July 2016 - mathfun@mailman.xmission.com

Re: [math-fun] Reentrant polygons with area zero
by Dan Asimov 30 Jul '16

30 Jul '16

Tom, can you please explain what the 2nd number on each line of your table means? And what it means when the leftmost number (which roots of unity you are using) is repeated on multiple lines? Dan -----Original Message----- >From: Tom Rokicki <rokicki(a)gmail.com> >Sent: Jul 30, 2016 11:03 AM >To: math-fun <math-fun(a)mailman.xmission.com> >Subject: Re: [math-fun] Reentrant polygons with area zero > >Here are the rotational symmetry searches I've done. >The 'no's are for exhaustive complete searches. >Searches that have not yet found solutions, but are >still running, are not listed. > >So far it looks like, for prime m, n has a >rotationally symmetrical solution of order m >if n is divisible by m^2. > > n sym solution found? > 9 3 yes >15 1 yes >15 3 no >15 5 no >21 1 yes >21 3 no >21 7 no >25 5 yes >27 3 yes >27 9 yes >33 11 no >35 5 no >35 7 no >39 13 no >45 9 no >45 15 yes >49 7 yes >51 17 no >57 19 no >63 21 yes >69 23 no >75 15 yes >75 25 no >81 27 yes > >(Sorry for spamming the list; I'll stop now.) > >-tom > >On Sat, Jul 30, 2016 at 10:28 AM, Tom Rokicki <rokicki(a)gmail.com> wrote: > >> Oddly, there appear to be a lot of n=15 solutions, but none of them have >> rotational symmetry (like the n=9 case did). >> >> -tom >> >> On Sat, Jul 30, 2016 at 9:42 AM, Tom Rokicki <rokicki(a)gmail.com> wrote: >> >>> 31 factorial is really large; it could be a very long night... :-) >>>> >>> >>> My CPU runs at 2.4 GHz. >>> >>> time-consuming) FullSimplify. How do you try to avoid getting false >>>> positives and false negatives? >>>> >>> >>> My philosophy is don't worry, be happy. Every additional vertex >>> introduces an error of at most 2^-51 or something like that, so I >>> check the final area against 0 with an epsilon of about 1e-8, and >>> then print the computed final area with the solution. If it's on the >>> order of 1e-15, I'm not certain it's zero, but close enough to report. >>> I'll let Bill and his posse do the hard bits of proving the result. >>> (Or I'll throw multiprecision arithmetic at it. Anyone want to write >>> a Mathematica procedure that takes a permutation, applies it >>> to the roots of unity and computes the area, using high-precision >>> math?) If multiprecision math says the area is < 1e-10000, I'm >>> inclined to view it as zero, at least provisionally. >>> >>> -tom >>> >>> -- >>> -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link >>> suppressed; ask me why] -- >>> >> >> >> >> -- >> -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link >> suppressed; ask me why] -- >> > > > >-- >-- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; >ask me why] -- >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

3 2

Re: [math-fun] Reentrant polygons with area zero
by Bill Gosper 30 Jul '16

30 Jul '16

James>http://faculty.uml.edu/jpropp/mathenchant/nonagon.pdf Nice. I think those are illegal in California. Use PossibleZeroQ. Tell us about any false positives. Alternatively, for n=9, e.g., In[478]:= I^Range[0, 4 - 4/9, 4/9] Out[478]= {1, (-1)^(2/9), (-1)^(4/9), (-1)^(2/3), (-1)^(8/9), -(-1)^(1/9), -(-1)^(1/3), -(-1)^(5/9), -(-1)^(7/9)} In[479]:= areaPolygon[%[[{1, 5, 3, 4, 8, 6, 7, 2, 9}]]] Out[479]= 1/2 (Cos[\[Pi]/18] + 1/2 Sqrt[3] Cos[\[Pi]/9] - 2 Cos[\[Pi]/18] Cos[\[Pi]/9] + 1/2 Sqrt[3] Cos[(2 \[Pi])/9] + 1/2 Sin[\[Pi]/9] - 2 Sin[\[Pi]/18] Sin[\[Pi]/9] + 1/2 Sin[(2 \[Pi])/9] - 2 Cos[(2 \[Pi])/9] Sin[(2 \[Pi])/9]) In[495]:= MinimalPolynomial@%479 // tim During evaluation of In[495]:= 0.241593,1 Out[495]= #1 & --rwg On 2016-07-30 04:34, James Propp wrote: > Tom, > > I'm assuming you want the "area" as defined by the winding rule, which >> means that some geometric areas may be counted multiple times. >> For instance, in a five-pointed star, the central pentagon, with winding >> number 2, counts double. Is this what you meant? > > > Yes, this is what I meant. > > >> Odd n=9 has the first solutions. >> >> Odd n=15 has many; here are some: >> >> 0 1 2 3 4 6 5 11 7 14 13 10 12 9 8 >> 0 1 2 3 4 6 5 12 8 7 14 11 13 10 9 >> 0 1 2 3 4 6 5 12 11 8 10 7 14 13 9 >> 0 1 2 3 4 6 5 12 14 11 7 13 10 9 8 >> 0 1 2 3 4 6 5 12 14 11 10 7 13 9 8 >> >> Odd n=21 has many; here are some: >> >> 0 1 2 3 4 5 6 7 8 16 10 15 14 13 12 20 19 18 17 11 9 >> 0 1 2 3 4 5 6 7 8 16 14 13 12 20 19 18 17 11 10 15 9 >> 0 1 2 3 4 5 6 7 8 16 15 13 12 20 14 19 18 17 11 10 9 >> 0 1 2 3 4 5 6 7 8 16 15 14 12 20 19 13 18 17 11 10 9 >> 0 1 2 3 4 5 6 7 8 16 15 14 13 18 12 20 19 17 11 10 9 > > > Thanks! > > >> So far I have not found any for any other odd values >> through n=31 (but I'll let them run overnight). > > > 31 factorial is really large; it could be a very long night... :-) He seems to be on n=33 ! > > One could mod out by a couple of dihedral group actions (one is geometric, > based on the cyclic ordering of the n points, the other is combinatorial, > based on the order in which the tour visits the points), but it's tricky > because if I'm not mistaken they act "from opposite sides" (I never learned > to like double cosets), and anyway, 31 factorial divided by 62 squared is > still enormous... > > Searching for symmetric solutions may well prove to >> find them more quickly than what I'm currently doing >> (just searching lexicographically). > > > Are any of your new solutions symmetrical? > > I should mention that my program found some of the n=9 solutions but missed > others because Mathematica wrongly computed their area to be nonzero (on > the order of 10^-16); the problem disappeared when I stuck in a (somewhat > time-consuming) FullSimplify. How do you try to avoid getting false > positives and false negatives? > > Jim

1 0

Re: [math-fun] Reentrant polygons with area zero
by Dan Asimov 30 Jul '16

30 Jul '16

Nice question! —Dan >For which n>2 does there exist a reentrant n-gon whose vertices are evenly >spaced on a circle and whose signed area (aka algebraic area) is zero? > >When n is even, there is an easy solution: draw a circuit from 1 to 2 to 3 >to ... to n/2-1 to n/2 to n to n-1 to n-2 to ... to n/2+1 to 1 (here I'm >numbering the n vertices cyclically). > >For n=3, it's trivially impossible to find such an n-gon, and I've checked >by brute force (using Mathematica) that there's no such n-gon when n=5 or >n=7. > >For n=9, there is such an n-gon (which I found through brute-force search): >http://mathenchant.org/nonagon.pdf >Mathematica assures me that >1/2 (-1/2 Sqrt[3] sin(\[Pi]/18)+1/2 sin(\[Pi]/9)-sin((2 \[Pi])/9)-2 >sin(\[Pi]/18) sin((2 \[Pi])/9)-1/2 cos(\[Pi]/18)-1/2 Sqrt[3] cos(\[Pi]/9)+2 >cos(\[Pi]/18) cos((2 \[Pi])/9)+2 sin(\[Pi]/9) cos(\[Pi]/9)) >simplifies to 0, which implies the claim. Is there a nice (preferably >trig-free) way to prove it? That is: is there a nice way to show that the >total area of the three scalene black triangles equals the total area of >the four nonscalene black triangles, maybe using dissection and >area-preserving linear transformations, or maybe using algebraic tricks >involving roots of unity? > >I suspect that when n is prime, there's no such polygon. Can any of you >find a proof? > >Also, can any of you construct such a polygon with n=15? It'd be nice if >the answer to my original question were "Precisely when n is composite".

1 0

Re: [math-fun] True or False, Log[Tan[t + p/4]]
by Dan Asimov 30 Jul '16

30 Jul '16

I like this proof method best of all. But there needs a workaround of the usual proof that the antiderivatives of 2 sec(2x) all differ by just a constant. That does not hold in this case, since each sigmoid of the original function can be slud up or down by an arbitrary amount. —Dan ----- f(x) = log(tan(x + pi/4)) f(0) = 0 and f'(x) = 2 sec(2x) is even, hence f(x) is odd. -----

2 1

Re: [math-fun] Robert Fano, 98, Dies -- Helped Develop Interactive Computers
by Dan Asimov 29 Jul '16

29 Jul '16

<< I'm guessing that it can't be that simple, so where am I going wrong? I'm not sure what question is being addressed. The one I asked was about describing *all* ways to interchange the points with the lines of the Fano plane such that incidence is preserved, not just to find some way to do it. But specifically regarding (Z/2Z)^3, I'm not sure there's a way to say that each line (1-dim subspace) has a unique plane (2-dim subspace). At least, I don't see a natural definition of a unique plane perpendicular to a given line (or vice versa). E.g., using the usual definition of perpendicularity determined by a 0 dot product with the usual definition of dot product, the vector (1,1,1) is perpendicular to the plane <(1,0,0), (0,1,0)> but also to the plane <(0,1,0), (0,0,1)>, and these two are clearly distinct. It's a good question, though: To what extent can standard geometrical concepts be defined naturally for vector spaces over finite fields? —Dan -----Original Message----- From: James Propp Sent: Jul 28, 2016 5:43 PM To: math-fun Cc: Dan Asimov Subject: Re: [math-fun] Robert Fano, 98, Dies -- Helped Develop Interactive Computers I know characteristic 2 can behave differently from characteristic 0, but in R^3, each line through the origin determines a unique plane through the origin (namely the one that's perpendicular); doesn't this also hold in (Z/2Z)^3? If so, can't you just projectivize to get the desired association between projective points and projective lines? I'm guessing that it can't be that simple, so where am I going wrong? Jim Propp On Thursday, July 28, 2016, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: << There is, incidentally, a type of duality between the points and lines of FP: If points are redefined as lines and the original lines redefined as points, it's still a Fano plane. I'm not very clear on how to best describe all incidence-preserving bijections between the set of points and the set of lines, i.e., ways of specifically interchanging each point with one and only one line to realize this duality. >> For integer i (mod 7) , label points and lines via points { i (mod 7) } and lines { i' = (i i+1 i+3) (mod 7) } , lines { i' (mod 7) } and points { i = (i' (i-1)' (i-3)' (mod 7) } ; spelled out in numbers, 0' = 013, 1' = 124, 2' = 235, 3' = 346, 4' = 450, 5' = 561, 6' = 602 , 0 = 064', 1 = 653', 2 = 542', 3 = 431', 4 = 320', 5 = 216', 6 = 105' . 1 / | \ 5 | 2 5' / 0 \ 1' / | \ 6 ----- 3 ----- 4 3' For much better pictures (but sadly no labels) see https://en.wikipedia.org/wiki/Fano_plane Transformations which leave the diagram unchanged are called "correlations". The Fano correlation group is a double cover of the usual symmetry group of 168 collineations, PGL(3,2) = PSL(2,7) : the extra symmetries comprise all collineations composed with any one duality, including f() : i -> -i' , i' -> -i . It is easy to verify (if I haven't fouled up again) that f() preserves collineations. Notice that it transforms this 7-cycle permutation of points into its inverse on lines (0 1 2 3 4 5 6 7) -> (0' 6' 5' 4' 3' 2' 1') ; therefore it exchanges two isomorphic conjugacy classes of PSL(2,7) , and is an outer automorphism. [ Essentially it is the only such, since Out(PSL(2, 7)) = Z_2 --- see https://en.wikipedia.org/wiki/Outer_automorphism_group --- contradicting the footnote at the bottom of the Fano_plane page, claiming that there are none. I can't offhand remember what "P Gamma L" stands for anyway: whatever, it's not what's written down there! ] A pleasant oddity from the Fano_plane page: if i is expressed in binary, converting from standard to Gray code yields a collineation of the plane! Fred Lunnon On 7/28/16, Dan Asimov <dasimov(a)earthlink.net> wrote: > So I wondered if this was the guy who discovered the Fano plane FP. > > (FP is the discrete projective plane that consists of 7 points and 7 lines > such that each point is contained in 3 lines and each line contains > 3 points. It's usually depicted as an equilateral triangle with all > 3 medians drawn in, plus a 7th "line" being its inscribed circle. > > FP can be obtained geometrically analogous to the ordinary projective plane > P^2, > by substituting the smallest finite field — F_2 — for the reals: > > Let V denote the vector space (F_2)^3. > > Define the points of FP to be the 1-dimensional subspaces of V, > and the lines of FP to be the 2-dimensional subspaces of V. > > (There is, incidentally, a type of duality between the points and lines of > FP: > If points are redefined as lines and the original lines redefined as points, > it's still a Fano plane. > > (I'm not very clear on how to best describe all incidence-preserving > bijections > between the set of points and the set of lines, i.e., ways of specifically > interchanging each point with one and only one line to realize this > duality.) > > > FP is particularly interesting in that its automorphism group > (permutations of its vertices that preserve incidence) is isomorphic > to the 2nd-smallest nonabelian simple group: the one of order 168. > > Turns out that the Fano who discovered this is Robert Fano's father, > the Italian mathematician Gino Fano. > > —Dan > > > ----- > FYI -- > > Bob Fano was my undergraduate advisor; his lovely daughter, Paola, worked at > BBN for a while. > > It's bizarre that the NYTimes makes such a big deal out of CTSS; I consider > Fano's work in information & coding theory (& whatever he did during WWII, > much of which may still be classified) his major accomplishments. He was > also an incredibly good teacher, and a wonderful advisor: he finally > convinced me that Kronecker may have been somewhat short-sighted ("God > created the integers..."), and got me to take probability. > > http://www.nytimes.com/2016/07/27/technology/robert-fano-98-dies-engineer-w… > > Robert Fano, 98, Dies; Engineer Who Helped Develop Interactive Computers > -----

2 1

Re: [math-fun] Interactive virtual planimeter
by Bill Gosper 29 Jul '16

29 Jul '16

It seems to me that planimeters are winding-numberish, based on the formula areaPolygon[L_List] := Total[MapThread[Im[#1\[Conjugate]*#2] &, {L, RotateLeft[L]}]]/2 whereas Mike's is sort of Riemann-Stieltjesish. Here's <http://gosper.org/planimeter.pdf> a crude Mathematica attempt at the former. Drag-undrag-drag makes points-linesegment-points. OOPS, bug in Mathematica's pdf renderer! The top illustration should look like gosper.org/planimeter.png , i.e. winding number mod 2. I'm not sure if Wolfram has a site for running toy programs, but this one seems short enough to be fairly portable. --rwg On 2016-07-26 10:57, Mike Stay wrote: > Here's a cheap one I just whipped up: > https://www.khanacademy.org/computer-programming/planimeter/5456307495501824 > > > On Tue, Jul 26, 2016 at 10:33 AM, James Propp <jamespropp(a)gmail.com> wrote: >> Does anyone know of a good interactive virtual planimeter that will tell >> you the signed area of a region that you've specified by dragging around a >> cursor? Even a version that only works for polygons (where the user clicks >> successive vertices) would be better than nothing. >> >> I've looked on the web, but I couldn't find what I need. Maybe one of you >> is inclined to build one? I would feature it (with credit, of course) in my >> upcoming Mathematical Enchantments essay about signed area (later this >> year). >> >> I want something that runs from a browser, so readers of my blog can just >> click on a link and a new tab will open. I am not interested in explaining >> the mechanics of how physical planimeters work (though I suppose I wouldn't >> be opposed to a webpage that explained that too, as long as the visuals >> associated with the wheels and arms didn't submerge the main message). I'd >> much rather have extra visuals that reinforce the concept of signed area. >> For instance, if the demo does just polygons, then every time a new vertex >> gets added, one could see a new triangle joining the new vertex, the >> previous vertex, and the origin, with shading or color to indicate whether >> it's oriented clockwise or counterclockwise, and with appropriate >> cancellation of positive and negative area being represented visually. >> >> It'd be especially nice if the demo handled reentrant polygons sensitively, >> so that regions encircled m times had a different color than regions >> encircled n times, when the integers m and n (positive, negative, or zero) >> are unequal. >> >> Jim Propp

2 1

Re: [math-fun] True or False, Log[Tan[t + π/4]]
by Bill Gosper 29 Jul '16

29 Jul '16

[is an odd function of t.] This is subtle. Mathematica plots the real part odd and the imaginary part even! gosper.org/reim ln tan.png Is Mathematica wrong? Julian writes "No. To make it an odd function, you need to make choices about branches of the composition of log and tan, while mathematica has a standard choice of branch for log which it is applying; this is the correct behavior." Thus igd(t) is so "darn strange" that it is odd but doesn't look it. Said more carefully, igd(2t) is not identically equal to log(tan(t+pi/4)). --rwg On 2016-07-27 21:53, rcs(a)xmission.com wrote: > Assuming you mean log(tan(t+pi/4)), this is igd(2t), > inverse gudermannian of 2t, which is an odd function > for small t. > To see this without appealing to special functions, > notice that (switching to degrees) > > tan(45+x) = cot(45-x) > > since the angles sum to 90. > Using cot == 1/tan, and applying the log, we get > > log(tan(45+x)) = - log(tan(45-x)). > > Rich > > ------- > Quoting Dan Asimov <dasimov(a)earthlink.net>: > >> is an odd function of t? >> ----- >> >> >> It's pretty darn strange, no doubt about it. >> >> ?Dan

1 0

Re: [math-fun] Robert Fano, 98, Dies -- Helped Develop Interactive Computers
by Dan Asimov 29 Jul '16

29 Jul '16

So I wondered if this was the guy who discovered the Fano plane FP. (FP is the discrete projective plane that consists of 7 points and 7 lines such that each point is contained in 3 lines and each line contains 3 points. It's usually depicted as an equilateral triangle with all 3 medians drawn in, plus a 7th "line" being its inscribed circle. FP can be obtained geometrically analogous to the ordinary projective plane P^2, by substituting the smallest finite field — F_2 — for the reals: Let V denote the vector space (F_2)^3. Define the points of FP to be the 1-dimensional subspaces of V, and the lines of FP to be the 2-dimensional subspaces of V. (There is, incidentally, a type of duality between the points and lines of FP: If points are redefined as lines and the original lines redefined as points, it's still a Fano plane. (I'm not very clear on how to best describe all incidence-preserving bijections between the set of points and the set of lines, i.e., ways of specifically interchanging each point with one and only one line to realize this duality.) FP is particularly interesting in that its automorphism group (permutations of its vertices that preserve incidence) is isomorphic to the 2nd-smallest nonabelian simple group: the one of order 168. Turns out that the Fano who discovered this is Robert Fano's father, the Italian mathematician Gino Fano. —Dan ----- FYI -- Bob Fano was my undergraduate advisor; his lovely daughter, Paola, worked at BBN for a while. It's bizarre that the NYTimes makes such a big deal out of CTSS; I consider Fano's work in information & coding theory (& whatever he did during WWII, much of which may still be classified) his major accomplishments. He was also an incredibly good teacher, and a wonderful advisor: he finally convinced me that Kronecker may have been somewhat short-sighted ("God created the integers..."), and got me to take probability. http://www.nytimes.com/2016/07/27/technology/robert-fano-98-dies-engineer-w… Robert Fano, 98, Dies; Engineer Who Helped Develop Interactive Computers -----

3 3

Re: [math-fun] True or False, Log[Tan[t + π/4]]
by Dan Asimov 28 Jul '16

28 Jul '16

is an odd function of t? ----- It's pretty darn strange, no doubt about it. —Dan

3 2

Re: [math-fun] Encyclopedia of math notation?
by Bill Gosper 28 Jul '16

28 Jul '16

Gary, I can't resist appealing to math-fun. I'm away from Knuth's AoCPs right now, but I think they have nice glossaries in the back. Likewise Concrete Math. There's https://en.wikipedia.org/wiki/List_of_mathematical_symbols some of which are arguable. (I just told Rohan "Hey, a brassiere" and showed him, partially concealed, the mention of the bra symbol. "(Yawn), where's the ket?") I've lost track of my review copy of Weisstein's monster The CRC Encyclopedia of Mathematics, Third Edition - 3 ... - CRC Press <https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uac…> but I don't remember it having a separate notation appendix, and I don't know how it would "alphabetize", say, a Legendre symbol. --rwg On Wed, Jul 27, 2016 at 9:38 AM, Gary Snethen <gsnethen(a)gmail.com> wrote: > Bill, > > I'm at a technical conference (SIGGRAPH) this week. When reading papers > and watching presentations, I occasionally come across unfamiliar notation. > > Are there any good resources for looking up formula notation and tracing > it back to a good explanation of the associated mathematics concept? > > As a mundane example, let's say someone came across a radical with a small > number above and to left of it. They could use such a resource to find > this notation and learn that it's the "nth root", perhaps with a link to a > page describing the concept: > > https://en.m.wikipedia.org/wiki/Nth_root > > A comprehensive resource would probably need to disambiguate the notation > based on the knowledge domain where it's being used. > > ---Gary > > > -- > Sent from my iPhone >

5 4