March 2018 - mathfun@mailman.xmission.com

Re: [math-fun] Infinitely long *chain-like* codes ?
by Henry Baker 29 Mar '18

29 Mar '18

Oops! I found reference to chain codes in the Gray code wiki page, but didn't change the subject header. At 05:02 PM 3/28/2018, Henry Baker wrote: >Most of the applications in the Wikipedia article on Gray codes >consider *fixed length* codes for some length n. > >One of the properties of binary gray codes is that substrings of them >contain all of the n-bit binary integers, but far more efficiently >than simply concatenating all 2^n of them. > >I'm interested in a slightly different *infinite* sequence of >bits, which contain these n-bit substrings as early as possible, >s.t., for any n, there is a function f(n) that tells my how >large of an initial substring of this infinite sequence I have >to generate in order to make sure that all 2^n binary integers >appear at least once. > >Furthermore, I may want to "tune" this infinite sequence in >order to change the statistics of how frequently all of the >different k-bit integers appear, how frequently all of the >(k+1)-bit integers appear, etc. > >Has anyone studied such sequences?

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Re: [math-fun] [MOG-group] non-MOG
by Brent Meeker 29 Mar '18

29 Mar '18

Dunno about that. It's hard to compare since the Wankel has all those face seal "rings" as well as the tip "rings". But Wankel's were notoriously fuel inefficient because of all the surface area through which they lost heat. I had an RX2 sedan though...nice car...fast for it's size. The problem was that once it blew a seal, it was going to cost more than I paid for the car to replace them. Brent On 3/28/2018 4:23 PM, Mickey wrote: > It seems to me the Wankel engine would have far les fiction loss than > either a radial or the Brickley. > STL Mike > > Sent from my iPhone > > On Mar 28, 2018, at 5:17 PM, Brent Meeker <meekerdb(a)verizon.net > <mailto:meekerdb@verizon.net>> wrote: > >> >> >> On 3/28/2018 2:26 PM, Duncan Charlton wrote: >>> Brent, >>> >>> He’s in the process of totally changing the look of his website so >>> the guts of the engine can no longer be seen there but I found this >>> site capture in the web archives from October 1st, 2016. Be sure to >>> allow Flash to load and then click to put the parts into motion. If >>> it looks like there might be a hidden or missing pivot point, that’s >>> because he didn’t want to make it too easy for someone to steal his >>> design. You’ll have to use your imagination to figure out what the >>> inboard end of the connecting rod is fastened to. >> >> I have no problem with that. What I wonder is what the H shaped >> links behind each piston are doing? They seem to add three more >> friction surfaces per piston. I'm afraid I'm less than impressed. >> Compare a radial engine >> >> <ecgbciaaihjghmhb.png> >> >> It has one bearing that turns through 360deg while all the others >> just oscillate through a small angle. Are radial engines >> significantly more fuel efficient?...no. First, most friction comes >> from the piston/cylinder interface, specifically at the compression >> rings. Second, friction is not 15% of loss. Even at low load it is >> only 9% and and at high load it's 3%. Here's from an MIT report >> based on a 1.6L four cylinder: >> >> <fabjgpemblmdmfed.png> >> >> Brent >> >>> >>> https://web.archive.org/web/20161001161418/http://brickleyengine.com/ >>> >>> Duncan >>> >>>> On Mar 28, 2018, at 2:46 PM, Brent Meeker <meekerdb(a)verizon.net >>>> <mailto:meekerdb@verizon.net>> wrote: >>>> >>>> Is there some cut away drawing or disassembled picture of how this >>>> engine is constructed? >>>> >>>> Brent >>>> >>>> On 3/28/2018 7:31 AM, Duncan Charlton wrote: >>>>> Not easily! It would need a set of transfer gears to carry the >>>>> drive to the gearbox, and it would sit pretty high (and sideways). >>>>> >>>>> Duncan >>>>> >>>>>> On Mar 27, 2018, at 9:39 PM, Tim Waller <btmfdchn(a)gmail.com >>>>>> <mailto:btmfdchn@gmail.com>> wrote: >>>>>> >>>>>> Greetings... Will it fit in the Unipolar? >>>>>> >>>>>> Tim(D.Y.M.) >>>>>> >>>>>> On Tuesday, March 27, 2018 at 4:46:17 AM UTC-7, Duncan wrote: >>>>>> >>>>>> Some of you may recall me mentioning my friend Mike >>>>>> Brickley’s engine. Here’s a video that shows how well >>>>>> balanced his 4-cylinder prototype is: >>>>>> https://www.youtube.com/watch?v=SjJegdFC54c&feature=youtu.be >>>>>> <https://www.youtube.com/watch?v=SjJegdFC54c&feature=youtu.be> >>>>>> >>>>>> He created an unusual bottom end in which there are only a >>>>>> few rotating bearings, which cuts friction and heat, >>>>>> improving fuel economy. Most of the components rock back and >>>>>> forth with wrist pin joints. He’s in the testing stage now >>>>>> after completing his hand-built prototype. >>>>>> >>>>>> His website shows a photo of the complete engine next to him >>>>>> (it’s the thing with the four square yellow air cleaners on >>>>>> it): http://brickleyengine.com <http://brickleyengine.com/> >>>>>> >>>>>> Duncan >>>>>> >>>>>> >>>>>> -- >>>>>> You received this message because you are subscribed to the >>>>>> Google Groups "MOG-group" group. >>>>>> To unsubscribe from this group and stop receiving emails from it, >>>>>> send an email to mog-group+unsubscribe(a)googlegroups.com >>>>>> <mailto:mog-group+unsubscribe@googlegroups.com>. >>>>>> For more options, visit https://groups.google.com/d/optout. >>>>> >>>>> -- >>>>> You received this message because you are subscribed to the Google >>>>> Groups "MOG-group" group. >>>>> To unsubscribe from this group and stop receiving emails from it, >>>>> send an email to mog-group+unsubscribe(a)googlegroups.com >>>>> <mailto:mog-group+unsubscribe@googlegroups.com>. >>>>> For more options, visit https://groups.google.com/d/optout. >>>> >>>> >>>> -- >>>> You received this message because you are subscribed to the Google >>>> Groups "MOG-group" group. >>>> To unsubscribe from this group and stop receiving emails from it, >>>> send an email to mog-group+unsubscribe(a)googlegroups.com >>>> <mailto:mog-group+unsubscribe@googlegroups.com>. >>>> For more options, visit https://groups.google.com/d/optout. >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "MOG-group" group. >>> To unsubscribe from this group and stop receiving emails from it, >>> send an email to mog-group+unsubscribe(a)googlegroups.com >>> <mailto:mog-group+unsubscribe@googlegroups.com>. >>> For more options, visit https://groups.google.com/d/optout. >> >> -- >> You received this message because you are subscribed to the Google >> Groups "MOG-group" group. >> To unsubscribe from this group and stop receiving emails from it, >> send an email to mog-group+unsubscribe(a)googlegroups.com >> <mailto:mog-group+unsubscribe@googlegroups.com>. >> For more options, visit https://groups.google.com/d/optout. > -- > You received this message because you are subscribed to the Google > Groups "MOG-group" group. > To unsubscribe from this group and stop receiving emails from it, send > an email to mog-group+unsubscribe(a)googlegroups.com > <mailto:mog-group+unsubscribe@googlegroups.com>. > For more options, visit https://groups.google.com/d/optout.

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Re: [math-fun] Hexaflexagon paper
by John Golden 27 Mar '18

27 Mar '18

> > I?m sort of a newcomer to hexaflexagons. Any thoughts about what sort of > paper works best? > > (I might just go to a stationer?s and ask for samples of various weights of > paper, assuming they have samples sitting around. But that may not be an > option in the 21st century.) > > Jim Propp I’ve had good luck with K-8 learners with copier weight paper. A little less durable, a little more flexible for turning.

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[math-fun] Hexaflexagon paper
by James Propp 27 Mar '18

27 Mar '18

I’m sort of a newcomer to hexaflexagons. Any thoughts about what sort of paper works best? (I might just go to a stationer’s and ask for samples of various weights of paper, assuming they have samples sitting around. But that may not be an option in the 21st century.) Jim Propp

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[math-fun] Collatz thoughts
by Dan Asimov 26 Mar '18

26 Mar '18

Three observations: 1) The Collatz mapping can be thought of as just going from odd integers to odd integers, via f(x) = 2K+1 for any odd integer x, where 3x+1 = 2^n * (2K+1) for integers n >= 0 and K >= 0. This seems much more natural than dividing by 2 only once per "time step". Then the Collatz conjecture can be phrased as saying that for odd x > 0, iterating f to get f(x), f(f(x)), ... must end up in the (1 ) cycle. 2) No reason we can't look at this for odd x < 0. In that case the cycles I know of are (-1 ), (-5 -7 ), and (-17 -25 -37 -55 -41 -61 -91 ). 3) It seems even more natural to look, more generally, at Collatz on rational numbers of the form x = odd/odd. Let x be such a rational. Then much as in 1) set f(x) = p/q in lowest terms where 3x+1 = 2^n * p/q for n, p, q integers with p and q odd. There seem to be lots of cycles, but they don't seem easy to find. And it's not clear if studying this more general situation will shed any light on the original conjecture. —Dan

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[math-fun] For a dance...
by Bill Gosper 26 Mar '18

26 Mar '18

On 2018-03-24 14:43, françois mendzina essomba2 via math-fun wrote: > Hello... > > I noticed that : > > atan(x)=(205*x^7+2184*x^5+5760*x^3+4096*x)/(36*x^8+820*x^6+4368*x^4+7680*x^2+4096)+sum(S(n),n=1..infinity); > > with > > S(m)=2^(-m-2)*x*sum(1/((2*k-1)^2*2^(-2*m-4)*x^2+1)-1/(k^2*2^(-2*m-2)*x^2+1),k,1,2^(m+1)); > > we deduce from it: > > 1) > > Pi= 2449/850 + 4*sum(S(n),n=1..infinity); > > with > > S(m)=2^(-m-2)*sum(1/((2*k-1)^2*2^(-2*m-4)+1)-1/(k^2*2^(-2*m-2)+1),k,1,2^(m+1)) ; > [. . .] Dr. Essomba: While In[11]:= Pi==2449/850+4*Sum[S@n,{n,\[Infinity]}] In[13]:= S@m_- >2^(-m-2)Sum[1/((2*k-1)^2*2^(-2*m-4)+1)-1/(k^2*2^(-2*m-2)+1),{k,2^(m+1)}] Out[14]= S[m_]->1/8 (2^-m+2 I (2 I \[Pi] Csch[2^(2+m) \[Pi]]+HarmonicNumber[(1-I) 2^(1+m)]-HarmonicNumber[(1+I) 2^(1+m)]-PolyGamma[0,1/2+(1-I) 2^(1+m)]+PolyGamma[0,1/2+(1+I) 2^(1+m)])) In[15]:= %11/.% Out[15]= \[Pi]==2449/850+4 \!\( \*UnderoverscriptBox[\(\[Sum]\), \(n\), \(\[Infinity]\)]\( \*FractionBox[\(1\), \(8\)]\ \(( \*SuperscriptBox[\(2\), \(-n\)] + 2\ I\ \((2\ I\ \[Pi]\ Csch[ \*SuperscriptBox[\(2\), \(2 + n\)]\ \[Pi]] + HarmonicNumber[\((1 - I)\)\ \*SuperscriptBox[\(2\), \(1 + n\)]] - HarmonicNumber[\((1 + I)\)\ \*SuperscriptBox[\(2\), \(1 + n\)]] - PolyGamma[0, \*FractionBox[\(1\), \(2\)] + \((1 - I)\)\ \*SuperscriptBox[\(2\), \(1 + n\)]] + PolyGamma[0, \*FractionBox[\(1\), \(2\)] + \((1 + I)\)\ \*SuperscriptBox[\(2\), \(1 + n\)]])\))\)\)\) In[16]:= N@% Out[16]= True is numerically plausible, I'm afraid that under US law, this makes you a PolyGamist. --rwg

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[math-fun] Khinchin - Collatz connection??
by Bill Gosper 25 Mar '18

25 Mar '18

There are 356033 integers within 41 Collatz (#/2 | (3#+1)/2) steps of 1. 356033/(4/3)^41 = 2.685357138639965 . . . 843774/(4/3)^44 = 2.684860516180538 . . . 63149973/(4/3)^59 = 2.685268450566904 . . . 112268898/(4/3)^61 = 2.685322969193732 But N@Khinchin = 2.685452001065306 (Thank Julian for the 4/3.) Does anyone recall an objection that the usual definition isn't really the expected geometric mean because its derivation assumes consecutive partial quotients are independent? Is there a "CorrectedKhinchin"? https://en.wikipedia.org/wiki/Khinchin%27s_constant and http://mathworld.wolfram.com/KhinchinsConstant.html say flatly No. (My Finch is on loan.) --rwg (EWW: In (8), the ln k belongs in the exponent.) Happy 8th, Gabriel!

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[math-fun] Avoiding Collisions in Outer Space
by Bill Gosper 25 Mar '18

25 Mar '18

Fred, if you are reading the following for the first time, I sorely regret missing out on your eruption of Irish adjectives: https://en.wikipedia.org/wiki/Project_West_Ford https://www.wired.com/2013/08/project-west-ford/ --rwg On 3/25/18, Allan Wechsler <acwacw(a)gmail.com> wrote: I didn't say I was in favor of this plan. I was trying, merely, to interpret the intention. On Sat, Mar 24, 2018 at 8:10 PM, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: You mean, having trashed the planet down here, we're going to start trashing it up there too? Brilliant plan ... WFL On 3/24/18, Allan Wechsler <acwacw(a)gmail.com> wrote: I'm not familiar with the proposal, but I think I can answer that question. The microsatellites will be very cheap, and mass-produced. Ambient conditions will certainly eventually cause them to fail. Using redundant circuits should be able to extend their usable lifetimes, but mostly, when they fail, send up another ten thousand.

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Re: [math-fun] Avoiding Collisions in Outer Space
by Keith F. Lynch 25 Mar '18

25 Mar '18

Henry Baker <hbaker1(a)pipeline.com> wrote: > The highway-in-the-sky approach works more-or-less for satellites in > exactly the same orbit, but keeping them equidistant may require > active maneuvers and hence fuel. Active maneuvers don't necessarily require fuel if the satellites are very light and in a fairly high orbit. Dump a can of Pringles (tm) in a circular orbit a few thousand miles up. Except they aren't potato chips, but computer chips. They're solar powered, and they have patches around their edges which can be turned dark or light so as to use sunlight to maneuver in both position and attitude. These satellite constellations are simultaneously communications satellites, directional antennas, computers, file servers, and cooling fins. They should be able to outcompete all ground-based computer and communications services except those that require low latency. See http://server-sky.com/ All credit goes to Keith Lofstrom, perhaps better known for his proposed launch loop. And now for some shameless name-dropping: http://KeithLynch.net/3keith.jpg is a photo of he and I sitting together on a couch in a hotel in Toronto. The third person is Keith Henson. I'm the only person on that couch who doesn't have a Wikipedia page about him.

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[math-fun] For a dance...
by françois mendzina essomba2 24 Mar '18

24 Mar '18

Hello... I noticed that : atan(x)=(205*x^7+2184*x^5+5760*x^3+4096*x)/(36*x^8+820*x^6+4368*x^4+7680*x^2+4096)+sum(S(n),n=1..infinity); with S(m)=2^(-m-2)*x*sum(1/((2*k-1)^2*2^(-2*m-4)*x^2+1)-1/(k^2*2^(-2*m-2)*x^2+1),k,1,2^(m+1)); we deduce from it: 1) Pi= 2449/850 + 4*sum(S(n),n=1..infinity); with S(m)=2^(-m-2)*sum(1/((2*k-1)^2*2^(-2*m-4)+1)-1/(k^2*2^(-2*m-2)+1),k,1,2^(m+1)) ; 2) Pi=139701*sqrt(3)/91504 + 3*sum(S(n),n=1..infinity); with S(m)=sqrt(3)*2^(-m-2)*sum(1/(3*(2*k-1)^2*2^(-2*m-4)+1)-1/(3*k^2*2^(-2*m-2)+1),k,1,2^(m+1)) ; 3) Pi=169189*sqrt(3)/96824 + 6*sum(S(n),n=1..infinity); with (2^(-m-2)*sum(1/(((2*k-1)^2*2^(-2*m-4))/3+1)-1/((k^2*2^(-2*m-2))/3+1),k,1,2^(m+1)))/sqrt(3);

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