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[math-fun] Complete quaternion soln: AX=XB
by Henry Baker 03 Oct '20

03 Oct '20

For the record, here's the complete solution. Let A,B,X be quaternions; A',B',X' are their conjugates. Here's some formulae we may need: S(A) = (A+A')/2 V(A) = (A-A')/2 A = S(A)+V(A) A' = S(A)-V(A) N(A) = AA'=A'A=|A|^2 N(V(A)) = N(A)-S(A)^2 V(A)^2 = -N(V(A)) A is a *unit* quaternion when N(A)=1. A is a 'pure' vector when S(A)=0. A is a 'pure unit vector' when N(A)=1 and S(A)=0. X=0 is the trivial solution of AX=XB. For X to have a nontrivial solution, N(AX)=N(A)N(X)=N(X)N(B)=N(XB) => N(A)=N(B). Also, since S(B) = S(1/X A X) = S(1/X (S(A)+V(A)) X) = S(1/X S(A) X + 1/X V(A) X) = S(S(A) X/X + 1/X V(A) X) = S(S(A) + 1/X V(A) X) = S(S(A)) = S(A) (Because A => (1/X V(A) X) merely rotates a pure vector into another pure vector w/o a scalar part.) So also for X to have a nontrivial solution, S(A)=S(B). Note also that N(A)=S(A)^2+N(V(A))=S(B)^2+N(V(B))=N(B), hence N(V(A))=N(V(B)). Let X = (N(V(A)) - V(A) V(B)) = |V(A)||V(B)| (1 - (V(A)/|V(A)|) (V(B)/|V(B)|)) (Because |V(A)|=|V(B)|, N(V(A))=|V(A)||V(B)|.) So this X is a real multiple of the X we found in the previous email when A,B were pure unit vectors. Indeed, V(A)/|V(A)|, V(B)/|V(B)| ARE pure unit vectors. Checking: AX = XB (S(A)+V(A)) X = X (S(B)+V(B)) S(A) X + V(A) X = S(B) X + X V(B) S(A) X + V(A) X = S(A) X + X V(B) V(A) X = X V(B) V(A) (N(V(A))-V(A)V(B)) = (N(V(A))-V(A)V(B)) V(B) V(A)N(V(A)) - V(A)^2 V(B) = N(V(A))V(B) - V(A)V(B)^2 N(V(A))V(A) - (-N(V(A))) V(B) = N(V(A)) V(B) - V(A) (-N(V(B))) N(V(A)) V(A) + N(V(A)) V(B) = N(V(A)) V(B) + N(V(B)) V(A) N(V(A)) (V(A)+V(B)) = N(V(A)) V(B) + N(V(A)) V(A) N(V(A)) (V(A)+V(B)) = N(V(A)) (V(B)+V(A)) V(A)+V(B) = V(B)+V(A) Check!

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Re: [math-fun] Trivial soln for AX=XB in quaternions
by Henry Baker 02 Oct '20

02 Oct '20

Hi Neil: Yes, a "pure" quaternion A has S(A)=(A+A')/2=0, while a "unit" quaternion A has AA'=N(A)=1 and a pure unit quaternion A has AA=-1. Proof: AA = (AxA-A.A) = 0-(AA') = -N(A) = -1. If AX=XB, then clearly *any* real multiple (including 0) of X will suffice: A(nX)=(nX)B n(AX)=n(XB) AX=XB In particular, n=-1 works, as does n=|X|. So once we find a solution X, then we can have a *unit* (i.e., 'normalize') solution X/|X|. At 07:58 PM 10/1/2020, Neil Bickford wrote: >Hi Henry, > >Just to clarify - in this context, does unit vector quaternion refer to >a quaternion > >xi + yj + zk > >such that > >x^2 + y^2 + z^2 = 1 > >(i.e. a quaternion that is unit and only has a vector part?) > >I haven't been able to find this in the literature yet - my best guess so far >is that because there are an infinite number of solutions (described below), >robotics applications seem to focus on the problem of finding a value of X >that minimizes a metric of error (e.g. least squares) in a system of >equations > >A_1 X = X B_1, >A_2 X = X B_2, >... > >It looks like many applications also consider the case where X, A, B, ... >are >general matrices, which could include translation (e.g. solutions involving >dual quaternions), as well as rotations that aren't odd multiples of pi/2 >(i.e. A, B not necessarily unit vector quaternions). > >For the infinite number of solutions, suppose X is a quaternion that >satisfies AX = XB, where A and B are nonzero quaternions. Then for any real >u, v, (defining the power of a quaternion in an appropriate way), > >Y = A^u X B^v is another solution to AY = YB, > >since > >A (A^u X B^v) B^-1 >= A^u A X B^-1 B^v >= A^u X B^v >= Y > >so AY = YB. > >In other words, if a unit quaternion solution to AX=XB exists, and A and B >are also unit quaternions (not necessarily unit vector quaternions) >(and assuming I haven't made any mistakes!), then there can be up to a >2-dimensional space of unit quaternion solutions. > >Best regards, > >--Neil Bickford

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[math-fun] a simple problem, but ?
by Simon Plouffe 02 Oct '20

02 Oct '20

Here is a simple problem I saw in one math marathon at the University of Orsay for youngsters, it is similar to the PUTNAM. If we have a*b + 7*a*c + 15*c = a*b*c then find all a,b,c that are primes that satisfies the equation. a,b,c are all >=2. By using brute force I can find 3 solution with small primes and 49 others with non-primes, 52 solutions in all. I am stuck with this problem, I do not see how to resolve this without using brute force. Normally, that problem should be feasible with simple means, I don't see how to find the 52 solutions and from that , the 3 triplets of primes. hint, a, b c are relatively small one of the solution is 2, 29, 2 for a, b, c, see below for the complete set (spoiler). here is the original paper : https://www.imo.universite-paris-saclay.fr/marathon/sept20.pdf Best regards, Simon Plouffe 1, 23, 23 1, 24, 12 1, 33, 3 1, 44, 2 2, 15, 30 2, 29, 2 3, 13, 13 3, 14, 7 3, 15, 5 3, 16, 4 3, 18, 3 3, 24, 2 4, 11, 44 5, 11, 11 5, 12, 6 5, 15, 3 5, 20, 2 6, 10, 20 6, 19, 2 8, 9, 72 9, 9, 27 9, 13, 3 10, 9, 18 10, 17, 2 12, 9, 12 12, 11, 4 15, 9, 9 15, 10, 5 15, 12, 3 15, 16, 2 16, 8, 128 17, 8, 68 18, 8, 48 19, 8, 38 20, 8, 32 21, 8, 28 21, 9, 7 23, 8, 23 25, 8, 20 27, 8, 18 30, 8, 16 30, 9, 6 30, 10, 4 30, 15, 2 35, 8, 14 39, 8, 13 45, 8, 12 45, 11, 3 55, 8, 11 75, 8, 10 75, 9, 5 135, 8, 9

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[math-fun] Trivial soln for AX=XB in quaternions
by Henry Baker 02 Oct '20

02 Oct '20

I assume that this solution is known, but I've not seen it before -- e.g., in Wikipedia, etc. I am mystified why not. Find quaternion X s.t. AX=XB, A,B unit vector quaternions. A,B are unit vector quaternions => AA = BB = -1. Consider X=(1-AB). A(1-AB)=(1-AB)B A-AAB = B-ABB A-(-1)B = B-A(-1) A+B = B+A (Check!) -X = AB-1 is also a solution: A(AB-1)=(AB-1)B AAB-A = ABB-B -B-A = -A-B (Check!) Example: A = J, B = (cos(2t) + I sin(2t)) J, X = 1-AB = 2*cos(t)*(cos(t) - I sin(t)). Note that X is actually a square root, as is typical for quaternion rotations. If we want to normalize X, simply divide by 2*cos(t). If X isn't normalized, then denote X' as the conjugate of X, and compute B = X'AX/(X'X); since X'X=|X|^2, we never need to compute |1-AB|. ----- The above is closely related to the following hack: To compute the square root of a complex number z = r*(cos(t)+i*sin(t)): Let sz = |z|+z = r*(1+cos(t)+i*sin(t)). Then sz^2 = 2*(1+cos(t))*r^2*(cos(t)+i*sin(t)) = 2*(1+cos(t))*r*z Thus sqrt(z) = sz/|sz| = sz/sqrt(2*r*(1+cos(t))). Note that we have utilized trig only to show that this hack works; we never have to calculate trig functions to compute the square root. This hack works because the points (0,|z|,z,|z|+z) form a rhombus, so |z|+z has 1/2 the angle of z. Computer graphics people also utilize this hack for quaternion square roots, so how come they don't compute (1-AB) for these rotations, as above?

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Re: [math-fun] Trivial soln for AX=XB in quaternions
by Dan Asimov 02 Oct '20

02 Oct '20

This seems like a very limited question, since if A^2 = -1, B^2 = -1 that means both A and B lie on the sphere S^2 = {a+bi+cj+dk ∊ H | a=0, b^2+c^2+d^2 = 1} of the quaternions H. And what is the general solution, in any case, of AX = XB ??? —Dan Henry Baker wrote: ----- I assume that this solution is known, but I've not seen it before -- e.g., in Wikipedia, etc. I am mystified why not. Find quaternion X s.t. AX=XB, A,B unit vector quaternions. A,B are unit vector quaternions => AA = BB = -1. Consider X=(1-AB). A(1-AB)=(1-AB)B A-AAB = B-ABB A-(-1)B = B-A(-1) A+B = B+A (Check!) -X = AB-1 is also a solution: A(AB-1)=(AB-1)B AAB-A = ABB-B -B-A = -A-B (Check!) Example: A = J, B = (cos(2t) + I sin(2t)) J, X = 1-AB = 2*cos(t)*(cos(t) - I sin(t)). Note that X is actually a square root, as is typical for quaternion rotations. If we want to normalize X, simply divide by 2*cos(t). If X isn't normalized, then denote X' as the conjugate of X, and compute B = X'AX/(X'X); since X'X=|X|^2, we never need to compute |1-AB|. -----

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Re: [math-fun] Three-dimensional flat torus composed of seven cubes each touching the other six?
by Dan Asimov 01 Oct '20

01 Oct '20

Yes, that's exactly (one way that) I solved this, and agrees with what Fred Lunnon wrote — and also Allan Wechsler had another solution that works in Hilbert space. Note that the Voronoi regions* of the integer points in R^3 are unit cubes. The way I came to this originally was from realizing that if we define the "3D plus sign" as the union of the 7 Voronoi cubes Plus = V(0) ∪ V(e_1) ∪ V(-e_1) ∪ V(e_2) ∪ V(-e_2) ∪ V(e_3) ∪ V(-e_3) (where {e_1,e_2,e_3} is the standard basis of R^3), then appropriate translates of Plus will tessellate R^3. One lattice of translations that works is generated by the vectors {(2,-1,0), (0,2,-1), (-1,0,2)}. I think the Voronoi region of this lattice is a (non-uniform) truncated octahedron, but I'm not sure yet. —Dan ————— * For a discrete point set X in R^n, the Voronoi region V(x) of x ∊ X is V(x) = {v ∊ R^n | ‖x-v‖ ≤ ‖v-y‖ for all y ∊ X} — the points of R^n at least as close to x as to any other y ∊ X. Scott Huddleston wrote: ----- Yes. Let L = { (a_1, a_2, a_3) | a_1 + 2a_2 + 3a_3 = 0 mod 7 } This also generalizes to d dimensions, using L = { (a_1, a_2, ..., a_d) | a_1 + 2a_2 + ... + d*a_d = 0 mod 2d+1 } I wonder what the Voronoi regions look like. On Tue, Sep 22, 2020 at 11:01 AM Dan Asimov <dasimov(a)earthlink.net> wrote: > Let L be any discrete subgroup of R^3 isomorphic to Z^3. > > The the quotient R^3/L is a "flat 3-torus", and it carries > a well-defined metric. Its metric can be be described as the > result of identifying opposite faces of a parallelepiped > (the bounded intersection of three slabs* in R^3) in the > obvious way. > > Of course the cubical 3-torus R^3/Z^3 is one example, but > a flat 3-torus can have surprising properties. Note that > the cubical 3-torus can be tessellated by 8 cubes each > touching three of the other seven along two common square > faces. > > > Puzzle: > ------- > Does there exist a flat 3-torus that can be tessellated by > seven cubes each touching the other six along a common face? > > —Dan > > > ————— > * A "slab" in R^3 is the closed region between two parallel planes. -----

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[math-fun] Three-dimensional flat torus composed of seven cubes each touching the other six?
by Dan Asimov 01 Oct '20

01 Oct '20

Let L be any discrete subgroup of R^3 isomorphic to Z^3. The the quotient R^3/L is a "flat 3-torus", and it carries a well-defined metric. Its metric can be be described as the result of identifying opposite faces of a parallelepiped (the bounded intersection of three slabs* in R^3) in the obvious way. Of course the cubical 3-torus R^3/Z^3 is one example, but a flat 3-torus can have surprising properties. Note that the cubical 3-torus can be tessellated by 8 cubes each touching three of the other seven along two common square faces. Puzzle: ------- Does there exist a flat 3-torus that can be tessellated by seven cubes each touching the other six along a common face? —Dan ————— * A "slab" in R^3 is the closed region between two parallel planes.

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[math-fun] Puzzle: Hilbert space coloring of the cubic tiling
by Dan Asimov 01 Oct '20

01 Oct '20

Let the coordinates of (infinite-dimensional) Hilbert space H be denoted as x_1, x_2, x_3, ..., x_n, .... Note that the integer points of H correspond to arbitrary sequences of integers *only finitely many of which* are non-zero. The cubic tiling of H has one cube for each integer point of H. Definition: ----------- The cube at the integer point x is "adjacent" to the cube at the integer point y exactly if y differs from x in precisely one coordinate, by the amount ±1 in that coordinate, and has all the other coordinates equal to those of x. Puzzle: ------- Prove that given infinitely many colors, indexed by the set of integers, the cubes of the cubic tiling of H can be colored so that each cube is adjacent to exactly one cube of each other color (and not adjacent to any cube of its own color). —Dan

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[math-fun] record polynomial for primes, 70 and 118 values.
by Simon Plouffe 01 Oct '20

01 Oct '20

Hello, I have been working on the idea of producing a polynomial that gives primes. I have found this expression, if c = 1.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\ 000000000000000000000000000000000000000000000000000000000000000000000000000000001556808232856463140\ 060524522424280772934429200613394089802373785244587288947728630209290312111238140972072983869588571\ 830870498647900351444778386216146076636515297094962921999030102707899655875083040665145525741443694\ 679398437806398341541385226365101056503219529413038357761603719148231623498218255233850813147568998\ 744093447755692241415245289626420200959624751354414736085310934015664680524107510074349656822480162\ 844410900000000000000000000006345174532890027939294089368385587928671551372984708501441585620418908\ 704346095361979911978459546006902805287612053866464609018035122943293118288150470557741929174121124\ 290158771911527759861726269584165754859588606741441445142853262133802425477033893100601759975564856\ 010884639189352574717896440249667645304974867391425986281906375105044471660347105158250104363991834\ 652263421461752834975705269597512556117680247632410969440028501070105083651713662403142442324953222\ 225461614957943694687384105246348009815340760743692542284884699931895728286412542979603098044884198\ 867262644327255465732046886424762403984621514353794473387936916290029911549331529954509625933626714\ 214956674379391784430822602413846681316909794084400273637943424856690207465817148161052758912779700\ 994744869116506018623229585002806564436851770494944385807359855293872240395791047188062073418411994\ 344296873461271952410865026038448423723192039177383879070574428090707165141606873722573750361413628\ 118939617030810117859967828092642486835827304995753842763548295039042203138474167878091109846423833\ 375302339937989596168347847772485744109449547853544023508997274222527142776574434316734525917951119\ 963956001777673781864527502261570125603162371258224293816247417631732394744100802942101015408972089\ 551902455081699086722777589571816095705739686859606854254953133788523318655544571643416019528317460\ 445413625249324735829 Then { c*n^599 } will produce 70 primes in a row. { } is the nearest integer function from n = 2 to 71. Pushing the argument further, we can compute another c c:=1.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\ 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\ 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\ 000000000000000000000000000000000000000000000000000000000000000000000000420808913403862852286037439\ 407207843280000454871709619989894140740709325325252736588820344022475410602235657175990218497430264\ 507047719507377755195716569424888020122974216275756432352375373324176830942254809073810657745081913\ 603888834668238535298596378009711467814200413802525979955404026071836372488037677429251875416844528\ 065861840386027884753763707042193371171062601284648726970657466685180064833875657919373385253912481\ 027773992391561048642338757691327360781626065347664616770332572923617311886082595296236132142036996\ 417631450188411788269401899358010244184959963090467560534373012929594222867337833389724761101476675\ 265940651881737969147941607553518894806698237842174373575406499207457956921830241573813363914109602\ 358518386220031857824704794430689335145845120578114984070186291502181713823988558402910508064322733\ 386664486830182828399818696846253725628343900000000000000000000000000000000000000000000000084175367\ 076263959877259166560848274081827502629329509901369386601926351191550840426061007096278515366265573\ 952742294088796839541191179807808635987650890468860605139264215146106576144341811285085551240005000\ 227988558319987250361379891704780432636060689056700626535379944795660666700454584856310206434590094\ 563862726354995211336248348977032053078689859896535496416742714456066139307160173240122478999781218\ 197906243111555919375402725879595014949858452214458311548245169558232561733182222402280681701375871\ 793674008212637679797212418300296261196701357686395359413115610049111887480023065235108355761846794\ 154242861989506511616703890355282922203837990655338427540314744201707762522718995260468275097894687\ 992658082255204558477493326827178073388363120958685523709670974720969561623848267686506891223902113\ 356465938801122508318610403664836118814508451633512182695560344508500322312812467848420047540700118\ 756996471899350068204423209609398912635469407815389181128071148265623479744625601961828151165458057\ 147404102417248338590636605285714210912430034480012999178228996360603231909255738065132728968316201\ 942832143187208766136927730107179702837907838280367350477002598486254090817302319631003233911930910\ 326471343720878750834773567638442701853950231270477449195094392125036285692593745404976928391598493\ 120102332751565874902239824210432449599913477786262806371486887301944497146993921610645697910752375\ 471592158629224985009940446389322350619119669969165095587552413272311756200137164763698462482237842\ 085305489252071338731483772934377614611471560053211589913596117355646404568766829712511199143649379\ 532868420835646628215546104322304046559478794811449490815559145536588643275429191044262263030906449\ 008533040348061447967481937845138943509232559583664683731050066160246941104648886339886131997547626\ 564120786805439643038360743772486321929905049774075239032391693139009562591896452809951937070106931\ 642757393826901443265817325728278062434238566608603447589227610334941793131104068971110670780632223\ 501992904396028760342780374806446000650380882469871510606798126329459070489028928886613221959394681\ 050313121338162899918203136695650902893260267066828147220878717494110655700034933423554789843042459\ 788022312976819102949452912553641155543969632606853848755713600338142981448378093170817347349572333\ 952846367305250612191724450242639374172385705018208921153877985055819331461598505096730761179995331\ 035421322004624870914956170836838658721047385074001034842012584599289337696078080593315480435697028\ 797774107860521215331382017992014682105116071846031465281846724781837631731709033583747490253551521\ 803421554038976456384199449968082545362824346045297834204279214326015137442652491792514695344393933\ 611548475720971484567438167471354078436388186768821886590160600734921902666008892973993959127619935\ 979220653987693039715708896453055494647408765542658440234527096823567561970767330545161066354821607\ 244431370845177234625939025174648851026905740292276726858715610711059412344054491438441794529481070\ 832085773331946632572075193269154720991959262777548164921202141769518765360369196200318903302241851\ 418383237112822078939390705199644185307459995247776965884156912985571438965918094406295259053539143\ 012557100340081332145565524386009208260466781433220908290351633667375248793342095089544887182663796\ 959414821279536006717126997713929475971075504116872033184698268145793823163522155260021118497275248\ 218585277043570493808327959903835073039657749082814079252504658007442530811406763989164961498255778\ 804250702976093118870182366132160978707672267041810251065555501606510564729290105611727839326557928\ 121443797072417029288704025364542736049982739791265476935477664530256067268014661044783782058238843\ 329500015341452854520505118062087287565683882017664 { c*n^1229 } is prime for n = 2 .. 119. There is probably a lower degree polynomial which would produce more primes , the one found here is the simplest possible expression. Details of this algorithm are here,. https://vixra.org/pdf/2009.0116v3.pdf Best regards, Simon Plouffe

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[math-fun] monkeys are clever than you think
by Simon Plouffe 28 Sep '20

28 Sep '20

Hello, here are 2 films about monkeys. https://www.futura-sciences.com/planete/actualites/zoologie-video-etonnante… https://www.regardecettevideo.fr/video/3966/l-homme-est-plus-intelligent-qu… They can memorize 19 numbers in 210 milliseconds. and more. best regards, Simon Plouffe

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