March 2014 - mathfun@mailman.xmission.com

[math-fun] calculating pi with puzzle pieces
by Steve Witham 14 Mar '14

14 Mar '14

Thinking about calculating pi... if only there were a way to do calculations by hand that was absolutely error-proof, so one wouldn't die having calculated ten thousand digits, only to have an error in digit 7195 detected after one's death. Well, you could calculate with appropriately-designed puzzle pieces. Think of the basic cell in a multiplication total factor A input & carry input ^ / \ + + factor B | | factor B output | | input + + \ / factor A v total output & carry output In decimal, there are 10,000 combinations of the four input digits (of which 9100 are possible). So, just make molds or cutting dies for all those combinations, with appropriate outputs of course. Then you need a frame, and pieces for the edge conditions. On the input and output edges, you could put raised type and print the thing when done. In fact you could design a set of pieces around the sides, or maybe wires, turning shafts or sliders, to convey all the inputs and outputs to a small area for printing. If you broke each digit into two pairs of binary-coded-decimal (BCD) bits, you would need four little pieces out of an alphabet of only 64, inside each cell. Then you would need to recode the result as BCD, but there's an issue there. The bit-pairs for factor A need to be interleaved with the bit-pairs for the carry, in order to get one of each to each upper-right little piece. I can imagine either using signal-crossing pieces, or handling the output of all combinations of (factor A, carry, total) with single pieces out of a set of 1000. But, a better way to distribute the factor inputs would be as crisscrossing bars lying under the pieces. Then there would still be 64 types of little piece, but only 100 BCD recoder shapes. --Steve "Did you check your calculations?" "I have my boards, right here!"

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[math-fun] Comments on "number wall" paper by Lunnon, and fast algorithms
by Warren D Smith 14 Mar '14

14 Mar '14

W. F. Lunnon : The Number-Wall Algorithm: an LFSR Cookbook, Journal of Integer Sequences 4 (2001) Article 01.1.1 https://cs.uwaterloo.ca/journals/JIS/VOL4/LUNNON/numbwall10.html J.H.Conway, R.K.Guy: The Book of Numbers, Springer 1996, see pages 85-89. A 2D square grid of numbers where any stencil of 5 adjacent numbers ...N... W..X..E obeys the relation X*X = N*S+E*W ...S... is a "number wall." These authors mainly are interested in the "mainstream case" where the wall contains a row of all 1s then the row immediately below is some (bi-infinite) integer sequence, then all wall entries below these can be completed using the stencil relation and will automatically be all-integer. (All entries above the all-1-row are automatically 0.) Except: actually sometimes the stencil relation tells you to divide 0/0 in which case it is not clear how to fill in that entry of the number wall. Lunnon defines all entries of the number wall as certain Toeplitz determinants -- in fact the stencil relation is exactly Lewis Carroll's determinant method applied to Toeplitz matrices. Lunnon then proves, he says for the first time, the fundamental theorems about how to fill in all number wall entries efficiently even in the 0/0 cases. Remarkably, 0 entries in mainstream number walls always form gXg square "windows" surrounded by 4g+4 nonzero entries forming the "inner rim" then 4g+12 entries forming the "outer rim" of the window. The 4 inner rim sides each automatically are geometric sequences. Lunnon gives fundamental theorems on page 11 relating inner and outer rim entries. On page 16 he explains how to use them to calculate a mainstream number wall. On page 23 he is somewhat confusing to the poor reader about computational complexity. To unconfuse: The stuff on page 16 allows one to calculate mainstream number walls row by row going downward, where the algorithm stores the most recent 3 rows; and also for each block of 0s in a row, it stores/knows which row of their window that block is. Then the calculation will compute an NxN chunk of the number wall in O(N^2) arithmetic ops and time, and using O(N) storage. This is optimal for any algorithm that computes the entire wall (not skipping any entries) and which has to store the input (top row). Other constructs which are equivalent modulo disguises to Lunnon's number wall include Heinz Rutishauser 1954's "quotient-difference algorithm," Peter Wynn's "epsilon algorithm" (and "rho algorithm"?), Shanks' sequence transformation, Aitken's delta-squared process, and values of "Pade tables" at specific x. Which of these is the "best" formulation is not immediately obvious but Lunnon's formulation has the advantages of preserving integrality and of obeying the amazing "window" laws. If the seed sequence obeys a linear recurrence of order k (e.g. the Fibonacci numbers obey one with k=2) then the number wall automatically will have exactly k+1 nonzero rows. The bottommost of them will automatically be a geometric sequence. The recurrence-defining polynomial J(x) can be computed using number-wall techniques but using polynomials as entries, not numbers (Lunnon shows) which is a rather inefficient way to find J(x), but it works. However, I point out the constant term J(0) in the scaled-to-be-monic version of J(x) can be computed faster; it is exactly (-1)^(k+1) times the growth ratio of the geometric sequence in the bottommost nonzero row of the number wall. Now if one wants to compute certain entries of the number wall but NOT all of them (skipping many entries is ok) then considerably faster algorithms than those Lunnon discusses, become possible. If you want to compute one entry m down in the number wall, you do not need m*m operations, you can do it using "ultrafast Toeplitz determinant" algorithms in O(m*log(m)^2) ops. And the entire J(x) polynomial (if degree=m) can be computed in this time bound e.g. R. P. Brent, F. G. Gustavson and D. Y. Y. Yun: Fast solution of Toeplitz systems of equations and computation of Pade approximants, J. Algorithms 1 (1980) 259-295 R. R. Bitmead and B. D. O. Anderson: Asymptotically fast solution of Toeplitz and related systems of linear equations, J. Linear Algebra Appl. 34 (1980) 103-116. Frank R. de Hoog: A new algorithm for solving Toeplitz systems of equations, J. Linear Algebra Appl. 88/89 (1987) 122-138. Victor Y. Pan: Structured Matrices and Polynomials: Unified Superfast Algorithms, Birkhauser 2001. These results of course are optimal except for log factors. It also is of interest to "skip ahead" in number walls, not only in the South direction, but also in the East direction. (Or West.) That is because Lunnon page 31 says the roots X[m] of the recurrence-defining-polynomial J(x) if all real and decreasing (i.e. iff all have distinct magnitudes) are given by X[m] = lim(n-->infinity) W[m, n+1] * W[m-1, n] / (W[m-1, n+1] * W[m,n]) where W[down, across] are the number-wall entries. If there are two roots X[m] and X[m+1] forming a complex conjugate pair then they will be roots of the quadratic W[m,n]/W[m-1,n] * x^2 - W[m,n+1]/W[m-1,n+1] * x + W[m,n+2]/W[m-1,n+2] = 0. An excellent application of this fact is a way to compute approximations to the roots of a given degree-k polynomial by using the polynomial to define a recurrence, using recurrence to compute seed sequence out far far to the East, then finally computing number wall entries going up to k South from there. The speed of convergence behaves roughly like |error| proportional to C^n where C=max(i,j with |X[i]|<|X[j]|)of |X[i]/X[j]| and n=how far East you walk. This is poor speed. However, if we were to make n=2^t be exponentially large, then the error would be like C^(2^t) which would be excellent speed ("quadratic convergence" as opposed to mere "linear convergence"). This can be accomplished via the following "fast skip ahead East" method. The linear recurrence defined by the input polynomial can be regarded as its kXk "companion matrix." Compute the (2^t)th power of that matrix using t consecutive matrix squarings, then use it to compute entries 2^t, 1+2^t, ..., 2*k+2^t of the sequence from its k starting entries. Now compute chunk of size about kXk of the number wall going South using Lunnon's usual method. In this way we get approximations to all the roots (and/or quadratic factors) of the input polynomial which have error of order C^(2^t), and the work (number of arithmetic operations) required is O( t * MatMult(k) + k*k ) i.e polynomial in k,t operations, and convergence is guaranteed. MatMult(k)=O(k^2.376). This is not that bad a rootfinding method, but I think it can be improved further, e.g. I think the matrix powering can be replaced by modular polynomial powering which would speed it up to O(t*PolyMult(k) + k*k) operations; PolyMult(k) = O(k*log(k)). Wynn's epsilon algorithm and rho algorithm are intended to extrapolate sequences to infinity under certain assumptions about their behavior (thus enabling, e.g. acceleration of series). The whole acceleration process for N terms of the sequence as input could be run, I would now claim, in O(N*log(N)^2) operations (via fast Toeplitz), as opposed to Wynn's processes based on number-wall-like tableaus, which had required order N^2. For example, http://en.wikipedia.org/wiki/Shanks_transformation shows the "Shanks transformation" acceleration method arises from solving NxN Toeplitz linear system, and E.J.Weninger in his review paper http://arxiv.org/abs/math.NA/0306302v1 discusses many other acceleration methods, which although Weniger never mentions it can also be done "ultrafast" using structured-matrix and/or "fast wall" techniques e.g. Levin's transform and the new transform in Weniger section 8.2 can be done using fast high order differencing via fast operator powering, and various other accelerators (Romberg integrator extrapolation, RIchardson extrapolation...) can be done fast using fast polynomial interpolation and multipoint evaluation algorithms. I'm not currently sure whether Brezinski's "theta algorithm" can be done ultrafast. These realizations about fast series/sequence acceleration/extrapolation, which I have not seen mentioned before (there are several books by Claude Brezinski on this topic -- do they mention ultrafast algorithms?), may have wide computational complexity implications: basically, slow-converging series & stuff, ARE fast-converging in the sense you can compute N terms and accelerate in O(N*log(N)^2) steps. In all of these arithmetic-op-counts I have ignored issues about inexact arithmetic. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Pi Day artwork
by Henry Baker 14 Mar '14

14 Mar '14

FYI -- http://www.theguardian.com/science/alexs-adventures-in-numberland/gallery/2…

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Re: [math-fun] avoidance of division (in English)
by Henry Baker 14 Mar '14

14 Mar '14

The practical problem is that putting off division typically blows up the sizes of the eventual numerators & denominators, so that the total number of bit operations actually increases. I keep running into this problem in many guises; most recently in trying to compute more-or-less exact intersections of facets in STL (3D printing) files. We need better methods for initially approximating data with rational numbers (or some other exact field) in such a way that subsequent operations don't explode in bit complexity. We already know that initially approximating data with -- e.g., single precision numbers -- could require triple, quadruple, or higher precision in intermediate results. Perhaps there is a way to _gradually increase the precision_ to keep the computation sane, but I'm not aware of any elegant way to do this. At 04:28 PM 3/13/2014, Warren D Smith wrote: >Volker Strassen: Vermeidung von Divisionen, >J.fur die reine und Angewandte Mathematik 264 (Jan 1973) 184-202 >https://eudml.org/doc/151394

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[math-fun] Calculus conundrum
by James Propp 13 Mar '14

13 Mar '14

I just invented this paradox, though I doubt I'm the first: "On the one hand, the indefinite integral of 2x+2 w.r.t. x is x^2 + 2x + C; on the other hand, putting u = x+1, we can write the integral as the integral of 2u w.r.t. u, which is u^2 + C = x^2 + 2x + 1 + C. Equating x^2 + 2x + C with x^2 + 2x + 1 + C we get 0 = 1." Here's my question for you guys: For those students who are stymied by this paradox, what would be a good hint? I've tried "Think about what indefinite integration IS", but that just seems to confuse them more. Is there a good hint that doesn't give away the game? Jim Propp

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[math-fun] avoidance of division (in English)
by Warren D Smith 13 Mar '14

13 Mar '14

Volker Strassen: Vermeidung von Divisionen, J.fur die reine und Angewandte Mathematik 264 (Jan 1973) 184-202 https://eudml.org/doc/151394 This paper is about evaluating (multivariate) polynomials with algorithms that include divisions, versus algorithms that only include ring operations +,-,*. Strassen upper bounds how much divisions can help. Theorem 2: If the polynomial degree is d, then a division-free algorithm exists with at most (d-1)*d/2 times the complexity of the division-using algorithm. But Strassen says it is unlikely this is optimal. Theorem 5: reduces (d-1)*d/2 to 4*d*log2(d) but where I think a further O(loglog(d)) factor really is needed, i.e. this should be O(d * logd * loglogd). That depends on your exact model of computation and exactly what you allow as operations. Sketch of how does Strassen's proof works: Assume your division-using algorithm has some known input point P which causes all the divisions to be by 1, i.e. not divisions at all. Now replace the inputs by P+x*(Y-P) where Y is the input you want, and x is an indeterminate. When x=1 we'll have the input we want. Now employ the division-using algorithm in the ring of formal power series in x. A division by a formal power series (1+a*x+b*x*x+...) can be done using ring-operations only -- and only order N*logN*loglogN of them if go out to degree N. Now at the end of the day since we know by assumption that the final output is polynomials of degree<=d, we know we did not need to use more than d+1 terms of each power series. Hence the complexity is multiplied by O(d*logd*loglogd) at most in converting the division-using to a division-free algorithm. This actually shows a bit more than Strassen's theorems 2 & 5 said. MY VERSION OF STRASSEN THEOREM: If the division-using algorithm employs D divisions, M multiplications, and A add/subtract operations, and has polynomial degree N, then a division-free algorithm exists employing O((D+M)*N*logN*loglogN + A*N) operations.

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[math-fun] Hypergeometric solution of the quintic
by Ahmad El-Guindy 13 Mar '14

13 Mar '14

Stillwell's paper `Eisenstein's Footnote' (link below) gives a lucid explanation of how Eisenstein (at the tender age of 14) obtained a solution y for y^5+y=x as an infinite series in x of the form y=x-x^5+(10*x^9)/2!-(14*15*x^13)/3!+.... Furthermore, the Bring-Jerrard Theorem asserts that a general quintic can be reduced to one of the above form, with x a radical expression of the coefficient of said quintic. http://web.a.ebscohost.com/ehost/pdfviewer/pdfviewer?sid=9dd1fb0a-49ef-42b5… Nonetheless, I don't think this gives the whole picture since the range of convergence for x isn't that big. Namely, the above series is exactly x*4F3(1/5, 2/5, 3/5, 4/5; 5/4, 2/4, 3/4|5^5*x^4/4^4), which converges (absolutely) for |x| less than or equal to (0.8)*(0.2)^(1/4), which is approximately 0.53499. So, if one is interested in solving y^5+y=1, say, then the series above won't converge. Is there an analytic continuation of the 4F3 that would be defined for large values of x and which would still give a solution of that quintic, or is there another way to get around that? Cheers, Ahmad

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[math-fun] hypergeometric solution of quintic
by Warren D Smith 12 Mar '14

12 Mar '14

>Stillwell's paper `Eisenstein's Footnote' (link below) gives a lucid >explanation of how Eisenstein (at the tender age of 14) obtained a solution > y for y^5+y=x as an infinite series in x of the form > > y=x-x^5+(10*x^9)/2!-(14*15*x^13)/3!+.... > ... > Nonetheless, I don't think this gives the whole picture since the range of > convergence for x isn't that big. Namely, the above series is exactly > x*4F3(1/5, 2/5, 3/5, 4/5; 5/4, 2/4, 3/4|5^5*x^4/4^4), which converges > (absolutely) for |x| less than or equal to (0.8)*(0.2)^(1/4), which is > approximately 0.53499. --incidentally, this is best possible. The discriminant of y^5 + y - x is 256+3125*x^4 which becomes 0 when |x|=4*5^(-5/4)=0.53499..., i.e. a double root y of y^5+y=x springs into existence for five different x with |x|=0.53499... via a bifurcation, whereas for all x with |x|<0.53499.... the root is isolated. Hence, the function itself becomes undefined (or more precisely is no longer uniquely defined) if |x|>0.53499... So, if one is interested in solving y^5+y=1, say, then the series above won't converge. Is there an analytic continuation of the 4F3 that would be defined for large values of x and which would still give a solution of that quintic, or is there another way to get around that? > 6. Re: hypergeometric solution of quintic (Warren D Smith) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Wed, 12 Mar 2014 18:07:59 +0100 > From: Joerg Arndt <arndt(a)jjj.de> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Lewis Carroll's determinant method and some > (new?) uses > Message-ID: <20140312170759.GB14741(a)jjj.de> > Content-Type: text/plain; charset=us-ascii > > * Fred Lunnon <fred.lunnon(a)gmail.com> [Mar 10. 2014 19:46]: >> On 3/10/14, Joerg Arndt <arndt(a)jjj.de> wrote: >> > * Warren D Smith <warren.wds(a)gmail.com> [Mar 10. 2014 07:22]: >> >> [...] >> > >> > For Toeplitz matrices, there should be FFT methods. >> >> Can you say a bit more about how this would work? > > I wish I could. It's many years ago I looked into that and > ended up with no notes, that indicates lack of success on my > side. > > I vaguely remember that one has to use a larger matrix that > somehow makes the problem look like the one for circulants. > No more than that, sorry. > > Best, jj > >> [...] > > > > ------------------------------ > > Message: 2 > Date: Wed, 12 Mar 2014 16:47:14 -0400 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] hypergeometric solution of quintic > Message-ID: > <CAAJP7Y0b2oa6ZfmXjkTW=1k_H_z+4AHMg_4=W5Hdn34Pw4YgZA(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > Of course 4F3(x) and indeed any pFq(x) with p<=q+1, is defined for > all complex x except for a set of measure zero, via analytic continuation. > And this particular 4F3 is of course an algebraic function which makes > that obvious. > > The solution of the quintic via this method is however nuts in the sense > that you will be far better off solving quintic by typical numerical > methods, than by > applying the "exact solution," i.e. you will obtain more digits > faster. Indeed the > best way to evaluate that 4F3 is by solving the quintic, not the other > way round. > > -- > Warren D. Smith > http://RangeVoting.org <-- add your endorsement (by clicking > "endorse" as 1st step) > > > > ------------------------------ > > Message: 3 > Date: Wed, 12 Mar 2014 15:14:01 -0600 > From: rkg <rkg(a)cpsc.ucalgary.ca> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] G4G11 > Message-ID: <alpine.LRH.2.03.1403121419550.26323(a)cpsc.ucalgary.ca> > Content-Type: text/plain; format=flowed; charset="US-ASCII" > > Dear all, > I've also (as well as Jim Propp) been > assembling something for the Gathering for > Gardner next week. This is honoring John Conway, > so the following, which may be distributed in any > way one wishes, is dedicated to him, with the > hope that he will complete The Triangle Book. > It would be flattering if he can use any of this. > > Guyration > is the combination of `quadration' and `twinning' > with Conway's `extraversion'. It yields a striking > generalization of the triangle, sometimes with as > many as 32 items appearing for the price of one. > > Quadration grants the same status to the orthocentre > of a triangle as it does to the vertices, so that > each of the four points is the orthocentre of the > triangle formed by the other three. It regards the > triangle as an orthocentric quadrangle, which now has > 4 vertices, 6 edges and 3 diagonal points. > > Twinning draws the perpendicular bisectors of each > of the six edges, producing a quadrangle of circumcentres, > congruent to the original quadrangle. Alternatively, the > two twins are related by reflexion in, or rotation > through 180 deg about, a common `centre' . > > Our triangle now has eight vertices, which are also both > circumcentres, and orthocentres. It has six pairs of > parallel edges which form three rectangles whose twelve > vertices lie on the `50-point circle' [9-point? > Euler-Feuerbach?] The other 38 points are the points of > contact with the 32 touch-circles and six points of contact > with the double-deltoid [Steiner, with the appearance of a > Star-of-David]. > > There are 12 edge-circles, 24 medial circles, 32 touch-circles, > 32 Gergonne points, 32 Nagel points, 144 Morley triangles, > 64 Pavillet tetrahedra, 256 radpoints, 384 guylines [haven't > checked if these are distinct; maybe they coincide in sets of > four], and much more. > > In case you'd like to do-it-yourself, here are the > makings of a picture of the 32 touch-circles > touching the 12 edges at 8 points, and each touching > the 50-point circle. > > Vertices V1 (36,51), V2 (-204,-77), V4 (132,-77), V8 (36,103) > Ve (-36,-51), Vd (204,77), Vb (-132,77), V7 (-36,-103) > > [hexadecimal subscripts; the 12 lines are ViVj with i,j in > {1,2,4,8} and in {e,d,b,7} = {14,13,11,7}.] > > The 32 circles are 16 \put(x,y){\circle{d}} and 16 with > (-x,-y) d, and the following centres (x,y) and diameters d. > > (20,-21) 112, (-92,-525) 896, (180,19) 192, (-252,115) 384, > (48,55) 24, (660,259) 1248, (20,99) 32, (88,-105) 104, > (16,63) 40, (-484,363) 1040, (60,91) 48, (-120,-209) 312, > (12,-5) 144, (216,63) 280, (-288,175) 504, (-84,-437) 720. > > 32 touch-circles touch the 12 edges at 8 points, > and each touches the 50-point circle. > > There are 3 or 4 largish files which can be sent to > interested individuals. R. > > > > > ------------------------------ > > Message: 4 > Date: Wed, 12 Mar 2014 14:37:04 -0700 > From: Henry Baker <hbaker1(a)pipeline.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] G4G11 > Message-ID: <E1WNqps-0002wj-Mv(a)elasmtp-kukur.atl.sa.earthlink.net> > Content-Type: text/plain; charset="us-ascii" > > Very cool! > > I wonder how this stuff "intersects" with a posting I made > almost 2 years ago regarding incircles/excircles. In > particular, an elegant quartic is formed by considering > the incenter and 3 excenters as roots of a complex > polynomial, where the triangle "lives" in the complex > plane. > > I'd be interested in getting the files. Thx. > > At 02:14 PM 3/12/2014, rkg wrote: >>Dear all, >> I've also (as well as Jim Propp) been >>assembling something for the Gathering for >>Gardner next week. This is honoring John Conway, >>so the following, which may be distributed in any >>way one wishes, is dedicated to him, with the >>hope that he will complete The Triangle Book. >>It would be flattering if he can use any of this. >> >> Guyration >>is the combination of `quadration' and `twinning' >>with Conway's `extraversion'. It yields a striking >>generalization of the triangle, sometimes with as >>many as 32 items appearing for the price of one. >> >>Quadration grants the same status to the orthocentre >>of a triangle as it does to the vertices, so that >>each of the four points is the orthocentre of the >>triangle formed by the other three. It regards the >>triangle as an orthocentric quadrangle, which now has >>4 vertices, 6 edges and 3 diagonal points. >> >>Twinning draws the perpendicular bisectors of each >>of the six edges, producing a quadrangle of circumcentres, >>congruent to the original quadrangle. Alternatively, the >>two twins are related by reflexion in, or rotation >>through 180 deg about, a common `centre' . >> >>Our triangle now has eight vertices, which are also both >>circumcentres, and orthocentres. It has six pairs of >>parallel edges which form three rectangles whose twelve >>vertices lie on the `50-point circle' [9-point? >>Euler-Feuerbach?] The other 38 points are the points of >>contact with the 32 touch-circles and six points of contact >>with the double-deltoid [Steiner, with the appearance of a >>Star-of-David]. >> >>There are 12 edge-circles, 24 medial circles, 32 touch-circles, >>32 Gergonne points, 32 Nagel points, 144 Morley triangles, >>64 Pavillet tetrahedra, 256 radpoints, 384 guylines [haven't >>checked if these are distinct; maybe they coincide in sets of >>four], and much more. >> >>In case you'd like to do-it-yourself, here are the >>makings of a picture of the 32 touch-circles touching the 12 edges at 8 >> points, and each touching >>the 50-point circle. >> >>Vertices V1 (36,51), V2 (-204,-77), V4 (132,-77), V8 (36,103) >> Ve (-36,-51), Vd (204,77), Vb (-132,77), V7 (-36,-103) >> >>[hexadecimal subscripts; the 12 lines are ViVj with i,j in >>{1,2,4,8} and in {e,d,b,7} = {14,13,11,7}.] >> >>The 32 circles are 16 \put(x,y){\circle{d}} and 16 with >>(-x,-y) d, and the following centres (x,y) and diameters d. >> >>(20,-21) 112, (-92,-525) 896, (180,19) 192, (-252,115) 384, >>(48,55) 24, (660,259) 1248, (20,99) 32, (88,-105) 104, >>(16,63) 40, (-484,363) 1040, (60,91) 48, (-120,-209) 312, >>(12,-5) 144, (216,63) 280, (-288,175) 504, (-84,-437) 720. >> >>32 touch-circles touch the 12 edges at 8 points, >>and each touches the 50-point circle. >> >>There are 3 or 4 largish files which can be sent to >>interested individuals. R. > > > > > ------------------------------ > > Message: 5 > Date: Wed, 12 Mar 2014 16:53:55 -0700 > From: Bill Gosper <billgosper(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] Name that octadecahedron > Message-ID: > <CAA-4O0FHTgCB5M=dNUbtwbPrcZCQf2U3F=TcksEEaGxUumw51A(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > http://gosper.org/IMG_0216.JPG > http://gosper.org/IMG_0217.JPG > --rwg > > > ------------------------------ > > Message: 6 > Date: Wed, 12 Mar 2014 20:03:48 -0400 > From: Warren D Smith <warren.wds(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] hypergeometric solution of quintic > Message-ID: > <CAAJP7Y0-pkny8zymJVMGg4a-jzaDB9XQZtJTFzq9sHW_109LnQ(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > On 3/12/14, Warren D Smith <warren.wds(a)gmail.com> wrote: >> Of course 4F3(x) and indeed any pFq(x) with p<=q+1, is defined for >> all complex x except for a set of measure zero, via analytic >> continuation. >> And this particular 4F3 is of course an algebraic function which makes >> that obvious. > > --and > http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQ/ > gives explicit formulas such as integral representations. > Pade and Borel summation should also work. > > > > ------------------------------ > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > End of math-fun Digest, Vol 133, Issue 17 > ***************************************** > -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] hypergeometric solution of quintic
by Warren D Smith 12 Mar '14

12 Mar '14

Of course 4F3(x) and indeed any pFq(x) with p<=q+1, is defined for all complex x except for a set of measure zero, via analytic continuation. And this particular 4F3 is of course an algebraic function which makes that obvious. The solution of the quintic via this method is however nuts in the sense that you will be far better off solving quintic by typical numerical methods, than by applying the "exact solution," i.e. you will obtain more digits faster. Indeed the best way to evaluate that 4F3 is by solving the quintic, not the other way round. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Name that octadecahedron
by Bill Gosper 12 Mar '14

12 Mar '14

http://gosper.org/IMG_0216.JPG http://gosper.org/IMG_0217.JPG --rwg

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