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[math-fun] Unlucky thirteen
by James Propp 24 Mar '18

24 Mar '18

#1. Puzzle: Show that for every positive integer n less than thirteen, there exists a Laurent polynomial p(x) with coefficients in {0,1,2,3,4} satisfying p(1/2) = p(1/3) = n. (Hint: Use an iterative construction n -> n+1.) #2. Problem: Show that this NOT the case for thirteen or any larger positive integer. (I don't know how to prove this.) If #1 seems tricky, here's a simpler version: Show that for every positive integer n there exists a Laurent polynomial p(x) with coefficients in {0,1,2} satisfying p(1) = p(2) = n. Equivalently, show that every positive integer n can be written as the sum of n powers of 2, each of which may appear at most twice (where "power of 2" means "2^k for some k in Z"). Jim Propp

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[math-fun] Curved space paradox
by Allan Wechsler 23 Mar '18

23 Mar '18

The following intuitive line of reasoning leads to a contradictory conclusion, and I don't know what's wrong with it. Obviously one of my assumptions or beliefs about spaces with constant Gaussian curvature is wrong, but I can't figure out which. Suppose we have a 3-dimensional space of constant, negative Gaussian curvature. For some large enough R, construct the locus of points at distance R from a fixed center C. This forms a 2-dimensional surface which is topologically an ordinary sphere. Select a point P on this sphere, and begin drawing circles of increasing radius r on the sphere centered at P. I mean r to be measured along the surface of the sphere, not straight through the hyperbolic 3-space in which the sphere is embedded. Now, this sphere must itself have constant Gaussian curvature; I don't have a fast proof but surely there is a symmetry argument that can establish it. Watch how the circumference of a circle, centered at P, changes with its radius r. For small r, it looks like 2 pi r. But as r gets larger, the negative curvature of the ambient hyperbolic space begins to make itself felt, and the circumference goes above 2 pi r, by some factor related to sinh. (If it doesn't, you didn't make R big enough -- that is exactly what I meant by "large enough R" above.) But later, as r becomes commensurate with R, the circumference rises to a maximum and begins to shrink again, because the sphere itself has finite circumference. On a surface of constant curvature, the circumference should either fall below 2 pi r, or it should exceed it forever. It should not do both. Surely I am not the first person to think these confusing thoughts. What is wrong with my reasoning?

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[math-fun] A Star Grazed Our Solar System 70,000 Years Ago
by Henry Baker 23 Mar '18

23 Mar '18

FYI -- This incident is particularly interesting, since there was a human population "bottleneck" at approximately the same time. The "bottleneck" is supported by DNA evidence, although its exact timing is difficult to determine. However, the purported cause of this bottleneck is quite controversial. https://www.rochester.edu/newscenter/scholz-star/ https://www.space.com/40043-star-grazed-our-solar-system-disrupted-orbits.h… A Star Grazed Our Solar System 70,000 Years Ago, and Early Humans Likely Saw It By Mike Wall, Space.com Senior Writer | March 21, 2018 11:50am ET https://en.wikipedia.org/wiki/Toba_catastrophe_theory "In 1993, science journalist Ann Gibbons posited that a population bottleneck occurred in human evolution about 70,000 years ago, and she suggested that this was caused by the [Toba] eruption. Biologist Michael R. Rampino of New York University and volcanologist Stephen Self of the University of Hawaii at Manoa support her suggestion. In 1998, the bottleneck theory was further developed by anthropologist Stanley H. Ambrose of the University of Illinois at Urbana-Champaign." https://en.wikipedia.org/wiki/Genetic_bottleneck "The controversial Toba catastrophe theory, presented in the late 1990s to early 2000s, suggested that a bottleneck of the human population occurred c. 70,000 years ago, proposing that the human population was reduced to perhaps 10,000Â30,000 individuals when the Toba supervolcano in Indonesia erupted and triggered a major environmental change. Parallel bottlenecks were proposed to exist among chimpanzees, gorillas, rhesus macaques, orangutans and tigers. The hypothesis was based on geological evidence of sudden climate change and on coalescence evidence of some genes (including mitochondrial DNA, Y-chromosome DNA and some nuclear genes) and the relatively low level of genetic variation in humans."

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[math-fun] Avoiding Collisions in Outer Space
by Henry Baker 20 Mar '18

20 Mar '18

An OpEd in this morning's NYTimes got me to thinking about the problem of non-intersecting orbits. https://www.nytimes.com/2018/03/19/opinion/space-race-regulation.html Avoiding Collisions in Outer Space By Yousaf Butt March 19, 2018 ----- There are multiple issues here, both practical & theoretical; I'm curious about the theoretical issues. In particular, I'm curious about satellites orbiting a perfectly spherical Earth (no gravitational anomalies), with no atmospheric drag, and no gravitational influences from the Moon. These conditions are intended to make satellite orbits as periodic as possible, so that hopefully zero fuel is required to avoid collisions. Most of you probably know that airplanes are restricted to various altitudes for various directions in order to cut down on the possibilities of "intersections". Furthermore, airplanes often use a "highway in the sky", wherein they follow one another at a fixed distance, just as if they were constrained to a physical roadway. But a simple altitude-sorting strategy won't work for satellites, even for those in perfectly circular orbits, because different circular altitudes produce different periods. 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. But the biggest problem is that altitude-sorting rules out interesting orbits -- e.g., transfer orbits and Molniya orbits. (Molniya orbits are highly eccentric orbits used for satellite broadcasting at high latitudes.) So-called satellite "constellations" enable a number of satellites *under common control* to orbit without intersections, but it isn't clear whether every satellite needs or wants to live within the restrictions of these mostly circular orbits. https://en.wikipedia.org/wiki/Satellite_constellation There must be some incredibly beautiful orbit constellations left to discover -- e.g., periodic constellations of *elliptical orbits* s.t. communications satellites can periodically get close enough to one another for exceedingly high bandwidth communications. But if some international agreements prematurely rule out such constellations, then we may never be allowed to use them.

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[math-fun] Isaac Newton quote (?)
by James Propp 20 Mar '18

20 Mar '18

I could've sworn that Isaac Newton, when asked how he got his solutions to math and physics problems, replied "By thinking upon them constantly", but I've tried entering this sentence (and semantically-equivalent variants) into Google, and I haven't been able to find it. Does anyone know the quote I'm mis-remembering? I should say I've done my due diligence (five minutes with a search engine) before daring to bother you all. Jim

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Re: [math-fun] Isaac Newton quote (?)
by Henry Baker 20 Mar '18

20 Mar '18

You might try finding Newton's original Latin (I think that Google knows & can translate Latin). BTW, Newton's "shoulders of giants" quote is originally from Bernard of Chartres ~ 1100AD. Newton probably liked & retweeted this quote at some point in his life. https://en.wikipedia.org/wiki/Bernard_of_Chartres At 11:39 AM 3/20/2018, James Propp wrote: >Thanks for the exact wording, Henry! > >I'm now finding the quote in various places, but I'm guessing that it's >apocryphal; all the references are to collections of quotations and >self-help books, none of which provide primary source information. > >Jim > >On Tue, Mar 20, 2018 at 2:27 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: > >> "You can see more by standing on the shoulders of giants, than by stepping >> on their toes." -- unknown (??) >> >> From goodreads.com: >> >> "Sir Isaac Newton was asked how he discovered the law of gravity. He >> replied 'by thinking about it all the time'" >> >> He also supposed said: >> >> "If I have ever made any valuable discoveries, it has been due more to >> patient attention, than to any other talent" >> >> --- >> We all learned while kids -- probably due to Walt Disney or equivalent -- >> that Newton discovered the law of gravity by sleeping under an apple tree! >> >> At 11:05 AM 3/20/2018, James Propp wrote: >> >I could've sworn that Isaac Newton, when asked how he got his solutions to >> >math and physics problems, replied "By thinking upon them constantly", but >> >I've tried entering this sentence (and semantically-equivalent variants) >> >into Google, and I haven't been able to find it. >> > >> >Does anyone know the quote I'm mis-remembering? >> > >> >I should say I've done my due diligence (five minutes with a search engine) >> >before daring to bother you all. >> > >> >Jim

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Re: [math-fun] Isaac Newton quote (?)
by Henry Baker 20 Mar '18

20 Mar '18

"You can see more by standing on the shoulders of giants, than by stepping on their toes." -- unknown (??) From goodreads.com: "Sir Isaac Newton was asked how he discovered the law of gravity. He replied 'by thinking about it all the time'" He also supposed said: "If I have ever made any valuable discoveries, it has been due more to patient attention, than to any other talent" --- We all learned while kids -- probably due to Walt Disney or equivalent -- that Newton discovered the law of gravity by sleeping under an apple tree! At 11:05 AM 3/20/2018, James Propp wrote: >I could've sworn that Isaac Newton, when asked how he got his solutions to >math and physics problems, replied "By thinking upon them constantly", but >I've tried entering this sentence (and semantically-equivalent variants) >into Google, and I haven't been able to find it. > >Does anyone know the quote I'm mis-remembering? > >I should say I've done my due diligence (five minutes with a search engine) >before daring to bother you all. > >Jim

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Re: [math-fun] Strip of squares in space puzzle (new & improved) (final version?)(finaler version)
by Dan Asimov 20 Mar '18

20 Mar '18

I am incapable of stating my puzzle correctly, even though I know exactly what I mean. Why is it so difficult for everyone else? 4th try?: ----- Each square Q is adjacent to exactly 2 others, Q- and Q+, such that each pair ((Q-, Q) and (Q, Q+) is perpendicular and shares one of 2 adjacent edges of Q, and Q- and Q+ lie on opposite sides of the plane of Q. ----- —Dan

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[math-fun] 1/7 etc.
by Bill Gosper 19 Mar '18

19 Mar '18

On 2018-03-19 08:35, James Propp wrote: > It’s nice to 89, in addition to being 10^2-10-1, is also a Fibonacci > number. > How many Fibonacci numbers are of the form n^2-n-1? > > Jim Propp In[26]:= Select[Fibonacci@Range@99999, squareq[4 # + 5] &] // tim During evaluation of In[26]:= 4767.121894,5 Out[26]= {1, 1, 5, 55, 89} In[24]:= squareq[n_Integer] := IntegerQ@√n Looks like a companion to 0,1,144. (Neither in OEIS) How about OEIS typographical finiteness conventions? E.g.: (Heegner) ..., 163. (Period means no more.) (Wieferich) 1093,3511,? means probably more (Pemutables) 991,1111111111111111111,11111111111111111111111.? maybe no more (Fibs) 4181,6765, fer sure more. --rwg > > On Monday, March 19, 2018, James Buddenhagen <jbuddenh(a)gmail.com> wrote: > >> Looks nicely fiby up till 90. (re: Allan Wechsler's 1/9899). >> >> On Sun, Mar 18, 2018 at 12:42 PM, Allan Wechsler <acwacw(a)gmail.com> wrote: >> >> > What do you think of 1/9899? In[2]:= RealDigits[1/9899] Out[2]= {{{1, 0, 1, 0, 2, 0, 3, 0, 5, 0, 8, 1, 3, 2, 1, 3, 4, 5, 5, 9, 0, 4, 6, 3, 6, 8, 3, 2, 0, 0, 3, 2, 3, 2, 6, 4, 9, 7, 6, 2, 6, 0, 2, 2, 8, 3, 0, 5, 8, 8, 9, 4, 8, 3, 7, 8, 6, 2, 4, 1, 0, 3, 4, 4, 4, 7, 9, 2, 4, 0, 3, 2, 7, 3, 0, 5, 7, 8, 8, 4, 6, 3, 4, 8, 1, 1, 5, 9, 7, 1, 3, 1, 0, 2, 3, 3, 3, 5, 6, 9, 0, 4, 7, 3, 7, 8, 5, 2, 3, 0, 8, 3, 1, 3, 9, 7, 1, 1, 0, 8, 1, 9, 2, 7, 4, 6, 7, 4, 2, 0, 9, 5, 1, 6, 1, 1, 2, 7, 3, 8, 6, 6, 0, 4, 7, 0, 7, 5, 4, 6, 2, 1, 6, 7, 8, 9, 5, 7, 4, 7, 0, 4, 5, 1, 5, 6, 0, 7, 6, 3, 7, 1, 3, 5, 0, 6, 4, 1, 4, 7, 8, 9, 3, 7, 2, 6, 6, 3, 9, 0, 5, 4, 4, 4, 9, 9, 4, 4, 4, 3, 8, 8, 3, 2, 2, 0, 5, 2, 7, 3, 2, 5, 9, 9, 2, 5, 2, 4, 4, 9, 7, 4, 2, 3, 9, 8, 2, 2, 2, 0, 4, 2, 6, 3, 0, 5, 6, 8, 7, 4, 4, 3, 1, 7, 6, 0, 7, 8, 3, 9, 1, 7, 5, 6, 7, 4, 3, 1, 0, 5, 3, 6, 4, 1, 7, 8, 1, 9, 9, 8, 1, 8, 1, 6, 3, 4, 5, 0, 8, 5, 3, 6, 2, 1, 5, 7, 7, 9, 3, 7, 1, 6, 5, 3, 7, 0, 2, 3, 9, 4, 1, 8, 1, 2, 3, 0, 4, 2, 7, 3, 1, 5, 8, 9, 0, 4, 9, 3, 9, 8, 9, 2, 9, 1, 8, 4, 7, 6, 6, 1, 3, 7, 9, 9, 3, 7, 3, 6, 7, 4, 1, 0, 8, 4, 9, 5, 8, 0, 7, 6, 5, 7, 3, 3, 9, 1, 2, 5, 1, 6, 4, 1, 5, 7, 9, 9, 5, 7, 5, 7, 1, 4, 7, 1, 8, 6, 5, 8, 4, 5, 0, 3, 4, 8, 5, 2, 0, 0, 5, 2, 5, 3, 0, 5, 5, 8, 6, 4, 2, 2, 8, 7, 0, 9, 9, 7, 0, 7, 0, 4, 1, 1, 1, 5, 2, 6, 4, 1, 6, 8, 0, 9, 7, 7, 8, 7, 6, 5, 5, 3, 1, 8, 7, 1, 9, 0, 6, 2, 5, 3, 1, 5, 6, 8, 8, 4, 5, 3, 3, 7, 9, 1, 2, 9, 2, 0, 4, 9, 7, 0, 1, 9, 9, 0, 1, 0, 0, 0}}, -3} >> > >> > >> > On Sun, Mar 18, 2018 at 2:26 PM, Thane Plambeck <tplambeck(a)gmail.com> >> > wrote: >> > >> > > i like the decimal expansion of 1/9801 >> > > >> > > On Sat, Mar 17, 2018 at 2:59 PM, Cris Moore <moore(a)santafe.edu> wrote: >> > > >> > > > >> > > > Indeed, 14+28+57 = 99… because 7 is a divisor of 10101... >> > > > and 07+69+23 = 99, because 13 is too. >> > > > >> > > > - Cris >> > > > >> > > > > On Mar 17, 2018, at 3:13 PM, Simon Plouffe < >> simon.plouffe(a)gmail.com> >> > > > wrote: >> > > > > >> > > > > >> > > > > >> > > > > Hello, >> > > > > >> > > > > we can remark that 142857 is symmetrical, 142 + 857 = 999 >> > > > > >> > > > > so one just has to remember half of the period in order to >> > > > > know all of the digits, >> > > > > 1/13 = 076923 , 076+923 = 999 too. >> > > > > >> > > > > this phenomena is true whenever p <> 2 or p <> 11 and >> > > > > the period is even. >> > > > > >> > > > > best regards, >> > > > > Simon Plouffe >> > > > > >> > > > > >> > > > > Le 2018-03-17 à 21:59, Cris Moore a écrit : >> > > > >> well, >> > > > >> >> > > > >> 1/13 = 0769/9997 = 0.0769 (1 + 0.0003 + 0.00000009 + …) = 0.0769 >> > 2307 >> > > … >> > > > >> >> > > > >> although this doesn’t quite break up the repetend 076923. You >> could >> > > > use the clumsier >> > > > >> >> > > > >> 1/13 = 076/988 = 0.076 (1 + 0.012 + 0.000144) = 0.076 + 0.000912 + >> > > > 0.0000109… = 0.076923… >> > > > >> >> > > > >> but this involves a bunch of carrying. >> > > > >> >> > > > >> Cris >> > > > >> >> > > > >>> On Mar 17, 2018, at 2:36 PM, James Propp <jamespropp(a)gmail.com> >> > > wrote: >> > > > >>> >> > > > >>> I just figured out for myself a probably well-known trick for >> > > > >>> deriving/remembering the decimal expansion of 1/7: 1/7 = 14/98 = >> > > > 14/(100-2) >> > > > >>> = .14/(1-.02) = .14 + .0028 + .000056 + .00000112 + ... = >> > .142857... >> > > > >>> >> > > > >>> >> > > > >>> Are there other examples where the repetend of the decimal >> > expansion >> > > > of 1/n >> > > > >>> in splits into blocks that are related to this sort of fashion? >> > > > >>> >> > > > >>> >> > > > >>> Jim Propp

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Re: [math-fun] 1/7 etc.
by Keith F. Lynch 19 Mar '18

19 Mar '18

James Propp <jamespropp(a)gmail.com> wrote: > How many Fibonacci numbers are of the form n^2-n-1? For positive Fibonacci numbers less than 2^63, none except 1, 5, 55, and 89. If I replace one or both minus signs with plus signs, there are still no solutions above 89 (though there are a few more below it). Similarly if I leave out the n or the 1 (leaving n^2), except that 144 is of course square.

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