May 2019 - mathfun@mailman.xmission.com

[math-fun] Quotient rings of the ring of Eisenstein integers
by Dan Asimov 21 May '19

21 May '19

The ring of Eisenstein integers E = Z[w], w = exp(2πi/3) form a nice commutative ring. It's "nice" because it's a discrete subring of the complex numbers C with additive group a rank-2 lattice in C. (When you identify any two points z, w of C whenever z - w in E, the quotient of C becomes a torus T with a very specific geometry.) Namely, T = the result of starting from a regular hexagon and then sewing together corresponding points on the three pairs of opposite edges. Anyhow, the ideals of the ring E are all principal, so of the extremely simple form J = a E, for some a in E. (The condition a^2 ≠ a ensures that the ideal is non-trivial.) Now consider the *quotient ring* S = E / a E. If a = K + L w, then the number of points in S is |S| = |a|^2 = K*K - K*L + L*L. These {K*K - K*L + L*L} are the same numbers represented by K*K + K*L + L*L when K, L range over all integers. Assume gcd(K,L) = 1. Then: What is the quotient ring E / a E ??? The additive portion is just Z / (K*K + K*L + L*L) Z, nothing surprising. Question: ----- What is the multiplicative structure? ----- E.g., for K + L w = 3 + 4w, |S| = 37. What is the best way to describe the multiplication table? I know how to figure it out, but am looking for a general formula for the multiplication table of E / (K + Lw) E. NOTE: The elements of the quotient ring S = E / a E are naturally identified with the hexagonal tiling of the torus T defined as T = C / a E by the Voronoi regions of the points of S in T, which is kind of cool. —Dan

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[math-fun] Jump and fill my square
by Éric Angelini 21 May '19

21 May '19

Hello MathFun The aim of this challenge is to fill a square (that has a 0 --zero-- in its upper-left corner) with integers. The best square is the one with the lowest sum of the said integers. Here is my personal best for the 2 x 2 square (sum = 11): 0.1 6.4 ... my best 3 x 3 has sum 63: 0.1.20 6.8.10 4.2.12 In order to fill a square, you have to start on the zero cell and jump: a) over zero cell (you land on 1 of the 4 adjacent cells – diagonal jumps are forbidden) b) from there, over 1 cell (you always have the choice to jump up, right, down or left if the landing cell is empty) c) from there, over 2 cells (same rules as above) d) from there, over 3 cells (same rules – at every step the size of the jump is increased by 1). etc. Here is my path two fill the 2 x 2 square, starting on 0 (R=right, D=down, U=up and L=left; the size of the jump is increased by 1 at every step, the 1st jump being of size zero): R D D U R L Here is my path to fill the 3 x 3 square -- same convention: R D R L D U L R L R U D L R D U D U R L Can you beat my personal records? And find records for bigger squares? (forgive me if this is old hat) (some illustrations here, on my personal weblog): https://cinquantesignes.blogspot.com/2019/05/jump-and-fill-my-square.html Best, É.

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Re: [math-fun] "Solid" geometry
by Dan Asimov 21 May '19

21 May '19

If we apply a linear transformation to the elliptical cone to get a round one, the sectioning plane goes to another plane cutting a conic section, so the inverse transformation shows the answer is Yes, since linear images of conic sections are still conic. —Dan ----- Is a section of an elliptical cone a Conic Section? -----

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[math-fun] "Solid" geometry
by Bill Gosper 21 May '19

21 May '19

Is a section of an elliptical cone a Conic Section? —rwg

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[math-fun] Nice little CF paradox
by Bill Gosper 20 May '19

20 May '19

Defn: "Regular CF" just means unit numerators. True or false?: Two consecutive 0s in a regular CF cancel each other out. True or false?: Tacking a 0 onto the end of a finite regular CF erases the (formerly) final term. Therefore, tacking on two 0s erases the last two terms! Or does the 2nd 0 erase the first before the first erases the previous? —rwg Note added in obfuscation: As we saw yesterday, three consecutive i <https://i.ytimg.com/vi/BGnoZ9ndYJc/maxresdefault.jpg> (=√-1) terms cancel each other out: In[151]:= FromContinuedFraction[{1, 2, 2, 2}] Out[151]= 17/12 %^2/2 289/288 In[152]:= FromContinuedFraction[{1, 2, 2, 2, I, I, I}] Out[152]= 17/12 In[153]:= FromContinuedFraction[{1, 2, 2, I, I, I, 2}] Out[153]= 17/12

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[math-fun] Nonquadratic surds with periodic continued fractions?
by Bill Gosper 20 May '19

20 May '19

Jeffrey Shallit's 1979 undergrad thesis https://cs.uwaterloo.ca/~shallit/Papers/thesispt1.pdf, https://cs.uwaterloo.ca/~shallit/Papers/thesispt2.pdf claims that "the" correct decomposition of the complex plane into unit-area regions is tilted √2 by 1/√2 dominoes including their two left edges, excluding their top- and bottommost points, placed with the midpoint of the lower left side on the representative lattice point. It may not be obvious, but periodic continued fractions with Gaussian integer terms can represent algebraic numbers beyond quadratic. Shallit's thesis claims that the correct complex floor function is complexfloor[z_] := Floor@z + Which[#1 + #2 < 1, 0, #2 > #1, I, True, 1] & @@ReIm@FractionalPart@z which he then uses to define complex continued fractions. Here is a "resumable" version, cCF[srd_, max_: 33] := NestWhileList[FullSimplify[1/(# - complexfloor@#)] &, srd, UnsameQ@## &, All, max] which lists the successive states of the computation, whose complexfloors are the actual terms. Moreover, leaving the last computed term unfloored, when interpreted as a finite CF, recovers the original "fractionand". E.g., for 2 i^⅓, In[220]:= cCF[2 Power[I, 1/3]] Out[220]= {2 (-1)^(1/6), 1/2 (1 + Sqrt[3]), 1 + Sqrt[3], 1/2 (1 + Sqrt[3])} cCF terminated: The 4th state equals the 2nd. Taking the floors, In[221]:= complexfloor /@ % Out[221]= {1 + I, 1, 2, 1} Mathematica denotes periodicity with extra braces: In[222]:= Append[Drop[%, -2], Take[%, -2]] Out[222]= {1 + I, 1, {2, 1}} In[223]:= FromContinuedFraction@% Out[223]= Root[16 - 4 #1^2 + #1^4 &, 4, 0] A quartic. In[224]:= ToRadicals@% Out[224]= Sqrt[2 (1 + I Sqrt[3])] In[225]:= FullSimplify@% Out[225]= 2 (-1)^(1/6) as hoped. For fancier algebraics, we would expect periodicity to fail. E.g., for the real root of Brillhart's famous x^3=8x+10, cCF emulates the usual Euclid process: In[126]:= Solve[x^3 == 8 x + 10] Out[126]= {{x -> Root[-10 - 8 #1 + #1^3 &, 1, 0]}, {x -> Root[-10 - 8 #1 + #1^3 &, 2, 0]}, {x -> Root[-10 - 8 #1 + #1^3 &, 3, 0]}} #1 is the real root: cCF[x /. %126[[1]]] //InputForm {Root[-10 - 8 #1 + #1^3 &, 1, 0], 1/(-3 + Root[-10 - 8 #1 + #1^3 &, 1, 0]), Root[-7 - 44 #1 - 66 #1^2 + 10 #1^3 &, 1, 0], Root[-10 - 144 #1 - 502 #1^2 + 119 #1^3 &, 1, 0], Root[-119 - 926 #1 - 1552 #1^2 + 1002 #1^3 &, 1, 0], . . .} In[246]:= complexfloor /@ % Out[246]= {3, 3, 7, 4, 2, 30, 1, 8, 3, 1, 1, 1, 9, 2, 2, 1, 3, 22986, 2, 1, 32, 8, 2, 1, 8, 55, 1, 5, 2, 28, 1, 5, 1, 1501790} But the other two roots are complex. In[212]:= cCF[x /. %126[[2]]] Out[212]= {Root[-10 - 8*#1 + #1^3 & , 2, 0], 1/((2 + I) + Root[-10 - 8*#1 + #1^3 & , 2, 0]), 1/(2*I + 1/((2 + I) + Root[-10 - 8*#1 + #1^3 & , 2, 0])), -2*I + 1/(2 + Root[-10 - 8*#1 + #1^3 & , 2, 0]), I + 1/((2 + I) + Root[-10 - 8*#1 + #1^3 & , 2, 0]), 1/(2*I + 1/((2 + I) + Root[-10 - 8*#1 + #1^3 & , 2, 0]))} It repeated! What is the CF? In[213]:= complexfloor /@ % Out[213]= {-2 - I, -2 I, -I, -I, -I, -I} Here is where Murphy steps in. There is no information in the triplet i,i,i ! In[247]:= MatrixPower[{{-I, 1}, {1, 0}}, 3] Out[247]= {{-I, 0}, {0, -I}} Shallit's CF algorithm, as I carelessy read it, usually doesn't converge! Most times, it eventually falls into a period 3 loop with hairier terms than Out[212] above, but their complexfloors are all -i! Can some hay be made from this looping? —rwg

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[math-fun] Covering sets and the game Spot It
by Ed Pegg Jr 16 May '19

16 May '19

The game Spot It Jr https://smile.amazon.com/SpotIt/dp/B0075XNGGK/ has 31 cards each with 6 of 31 animals. Every pair of cards shares 1 animal. This is a clever usage of the {31,6,2} cyclic difference set. Here are the first few difference sets. {{{7, 3}, {1, 2, 4}}, {{13, 4}, {0, 1, 3, 9}}, {{21, 5}, {3, 6, 7, 12, 14}}, {{31, 6}, {1, 5, 11, 24, 25, 27}}, {{57, 8}, {0, 1, 6, 15, 22, 26, 45, 55}}, {{73, 9}, {0, 1, 12, 20, 26, 30, 33, 35, 57}}, {{91, 10}, {0, 2, 6, 7, 18, 21, 31, 54, 63, 71}}, {{133, 12}, {1, 10,11, 13, 27, 31, 68, 75, 83, 110, 115, 121}}, {{183, 14}, {1, 13, 20, 21, 23, 44, 61, 72, 77, 86, 90, 116, 122, 169}} What if there are 7 symbols per card? The best possible is 40. And a solution for that was found by Alessandro Jurcovich. C(40,7,2) = 40 https://ljcr.dmgordon.org/cover/show_cover.php?v=40&k=7&t=2 What about 11, 13, 15, 16? I don't know, and the La Jolla Covering repository doesn't help. What are the lower bounds for these? Have solutions been found? I think this would be a good sequence for OEIS. --Ed Pegg Jr

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Re: [math-fun] Is the triangular array a disjoint union of proper similar ones?
by Dan Asimov 16 May '19

16 May '19

You are right, Jim. Actually, it's a kind of fun puzzle to find a partition E = U E_j of E as the disjoint union of cosets E_j of distinct |a_j|, i.e., the magnification factors |a_j| satisfy : j ≠ k => |a_j| ≠ |a_k|. Of course, the densities d_j = 1 / |a_j|^2 will be numbers of the form 1/K*K + K*L + L*L) for some K,L in Z, such that (*) Sum d_j = 1. Conjecture: Maybe a converse is true (?): ----- Whenever a set {d_j} of rationals of the form 1/(K*K + K*L + L*L) sums to 1, then there exists a partition of E into E_j as in (*). ----- —Dan Jim Propp wrote: ----- Dan, did you mean to specify that the dilation factors |a_j| should all be distinct? Because if you don’t require this, it seems to me that there are numerous examples; e.g., for a fixed a, take all the distinct cosets of a E. Jim Propp On Wed, May 15, 2019 at 6:58 PM Dan Asimov <dasimov(a)earthlink.net> wrote: > Let w = primitive cube root of unity, and let > > E = {P(w) in C | P is an integer polynomial} > > be the triangular array in question, where C = complex plane, > aka the Eisenstein integers. > > Definition: > ----- > A proper triangular array (PTA) is any proper subset A of E that is > similar to E. > ----- > > Any PTA A is of form > > A = a E + b > > for a and b any elements of E with |a| > 1. > > Question: > ----- > Does there exist a collection {A_j | j in J} of disjoint PTA's whose > union is E ??? > ----- > > ... where J is any index set -----

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[math-fun] Is the triangular array a disjoint union of proper similar ones?
by Dan Asimov 15 May '19

15 May '19

Let w = primitive cube root of unity, and let E = {P(w) in C | P is an integer polynomial} be the triangular array in question, where C = complex plane, aka the Eisenstein integers. Definition: ----- A proper triangular array (PTA) is any proper subset A of E that is similar to E. ----- Any PTA A is of form A = a E + b for a and b any elements of E with |a| > 1. Question: ----- Does there exist a collection {A_j | j in J} of disjoint PTA's whose union is E ??? ----- ... where J is any index set —Dan

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[math-fun] US Opposition to teaching of Arabic Numerals at 56%
by Ed Pegg Jr 15 May '19

15 May '19

https://pbs.twimg.com/media/D6TskeIXoAAnZHL.png

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