June 2015 - mathfun@mailman.xmission.com

[math-fun] Geometric significance of algebraic discriminant ?
by Henry Baker 02 Jul '15

02 Jul '15

I recall learning a bunch of algebra about the discriminant, which becomes zero when there are coincident roots. https://en.wikipedia.org/wiki/Discriminant Has someone come up with geometric insights about this particular formula ? In the case of a quadratic, the formula is (x1-x2)^2, but this isn't the real number |x1-x2|^2. Perhaps the norm of the discriminant (DD*) is more important? What about the discriminant of the cubic ? Shouldn't this say something interesting about the triangle in the complex plane formed by the roots?

7 8

Re: [math-fun] Fano Plane puzzle
by Bill Gosper 01 Jul '15

01 Jul '15

The Sierra Club has you beaten. They've quoted the volume of nuclear waste in cubic liters. -- Gene Why do only bad things come in cubic liters? I saw a GreenPeace flyer celebrating the sabotage of a drainpipe dumping "thousands of cubic liters of pharmaceutical waste" into the ocean. Unfortunately, the photo showed only three-dimensional damage, leaving hundreds of square gallons unspilled, resulting in attacks by fifteen pound purple jellyfish: http://sanfrancisco.cbslocal.com/2015/06/12/big-purple-sea-slugs-wash-ashor… --rwg From: rwg <rwg(a)sdf.org> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Wednesday, June 24, 2015 6:28 PM Subject: Re: [math-fun] Fano Plane puzzle On 2015-06-24 17:07, Dan Asimov wrote: On Jun 24, 2015, at 3:22 PM, David Wilson <davidwwilson(a)comcast.net> wrote: Sadly, it would difficult to make an office toy in R^6. Not at all. In R^6 they have even better machine shops than we do. —Dan And the wealth to operate them. Like I was just telling the kids, On Wed, Jun 24, 2015 at 7:19 AM, Bill Gosper <billgosper(a)gmail.com> wrote: Subj: Scrooge McDuck was said to have three cubic acres of money. How much is that in square gallons? In[169]:= UnitConvert[Quantity[3, "Acres"]^3, Quantity["Gallons"]^2] Out[169]= Quantity[95610810544128000000000000000000000000000000000000000/ 765615812545937377500183749853000049, ("Gallons")^2] Oops, that's 27 cubic acres! Should have been In[170]:= UnitConvert[Quantity[3, "Acres"^3], Quantity["Gallons"]^2] Out[170]= Quantity[10623423393792000000000000000000000000000000000000000/ 765615812545937377500183749853000049, ("Gallons")^2] I hate it when ducks exaggerate. --Bill -----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com <mailto:math-fun-bounces@mailman.xmission.com>] Dan Asimov Sent: Wednesday, June 24, 2015 3:30 PM But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.

7 16

[math-fun] projective plane embeddings
by Warren D Smith 30 Jun '15

30 Jun '15

> Subject: Re: [math-fun] Fano Plane puzzle (Fred Lunnon) > Message-ID: > <CAN57YqvtqfqOvaYUjHraaUeXdSdch6Fks3xXAiNnA61wbdjhkw(a)mail.gmail.com> > Content-Type: text/plain; charset=UTF-8 > > If each projective line meets n+1 points, you can always embed lines as > (n-1)-spheres in n-space; and there is so much freedom in choosing point > coordinates that you can surely then constrain all radii to unity in > principal. > > If you want to reduce the number of dimensions, it becomes much harder: > the Heawood embedding in 2-space was only solved in 2009 by Gerbracht. > --well as I'd explained in a previous post, any bipartite graph can be embedded as a unit distance graph in 4 dimensions so projective planes can be embedded in 4D using unit-radius spheres as the "lines." Lunnon's claim here about general n was therefore obsolete before issuance because n=4 suffices. However, the interesting question then is: for which projective planes does n=2 or n=3 suffice? One might suspect based on crude degree-of-freedom counting that only a finite number of projective planes can be embedded in 2D or 3D. However, clearly such crude arguments are not actually valid! Furthermore, my 4D solution could be considered undesirable since although all the spheres are different, the N points all happen to lie on a common circle which is the intersection of all N spheres. So then you might want to go to 5D, or whatever... in 5D the analogous solution has all points lying on a common surface-dimension-2 sphere which is the intersection of the spheres...

1 0

Re: [math-fun] Fano Plane puzzle (Fred Lunnon)
by Warren D Smith 29 Jun '15

29 Jun '15

Can you do the same sort of 3D (or 2D) embedding via circles for other cyclic projective planes? The next one is {0,1,3,9}+m mod 13.

2 1

[math-fun] Wheels
by James Propp 29 Jun '15

29 Jun '15

Thanks to all of you who steered me towards Gardner's article "Wheels" (reprinted as the first chapter of "Wheels, Life and Other Mathematical Amusements"). It's rare for Gardner to get anything wrong, but he may have done that in one or two places in this piece; what do you all think? Speciifcally: (1) Gardner writes (in the second to last paragraph of the article proper, on page 8): "There are varied and perplexing problems that involve non-circular "wheels." For example, suppose a square wheel rolls without slipping on a track that is a series of equal arcs, convex sides up. What kind of curve must each arc be to prevent the center of the wheel from moving up and down? (In other words, the wheel's center must travel a straight horizontal path.) The curve is a familiar one and, amazingly, the same curve applies to similar tracks for wheels that are regular polygons with any number of sides." But my understanding is that when the polygon is an equilateral triangle, the inverted catenary curve doesn't work: the corner of the two triangle tries to cut into the catenary arc that it's purportedly moved beyond. (2) Gardner writes (in the Answers section): "If a wheel is an irregular convex polygon, the track must have arcs that are differently shaped catenaries, one for each side of the wheel." But will that work for EVERY convex polygon? I'm guessing "no", if only because of triangles and things with similar sharp corners. But my intuition tells me that for a generic convex polygon, the answer is "no" (unless you discontinuously change what you mean by the "center" of the wheel as you roll it). I looked at "Rockers and Rollers" ( http://steiner.math.nthu.edu.tw/disk5/js/cardioid/6.pdf) but it seems to me that at best, Robison proves that the only curve that might work is one made of catenaries; he does not explore conditions under which this necessary condition is also sufficient. Jim Propp

1 0

[math-fun] A math-y US Supreme Court decision
by Thane Plambeck 29 Jun '15

29 Jun '15

A friend pointed out this recent USA Supreme Court decision. http://www.supremecourt.gov/opinions/14pdf/no5orig_d18e.pdf Exhibit A is titled "The Location of the Fixed Offshore Boundary Between the United States and California that is Parallel to the Coastline of Mainland California." It's over 100 pages long. Anyone up for using Exhibit A to calculate the official length of the Pacific ocean-facing border of mainline California? I tried a quick and dirty import into Mathematica, but it failed. To see a map of it, scroll to the very end. Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

2 1

Re: [math-fun] Leap Frog (fwd)
by rkg 27 Jun '15

27 Jun '15

To save duplicated effort, here are two messages from Richard Nowakowski, who has used Aaron Siegel's CG-Suite to analyze Leap Frog as far as n = 11. R. ---------- Forwarded message ---------- Date: Fri, 26 Jun 2015 19:15:09 +0000 From: Richard Nowakowski <R.Nowakowski(a)Dal.Ca> To: rkg <rkg(a)ucalgary.ca> Subject: Re: Leap Frog Richard, if I understood the game correctly then CGSuite gives: [0] = * [0,0] = 0 [0,0,0] = * [0,0,0,0] = 0 [0,0,0,0,0] = * [0,0,0,0,0,0] = 0 [0,0,0,0,0,0,0] = *2 [0^8] = 0 [0^9] = * [0^10] = 0 The options are: n=1, 0 n=2, all * n=3, all 0 n=4, [1,0,0,0]=[0,0,0,1] =* other two are *2 n=5, all 0 n=6, all * n=7, all * except [0,0,0,1,0,0,0] = *2 n=8, all * n=9 all *2 except [1,0^8], [0^4,1,0^4], [0^8,1] which are 0 n=10, all * I'm waiting on n=11 and n=12. No insights though. Second message: The game is `symmetric' so I'll only give half the options. n=11 is * but [1,0^{10}] = [0,1,0^9] = [0^5,1,0^5] = 0 [0,0,1,0^8] = *2 [0,0,0,1,0^7] = *6 [0^4,1,0^6]=*5 Cheers, Richard. ________________________________________ From: rkg <rkg(a)ucalgary.ca> Sent: Friday, June 26, 2015 3:12 PM To: math-fun Subject: Leap Frog To whom it may concern. Here is a game, Leap Frog, which may not be original. It is played on a strip of squares. A move is to place a bean on an unoccupied square. So far we have one of the many manifestations of She-Loves-Me-She-Loves-Me-Not, but there is a second type of move. A player may jump a bean onto an unocupied square, provided there is at least one intervening square and all such intervening squares are occupied by beans. To ensure that the Ending Condition is satisfied, we will say that such jumping moves may not be reversed. If you make that ``may not be IMMEDIATELY reversed'' or ``may not be reversed by the opposite player'' then these versions may be of interest, but I suspect that ties occur for n > 5. For 0 < n < 5 the game is equivalent to S-L-M-S-L-M-N, since jumping moves need not arise. For example, for n = 4, 1 & 3 are good replies to one another, as are 2 & 4. n = 5 is an N-position. For example, placing a bean on square 3 wins. The moves 1 & 5 are good replies to each other, while 2 (or 4) can be met by 1 (or 5), since the jumping moves 2 to 4 or 1 to 4 are met by the jump 3 to 5. n = 6 is beginning to get difficult by hand. Over to you! Fame & fortune await a complete analysis! R.

2 2

[math-fun] alleged impressive fiber-optic accomplishment
by Warren D Smith 27 Jun '15

27 Jun '15

NY Times today (26 June 2015) says Nikola Alic and others have successfully sent optical signals thru an optical fiber 7456 miles long, WITHOUT any "repeaters" along the way. Supposedly, the method involves understanding what kinds of transformations happen to the signal during transit, and pre-applying an approximately-inverse transformation. That causes the signal to be maximally easy to detect at the receiving end. E. Temprana, E. Myslivets, B.P.-P. Kuo, L. Liu, V. Ataie, N. Alic, S. Radic: Overcoming Kerr-induced capacity limit in optical fiber transmission Science 348,6242 (26 June 2015) 1445-1448 http://www.sciencemag.org/content/348/6242/1445 As far as I understand (this may have been hype, but probably not) semiconductor chip manufacturers trying to optically-lithograph ultrafine patterns, have for years all done a similar trick. That is, (a) light is transmitted thru an optical mask defining the pattern, then (b) hits photoresist on the chip, whch (c) polymerizes, then (d) chemical etching or whatever is used to generate the pattern on the chip. This process abcd causes the pattern you get to differ from the pattern on the mask via some transformation which is somewhat understood, and can be, and is, approximately inverted. Hence the picture they draw on the mask intentionally is NOT the same as the intended pattern on the chip. This is actually a harder mathematical problem since 2 or 3 or 4 dimensional, whereas the fiber problem presumably is only 1 or 2 dimensional.

3 2

[math-fun] Circular motion
by James Propp 27 Jun '15

27 Jun '15

Here are some questions related to my next blog post: 1) Who first came up with the idea of a square-wheeled bicycle/tricycle, and who first built one? 2) Did Martin Gardner ever discuss it? (And if so, where?) 3) Did Gardner ever discuss the fact that when a flange-wheeled train ( https://en.wikipedia.org/wiki/Train_wheel) is speeding down a track, a part of the wheel is actually going backwards? 4) Did Gardner ever discuss the following puzzle?: "Rotate one dime around another, without slipping. How many turns does the rolling dime make as it goes around once?" 5) Did Gardner ever discuss the fact that there are only 23 hours and 56 minutes in a day, or the related fact that in the course of a year, the Earth spins on its axis 366.25 times? 6) Did Gardner ever discuss the following puzzle?: "Cut a round hole into a piece of plastic so that the diameter of the hole is twice the diameter of a dime, and roll a dime around the inside of the hole without slipping. What path is traveled by a point on the perimeter of the dime?" 7) Does anyone know of a really elementary solution to the preceding puzzle? (Mine using analytic geometry. I could recast it in terms of vectors. But I'm hoping for a more accessible solution.) 8) Does the answer to the preceding puzzle have applications that involve converting linear motion to circular motion and vice versa, e.g. in steam-drive trains? 9) Did Gardner ever discuss other puzzles about circular motion, in the same vein as the ones given above? (I know about the two-screws puzzle, but no others come to mind.) 10) Do any of you know of any other puzzles of this kind that might belong in the same post? (No curves-of-constant-width stuff, please; that'll be a whole different blog-post.) I plan to send a draft of the essay to math-fun in a week or so. Thanks, Jim Propp

9 12

Re: [math-fun] Circular motion
by Henry Baker 26 Jun '15

26 Jun '15

Towards the end of the 19th century, there were a large number of patents filed regarding translating various forms of motion into other forms. Many would probably be regarded as trivial by mathematicians today, but many are quite clever, and involve interesting mathematical principles. I haven't tried to DuckDuckGo these patents, but perhaps it might be interesting to do so. BTW, back then, you actually had to produce a working mechanical model of your invention, so a lot of these inventions exist in the Patent Office or in the Smithsonian. PS, in those days you also had to send the Copyright Office copies of your works, which I think was a terrific idea, since many works exist *only* in the form that was saved by the Copyright Office -- e.g., some early film clips exist as sequences of B/W images on paper. The nitrate original film subsequently disintegrated, burned, or converted into WWI ammunition. At 10:35 AM 6/26/2015, James Propp wrote: >Here are some questions related to my next blog post: > >1) Who first came up with the idea of a square-wheeled bicycle/tricycle, >and who first built one? > >2) Did Martin Gardner ever discuss it? (And if so, where?) > >3) Did Gardner ever discuss the fact that when a flange-wheeled train ( >https://en.wikipedia.org/wiki/Train_wheel) is speeding down a track, a part >of the wheel is actually going backwards? > >4) Did Gardner ever discuss the following puzzle?: "Rotate one dime around >another, without slipping. How many turns does the rolling dime make as it >goes around once?" > >5) Did Gardner ever discuss the fact that there are only 23 hours and 56 >minutes in a day, or the related fact that in the course of a year, the >Earth spins on its axis 366.25 times? > >6) Did Gardner ever discuss the following puzzle?: "Cut a round hole into a >piece of plastic so that the diameter of the hole is twice the diameter of >a dime, and roll a dime around the inside of the hole without slipping. >What path is traveled by a point on the perimeter of the dime?" > >7) Does anyone know of a really elementary solution to the preceding >puzzle? (Mine using analytic geometry. I could recast it in terms of >vectors. But I'm hoping for a more accessible solution.) > >8) Does the answer to the preceding puzzle have applications that involve >converting linear motion to circular motion and vice versa, e.g. in >steam-drive trains? > >9) Did Gardner ever discuss other puzzles about circular motion, in the >same vein as the ones given above? (I know about the two-screws puzzle, but >no others come to mind.) > >10) Do any of you know of any other puzzles of this kind that might belong >in the same post? (No curves-of-constant-width stuff, please; that'll be a >whole different blog-post.) > >I plan to send a draft of the essay to math-fun in a week or so. > >Thanks, > >Jim Propp

1 0