a positive integer and r in the halfopen interval (0,a], and let s:=ar, in the halfopen interval [0,a). Then if n=1,..,q we have L(n,a,b)={a,bna} and, indeed, the union of L(n,a,b) and L(n1,a,b) has cardinality at most three. So suppose that n>q. Then I claim that this union equals the union of L(m,s,a) and L(m1,s,a) for some m at least 1 and at most nq. Since P(m) holds by assumption, we are then done. To get a feeling why the claim is true, look at the situation after q steps: the circle of length b has been cut into q pieces of length a and an additional piece of length r. Now concentrate first on those steps among the ones that follow that hit the first piece of length a; roughly 1 in every q steps does so, and it hits that piece at the positions s, 2s, 3s, ... (all modulo a). So in fact you are stepping along a circle of length a with constant step size s. But what about the remaining pieces of size a? They are all, onebyone, following the pattern in the first piece of size a, until all have been "updated" and you step back onto the firstor, indeed, onto the piece of size r; what's with that? Well, it's easy to see that that piece behaves exactly like the rtail of the apieces: when it is hit, the apieces are consecutively hit at the corresponding spot in their rtail. In rounds when the rbit is not hit, the apieces are consecutively hit in their sheads. We conclude that after every step hitting the qth apiece the set of lengths is exactly L(m,s,a) for some m. One step earlier it is L(m1,s,a) union L(m,s,a) because the qth apiece has not yet been updated, and one step later L(m,s,a) union L(m+1,s,a), etc. This proves the claim, and hence the theorem. This process can be formalised, of course. But perhaps more interesting is: are there higherdimensional analogues on tori or other compact Lie groups? Several years after I put this item online, a version of this on tori was proved by Henk Don; see this paper. A more efficient way to pack (a kind of) TangramYou probably know the classical puzzle of tangram. Somehow we obtained a variant of the game, containing two more small triangles; see the pictures below. It came packed in a square box, assembled as on the left. Note that if the short side of the smallest triangle is 1, then the sides of the square filled by the puzzle, after moving the pieces tightly together, have length 3, so that the square has area 9. But my fatherinlaw discovered that this is not quite the most efficient way to pack the puzzle. Indeed, consider the rectangle on the right. Its horizontal side has length 2, and its vertical side has length 1+sqrt(2)+2 (the lengths of the pieces on the left border, from top to bottom), which is approximately 4.414. So the area of this rectangle is approximately 2*4.414=8.83, which is smaller than 9! Funny, right? What's wrong here? 