January 2012 - mathfun@mailman.xmission.com

[math-fun] path invariant surprises #2 & 3
by Bill Gosper 25 Jan '12

25 Jan '12

>From the four dimensional "Gauss" system in pathiart.pdf, the N matrix becomes [ (n + i) (n + j) ] [ --------------- 1 ] N(i, j, k, n) := [ (n + 1) (n + k) ] [ ] [ 0 1 ] after the transformation k -> k-i-j and a "crosstie" by 1/n. But shouldn't this be the five dimensional [ (n + i) (n + j) ] [ --------------- 1 ] N(h, i, j, k, n) := [ (n + h) (n + k) ] [ ] [ 0 1 ] ? I thought not, because h is redundant--it can simply be an arbitrary constant, shifted away by substituting n-h+1 for n, k+h-1 for k, j+h-1 for j, and i+h-1 for i, purely within the 4D system. (Merely shifting variables by arbitrary constants preserves path invariance.) Without h, it is still possible to coordinate transform any of the four factors in N to the form (x n + y k + z j + w i + a), where a is arbitrary and x, y, z, and w are arbitrary integers, by a sequence like n -> p n - q k ... + b, k -> r k + u j +..., j -> s k + t j + ... . But it's not always obvious or convenient. For example, pathiart.pdf shows how coordinate-transform and specialize [ (n + i) (n + j) ] [ --------------- 1 ] N(i, j, k, n) := [ (n + 1) (n + k) ] [ ] [ 0 1 ] to a Ϛ(2) system, with [ 2 ] [ (n + k) ] N(k, n) := [ ------------------------- 1 ] [ (n + k + 1) (n + 2 k + 1) ] [ ] [ 0 1 ] and K(k,1) giving a 3F2[1/4]. This is easy: k->k+1, j->0, i->0, n->n+k. Suppose instead we want to see what sort of K matrix we get from specializing N to [ 2 ] [ n ] [ -------- 1 ] N(k, n) := [ 2 ] [ (n + k) ] [ ] [ 0 1 ] This makes a nice puzzle: (Ignoring the other matrices) How do we get here from N(i, j, k, n)? Try it! The legal moves are n -> n+ IntegerLinearCombination(i,j,k) n -> -n, which also inverts the n matrix! i,j,k -> IntegerLinearCombinations(i,j,k) i or j -> constant. Here's one solution: n->n-1, i->i+1, j->j+1, k->k+1, j->j+i,k->k+i,n->n-i, i->i-k, i->0, j->0 . Not too obvious. Incidentally, this gives a K(k,1) essentially identical to the former one, but substantially simpler matrices for K(k,n) with n=2 (for Ϛ(2)-1), n=1/2 (odd terms of Ϛ(2)), n=1/4 (Catalan's constant) etc. But suppose we had the redundant system with h: [ (n + i) (n + j) ] [ --------------- 1 ] N(h, i, j, k, n) := [ (n + h) (n + k) ] [ ] [ 0 1 ] All we'd need is h->k, i->0, j->0. So how do we get h? I always shifted it into N(i,j,k,n) as an arbitrary constant: N(n+h-1,i-h+1,j-h+1,k-h+1) and likewise for I, J, and K. Then noting that the whole system was symmetric in h and k, redefining K(h,i,j,k,n):=K(i,j,k,n) etc, then manually grafting in H(h,i,j,k,n) := K(k,i,j,h,n). But today's first surprise is a much simpler way to do this: Simply define H(h,i,j,k,n):=the 2x2 identity matrix, and K(h,i,j,k,n):=K(i,j,k,n) etc., even though there is no h! Now use the formal coordinate transformer to n->n+h, i->i-h, j->j-h ,k->k-h and H is magically the right thing! The transformation machinery can boost the dimension of the grid. The same trick works to add the index f to the Rosetta system: Graft on F(f,g,h,i,j,k,n):=3x3 identity, then g->g+1-f, h->h+1-f, i->i+1-f, j->j+1-f,k+1-f, n->n+f-1, is almost right, but changes the 1,3 element of N from n to n+f-1, which we fix with a crosstie by 1 f-1 0 0 1 0 0 0 1 Now the final surprise: Without h, transformations like k -> 2k are legal, but irreversible, at least without a fancy Abramov-style recurrence solver. But with h, if we have [ (n + i) (n + j) ] [ ----------------- 1 ] N(h, i, j, k, n) := [ (n + h) (n + 2 k) ] [ ] [ 0 1 ] we can change the 2k back to k: n->n-h, i->i+h, j->j+h, h->h+k, n->n+h, i->i-h, j->j-h, and we've (reversibly) undone the irreversible! Equally easy would be 69k -> k. (Except the K matrix would already be intractable.) --rwg (Resending "staggering" predecessor to remedy a crucial omission:) On Tue, Jan 24, 2012 at 1:51 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > (Over the last fortnight, we have discovered or rediscovered three > path-invariance surprises which I hadn't reported, thinking no one would > care.) > > Products of 3x3 matrices don't typically compute sums unless they're > upper-triangular: > > r a b > 0 s c > 0 0 t > > (where we usually scale out the t), producing ordinary sums in elements > 1,2 and 2,3, > and a triangular double sum in 1,3. If a path-invariant system is upper > triangular, we > can discard any (same) number of top rows and left columns, and still be > path-invariant. > Furthermore, by "cross-tying" (as in pathiart.pdf) a diagonal matrix and > an auxiliary > matrix of the form > > 1 d 0 > 0 1 0 > 0 0 1 > > it is often possible to annihilate the 1,2 elements in a 3x3 system, > > r 0 b > 0 s c > 0 0 t > > giving you a second 2x2 system by discarding the middle row and column. > Also, if one can annihilate this 1,2 element in a non-triangular system, > getting > > r 0 b > a s c > 0 0 t > > then you get upper-triangular simply by cross-tying with > > 0 1 0 > 1 0 0 , > 0 0 1 > > interchanging the left two columns and top two rows. > > Since, without specialization, only one of the six matrices in the 3F2 > Rosetta > system is upper triangular, the identities you get from closing contours > relate > recurrences usually inexpressible as traditional sums and products, > although > they provide valuable accelerations for the 3F2s computed by the > (triangular) > N matrix. (And can relate continued fractions.) > > Before venturing into 3x3s, I spent years puzzling over > why the N matrix for my 2x2 Dixon system was N(...,k,n) := > > [ (n - h) (n - i) (n - j) ] > [ ----------------------------------------- 1 ] > [ (n + 2 k + h) (n + 2 k + i) (n + 2 k + j) ] > [ ] > [ 0 1 ] > > and not > > [ (n - h) (n - i) (n - j) ] > [ ----------------------------------- 1 ] > [ (n + k + h) (n + k + i) (n + k + j) ] > [ ] > [ 0 1 ] > > From the latter, I could get the (predictably messy) former simply by > redefining > K(k) :=K(2*k).K(2k+1), pairwise grouping the K matrices. I yearned for the > (presumably more elegant) latter, for making neater identities. > > But the ***K(k) product of the*** > former computes pFp-1[-1/27]. This would require the latter to > compute qFq-1[i/sqrt(27)], clearly incompatible with the N matrix! > > But the 3x3 Rosetta permits *any* pattern of integer coefficients in any > of the > six numerator and denominator factors, e.g. (n+6g-9h...).../(n-2j+8k)..., > so > let's insist on (n-h).../(n+k+h)..., and see why K(k) only computes a sum > when > grouped pairwise. It's of the form > > 0 r a > s 0 b ! > 0 0 1 > > With r and s quite dissimilar. Setting h=i=j=0 for displayability, K(k):= > > [ 3 (n - 1) (2 n + 5 k - 2) 2 ] > [ 5 5 (------------------------- + k ) ] > [ k 10 ] > [ 0 - ------------------- ---------------------------------- ] > [ 2 3 2 ] > [ 18 (k - -) (n + k) 9 (k - -) ] > [ 3 3 ] > [ ] > [ n - 1 ] > [ 3 2 (----- + k) ] > [ 2 k 2 ] > [ ------------------ 0 ------------- ] > [ 1 3 1 ] > [ 3 (k - -) (n + k) 3 (k - -) ] > [ 3 3 ] > [ ] > [ 0 0 1 ] > > Notice -1/27 = (-1/18)(2/3). > > (if idiot GMail ruined this as usual, just delete the linebreaks that don't > immediately precede [.) > > So instead of elements 1,3 and 2,3 of the product being > > a(0) + r(0) a(1) + r(0) r(1) a(2) + r(0) r(1) r(2) a(3) +... > and > b(0) + s(0) b(1) + s(0) s(1) b(2) + s(0) s(1) s(2) b(3) +... > > we get > > a(0) + r(0) b(1) + r(0) s(1) a(2) + r(0) s(1) r(2) b(3) +... > and > b(0) + s(0) a(1) + s(0) r(1) b(2) + s(0) r(1) s(2) a(3) +... > > Forcing the above K(k) into conventional sum notation requires K(2k) > K(2k+1), > doubling the size of the expressions in the left two columns, and putting > a LARGE > irreducible quintic and sextic in the right column. > > Well, it staggered ME, anyway. > --rwg > Two other little surprises pending. >

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[math-fun] Staggering sums
by Bill Gosper 24 Jan '12

24 Jan '12

(Over the last fortnight, we have discovered or rediscovered three path-invariance surprises which I hadn't reported, thinking no one would care.) Products of 3x3 matrices don't typically compute sums unless they're upper-triangular: r a b 0 s c 0 0 t (where we usually scale out the t), producing ordinary sums in elements 1,2 and 2,3, and a triangular double sum in 1,3. If a path-invariant system is upper triangular, we can discard any (same) number of top rows and left columns, and still be path-invariant. Furthermore, by "cross-tying" (as in pathiart.pdf) a diagonal matrix and an auxiliary matrix of the form 1 d 0 0 1 0 0 0 1 it is often possible to annihilate the 1,2 elements in a 3x3 system, r 0 b 0 s c 0 0 t giving you a second 2x2 system by discarding the middle row and column. Also, if one can annihilate this 1,2 element in a non-triangular system, getting r 0 b a s c 0 0 t then you get upper-triangular simply by cross-tying with 0 1 0 1 0 0 , 0 0 1 interchanging the left two columns and top two rows. Since, without specialization, only one of the six matrices in the 3F2 Rosetta system is upper triangular, the identities you get from closing contours relate recurrences usually inexpressible as traditional sums and products, although they provide valuable accelerations for the 3F2s computed by the (triangular) N matrix. (And can relate continued fractions.) Before venturing into 3x3s, I spent years puzzling over why the N matrix for my 2x2 Dixon system was N(...,k,n) := [ (n - h) (n - i) (n - j) ] [ ----------------------------------------- 1 ] [ (n + 2 k + h) (n + 2 k + i) (n + 2 k + j) ] [ ] [ 0 1 ] and not [ (n - h) (n - i) (n - j) ] [ ----------------------------------- 1 ] [ (n + k + h) (n + k + i) (n + k + j) ] [ ] [ 0 1 ] >From the latter, I could get the (predictably messy) former simply by redefining K(k) :=K(2*k).K(2k+1), pairwise grouping the K matrices. I yearned for the (presumably more elegant) latter, for making neater identities. But the former computes pFp-1[-1/27]. This would require the latter to compute qFq-1[i/sqrt(27)], clearly incompatible with the N matrix! But the 3x3 Rosetta permits *any* pattern of integer coefficients in any of the six numerator and denominator factors, e.g. (n+6g-9h...).../(n-2j+8k)..., so let's insist on (n-h).../(n+k+h)..., and see why K(k) only computes a sum when grouped pairwise. It's of the form 0 r a s 0 b ! 0 0 1 With r and s quite dissimilar. Setting h=i=j=0 for displayability, K(k):= [ 3 (n - 1) (2 n + 5 k - 2) 2 ] [ 5 5 (------------------------- + k ) ] [ k 10 ] [ 0 - ------------------- ---------------------------------- ] [ 2 3 2 ] [ 18 (k - -) (n + k) 9 (k - -) ] [ 3 3 ] [ ] [ n - 1 ] [ 3 2 (----- + k) ] [ 2 k 2 ] [ ------------------ 0 ------------- ] [ 1 3 1 ] [ 3 (k - -) (n + k) 3 (k - -) ] [ 3 3 ] [ ] [ 0 0 1 ] Notice -1/27 = (-1/18)(2/3). (if idiot GMail ruined this as usual, just delete the linebreaks that don't immediately precede [.) So instead of elements 1,3 and 2,3 of the product being a(0) + r(0) a(1) + r(0) r(1) a(2) + r(0) r(1) r(2) a(3) +... and b(0) + s(0) b(1) + s(0) s(1) b(2) + s(0) s(1) s(2) b(3) +... we get a(0) + r(0) b(1) + r(0) s(1) a(2) + r(0) s(1) r(2) b(3) +... and b(0) + s(0) a(1) + s(0) r(1) b(2) + s(0) r(1) s(2) a(3) +... Forcing the above K(k) into conventional sum notation requires K(2k) K(2k+1), doubling the size of the expressions in the left two columns, and putting a LARGE irreducible quintic and sextic in the right column. Well, it staggered ME, anyway. --rwg Two other little surprises pending.

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[math-fun] New algorithm SMODA for integer factoring (apparently beats or ties world record asymptotic runtime formula)
by Warren Smith 24 Jan '12

24 Jan '12

Dear factorers. I invented a new algorithm I call SMODA for factoring integers. It requires a certain database to be precomputed once -- and this database is independent of N, the number being factored. You can read the paper here: http://dl.dropbox.com/u/3507527/SmoothFactor.html Once the database is available, my nonrigorous complexity analysis suggests that SMODA will factor integers N in runtime of order exp( (logN)^(1/3+o(1)) ) i.e. of the same or better order as the current champion the "number field sieve." SMODA is however quite different and a lot simpler than NFS. I then was wondering if somehow SMODA and NFS could be hybridized to get something better than either e.g. perhaps lowering the "1/3" to "1/4." However, I failed to accomplish that and the paper ends with a tentative argument that this runtime formula is actually optimal! You might want to consider this hybridization possibility yourself, since I do not currently feel tremendous confidence in my understanding of NFS (especially compared to the knowledge levels of some of the recipients) nor in my optimality conjecture. SMODA has not been implemented. I suspect it will outperform NFS. This is part I of a two-part paper. Part II is not yet ready for you to read, but it concerns how to build the database and some computer experiments I've been doing about that. [Perhaps there also should be a part III about actually implementing and testing SMODA. However, I do not currently have what it takes to do that. There is some worry that I am insane and SMODA is garbage. If so, please inform me of your concerns as soon as possible. However, I've now checked it enough so my current belief is I am not insane and SMODA is for real. If nobody can decimate it in some theoretical manner the next step would be experiments.] -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Approximating the Gamma function via odd-symmetric series without use of log or exp
by Warren Smith 23 Jan '12

23 Jan '12

Here are the Chebyshev series for the two difference functions I discussed in the two half-width real x-intervals discussed. F := (x) -> 1/GAMMA(x + 1/2); G := (x) -> 1/GAMMA(x+1); q1 := (x) -> (F(x) - F(-x))/x; q2 := (x) -> (G(x) - G(-x))/x; q1(x) = a0 + a2*T2(4x) + a4*T4(4x) + a6*T6(4x) + ... Chebyshev coefficients of q1(x) for |x|<=1/4 (too many sig figs): n a[n] 0 2.18822857229909830468416902185791956060390956512732610220916600781336524039221351614053661977 2 -0.273254664514069887739896092160782999307994656871336958404374793160797168865425760349942762e-1 4 0.298843558421361874292283480521214547841448396931759631063620088816173984493887620815045728e-4 6 0.115114573208780641806719170236838336056759053167834329543704098768549342425490865814478718e-6 8 -0.24596221043229878207737975051549880236147396698687820663893052392755238157647809614502267e-9 10 0.700903301236630192987360690303673696995814846727736386728059810535958517989594944216083e-13 12 0.1261660315051917112681936532323335993717963292755059998170626330008128274587874270179e-15 14 -0.735303620333152064004118473781548822474805015615497130102796146688323107360418741e-19 16 -0.148580315319405091733312214886224961213524515441229540845127685713957418126987e-22 18 0.156722821422181051333643262566131246642066989884915231128910055564051822536e-25 20 -0.289556822227957635447323026158405881619161468803372482129382974179845e-31 22 -0.14583581654088669266132166903935664061380995038478086507246363893226e-32 24 0.919144989701256450086199852156777444804522833664634040713852193e-37 26 0.752668142932638434388006341797133201199276768708658410356828e-40 28 -0.6131920222632910048942474544299913417053885888724490223e-44 30 -0.24584298938594666516314771402732888658399133619611947e-47 32 0.196824757391719781666884661682625968030834861808e-51 34 0.551217673024981009660703211364678071294336715e-55 36 -0.37254557177220635202040279728156859712691e-59 38 -0.8900502016985072541654131882688630657e-63 40 0.441983214718463044434018665176878e-67 Chebyshev coefficients of q2(x) for |x|<=1/4 (too many sig figs): 0 1.15168363865576457705814530189465293229955060141593563368689524377854453753257249472725484500 2 -0.278835019474058038032956993831997136012779702403009023299260927859084781612456532789004573e-2 4 -0.40549257019888606583593775275055759936819477357423917507549633776139962493978575979661184e-4 6 0.1097384304887659151388140595412942905507915514982821715429963111267035066105692423482352986e-6 8 -0.52063341146368135636242068873450151617607660317907541879673101716830274615902745052075940e-10 10 -0.7421627199946726954059060253533201627802522519778046414777521440945128601179737263758490e-13 12 0.660267276982836929347795384431999632164757028142010788307875489843105082313448064188e-16 14 0.52942317528942885079120371250360945162626967557090543388947765597955864045772849e-20 16 -0.167554813606803522917419567307197359356327764959597352006173936168385154713923e-22 18 0.17633333649446838102944788403535051137759024643942454988967735664720308007e-26 20 0.176314055714521646530599912070028050509839133705044662115599052049549708e-29 22 -0.29231757438826628324200236073984754079659038161272527028043683314940e-33 24 -0.9956920708938751225814610094553410549308448163967976079073297640e-37 26 0.187916818188481602140697618536266190071442938779569508334821e-40 28 0.352535312621740588265806815750559184380716955178170093798e-44 30 -0.66654916712880469315737878543401886078998130564459950e-48 32 -0.869604314728724930195939309153741276286429652258e-52 34 0.147695558683492868605235261902054118647388149e-55 36 0.159139215387692683697334870224606153295582e-59 38 -0.2174748996700316207119288633926217464e-63 40 -0.221510015046556441186755906516959e-67 As you can see, the speed of convergence is pretty spectacular; 12 terms in each series will give you error<10^(-37). Were Gosper to Rational-Remez these it should become slightly better still. In MAPLE, you allegedly can get it to find rational Remez best approximation of degree 7/degree 5 by with(numapprox); wght := (x) -> 1; minimax(q1, -1/4 - 10^(-87)..1/4 + 15.4632*10^(-88), [7, 5], wght, 'maxerror' ); #notice I have used a slight random enlargement of the #interval -1/4, 1/4 to prevent MAPLE #from activating some bullshit... but then it just gives different kind of failure but in fact, I have never managed to make MAPLE actually succeed after several tries, gave up. MAPLE has a lot of bugs.

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Re: [math-fun] 6F5[-1/48]
by Bill Gosper 23 Jan '12

23 Jan '12

The arboricidal version is Gosper, R. Wm., Strip Mining in the Abandoned Orefields of Nineteenth Century Mathematics, Computers in Mathematics, (D. Chudnovsky & R. Jenks, eds.), Lecture Notes in Pure and Applied Mathematics, Vol 125 (1990), p282. but they printed my earliest draft instead of one of the many revisions I sent during the long publication delay. Better: http://www.tweedledum.com/rwg/stanfordn1.pdf http://www.tweedledum.com/rwg/stanfordn2.pdf http://www.tweedledum.com/rwg/stanfordn3.pdf http://www.tweedledum.com/rwg/stanfordn4.pdf (The machine I TeXed them on was too microcephalic to typeset more than a handful of pages per file! And it couldn't import graphics--there are some missing lines and a curve that were drawn by hand.) --rwg JPropp> Bill, what's the best reference to give people on path invariant matrix products? I want to mention this to Alexei Borodin at MIT, since it seems relevant to some of the work he's done. Jim Propp > Fooling with more path invariant matrices last nite, the kids & I found> HypergeometricPFQ[{1/2 - a, 1 - a, 1/2 + a, 1 + a,> 23/14 + (2 a)/7 - 1/7 Sqrt[1 - 3 a - 3 a^2],> 23/14 + (2 a)/7 + 1/7 Sqrt[1 - 3 a - 3 a^2]}, {5/4, 3/2, 7/4,> 9/14 + (2 a)/7 - 1/7 Sqrt[1 - 3 a - 3 a^2],> 9/14 + (2 a)/7 + 1/7 Sqrt[1 - 3 a - 3 a^2]}, -(1/48)] ==> (2 3^(3/2 + a) Sin[(a \[Pi])/3])/(a (11 + 12 a + 4 a^2))>> I don't recall any F[-1/48]. Julian found 27 a which rationalize the> parameters, e.g.,> HypergeometricPFQ[{46/43, 40/43, 49/86, 37/86, 883/602, 153/86}, {3/2,> 5/4, 7/4, 281/602, 67/86}, -1/48] ==> 159014*3^(37/86)*Sin[Pi/43]/18827>> There were three 3F2[-1/4] Correction: 3F2[-1/48] (which Macsyma could do), one 4F3,> and no 5F4s.>> Along the way, he found the minor simplification puzzle> 1 == 10*Sqrt[Pi]^3/(27*2^(2/3)*(1/3)!^2*(5/6)!*Sqrt[3])>> and the unlikely looking> 3*((4*Sqrt[3]*I + 1)/Sqrt[1 - I/4/Sqrt[3]] - (4*Sqrt[3]*I -> 1)/Sqrt[I/4/Sqrt[3] + 1]) -> 14*3^(1/4)*((47*Sqrt[3]*I + 24)/> Sqrt[4*Sqrt[3] - I]^5 - (47*Sqrt[3]*I - 24)/Sqrt[I + 4*Sqrt[3]]^5) ==> 121*Sqrt[Pi]^5/(288*2^(1/3)*(1/6)!*(1/4)!^2*(11/12)!^2*Sqrt[3])>> We also accelerated Dixon's thm to get a three-parameter 7F6[-1/4]:> HypergeometricPFQ[{a, b, c,> 2*a - 2*b, -c - b + 2*a, (r + 3*c - b + 4*a)/10 +> 1, (-r + 3*c - b + 4*a)/10 + 1}, {c - b + 1, (1 - b)/2 + a, -b/2 +> a + 1, c + b - a + 1, (-r + 3*c - b + 4*a)/> 10, (r + 3*c - b + 4*a)/10}, -1/> 4] == (2*a - b)!*2^(2*b - 2*a)*(c - b)!*(c + b - a)!*> Sqrt[Pi]/(a!*(-b + a - 1/2)!*c!*(c - a)!)>> r-> Sqrt[(3*c - b)^2 - 8*a*(2*c + b - 2*a)]>> a rational case of which is> HypergeometricPFQ[{3/2 + a/2, 2 + a/2, 1 - c, 1 + a + 2 c,> 1 + a/2 - c - (2 c)/a, 8/5 + a/5 - (2 c)/5 - (6 c)/(5 a),> 1 + a/2 + 2 c + (2 c)/a}, {1 + a/2, 3/2 + a/2 + c/2, 2 + a/2 + c/2,> 1 + a/2 - (2 c)/a, 3/2 - 2 c - (2 c)/a,> 3/5 + a/5 - (2 c)/5 - (6 c)/(5 a)}, -(1/4)] == (> 2^(-1 - a - 2 c)> Sqrt[\[Pi]] (2 + a + c)! (a/2 - (2 c)/a)! (1/2 - 2 c - (2 c)/> a)!)/(((3 + a)/> 2)! (a/2 + c)! (-(1/2) - c - (2 c)/a)! (1 + a/2 - c - (2 c)/a)!)> --rwg

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Re: [math-fun] Gosper's "path invariance" in matrix product land
by N. J. A. Sloane 23 Jan '12

23 Jan '12

Warren said: "Now that I finally see Gosper's ideas, wow!" Me: that's exactly what I said after hearing Bill give this talk at an AMS meeting in Boston some 30 years ago. I thought (and still think) that it was one of the greatest talks I've ever heard. I also felt that the audience didn't sufficiently appreciate it - it was too extraordinary. Neil

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Re: [math-fun] bending M.C.Escher "Circle Limit IV".
by Stuart Anderson 23 Jan '12

23 Jan '12

Some details from Vladimir; It is the same 3D bending of 2D hyperbolic symmetry group generated by reflections in the sides of quadrilateral I've described in http://bulatov.org/math/1107 All the tiles are the same and fit perfectly and can be mapped to each other using Moebius transformation. Angles and devils belongs to 2 different classes though. All the circular holes in the holed tilings are equivalent (including the largest hole with the center at infinity - outside of the disk) On Tue, Jan 24, 2012 at 5:26 AM, <math-fun-request(a)mailman.xmission.com>wrote: > Send math-fun mailing list submissions to > math-fun(a)mailman.xmission.com > > To subscribe or unsubscribe via the World Wide Web, visit > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > or, via email, send a message with subject or body 'help' to > math-fun-request(a)mailman.xmission.com > > You can reach the person managing the list at > math-fun-owner(a)mailman.xmission.com > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of math-fun digest..." > > > > 2. bending M.C.Escher "Circle Limit IV". (Stuart Anderson) > > > > > ------------------------------ > > Message: 4 > Date: Mon, 23 Jan 2012 16:44:36 +0000 > From: Fred lunnon <fred.lunnon(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] bending M.C.Escher "Circle Limit IV". > Message-ID: > <CAN57YquTA=LVrv+gASRnEp+S2mgP4dQcLhexgGf2tLU3mOtTeg(a)mail.gmail.com > > > Content-Type: text/plain; charset=ISO-8859-1 > > I'm intrigued by the blank circles in images on right-hand side: > it looks as if he has some way of mapping multiple hyperbolic > infinities to a lattice in the plane, suggesting a (2-D) space > peppered with black holes of varying masses. > > If the tiles were now blown up back to congruence again, what > shape does this space have: eg. is it describable by some simple > metric? > > Fred Lunnon > > > > ------------------------------ > > Message: 6 > Date: Mon, 23 Jan 2012 18:26:09 +0000 > From: Veit Elser <ve10(a)cornell.edu> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] bending M.C.Escher "Circle Limit IV". > Message-ID: <37BDD28D-9961-4189-8965-B00B67227F18(a)cornell.edu> > Content-Type: text/plain; charset="us-ascii" > > The cover art for this book by Hermann Weyl > > http://www.amazon.com/Open-World-Hermann-Weyl/dp/0918024706 > > shows the surface of the globe conformally mapped to a flat torus -- the > singularities > are cleverly hidden in the oceans. > > Veit > > On Jan 23, 2012, at 11:44 AM, Fred lunnon wrote: > > > I'm intrigued by the blank circles in images on right-hand side: > > it looks as if he has some way of mapping multiple hyperbolic > > infinities to a lattice in the plane, suggesting a (2-D) space > > peppered with black holes of varying masses. > > > > If the tiles were now blown up back to congruence again, what > > shape does this space have: eg. is it describable by some simple > > metric? > > > > Fred Lunnon > > > >an.xmission.com <math-fun(a)mailman.xmission.com> > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > >

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Re: [math-fun] Approximating the Gamma function via even-symmetric series without log or exp
by Bill Gosper 23 Jan '12

23 Jan '12

> From: Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>>> Wow, this Remezes <http://www.tweedledum.com/rwg/recipoddremez.png> to> mu=2*10^-36 with 14 parameters! Oop, more like 3*10^-36 > --rwg But noting |a[15]/a[1]| > |b[12]/1|, switch a[17] for b[12] <http://www.tweedledum.com/rwg/a17b10oddrecip.png> (which is a fairly notorious placebo anyway). μ = .6 * 10^-36 ! wds> --there is little need to Remez a fast-converging Chebyshev series since it already is Remezed to decent accuracy. >Oh, but I see, Gosper is not making a polynomial, but rather a rational approximation. The way he did it, though, he's Remezing something different -- really you want the approximation error of the rational function to be equiripple, while Gosper is making something else be equiripple. [That probably cost Gosper 50% more error, or something?] There are known convergent algorithms that do this right. I think we just needed a different weight function. (Although this produced a lovely plot, I'm completely confused by NSolve producing several solutions, with the only desirable one claiming μ = 0. By the way, I also disunderstand your half integer offset.) Despite the anomalies, I'll stick my neck out and claim this will be hard to beat. With two surplus digits of precision, you might hope we could punt b[10], but its > a[17]. --rwg wds> Thinking about it, the difference-based method I proposed (odd sym) seems superior to the sum-based (even sym) one because later when solving a*b=p, a+b=s for a,b... with the sum you will need to do a bad subtraction which loses a lot of significant digits in cancellation sqrt(s*s/4 - p) is subtracting two nearly equal quantities while with difference a-b=d that issue does not arise: sqrt(d*d/4+p) Also to correct what I said before the solution of a*b = p, a-b = d is: 1/a = 1/ [ -d/2 +- sqrt(d*d/4+p) ] = [ d/2 +- sqrt(d*d/4+p) ] / p where the last expression had been erroneously reciprocated. Also, Gosper earlier had the idea, which is still a good one with this approach, that he could get symmetry for the Gamma function around x=1/2, OR get it around x=0, either way, and thus make a PAIR of symmetric approximants now each with half the interval length. That is, the interval is now 1/2 wide. The two functions you then need to approximate are d1(x) and d2(x) where |x|<=1/4... or perhaps you prefer to approximate d1(x)/x and d2(x)/x to keep the relative error small near x=0...: F(x) = 1/GAMMA(x+1/2). F(x) * F(-x) = cos(Pi*x)/Pi d1(x) = F(x) - F(-x). G(x) = 1/(x!) G(x) * G(-x) = sin(Pi*x) / (Pi*x) d2(x) = G(x) - G(-x) . d1(0) = d2(0) = 0. d1(1/4) = -d1(-1/4) = (1/2)*(2*Pi-sqrt(2)*GAMMA(3/4)^2)/(GAMMA(3/4)*Pi) = 0.54023327626805366671714040736252461351838415204125 d2(1/4) = -d2(-1/4) = (sqrt(8)*GAMMA(3/4)^2-Pi)/(GAMMA(3/4)*Pi) = 0.28721371222257427636269621260241438575441350426278 d1(1/2) = 1 d2(1/2) = Pi^(-1/2) d1'(0) = (2*gamma+4*ln(2))/Pi^(1/2) = 2.2155838077457420463420706011273861532011964034195 d2'(0) = 2*gamma = 1.1544313298030657212130241801648048620843186718798

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[math-fun] bending M.C.Escher "Circle Limit IV".
by Stuart Anderson 23 Jan '12

23 Jan '12

Vladimir Bulatov <https://plus.google.com/u/0/109509141493915423605> - 7:18 AM <https://plus.google.com/u/0/109509141493915423605/posts/D1DaC5bjnZY> - Public [image: Vladimir Bulatov's profile photo]Vladimir Bulatov<https://plus.google.com/u/0/109509141493915423605>originally shared this post: Four stages of bending M.C.Escher "Circle Limit IV". The art work for this one was hellishly hard to simulate. And I even didn't reproduce M.C.Escher original woodcut detail - half of the angels and half of devils are actually looking back. http://bulatov.org/math/1201/ shared by Stuart Anderson

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[math-fun] Approximating the Gamma function via even-symmetric series without log or exp
by Warren Smith 23 Jan '12

23 Jan '12

> From: Bill Gosper <billgosper(a)gmail.com> > Wow, this Remezes <http://www.tweedledum.com/rwg/recipoddremez.png> to > mu=2*10^-36 with 14 parameters! > --rwg --there is little need to Remez a fast-converging Chebyshev series since it already is Remezed to decent accuracy. Oh, but I see, Gosper is not making a polynomial, but rather a rational approximation. The way he did it, though, he's Remezing something different -- really you want the approximation error of the rational function to be equiripple, while Gosper is making something else be equiripple. [That probably cost Gosper 50% more error, or something?] There are known convergent algorithms that do this right. Thinking about it, the difference-based method I proposed (odd sym) seems superior to the sum-based (even sym) one because later when solving a*b=p, a+b=s for a,b... with the sum you will need to do a bad subtraction which loses a lot of significant digits in cancellation sqrt(s*s/4 - p) is subtracting two nearly equal quantities while with difference a-b=d that issue does not arise: sqrt(d*d/4+p) Also to correct what I said before the solution of a*b = p, a-b = d is: 1/a = 1/ [ -d/2 +- sqrt(d*d/4+p) ] = [ d/2 +- sqrt(d*d/4+p) ] / p where the last expression had been erroneously reciprocated. Also, Gosper earlier had the idea, which is still a good one with this approach, that he could get symmetry for the Gamma function around x=1/2, OR get it around x=0, either way, and thus make a PAIR of symmetric approximants now each with half the interval length. That is, the interval is now 1/2 wide. The two functions you then need to approximate are d1(x) and d2(x) where |x|<=1/4... or perhaps you prefer to approximate d1(x)/x and d2(x)/x to keep the relative error small near x=0...: F(x) = 1/GAMMA(x+1/2). F(x) * F(-x) = cos(Pi*x)/Pi d1(x) = F(x) - F(-x). G(x) = 1/(x!) G(x) * G(-x) = sin(Pi*x) / (Pi*x) d2(x) = G(x) - G(-x) . d1(0) = d2(0) = 0. d1(1/4) = -d1(-1/4) = (1/2)*(2*Pi-sqrt(2)*GAMMA(3/4)^2)/(GAMMA(3/4)*Pi) = 0.54023327626805366671714040736252461351838415204125 d2(1/4) = -d2(-1/4) = (sqrt(8)*GAMMA(3/4)^2-Pi)/(GAMMA(3/4)*Pi) = 0.28721371222257427636269621260241438575441350426278 d1(1/2) = 1 d2(1/2) = Pi^(-1/2) d1'(0) = (2*gamma+4*ln(2))/Pi^(1/2) = 2.2155838077457420463420706011273861532011964034195 d2'(0) = 2*gamma = 1.1544313298030657212130241801648048620843186718798

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