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[math-fun] onion of revolution?
by Marc LeBrun 17 May '04

17 May '04

Hold one end of a long string of beads and get it twirling around a vertical axis with a vigorous stirring motion (like in some cowboy lasso tricks, but without the noose loop). As the speed increases the middle of the string pulls outward, drawing the lower end up, until the bottom part looks horizontal. Since the top is basically fixed it makes a cusp, and overall the profile the whole string traces in the air is shaped something like an onion, oriented cusp-up, like the Mogul-style onion domes seen on many buildings. At high speeds the string gets sharply bent, like a hairpin. Can you give an expression describing the "onion curve" the string forms at a given speed? What path does the string's bottom end trace as a function of the speed of rotation?

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[math-fun] lego robot solves rubik's cube
by Chris Landauer 17 May '04

17 May '04

------- Forwarded Message From: Brian Behlendorf <brian(a)collab.net> Date: Mon, 17 May 2004 13:08:31 -0700 (PDT) As a friend said, "this raises the bar." http://jpbrown.i8.com/cubesolver.html Brian ------- End of Forwarded Message

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[math-fun] possible new Mersenne prime
by Richard Schroeppel 15 May '04

15 May '04

The GIMPS project has another report of a new Mersenne prime. If confirmed, this would be the 41st known. George Woltman is doing some preliminary checking to exclude the type of false report that happened a few months ago. After that, the full check takes a couple of weeks. The full check involves rerunning the Lucas-Lehmer test on different hardware and software. The preliminary check just rechecks the last few thousand steps in the LL test, rerunning from a saved intermediate residue. The recent false report is believed to have been caused by a "just at the wrong time" hardware failure, creating an accidental 0 residue. Rerunning the last few LL steps excludes this case. Rich rcs(a)cs.arizona.edu

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RE: [math-fun] FW: funny limit to well known Generating Function
by Meeussen Wouter (bkarnd) 07 May '04

07 May '04

Ah, the 'real' rwg stood up. My Respects, Sir! The following generalisation pops up in my mind: ... can an algorithm be described that similarly simplifies any ratio of two (order n)polynomes, given that their coefficients of x^k in numerator and denominator are (* of closed form *) :: replace (* .. *) with "a non-recursive expression of n and k" as needed. Or will it always remain a job to be done by hand & eye? Application (for the mathematicaly underprivileged): There are loads of sequences whose GF can be approached by such f_n(x)/g_n(x) ratios. They occur as (I - x A)^-1 where I is the appropriate Identity matrix, and A is a substitution matrix symbolising for example (in the case at hand) the substitution game k->(1,2,...,k,k+1) that generates the Catalans. > -----Original Message----- > From: R. William Gosper [mailto:rwg@tc.spnet.com] > Sent: vrijdag 7 mei 2004 7:27 > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] FW: funny limit to well known Generating > Function > > > Wouter asked > >> f[x_] := Sum[(-1)^(n-j) Binomial[j,n-j] x^(n-j),{j,1,n}] / > Sum[(-1)^(n+2-j) Binomial[j,n+2-j] > >x^(n+2-j),{j,1,n+2}] > >> produces a well known Generating Function for the limit > n->Infinity. > > >> Q: What is needed to force Limit[f[x], n->Infinity] into > submission? > >> it equals (1-2x - Sqrt[1 - 4*x]) / (2 x^2) > > > >> Mathematica's package Calculus`Limit` doesn't feel like > tackling it. > >> Should it? > > and I blathered > > >Both Mma and Macsyma need a little help here. Just send the > sum limits > >to infinity first. It might also help to start from j=0, > which you may > >have intended. If not, Mma may need you to "manually" > subtract the j=0 > >term. Then you can even start from j=-inf. The resulting > sums are simple > >algebraics(n), and n->inf is a snap. > > Bull. This isn't even a limit problem. But it exposes > several defects > in my personally enhanced Macsyma. All that should have been > necessary > was (declare(n,integer), hypersimp(sum(...))). Instead, I have to > sum from floor(n/2) instead of 1: > (c225) > (declare(n,integer),sum((-1)^(n-j)*binomial(j,n-j)*x^(n-j),j,f > loor(n/2),n) > /sum((-1)^(n+2-j)*binomial(j,n+2-j)*x^(n+2-j),j,floor(n/2),n+2)) > > n > ==== > \ n + j n - j > > binomial(j, n - j) (- 1) x > / > ==== > n > j = floor(-) > 2 > (d225) --------------------------------------------------------- > n + 2 > ==== > \ n + j n - j + 2 > > binomial(j, n - j + 2) (- 1) x > / > ==== > n > j = floor(-) > 2 > > then manually reverse the summation direction, > (c226) rangereverse(%) > > n > floor(-) - 1 > 2 > ==== > \ n + j n - j > > binomial(j, n - j) (- 1) x > / > ==== > j = n + 1 > (d226) --------------------------------------------------------- > n > floor(-) - 1 > 2 > ==== > \ n + j n - j + 2 > > binomial(j, n - j + 2) (- 1) x > / > ==== > j = n + 3 > > and then manually change the summation variable: > (c227) linchange(%,n+1-j) > > n > n - floor(-) > 2 > ==== j > \ binomial(n - j, j) x > > (- ---------------------) > / j > ==== (- 1) > j = 0 > (d227) ---------------------------------------------- > n > n - floor(-) > 2 > ==== j + 2 > \ binomial(n - j, j + 2) x > > (- -----------------------------) > / j > ==== (- 1) > j = - 2 > > Finally, > (c228) hypersimp(%) > > 4 > (d228) -------------------- > 2 > (sqrt(1 - 4 x) + 1) > > I assume this is the square of what you'd have gotten with f(n)/f(n+1) > instead of f(n)/f(n+2). > --rwg > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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[math-fun] FW: funny limit to well known Generating Function
by Meeussen Wouter (bkarnd) 06 May '04

06 May '04

> hi all, > > (I tried to send this to rwg(a)tc.spnet.com but it bounced. Bill, where are you?) > > > f[x_] := Sum[(-1)^(n-j) Binomial[j,n-j] x^(n-j),{j,1,n}] / Sum[(-1)^(n+2-j) Binomial[j,n+2-j] x^(n+2-j),{j,1,n+2}] > produces a well known Generating Function for the limit n->Infinity. > > Q: What is needed to force Limit[f[x], n->Infinity] into submission? > it equals (1-2x - Sqrt[1 - 4*x]) / (2 x^2) > > Mathematica's package Calculus`Limit` doesn't feel like tackling it. > Should it? > > > Wouter. =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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[math-fun] Re: From Dan Asimov to Richard re: Divisor chains
by Richard Guy 05 May '04

05 May '04

Dan raises a question which has already been anticipated by some funsters. Many of you will now have several definitions. I (&/or Paul Vaderlind) originally meant `chain of order n beginning with n', but others have noted that they could begin with any number > 1. There is a question about `cyclic' chains (necklaces?). These have been defined as those whose first member divides n(n+1)/2, though that seems even more artificial. No chain can be truly cyclic, though one can ask for chains, or even necklaces, in which each member divides the sum of the previous k members. This is not possible for k = 1 or 2 [except for 4 2 3 1 -- which won't quite make a necklace, 'cos 2 doesn't divide 4+1], but perhaps for some larger k ? [for the first k members, `divisor of sum of all previous members'] Example of chain or necklace, or proof of impossibility ? R. On Tue, 4 May 2004, Daniel Asimov wrote: > Richard, [snip] > Do you mean to define a divisor chain or order n as necessarily > beginning with n ??? > > (That may lead to a trickier class of questions, but it seems a > bit artificial as a definition.) [snip] > > Dan Asimov > Visiting Scholar > Math Dept. > Univ. of Calif. > Berkeley, California > (510) 524-7789 >

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[math-fun] The Honors Class
by Thane Plambeck 05 May '04

05 May '04

I enjoyed this introductory quotation and first paragraph from Ben H. Yandell's excellent "The Honors Class: Hilbert's Problems and their Solvers" (AK Peters, 2002) * ** "That I have been able to accomplish anything in mathematics is really due to the fact that I have always found it so difficult." --- David Hilbert My mathematical readers know how to read this book. They will read it the way they read mathematics books. If they don't understand something (a state they are used to), they will keep reading in the hope tha they will understand the next thing. They will skip sections that don't interest them. So this is to my nonmathematical readers. If you are reading for the story, keep reading if you don't understand something. Skip a bit if you want---the biographical narrative will pick up again. Pretend you are reading Moby Dick and you've come to another section on whaling.... Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org

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Re: A proposal Re: [math-fun] emails about div. chains
by N. J. A. Sloane 04 May '04

04 May '04

personally i think Dan's idea is not so great! i hate having to use the web to get things that used to come in the mail the nice thing about a mailing list is that it comes to you, not the other way round Neil

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A proposal Re: [math-fun] emails about div. chains
by Daniel Asimov 04 May '04

04 May '04

Neil writes: << apparently some people have been posting to the math fun list when they should have been posting to the seqfan list let's keep the sequence stuff on the seq fan list , Ok? >> I very much sympathize with Neil's plight. Personally, I don't post to the seqfan list (or number theory, or Life), because I'm not sufficiently interested in these topics to want to get all the e-mail traffic they generate. I usually figure that if something's sufficiently interesting, it'll show up on math-fun eventually. ------------------------------------------------------------------- Here's a feasible though utopian solution that accommodates both points of view: Ideally, all these lists could be web based. Like Usenet, when somebody posts to several lists, that posting is marked as "read" for a given individual as soon as they read it on any list -- so they never have to step in the same article twice. Is there any possibility that the various limited-access math lists could be collected under some (perhaps web-based) Usenet-like meta-software that ensures no one has to inadvertently read a post twice? --Dan

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[math-fun] How Many Crossword Puzzles Are There?
by Richard Schroeppel 04 May '04

04 May '04

My friend Bob Baillie <rjbaillie(a)frii.com> asks > Has anyone ever estimated how many crossword puzzles of size N by N there are, where the entries are words in English? I assume that the answer for very large N must be exponential in the area of the crossword. Assume we can make two KxK tiles with the same boundary conditions but differeing internally, and that all the necessary fixups can be made on the outer edge of the pattern. Similar questions could be asked for a randomly generated language with English statistics, and for other real languages. But this is not really the question he's asking. Rich

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