July 2010 - mathfun@mailman.xmission.com

Re: [math-fun] CF closed forms Mea gufa!
by Bill Gosper 30 Jul '10

30 Jul '10

dear Bill, among the crowd of those silently checking every line you produce, and generally nodding in approval before passing on to other issues, some, as I, are more outspoken (having less to say): in your mail dd. Mon, Jul 26, 2010 at 2:48 PM you uttered:> ContinuedFractionK[d n + e, a n + b, {n, 0, Infinity}],>> where ContinuedFractionK[num[n],den[n],{n,0,inf}] :=> num[0]+den[0]/(num[1]+den[1]/... , throwing me slickly off the scent for days, since it should have been:> num[0]/(den[0]+num[1]/(den[1]+... , <ARGH! Apologies all. Thanks for catching this. in which the 'num' and 'den' are a bit of a give-away, admittedly. After I got that right, no thanks to the sparse documentation of Mma 7.0 on the web, I kept my peace since all you wrote checked out numerically. Except for: ContinuedFractionK[a*n + b, 1/4*-a^2*n^2 + d*n + e, {n, 0, Infinity}] <ARGH!! I meant <ContinuedFractionK[1/4*-a^2*n^2 + d*n + e, a*n + b,{n, 0, Infinity}] <Mea gufa. Mea gufa. Mea Macsyma gufa. <--rwg == (2*a*e*HypergeometricPFQ[{1 - (2*d)/a^2 - (2*Sqrt[d^2 + a^2*e])/a^2, 1 - (2*d)/a^2 + (2*Sqrt[d^2 + a^2*e])/a^2}, {}, -(a^2/(-a^2 + 2*a*b + 4*d))])/ ((-a^2 + 2*a*b + 4*d)*HypergeometricPFQ[{-((2*d)/a^2) - (2*Sqrt[d^2 + a^2*e])/a^2, -((2*d)/a^2) + (2*Sqrt[d^2 + a^2*e])/a^2}, {}, -(a^2/(-a^2 + 2*a*b + 4*d))]) I plugged in ru=Thread[{a,b,d,e}->{3/2,1,2,3}] and I got LHS 0.295876155510292644314878101041614639797922314028 against RHS 1.32599373379934479107782121176450331018758562578 even with the 'gospersRule' enforced on the beheaded 'ypergeometricPFQ' (so as not to allow Mma any built-in transformation into 'HypergeometricU' thingies. gospersRule= ypergeometricPFQ[{a_, b_}, {Sequence[]}, z_] -> (Gamma[a - b]*HypergeometricPFQ[{b}, {b - a + 1}, -z^(-1)]*(-z^(-1))^b)/Gamma[a] + (HypergeometricPFQ[{a}, {-b + a + 1}, -z^(-1)]*Gamma[b - a]*(-z^(-1))^a)/Gamma[b] For the legion who do not wield the mighty Mma7.0 weapons, but make do with trusted old Mma4.0, here is the Gospel-Checker:http://users.telenet.be/Wouter.Meeussen/Gosper_CF_3.nb Wouter.

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[math-fun] CF closed forms
by Bill Gosper 30 Jul '10

30 Jul '10

[Sorry for the wrapping on prev msg. I *thought* I followed Mike Stay's no-wrap recipe. You should be able to repair it merely by deleting the bogus linebreaks,] So quadratic/linear gives 2F1s and linear/linear gives 1F1s. And constant/linear gives 0F1 ContinuedFractionK[e, a n + b, {n, 1, Infinity}] == b e e Hypergeometric0F1[2 + -, --] a 2 a ------------------------------------, b e (a + b) Hypergeometric0F1[1 + -, --] a 2 a and a pattern emerges. But not quite. Quadratic/linear goes confluent when c=-a^2/4: ContinuedFractionK[ a*n + b, (1/4)*(-a^2)*n^2 + d*n + e, {n, 0, Infinity}] -> (2*a*e* HypergeometricPFQ[{1 - (2*d)/a^2 - (2*Sqrt[d^2 + a^2*e])/a^2, 1 - (2*d)/a^2 + (2*Sqrt[d^2 + a^2*e])/a^2}, {}, -(a^2/(-a^2 + 2*a*b + 4*d))])/((-a^2 + 2*a*b + 4*d)* HypergeometricPFQ[{-((2*d)/a^2) - (2*Sqrt[d^2 + a^2*e])/a^2, -((2*d)/a^2) + (2*Sqrt[d^2 + a^2*e])/a^2}, {}, -(a^2/(-a^2 + 2*a*b + 4*d))]) But, another scandal! In[62]:= HypergeometricPFQ[{.6, .9}, {}, .1] Out[62]= ComplexInfinity I.e., 7.0 thinks that 2F0 is infinite because its series doesn't converge! In fact it's two 1F1s: In[66]:= Rule[HypergeometricPFQ[List[a_, b_], List[Sequence[]], z_], ((Gamma[a - b]* HypergeometricPFQ[List[b], List[b - a + 1], -1/z]*(-1/z)^b)/(Gamma[ a])) + ((HypergeometricPFQ[List[a], List[-b + a + 1], -1/z]* Gamma[b - a]*(-1/z)^a)/(Gamma[b]))] Out[66]= HypergeometricPFQ[{a_, b_}, {}, z_] -> 1 a 1 (-(-)) Gamma[-a + b] HypergeometricPFQ[{a}, {1 + a - b}, -(-)] z z --------------------------------------------------------------- + Gamma[b] 1 b 1 (-(-)) Gamma[a - b] HypergeometricPFQ[{b}, {1 - a + b}, -(-)] z z -------------------------------------------------------------- Gamma[a] (A&S 13.1.10) Specializing, we get Bessels from quadratic/linear: 2 1 2 2 1 n 1 2 2 ContinuedFractionK[- (-1 ) n - ---- + -- 1 (-1 + 4 0 ), 1 n + 1 (1 + I z), {n, 0, Infinity}] == 4 4 16 ((-1 - 2 I z) BesselJ[0, z] + 2 z BesselJ[1, z] + (I - 2 z) BesselY[0, z] - 2 I z BesselY[1, z]) / (4 BesselJ[0, z] - 4 I BesselY[0, z]) 1^2? 0^2?? Like I said, ContinuedFractionK needs work. I wouldn't be surprised to find a localized term transformation that made this linear/linear. --rwg Wow, what's the real(imag)part of a CF with complex terms?

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[math-fun] Linear/linear CF closed form
by Bill Gosper 29 Jul '10

29 Jul '10

Poorly tested and recklessly derived, but it sure looks like d + e a (a + b) + d d e Hypergeometric1F1Regularized[-----, -------------, --] d 2 2 a a -------------------------------------------------------- = e a b + d d a Hypergeometric1F1Regularized[-, -------, --] d 2 2 a a ContinuedFractionK[d n + e, a n + b, {n, 0, Infinity}], where ContinuedFractionK[num[n],den[n],{n,0,inf}] := num[0]+den[0]/(num[1]+den[1]/... , and Hypergeometric1F1Regularized[a,b,z] := 1F1/Gamma[b]. Mma seems not to know this CF formula. Given the three-term recurrences for 1F1. the form of this result is not surprising. But starting from the recurrences, it's not obvious how to get pure polynomials, vs rational functions, in the CF terms. This is too nice to be new: y e gama(z + 1, y) z ----------------- = ----------------------------------- - 1 z y y z - y + --------------------------- 2 y z - y + 1 + --------------- 3 y z - y + 2 + --- . . . where gama is little ("lower") gamma := int_0^y. But then why isn't it in A&S, MathWorld, or Wikipedia? Hmm, the latter ascribes to Gauss a CF with alternating terms that might be two-for-one equivalent to the above. Also, for upper Gamma (:= int_y^oo) they give something very similar, but with quadratic numerators. In Wolframese, (E^y*Gamma[1 + z, 0, y])/y^z == -1 + z/(-y + z + ContinuedFractionK[n*y, n - y + z, {n, 1, Infinity}]) y E Gamma[1 + z, 0, y] --------------------- == -1 + z y z ------------------------------------------------------------- -y + z + ContinuedFractionK[n y, n - y + z, {n, 1, Infinity}] Good luck testing this numerically--ContinuedFractionK needs work. These are presumably limiting cases of K(quadratic/linear) = 2F1/2F1, which I'll now pursue. --rwg

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[math-fun] Interleavings of three arithmetic progressions
by Allan Wechsler 26 Jul '10

26 Jul '10

Some of you may recall my interest in the sequence 1, 2, 4, 8, 14, 24, 36 ... of the number of ways it is possible to interleave N elements chosen from two non-intersecting arithmetic progressions. Another way to view this is as (closely related to) the number of rasterized straight line segments with N+1 pixels, where we light up a pixel if a geometric line intersects its rectangle of screen real estate. The sequence is in OEIS as A005598 ( http://www.research.att.com/~njas/sequences/A005598) and is apparently originally due to Leendert "Leo" Dorst, who wrote about it for his doctoral research, first mentioning it in 1984. Oddly, the corresponding sequence for three arithmetic progressions is, perhaps, not in OEIS yet. The alternative formulation involves polycube models of three-dimensional line segments. Unless I've made a mistake in my enumerations, the sequence begins 1, 3, 9, 27, 75, 189 ... and OEIS has nothing with those elements. Unfortunately, my calculations are all with pen and paper; I haven't written any code to do the counting. So it's quite possible that I blundered, and that OEIS has the sequence after all. Can another funster verify my work? (I have one more element, which I'm holding back to avoid the power of suggestion.) There is an obvious generalization to any number of progressions/dimensions.

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Re: [math-fun] Interleavings of three arithmetic progressions
by Dan Asimov 26 Jul '10

26 Jul '10

Bill T. wrote: << Interleavings of two translate to words reperesemting simple curves on a punctured torus. I think each one is also valid when extended in this case. The lex-min cyclic permutations are associated <--> their slope. Tjere are O(n^2) slopes admitting O(n) cyclic permutations. -- is there a good logical analysis of the 3D simplification base = slope for interleavings of 3? You can think of in terms of the cynical subdivision, as seen from the origin. >> That seems a little jaundiced to me. But I was naively thinking of the torus also, perhaps in a different way: Given three real numbers a,b,c > 0, linear independent of each other and arbitrary reals a_0, ab_0, c_0, consider the line in R^3 [thought of as tiled by unit cubes] F(t) = (at + a_0, bt + b_0, ct + c_0)) Then F(t) enters successive unit cubes via faces perpendicular to the x_1, x_2, or x_3 axis in a specific order, which yields a sequence of 1's, 2's and 3's. Then: Which sequences of 1's, 2's, 3's are possible from any ordered triple of a,b,c as above? Or in general, which sequences of 1's through n's are possible ? (Or is this equivalent to the original question?) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] A Scientist Takes On Gravity
by Steve Witham 24 Jul '10

24 Jul '10

>From: Mike Stay <metaweta(a)gmail.com> > >On Wed, Jul 21, 2010 at 11:08 PM, Steve Witham <sw(a)tiac.net> wrote: >>> Subject: Re: [math-fun] A Scientist Takes On Gravity >> >> Frederick W Kantor's 1977 book _Information Mechanics_ says that >> energy must correspond to information, and then works out consequences > >That can't be right--they have different units! Doesn't that depend what units you think a bit is in? >Perhaps energy >corresponds to information *at a given temperature*. Landauer's >principle says the energy required to erase a bit is > > E >= k_B T ln 2 Since Landauer is talking about erasing a bit, while Kantor considers information conserved, I don't think these relate, or maybe they relate like available vs. total energy. I'm pretty sure Kantor's bits (unlike Verlinde's) are not about thermodynamics but about grain in the possible states of the underlying system. Maybe you can make some sense of his math & use of units here, I can't: http://www.gravityresearchfoundation.org/pdf/awarded/1974/kantor.pdf --Steve

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[math-fun] abstrusa yet again
by Bill Gosper 23 Jul '10

23 Jul '10

Mike Hirschhorn sent me two unpublished drafts (which I'll post, pending his permission) proving with only the triple product identity a result equivalent to theta[3](a,q)^2*theta[3](b,q)^2-theta[4](a,q)^2*theta[4](b,q)^2 = theta[2](0,q)^2*theta[2](b-a,q)*theta[2](b+a,q) which specializes to the aequatio abstrusa when a=b=0. This was a special case of the quadrivariate theta[3](x-w,q)*theta[3](x+w,q)*theta[3](z-y,q)*theta[3](z+y,q) = theta[2](y-w,q)*theta[2](y+w,q)*theta[2](z-x,q)*theta[2](z+x,q) +theta[4](y-x,q)*theta[4](y+x,q)*theta[4](z-w,q)*theta[4](z+w,q) which comes from the easily remembered hexa(nona?)variate 3x3 determinant 0 = | theta (x-u,q) theta (x+u,q) theta (x-v,q) theta (x+v,q) theta (x-w,q) theta (x+w,q) | | r r r r r r | | | | theta (z-u,q) theta (z+u,q) theta (z-v,q) theta (z+v,q) theta (z-w,q) theta (z+w,q) | | s s s s s s | | | | theta (y-u,q) theta (y+u,q) theta (y-v,q) theta (y+v,q) theta (y-w,q) theta (y+w,q) | | t t t t t t | equivalent to one Rich proved in arxiv.org/pdf/math/0703470 . This determinant subsumes a bleepload of special cases in Whittaker&Watson, making me wonder if they knew it. A conceivable reason to omit it is difficulty of usage. E.g., how would you specialize it to Mike's result, especially without the intermediate clue? This is also a computer algebra issue. --rwg

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[math-fun] Quantum time machine 'allows paradox-free time travel'
by Henry Baker 22 Jul '10

22 Jul '10

FYI -- http://www.telegraph.co.uk/science/7904712/Quantum-time-machine-allows-para… Quantum time machine 'allows paradox-free time travel' Quantum physicists at the Massachusetts Institute of Technology believe it is possible to create a time machine which could affect the past without creating a "grandfather paradox". By Tom Chivers Published: 3:09PM BST 22 Jul 2010 Scientists have for some years been able to 'teleport' quantum states from one place to another. Now Seth Lloyd and his MIT team say that, using the same principles and a further strange quantum effect known as 'postselection', it should be possible to do the same backwards in time. Lloyd told the Technology Review: "It is possible for particles (and, in principle, people) to tunnel from the future to the past." Postselection is a vital part of the nascent science of quantum computing. In traditional computing, if a user needs to determine which set of variables in an equation leads to the answer being true, the computer must try every combination until it hits upon one that works. In quantum computing, due to the weird parallel behaviour of subatomic particles, it seems to be possible to simplify the procedure by running all possible variations simultaneously, and selecting only the combinations that make the answer true. Professor Lloyd and his team say that, by combining teleportation and postselection, it would be possible to carry out the quantum teleportation effect in reverse; that is, to decide after the teleportation what the quantum state must have been before it. This works as postselection allows you to dictate which quantum states can be teleported, limiting what state it can have been in before the teleportation. The state of the particle post-teleportation has therefore, in effect, travelled back in time. Dr Richard Low, a quantum computing scientist from the University of Bristol, says: "You could think of it as postselection affecting the history of the particle, sending the state back in time." Unlike previous theories of teleportation, this apparently avoids the "grandfather paradox" - or, to Back to the Future fans, the Marty McFly problem. If you go back and change time, and accidentally end up killing your own grandparent, you create a paradox - you will not be born, so you cannot go back and affect time. Even with subatomic particles, this is still a problem: upon travelling back in time, the particle could somehow destroy its earlier self or move it, thus preventing it from travelling. However, because of the probabilistic nature of quantum mechanics, Prof Lloyd's method seems to avoid this. Anything caused by the time travel must have had a finite probability of happening anyway, so paradoxical impossibilities are out. Further, this time travel method does not involve bending spacetime, unlike other proposed systems. At the moment, the only conditions known that would bend spacetime sufficiently exist in black holes, which would be impractical at best. It is a controversial theory, to say the least. Some physicists claim that the apparently impossible things implied by postselection prove that it cannot work, which would destroy Prof Lloyd's theory before it got off the ground. The theory is unlikely to lead to DeLorean time machines or anything remotely similar. Instead, Prof Lloyd and his team hope that it can advance our understanding of physics: "Our hope is that this theory may prove useful in formulating a quantum theory of gravity," he says. Prof Lloyd's paper, The quantum mechanics of time travel through post-selected teleportation, is published in arxiv.org - Quantum Physics.

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Re: [math-fun] A Scientist Takes On Gravity
by Mike Stay 22 Jul '10

22 Jul '10

On Wed, Jul 21, 2010 at 11:08 PM, Steve Witham <sw(a)tiac.net> wrote: >> Subject: Re: [math-fun] A Scientist Takes On Gravity > > Frederick W Kantor's 1977 book _Information Mechanics_ says that > energy must correspond to information, and then works out consequences That can't be right--they have different units! Perhaps energy corresponds to information *at a given temperature*. Landauer's principle says the energy required to erase a bit is E >= k_B T ln 2 > of that in ways I didn't understand when I had a copy. But I > do remember his explaining gravity in terms of information: > > http://books.google.com/books?ei=AtFHTLjPN4H98AaX-NiaBQ&ct=result&id=VaKBAA… > > "That is, energy and information are released when two objects are > brought nearer each other. From this, I next construct a picture > for gravity." > > Here's a paper from 1974 introducing information mechanics: > http://www.gravityresearchfoundation.org/pdf/awarded/1974/kantor.pdf > > --Steve > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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Re: [math-fun] RE : RE : King walk on a square grid to write 0 to 100
by Michael Kleber 22 Jul '10

22 Jul '10

On Wed, Jul 21, 2010 at 6:20 PM, Eric Angelini <Eric.Angelini(a)kntv.be>wrote: > > I have 20 cells: > 29378 > 78154 > 60291 > 54360 > This might be improved, I guess. > Well, each digit must appear twice (because it all nine other digits as neighbors, and a single location has only eight neighbors). So there's no way to use fewer than 20 cells. The 4x5 rectangle has 4*4 + 3*5 + 4*6 = 55 neighboring pairs, and you need 45 of them. So it's possible that another arrangement of 20 cells could solve the puzzle with fewer than ten unneeded pairings, I suppose... --Michael > Best, > E. > > > -------- Message d'origine-------- > De: math-fun-bounces(a)mailman.xmission.com de la part de Robert Munafo > Date: jeu. 22/07/2010 02:02 > À: math-fun > Objet : Re: [math-fun] RE : King walk on a square grid to write 0 to 100 > > Thanks for the clarification Eric. So it's a word search puzzle where you > need to find the 10 single digits, plus 45 two-digit pairs where the two > digits differ. The need for 45 pairs places a lower limit on the size of > the > grid, since each vertical, horizontal and diagonal line is one unique pair. > > I thought about it for a bit, and I didn't improve on the 4x8 grid size but > at least found a more orderly pattern. Here I have deliberately left > redundant digits in (such as the upper-left 1) to show the pattern. Using a > mono-space font, the lines represent pairs of adjacent digits: > > 1-2-3-4-5-6-7-8 > |x|x|x|x|x|x|x| > 4-5-6-7-8-9-0-1 > |x|x|x|x|x|x|x| > 8-9-0-1-2-3-4-5 > |x|x|x|x|x|x|x| > 1-2-3-4-5-6-7-8 > > > On Wed, Jul 21, 2010 at 19:24, Eric Angelini <Eric.Angelini(a)kntv.be> > wrote: > > > > > Hello Robert, > > sorry to be unclear, > > the "words" alone have to be found, > > not the path linking them. > > So you can find "0", of course, > > "1", "2", etc. And "10", "11", "12"... > > But for "19" (in my example) you > > have to add an extra digit "1"... > > Hope I'm more clear now, > > E. > > > > > What would be a "minimal king's-tour spelling matrix" showing all > > integers > > > from 0 to 100? > > > > > > Double digits can be accomodated by the rule that a cell may be counted > > > twice. > > > > > > The attempt below is for sure not minimal (minimal = quantity of cells > = > > > 24, here): > > > > > > 9 2 3 4 7 8 9 > > > 4 7 8 0 5 3 . > > > 0 6 1 2 9 1 . > > > . 5 4 3 6 4 . > > > > -- > Robert Munafo -- mrob.com > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > -- Forewarned is worth an octopus in the bush.

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