- mathfun@mailman.xmission.com - mailman.xmission.com

[math-fun] Paradoxical decomposition of the plane or sphere?
by Dan Asimov 07 Aug '20

07 Aug '20

Does there exist a decomposition of the plane into three disjoint sets A, B, C such that all three are isometric to each other but also A is isometric to the union B + C ??? Ideally each set would be connected. Also measurable, but I would be surprised if measurable is possible. And ideally, without needing the Axiom of Choice. How about the same thing but with some points omitted: A + B + C = R^2 - Z where Z is a set of measure 0 ??? And if not the plane, how about the sphere S^2 ??? —Dan

4 3

Re: [math-fun] Paradoxical decomposition of the plane or sphere?
by Dan Asimov 07 Aug '20

07 Aug '20

For the sphere, yes. But for the plane, oo + oo = oo, so it's not immediately obvious why measurability must be ruled out. —Dan Allan Wechsler wrote: ----- Maybe I'm not understanding, but I think we can prove right away that the measurability requirement is not attainable. Two congruent measurable sets, after all, must have the same measure. And I think that if A is congruent to the union of the disjoint sets B and C, its measure must be at least |B| + |C|. But maybe I am forgetting some horrible pathological case. -----

1 0

[math-fun] What do you mean, "Fake news"?
by Bill Gosper 06 Aug '20

06 Aug '20

Out[364]= Gamma[1/4]==(1+√√5)^(3/2) √(5 (1+√2) (√2+√√5)) (1+√5)^(11/4) √√(3+√10) E^(-65π/3)π^(3/4) Sinh[20π]/2^(11/16) In[372]:= N[%364,105] Out[372]=3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 ==3.62560990822190831193068515586767200299516768288006546743337799956991924353872912161836013672338430036147 —rwg

6 7

[math-fun] Minimizing products of differences of powers in the same base
by Andres Valloud 05 Aug '20

05 Aug '20

Hi, I'm looking for some references to the problem of minimizing products of the form \prod_{j=0 \neq k}^d |b^k - b^j| where k is chosen between 0 and j, for a fixed base b. It looks like, for sufficiently large b > 1, minimizing this product requires choosing k = d, unless b gets close to 1 in which case eventually k must be chosen to be e.g. floor(d/2). If 0 < b < 1, the reverse process occurs: k must be chosen to be 0, unless b is really close to 1 in which case k must become e.g. floor(d/2). I'd like to believe there's literature on this subject, however searching for things like "minimize the product of differences of powers" returns material that is only related to the search query by the fact that the terms appear in the search results. Looking at the expression with Wolfram, like this, https://www.wolframalpha.com/input/?i=%5Cprod_%7Bj%3D0%7D%5Ed+%28b%5Ek+-+b%… returns a reference to the q-Pochhammer symbol. Digging through that, I was not able to immediately find a way to minimize the product or show which factor has to be removed from the product. I also tried making b = e, applying ln to the product, then approximating the sum by the integral. The resulting integral (courtesy of Wolfram Alpha) relies on Spence's dilogarithm function Li_2. After some approximations evaluating the integral, it looks like the product as a whole is about e^O(d). However, I do not immediately see how to minimize the product by choice of k depending on b. Do you have any suggestions on where to look? TIA! Andres.

1 0

[math-fun] Sum of consecutive terms = shortcut
by Éric Angelini 01 Aug '20

01 Aug '20

Hello Math-Fun Instead of writing: 1+2+3+4+5+6+7+8+9+10+11+12+13 +14+15+16+17+18+19+20+21+22+23 +24+25+... I could’ve written the « shortcut »: 3+7+11+24+46+62+37+41+94+... as 3=1+2 7=3+4 11=5+6 24=7+8+9 46=10+11+12+13 62=14+15+16+17 37=18+19 41=20+21 94=22+23+24+25 ... The terms 3,7,11,24,46,62,37,41,94,... are numbers that can be expressed as the sum of k>1 consecutive integers _in only one way_ (they belong to A038550). The terms were always chosen to be the smallest one that could belong to the « shortcut » sequence. One needs to backtrack sometimes as some chunks of consecutive integers cannot be « shortcut » (like the chunk that starts with 16+17+18... if I’m right). This « backtrack » remark applies to the lexicographically earliest « shortcut » tentative: 3,7,11,24,46,29,... which seems impossible to extend. Best, É. Catapulté de mon aPhone

1 0

[math-fun] 1 spiral + 10 diagonals = 1 sequence
by Eric ANGELINI 30 Jul '20

30 Jul '20

Hello Math-Fun (this is an attempt to send a message to the mailing list via my new gmail address I've subscribed in order to avoid warning messages that people sometimes get from me; the hereunder message will be more readableif you click this link: http://cinquantesignes.blogspot.com/2020/07/1-spiral-10-diagonals-1-sequenc… ) > Pick any digit zero and start a clockwise spiral on this infinite square lattice (streaked by the same succession of ten different diagonals, one digit per square); the resulting sequence *S* is the lexicographically earliest of distinct positive terms: 012569874301256987430125698743012569874301256987430125698743 125698743012569874301256987430125698743012569874301256987430 256987430125698743012569874301256987430125698743012569874301 569874301256987430125698743012569874301256987430125698743012 698743012569874301256987430125698743012569874301256987430125 987430125698743012569874301*2**5*6987430125698743012569874301256 874301256987430125698743012*569874301*256987430125698743012569 743012569874301256987430125*698743012*569874301256987430125698 430125698743012569874301256*987430125*698743012569874301256987 301256987430125698743012569*8743**0**1256*987430125698743012569873 012569874301256987430125698*74301*2*569*874301256987430125698743 125698743012569874301256987*430125698*743012569874301256987430 256987430125698743012569874*301256987*430125698743012569874301 569874301256987430125698743*012569874*301256987430125698743012 698743012569874301256987430125698743012569874301256987430125 987430125698743012569874301256987430125698743012569874301256 874301256987430125698743012569874301256987430125698743012569 743012569874301256987430125698743012569874301256987430125698 430125698743012569874301256987430125698743012569874301256987 301256987430125698743012569874301256987430125698743012569873 ... *S* = *0*, *1*, *2*, *10*, *3*, *4*, *30*, *12*, *5*, *6*, *52*, *103*, *47* , *8*, *7*, *430*, *125*, *69*, *89*, *65*, *210*, *34*, *78*, *9*, *698*, *74*, *301*, *25*, *6987*, *478*, *96*, *5210*, *347*, *896*, *525*, ... Any other order of the ten diagonals will not produce a lexicographically earliest sequence of distinct positive terms (I guess). Best, É.

1 0

Re: [math-fun] High schoolers: simplify this expression
by Henry Baker 27 Jul '20

27 Jul '20

This method isn't for high schoolers, but should be doable for undergrads: Find the cube root of 10+sqrt(108). (This method works for all(?) fields Q(sqrt(D)), not just D=3.) 1. Find the minimal polynomial for 10+sqrt(108). (%i1) (x-(10+sqrt(108)))*(x-(10-sqrt(108))),expand; 2 (%o1) x - 20 x - 8 2. Produce the companion matrix for this minimal polynomial. (%i2) polytocompanion(%,x); [ 0 8 ] (%o2) [ ] [ 1 20 ] Its determinant is the 'norm' of 10+sqrt(108), which by inspection is -8. Note that -8 = (-2)^3; i.e., -8 is a perfect cube. The general 2x2 companion matrix looks like: [ 0 - Determinant ] (%o4) [ ] [ 1 Trace ] Let T=Trace and N=Determinant: (%i5) %,Trace=T,Determinant=N; [ 0 - N ] (%o5) [ ] [ 1 T ] Cube this matrix: (%i6) %^^3,factor; [ 2 ] [ - N T - N (T - N) ] (%o6) [ ] [ 2 2 ] [ T - N T (T - 2 N) ] 'Companionize' this matrix by producing the companion matrix with the same characteristic polynomial as our cubed matrix: (%i7) polytocompanion(charpoly(%,z),z),factor; [ 3 ] [ 0 - N ] (%o7) [ ] [ 2 ] [ 1 T (T - 3 N) ] Equate this matrix with our integer matrix above, and extract equations in N,T: (%i8) list_matrix_entries(col(%-matrix([0,8],[1,20]),2)); 3 2 (%o8) [- N - 8, T (T - 3 N) - 20] Eliminate N and factor: (%i9) eliminate(%,[N]),factor; 2 2 (%o9) [- (T - 2) (T + 2 T + 4) (T + 2 T + 10) 4 3 2 (T - 2 T - 6 T - 20 T + 100)] The simplest integer answer is T=2 and N=-2. So (10+sqrt(108))^(1/3) = (T/2) + sqrt((T/2)^2 - N) = 1 + sqrt( 1^2 - -2) = 1 + sqrt(3) At 10:43 AM 7/10/2020, Henry Baker wrote: >True. > >Now do it like a high schooler... > >At 09:17 AM 7/10/2020, Simon Plouffe wrote: >>Hello, >> >>the expression is simplified in a fraction of a second (sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3); to 2 by Maple. >> >>Best regards, >> >>Simon Plouffe >> >>Le ven. 10 juil. 2020 Ã 17:47, Henry Baker <hbaker1(a)pipeline.com> a ÃÂ©crit : >>> (sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3) >>> >>> [I thought this was a pretty cool problem that came up on the Maxima email list.]

2 1

Re: [math-fun] Small chunks of letters
by Éric Angelini 26 Jul '20

26 Jul '20

Thanks Allan! (and this is the first time I encounter the « single token » concept — very instructive!-) Here is, I guess, the longest (though short) seq of terms counting the consonants used so far: THRee FiVe eiGHT TWeLVe SiXTeeN TWeNTy TWeNTySiX THiRTy THiRTySeVeN FoRTyoNe FoRTySeVeN FiFTyoNe FiFTySeVeN SiXTyTWo SiXTyeiGHT SeVeNTyFouR stop Best, É. > Le 26 juil. 2020 à 16:04, Allan Wechsler <acwacw(a)gmail.com> a écrit : > > This sequence is the fixed point of the morphism 1 -> 12; 2 -> 3; 3 -> > "41"; "41" -> 2122. Here, "41" must be treated as a single token. > > Limits of such morphisms are not usually periodic. The Thue-Morse sequence > is a famous example. > >>> On Sat, Jul 25, 2020, 4:24 PM Éric Angelini <eric.angelini(a)skynet.be> wrote: >> ONE TWO THREE FOUR ONE TWO ONE TWO TWO THREE ONE TWO THREE THREE FOUR ONE >> ONE TWO ... >> Insert a dot after each vowel: >> O.NE. TWO. THRE.E. FO.U.R O.NE. TWO. O.NE. TWO. TWO. THRE.E. O.NE.TWO. >> THRE.E. THRE.E. FO.U.R O.NE. O.NE. TW.O ... >> The size of the successive chunks is given >> by the sequence itself. >> Question: does the said sequence enter into >> a loop? >> Best, >> É. >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0

[math-fun] Small chunks of letters
by Éric Angelini 26 Jul '20

26 Jul '20

ONE TWO THREE FOUR ONE TWO ONE TWO TWO THREE ONE TWO THREE THREE FOUR ONE ONE TWO ... Insert a dot after each vowel: O.NE. TWO. THRE.E. FO.U.R O.NE. TWO. O.NE. TWO. TWO. THRE.E. O.NE.TWO. THRE.E. THRE.E. FO.U.R O.NE. O.NE. TW.O ... The size of the successive chunks is given by the sequence itself. Question: does the said sequence enter into a loop? Best, É.

2 1

[math-fun] More small chunks of letters
by Éric Angelini 25 Jul '20

25 Jul '20

TWO ONE TWO THREE FOUR ONE TWO ONE TWO TWO THREE ONE TWO THREE THREE FOUR ONE ONE... Insert now a dot _ before_ each vowell: the successive sizes of the chunks of letters are given by the sequence itself: TW.O .ON.E TW.O THR.E.E F.O.UR .ON.E TW.O .ON.E TW.O TW.O THR.E.E .ON.E TW.O THR.E.E THR.E.E F.O.UR .ON.E .ON.E ... Again: does the sequence enter at some point into a loop? Best, É.

1 0