November 2015 - mathfun@mailman.xmission.com

[math-fun] The Paintball Party Problem
by James Propp 16 Nov '15

16 Nov '15

2n kids are at a paintball party, and a mathematician parent is assigned the task of coming up with a sequence of 2n-1 ways of dividing the kids into two equal-sized teams, with the property that each kid gets to play on the same team with each other kid exactly n-1 times. For what values of n is this possible? For n=2, there is an obvious solution: {1,2 | 3,4}, {1,3 | 2,4}, {1,4 | 2,3} (here I am using notation that I hope is transparent). For n=3, there is no solution, as a brute-force approach discloses. (Unless I made a mistake.) For n=4, there is a solution based on the vector space F_2^3, where F_2 is the field with two elements. Specifically, associate the eight players with the eight elements of F_2^3 = {(x,y,z): x,y,z in F_2} in some arbitrary fashion, and then use the seven linear functions f(x,y,z) = x, y, z, x+y, x+z, y+z, and x+y+z to divide F_2^3 into two subsets of size 4 according to whether f(x,y,z) is 0 or 1. So I ask, for what values of n can a satisfactory rotation pattern of length 2n-1 be constructed? (This may be a known question in design theory, but I thought I'd ask you-all before I post to MathOverflow.) The case n=4 arose yesterday at my son's paintball birthday party, and I came up with the F_2^3 solution in the nick of time. Some extra issues arose that compromised the success of the design (as measured by my son's satisfaction), and it turns out that a different ordering of the seven linear functions would have led to a better rotation pattern, but I'll get into that later. Jim Propp

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Re: [math-fun] Better Dragon coruscator
by Bill Gosper 14 Nov '15

14 Nov '15

Fred: Better. But I still have no idea what principle is being demonstrated .... WFL Just illustrating draconic anatomy. The dyadic construction, the fivefold partitioning of the area, and the abundance of triple points. Especially if you have a pausable gif player. If it's reversible to boot, you only need one of these: gosper.org/coruscf1.gif gosper.org/1fcsuroc.gif --rwg I've doubtless overrepeated the anecdote of Mandelbrot on the way to the airport asking me the dimension of the Dragon's boundary, having seen one for the first time at Knuth's house only days before. I told him I'd call him with the answer after working it out from the recursive graphics description we used to correct the base i-1 figure in AoCP II. I got a messy cubic with a messier solution. He asked what was the cubic. I told him. "Yes, that's right." "You already knew? That's not your kind of formula! How did you get it?" "I looked how the wiggles fit into each other."

1 0

[math-fun] Better Dragon coruscator
by Bill Gosper 14 Nov '15

14 Nov '15

gosper.org/corrusc3.gif --rwg

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[math-fun] Draft of November essay
by James Propp 13 Nov '15

13 Nov '15

A draft of my November essay (a review of the picture book "The Boy Who Loved Math") is now previewable at https://mathenchant.wordpress.com?p=259&shareadraft=56464ea248cd1 If you use that link to comment, please be sure to include your name in your comment, so that I'll know who you are if I have any follow-up questions about your comments. (Unless you absolutely want to comment anonymously for some reason.) Thanks, Jim Propp

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[math-fun] Microsoft releasing quantum computing simulator
by Henry Baker 13 Nov '15

13 Nov '15

FYI -- "The simulator is 30 qubits, which requires 32 gigabytes of memory. That means it can run on a high-end laptop or desktop. Every qubit added, he warns, will *double* the memory requirement." http://fortune.com/2015/11/13/microsoft-quantum-computing-simulator/?xid=ya… Microsoft Simulator Brings Quantum Computing One Step Closer to the Masses by Barb Darrow @FortuneMagazine 9:11 AM EST On Friday, Microsoft is releasing simulation software that it says will let academics, scientists, or even do-it-yourself eggheads simulate quantum computing on their laptops. The promise of quantum computing, which breaks the nuts-and-bolts of computing down to the sub-atomic level, is that it can solve problems that go far beyond the capabilities of even todayÂs most powerful computers. The current generation of computers represent all data as ones and zeros. ItÂs all a very binary, on-or-off proposition. By relying on smaller particles, like photons or electrons, a quantum computer would be able to look at data that can hold several contradictory states at the same time. For instance, instead of those ones and zero, a quantum computer would deal with quantum bits (also known as qubits), which would accommodate multiple states. ItÂs a bit complicated for mere mortals to understand, but what all that means is that quantum computers should be able to calculate certain problemsÂmodeling molecules for exampleÂ-much, much faster than their transistor-based forebears. One problem with all of this work is itÂs still largely theoretical. Scientists think it will work, but the gear required to run these calculations isnÂt here yet. ThatÂs because to keep all those wee subatomic particles in a stable state, the equipment has to be very, very coldÂ100 times colder than the temperature in outer spaceÂaccording to Dave Wecker, chief architect of the MicrosoftÂs quantum team, also known as QuArC. As if thatÂs not difficult enough, itÂs also not clear how best to state the problems that need to be solved by quantum computers, or how to interpret those results since this will be a whole new qubit-inspired world. ThatÂs something the simulator can help researchers get their minds around. The software, which Microsoft has been using in house for some time, is called Language-Integrated Quantum Operations, or LIQUi|> (and no, thatÂs not a typo). LIQ stands for language integrated quantum, says Wecker. The vertical bar stands for ÂketÂ which is tech speak for the quantum state. U is the operation performed on the quantum state and the Âgreater thanÂ bracket is simply a bracket, Wecker says. As an example of a problem that a quantum computer could solve, Wecker cites the creation of high-temperature superconductors. ÂWe lose electricity on our transmission lines. Superconductors would have no losses, but no one can build a super conductor at anything near room temperature. We could model than on a quantum computer and at least have the hope of solving that problem,Â Wecker says. Ditto the creation of organic batteries, which wouldnÂt have to rely on expensive and polluting heavy metals. ÂWe know we can build them but which molecules do we use? There are millions. A cloud of quantum computers could test them all out.Â Drug modeling and testing is another potentially huge use for these computers, which explains why companies like Microsoft, Google, IBM, and others are investing in the field. D-Wave Systems, has its own version of a quantum processor, one of which was just sold to Los Alamos National Laboratory. There are also other quantum computing simulators on the market, including one from Google for example. Wecker says the differentiator here is this is industrial strength and academics or researchers can extend it, adding their own quantum calculations as needed. The simulator is 30 qubits, which requires 32 gigabytes of memory. That means it can run on a high-end laptop or desktop. Every qubit added, he warns, will double the memory requirement. The code means some simulations can run locally, or if you have bigger requirements, you can turn to the cloud of your choiceÂsuch as Microsoft Azure, Amazon Web Services, or Google Cloud PlatformÂ-he adds, there are no restrictions on use. The executable code, along with documentation, will be available on the Github code repository for download. Microsoft will talk more about the simulator next week at Supercomputing 2015 in Austin.

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[math-fun] about some continued fractions related to sqrt(5).
by Simon Plouffe 13 Nov '15

13 Nov '15

Hello funsters, here is something which is a puzzle to me, I was exploring numbers like Fibonacci(k)/(Lucas(k-n)*((1+sqrt(5))/2)^n, where n is small and k >> 1. Here is the odd thing : if you expand for example the number 5193981023518027157495786850488117/7177905237579946589743592924684178/(1/2+1/2*5^(1/2))^2 into a continued fraction, the surprise comes from the partial quotients of that expansion. It is quite chaotic. The maximal value being 83364870763649235403921261388869364666045817819140268784224747492762, What is this ? how come a simple number like a/b*sqrt(5) has a c.frac expansion which such values ? I thought that approximations of a number like sqrt(5) could not be like that. A quick examination shows that the size of these numbers (the maximal value of the c.f expansion) will be like Fibonacci(k)^2 (if n is small). In this example we have, n = 163 and n = 2. Can someone tell me how is this possible ? I really don't see a general formula, since for some values of n and k, the behavior of the c.f. is quite <normal> with no high values, what are the conditions to have the maximal value ? I made some programs to analyze this and found only bizarre examples. Best regards, ps : I am back on the math-fun list after a quick absence. Simon Plouffe

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[math-fun] Plouffe's Fibonnacci-Lucas-Golden numbers
by Warren D Smith 13 Nov '15

13 Nov '15

Simon Plouffe: I was exploring numbers like Fibonacci(k)/(Lucas(k-n)*g^n) where n is small and k >> 1 and g=(1+sqrt(5))/2 is the golden ratio. WDS: Plouffe mistyped the above formula, as is evident from his mismatched parentheses. I have written what I am guessing he intended. Plouffe: The continued fractions of such numbers contain large partial quotients. WDS: Plouffe's example, rewritten, is 5193981023518027157495786850488117 / (3588952618789973294871796462342089* (3+sqrt(5))) = 0.27639... = [0; 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 83364870763649235403921261388869364666045817819140268784224747492762, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 2, 8, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 4, 1, 2, 6, 2, 1, 13, 1, 7, 3, 1, 3, 7, 1, 13, 1, 7, 3, 1, 3, 7, 1, 13, 1, 7, 3, 1, 3, 7, 1, 13, 1, 7, 3, 1, 3, 7, 1, 13, 1, 7, 3, 1, 3, 7, 1, 13, 1, 7, 3, 1, 3, 7, 1, 13, 1, 7, 3, 1, 3, 7, 1, 13, 1, 7, 3, 1, 3, 7, 1, 12, 1, 1, 1, 5, 1, 2, 4, 2, 4, 2, 1, 5, 1, 1, 1, 11, 1, 1, 1, 5, 1, 2, 4, 2, 4, 2, 1, 5, 1, 1, 1, 11, 1, 1, 1, 5, 1, 2, 4, 2, 4, 2, 1, 5, 1, 1, 1, 11, 1, 1, 1, 5, 1, 2, 4, 2, 4, 2, 1, 5, 1, 1, 1, 11, 1, 1, 1, 5, 1, 2, 4, 2, 4, 2, 1, 5, 1, 1, 1, 11, 1, 1, 1, 5, 1, 2, 4, 6, 4, 7, 1, 6, 2, 2, 1, 31, 1, 2, 2, 6, 1, 7, 4, 6, 4, 7, 1, 6, 2, 2, 1, 31, 1, 2, 2, 6, 1, 7, 4, 6, 4, 7, 1, 6, 2, 2, 1, 31, 1, 2, 2, 6, 1, 7, 4, 6, 4, 7, 1, 6, 2, 2, 1, 31, 1, 2, 2, 6, 1, 5, 1, 2, 4, 3, 1, 3, 4, 2, 1, 5, 1, 4, 1, 2, 2, 2, 1, 1, 1, 29, 1, 1, 1, 2, 2, 2, 1, 4, 1, 5, 1, 2, 4, 3, 1, 3, 4, 2, 1, 5, 1, 4, 1, 2, 2, 2, 1, 1, 1, 29, 1, 1, 1, 2, 2, 2, 1, 4, 1, 5, 1, 2, 4, 3, 1, 3, 4, 2, 1, 5, 1, 4, 1, 2, 2, 2, 1, 1, 1, 29, 1, 1, 1, 2, 2, 2, 1, 13, 1, 5, 1, 7, 4, 4, 2, 4, 4, 7, 1, 5, 1, 13, 1, 2, 2, 7, 1, 1, 1, 78, 1, 1, 1, 7, 2, 2, 1, 13, 1, 5, 1, 7, 4, 4, 2, 4, 4, 7, 1, 5, 1, 13, 1, 2, 2, 7, 1, 1, 1, 78, 1, 1, 1, 7, 2, 2, 1, 13, 1, 5, 1, 7, 4, 4, 6, 4, 4, 2, 1, 5, 1, 5, 1, 1, 1, 12, 1, 2, 6, 7, 1, 1, 1, 1, 1, 76, 1, 1, 1, 1, 1, 7, 6, 2, 1, 12, 1, 1, 1, 5, 1, 5, 1, 2, 4, 4, 6, 4, 4, 2, 1, 5, 1, 5, 1, 1, 1, 12, 1, 2, 6, 7, 1, 1, 1, 1, 1, 76, 1, 1, 1, 1, 1, 7, 6, 2, 1, 12, ... eventually periodic] The Fibonacci numbers: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765,... each is sum of 2 preceding. The Lucas numbers: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349,... each is sum of 2 preceding. So what is going on here? First of all, every number of Plouffe's class is a quadratic irrational, hence has an ultimately-periodic continued fraction. So the only thing that is interesting is the initial segment of the CF before the period begins. That initial segment contains a long string of 1s basically because the Plouffe numbers when k is large yield very good approximations of the golden ratio. But the funny thing is, it then looks like a periodic sequence begins, but then it isn't what you thought, then it looks like a longer periodic sequence begins, but it again isn't what you thought, and so on for quite a while. So it is as though Plouffe's quadratic irrational, is very well-approximated by a sequence of less-complex quadratic irrationals. Well, that actually is (of course) true, although it may not be obvious which precise subsequence among them are good enough to cause this behavior. On 11/12/15, math-fun-request(a)mailman.xmission.com <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 > https://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..." > > > Today's Topics: > > 1. about some continued fractions related to sqrt(5). (Simon Plouffe) > 2. Re: about some continued fractions related to sqrt(5). (Mike Stay) > 3. Re: about some continued fractions related to sqrt(5). > (Dan Asimov) > 4. Re: about some continued fractions related to sqrt(5). > (Dan Asimov) > 5. Re: about some continued fractions related to sqrt(5). > (Dan Asimov) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Thu, 12 Nov 2015 20:30:14 +0100 > From: Simon Plouffe <simon.plouffe(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] about some continued fractions related to sqrt(5). > Message-ID: <5644E8C6.1020109(a)gmail.com> > Content-Type: text/plain; charset=utf-8; format=flowed > > > Hello funsters, > > here is something which is a puzzle to me, > > I was exploring numbers like Fibonacci(k)/(Lucas(k-n)*((1+sqrt(5))/2)^n, > where n is small and k >> 1. > > Here is the odd thing : if you expand for example the number > > 5193981023518027157495786850488117/7177905237579946589743592924684178/(1/2+1/2*5^(1/2))^2 > into a continued fraction, the surprise comes from the partial quotients > of that expansion. It is quite chaotic. > The maximal value being > 83364870763649235403921261388869364666045817819140268784224747492762, > > What is this ? how come a simple number like a/b*sqrt(5) has a c.frac > expansion which such > values ? I thought that approximations of a number like sqrt(5) could > not be like that. > > A quick examination shows that the size of these numbers (the maximal > value of the c.f expansion) > will be like Fibonacci(k)^2 (if n is small). In this example we have, > > n = 163 and n = 2. > > Can someone tell me how is this possible ? > > I really don't see a general formula, since for some values of n and > k, the behavior of > the c.f. is quite <normal> with no high values, what are the conditions > to have the > maximal value ? I made some programs to analyze this and found only > bizarre examples. > > Best regards, > > ps : I am back on the math-fun list after a quick absence. > > Simon Plouffe > > > > > ------------------------------ > > Message: 2 > Date: Thu, 12 Nov 2015 12:07:15 -0800 > From: Mike Stay <metaweta(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] about some continued fractions related to > sqrt(5). > Message-ID: > <CAKQgqTavya1y3Uci+r8iNL1FV-cTQn5Va6xjHJku33fGYeFDpQ(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > The 163 and large terms immediately makes me think of the "exotic" > continued fraction for the real root of x^3 - 8x - 10 = 0. > > On Thu, Nov 12, 2015 at 11:30 AM, Simon Plouffe <simon.plouffe(a)gmail.com> > wrote: >> >> Hello funsters, >> >> here is something which is a puzzle to me, >> >> I was exploring numbers like Fibonacci(k)/(Lucas(k-n)*((1+sqrt(5))/2)^n, >> where n is small and k >> 1. >> >> Here is the odd thing : if you expand for example the number >> >> 5193981023518027157495786850488117/7177905237579946589743592924684178/(1/2+1/2*5^(1/2))^2 >> into a continued fraction, the surprise comes from the partial quotients >> of that expansion. It is quite chaotic. >> The maximal value being >> 83364870763649235403921261388869364666045817819140268784224747492762, >> >> What is this ? how come a simple number like a/b*sqrt(5) has a c.frac >> expansion which such >> values ? I thought that approximations of a number like sqrt(5) could not >> be >> like that. >> >> A quick examination shows that the size of these numbers (the maximal >> value >> of the c.f expansion) >> will be like Fibonacci(k)^2 (if n is small). In this example we have, >> >> n = 163 and n = 2. >> >> Can someone tell me how is this possible ? >> >> I really don't see a general formula, since for some values of n and k, >> the >> behavior of >> the c.f. is quite <normal> with no high values, what are the conditions >> to >> have the >> maximal value ? I made some programs to analyze this and found only >> bizarre examples. >> >> Best regards, >> >> ps : I am back on the math-fun list after a quick absence. >> >> Simon Plouffe >> >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://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 > > > > ------------------------------ > > Message: 3 > Date: Thu, 12 Nov 2015 12:27:01 -0800 > From: Dan Asimov <dasimov(a)earthlink.net> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] about some continued fractions related to > sqrt(5). > Message-ID: <4A46CFC5-64BD-4738-8B8D-A89E8E20C4B9(a)earthlink.net> > Content-Type: text/plain; charset=utf-8 > > Mike's suggestion may be prescient, as the real root of > > x^3 - 8x - 10 = 0 > > is > > K = (5-q)^(1/3) + (5+q)^(1/3) > > where > > q = sqrt(163/27) = sqrt(6 + 1/27) > > ?Dan > > >> On Nov 12, 2015, at 12:07 PM, Mike Stay <metaweta(a)gmail.com> wrote: >> >> The 163 and large terms immediately makes me think of the "exotic" >> continued fraction for the real root of x^3 - 8x - 10 = 0. >> >> On Thu, Nov 12, 2015 at 11:30 AM, Simon Plouffe <simon.plouffe(a)gmail.com> >> wrote: >>> >>> Hello funsters, >>> >>> here is something which is a puzzle to me, >>> >>> I was exploring numbers like Fibonacci(k)/(Lucas(k-n)*((1+sqrt(5))/2)^n, >>> where n is small and k >> 1. >>> >>> Here is the odd thing : if you expand for example the number >>> >>> 5193981023518027157495786850488117/7177905237579946589743592924684178/(1/2+1/2*5^(1/2))^2 >>> into a continued fraction, the surprise comes from the partial quotients >>> of that expansion. It is quite chaotic. >>> The maximal value being >>> 83364870763649235403921261388869364666045817819140268784224747492762, >>> >>> What is this ? how come a simple number like a/b*sqrt(5) has a c.frac >>> expansion which such >>> values ? I thought that approximations of a number like sqrt(5) could not >>> be >>> like that. >>> >>> A quick examination shows that the size of these numbers (the maximal >>> value >>> of the c.f expansion) >>> will be like Fibonacci(k)^2 (if n is small). In this example we have, >>> >>> n = 163 and n = 2. >>> >>> Can someone tell me how is this possible ? >>> >>> I really don't see a general formula, since for some values of n and k, >>> the >>> behavior of >>> the c.f. is quite <normal> with no high values, what are the conditions >>> to >>> have the >>> maximal value ? I made some programs to analyze this and found only >>> bizarre examples. >>> >>> Best regards, >>> >>> ps : I am back on the math-fun list after a quick absence. >>> >>> Simon Plouffe > > > > ------------------------------ > > Message: 4 > Date: Thu, 12 Nov 2015 12:32:03 -0800 > From: Dan Asimov <asimov(a)msri.org> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] about some continued fractions related to > sqrt(5). > Message-ID: <B66BFE30-3B45-4A90-A173-3F5A43A25868(a)msri.org> > Content-Type: text/plain; charset=utf-8 > > P.S. Whose continued fraction to 100 terms is: > > [3; 3, 7, 4, 2, 30, 1, 8, 3, 1, 1, 1, 9, 2, 2, 1, 3, 22986, 2, 1, > > 32, 8, 2, 1, 8, 55, 1, 5, 2, 28, 1, 5, 1, 1501790, 1, 2, 1, 7, 6, 1, > > 1, 5, 2, 1, 6, 2, 2, 1, 2, 1, 1, 3, 1, 3, 1, 2, 4, 3, 1, 35657, > > 1, 17, 2, 15, 1, 1, 2, 1, 1, 5, 3, 2, 1, 1, 7, 2, 1, 7, 1, 3, > > 25, 49405, 1, 1, 3, 1, 1, 4, 1, 2, 15, 1, 2, 83, 1, 162, 2, 1, 1, 1, ...] > > ?Dan > > >> On Nov 12, 2015, at 12:27 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: >> >> Mike's suggestion may be prescient, as the real root of >> >> x^3 - 8x - 10 = 0 >> >> is >> >> K = (5-q)^(1/3) + (5+q)^(1/3) >> >> where >> >> q = sqrt(163/27) = sqrt(6 + 1/27) >> >> ?Dan >> >> >>> On Nov 12, 2015, at 12:07 PM, Mike Stay <metaweta(a)gmail.com> wrote: >>> >>> The 163 and large terms immediately makes me think of the "exotic" >>> continued fraction for the real root of x^3 - 8x - 10 = 0. >>> >>> On Thu, Nov 12, 2015 at 11:30 AM, Simon Plouffe <simon.plouffe(a)gmail.com> >>> wrote: >>>> >>>> Hello funsters, >>>> >>>> here is something which is a puzzle to me, >>>> >>>> I was exploring numbers like >>>> Fibonacci(k)/(Lucas(k-n)*((1+sqrt(5))/2)^n, >>>> where n is small and k >> 1. >>>> >>>> Here is the odd thing : if you expand for example the number >>>> >>>> 5193981023518027157495786850488117/7177905237579946589743592924684178/(1/2+1/2*5^(1/2))^2 >>>> into a continued fraction, the surprise comes from the partial >>>> quotients >>>> of that expansion. It is quite chaotic. >>>> The maximal value being >>>> 83364870763649235403921261388869364666045817819140268784224747492762, >>>> >>>> What is this ? how come a simple number like a/b*sqrt(5) has a c.frac >>>> expansion which such >>>> values ? I thought that approximations of a number like sqrt(5) could >>>> not be >>>> like that. >>>> >>>> A quick examination shows that the size of these numbers (the maximal >>>> value >>>> of the c.f expansion) >>>> will be like Fibonacci(k)^2 (if n is small). In this example we have, >>>> >>>> n = 163 and n = 2. >>>> >>>> Can someone tell me how is this possible ? >>>> >>>> I really don't see a general formula, since for some values of n and k, >>>> the >>>> behavior of >>>> the c.f. is quite <normal> with no high values, what are the conditions >>>> to >>>> have the >>>> maximal value ? I made some programs to analyze this and found only >>>> bizarre examples. >>>> >>>> Best regards, >>>> >>>> ps : I am back on the math-fun list after a quick absence. >>>> >>>> Simon Plouffe >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > ------------------------------ > > Message: 5 > Date: Thu, 12 Nov 2015 14:35:25 -0800 > From: Dan Asimov <dasimov(a)earthlink.net> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] about some continued fractions related to > sqrt(5). > Message-ID: <169CB4ED-D4EC-4131-8A69-B6629F3CC329(a)earthlink.net> > Content-Type: text/plain; charset=utf-8 > > No specific idea, but it's interesting to consider continued fractions > as a dynamical system: > > F: (0,1) ?> (0,1) > > via > > F(x) = 1/x - floor(1/x) > > Then as Gauss discovered and it's fun to prove, there is a measure M on > (0,1) > that is preserved by F. > > Namely, M([a,b)) := (1/ln(2)) Integral_{a <= x < b} dx/(1+x). > > The factor of 1/ln(2) is just to make the total measure come out to be 1. > > ( > F preserves the measure M in the sense that > > M(Finv(([a,b)) = M([a,b)) for all intervals [a,b) \sub (0,1). > ) > > > (This is discussed in the incredibly enjoyable book by Mark Kac, > "Statistical Independence in Probability, Analysis and Number Theory", > 1959.) > > With respect to a preserved measure (having total measure = 1), a dynamical > system > can have a number of interesting properties: > > 1. Ergodicity. This means that any sets that are invariant under the > dynamical system > have measure 0 or 1. > > The dynamical system F: (0,1) ?> (0,1) is in fact ergodic. > > IF a dynamical system > > 2. Mixing. This means that for any two measurable sets A, B of measures > > 0, > the intersection > > B \int [the image of A under the dynamical system] > > approaches having its measure = the product of [the measure of A]* [the > measure of B], > as time ?> oo. > > This can be interpreted as saying that for any two events A and B of > positive measure, > the image of A under the dynamical system approaches being an independent > event > with respect to the event B, as time ?> oo. > > I don't know if this holds for F: (0,1) ?> (0,1). > > Does it? > > Remark: It's easy to see that mixing implies ergodicity, but not > conversely. > > It's less easy to find an example of a dynamical system that is mixing > (but it's a reasonable exercise). > > ?Dan > > >> On Nov 12, 2015, at 11:30 AM, Simon Plouffe <simon.plouffe(a)gmail.com> >> wrote: > ... >> here is something which is a puzzle to me, >> >> I was exploring numbers like Fibonacci(k)/(Lucas(k-n)*((1+sqrt(5))/2)^n, >> where n is small and k >> 1. >> >> Here is the odd thing : if you expand for example the number >> >> 5193981023518027157495786850488117/7177905237579946589743592924684178/(1/2+1/2*5^(1/2))^2 >> into a continued fraction, the surprise comes from the partial quotients >> of that expansion. It is quite chaotic. >> The maximal value being >> 83364870763649235403921261388869364666045817819140268784224747492762, >> >> What is this ? how come a simple number like a/b*sqrt(5) has a c.frac >> expansion which such >> values ? I thought that approximations of a number like sqrt(5) could not >> be like that. >> >> A quick examination shows that the size of these numbers (the maximal >> value of the c.f expansion) >> will be like Fibonacci(k)^2 (if n is small). In this example we have, >> >> n = 163 and n = 2. >> >> Can someone tell me how is this possible ? >> >> I really don't see a general formula, since for some values of n and k, >> the behavior of >> the c.f. is quite <normal> with no high values, what are the conditions to >> have the >> maximal value ? I made some programs to analyze this and found only >> bizarre examples. >> >> Best regards, >> >> ps : I am back on the math-fun list after a quick absence. > > > ------------------------------ > > Subject: Digest Footer > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > ------------------------------ > > End of math-fun Digest, Vol 153, Issue 21 > ***************************************** > -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] ln(x^2+1)
by Erich Friedman 12 Nov '15

12 Nov '15

this morning i found that for f(x)=ln(x^2+1), that the (4n+2)th derivatives all vanish at x=1. it took me an hour to figure out why, using geometric series, but surely there must be an easier way to see this. does anyone have an elegant proof of this? erich

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Re: [math-fun] ln(x^2+1) -- correction
by Adam P. Goucher 12 Nov '15

12 Nov '15

[Actually, I made a slight typo in the previous e-mail. But this one is slicker since it doesn't require differentiation of log.] log(x^2 + 1) = log(x + i) + log(x - i), so it suffices to show that the (4n + 2)th derivatives of log(x + i) are all pure-imaginary when x = 1. Let x = 1 + y, so we are interested in the Taylor series of: log(y + (1 + i)) about the point y = 0. We want to show that the (4n + 2)th coefficients are all pure-imaginary. This follows from the claim that the coefficient in y^n of the series expansion of log(y + a) is [a real multiple of] a^-n, which is true by dimensional analysis. Sincerely, Adam P. Goucher > Sent: Thursday, November 12, 2015 at 3:24 PM > From: "Erich Friedman" <erichfriedman68(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] ln(x^2+1) > > this morning i found that for f(x)=ln(x^2+1), that the (4n+2)th derivatives all vanish at x=1. it took me an hour to figure out why, using geometric series, but surely there must be an easier way to see this. does anyone have an elegant proof of this? > > erich > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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[math-fun] 24-hour math hotline
by Henry Baker 11 Nov '15

11 Nov '15

Siri for the math-serious ? It would be fun to call this hotline with some really hard math problems & record their answers. http://solvex4u.com/

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