October 2016 - mathfun@mailman.xmission.com

Re: [math-fun] mult. group of GF(p) uniformly distributed?
by Henry Baker 15 Oct '16

15 Oct '16

"0" isn't a member of the "multiplicative *group* of GF(p)", so there is no off-by-one problem. I guess since x->g^x is 1-1, it maps uniform distributions to uniform distributions. But I guess this reasoning only works for finite (countable??) sets. At 03:12 PM 10/14/2016, Allan Wechsler wrote: >I did misunderstand. > >However, I think we have an off-by-one problem: g^k will never, in fact, take on the value 0. > >It will be equally distributed over the nonzero values, of which there are only (p-1). > >On Fri, Oct 14, 2016 at 6:02 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: >> But 2 is not a generator of [0,6] = Z/7. For a generator g of Z/p, the powers >> g^k will be all distinct for 0 <= k <= p-1; hence {g^k | 0 <= k < p} = Z/p. >> >> So g^X is in fact uniform on Z/p when X is. >> >> --Dan >> >> On Oct 14, 2016, at 2:25 PM, Allan Wechsler <acwacw(a)gmail.com> wrote: >> >> Let p=7, g=2. For X in [0,6], g^X takes on the values 1, 2, 4, 1, 2, 4, 1. >> This isn't uniform: it takes the value 1 thrice, the values 2 and 4 twice, >> and 3, 5, and 6 never. >> >> Am I misunderstanding your question? >> >> On Fri, Oct 14, 2016 at 4:08 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: >>> Is this known or conjectured with a conjecture name? >>> >>> If X is a uniform random distribution over [0,p-1], p prime, >>> is g^X a uniform random distribution, for a generator g ?

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[math-fun] Fwd: PolyhedronData["TriangularHebesphenorotunda", "Faces"]
by Bill Gosper 14 Oct '16

14 Oct '16

---------- Forwarded message ---------- From: Bill Gosper <billgosper(a)gmail.com> Date: Fri, Oct 14, 2016 at 12:20 PM Subject: PolyhedronData["TriangularHebesphenorotunda", "Faces"] To: Wolfram Technical Support <support(a)wolfram.com> <Johnson solid png censored courtesy xmission.com> (PolyhedronData["TriangularHebesphenorotunda"] default view hides C3 symmetry.) GraphicsComplex[{{-0.9601329154398749, 0.715228827607169, -0.03894034047821841}, {-0.8187282992850485, 0.1613051767267678, -0.8594117437126204}, ...}] Bleahh. GraphicsComplex[{{-1, 0, 1/32 (Sqrt[3] + 5 Sqrt[15])}, {-(1/2), -(1/(2 Sqrt[3])), -(( 45 + Sqrt[5])/(32 Sqrt[3]))}, {-(1/2), -(Sqrt[3]/2), 1/32 (Sqrt[3] + 5 Sqrt[15])}, {-(1/2), Sqrt[3]/2, 1/32 (Sqrt[3] + 5 Sqrt[15])}, {-(1/2), (2 + Sqrt[5])/( 2 Sqrt[3]), -((13 + Sqrt[5])/(32 Sqrt[3]))}, {0, 1/Sqrt[ 3], -((45 + Sqrt[5])/(32 Sqrt[3]))}, {0, -((3 + Sqrt[5])/( 2 Sqrt[3])), (-29 + 15 Sqrt[5])/(32 Sqrt[3])}, {1/ 2, -(1/(2 Sqrt[3])), -((45 + Sqrt[5])/(32 Sqrt[3]))}, {1/ 2, -(Sqrt[3]/2), 1/32 (Sqrt[3] + 5 Sqrt[15])}, {1/2, Sqrt[3]/2, 1/32 (Sqrt[3] + 5 Sqrt[15])}, {1/2, (2 + Sqrt[5])/( 2 Sqrt[3]), -((13 + Sqrt[5])/(32 Sqrt[3]))}, {1, 0, 1/32 (Sqrt[3] + 5 Sqrt[15])}, {1/4 (-3 - Sqrt[5]), 1/12 (Sqrt[3] - Sqrt[15]), -((13 + Sqrt[5])/(32 Sqrt[3]))}, {1/ 4 (-3 - Sqrt[5]), (3 + Sqrt[5])/(4 Sqrt[3]), (-29 + 15 Sqrt[5])/( 32 Sqrt[3])}, {1/4 (-1 - Sqrt[5]), -((5 + Sqrt[5])/( 4 Sqrt[3])), -((13 + Sqrt[5])/(32 Sqrt[3]))}, {1/ 4 (1 + Sqrt[5]), -((5 + Sqrt[5])/(4 Sqrt[3])), -((13 + Sqrt[5])/( 32 Sqrt[3]))}, {1/4 (3 + Sqrt[5]), 1/12 (Sqrt[3] - Sqrt[15]), -((13 + Sqrt[5])/(32 Sqrt[3]))}, {1/ 4 (3 + Sqrt[5]), (3 + Sqrt[5])/(4 Sqrt[3]), (-29 + 15 Sqrt[5])/( 32 Sqrt[3])}}, Polygon[{{14, 5, 6, 2, 13}, {5, 14, 4}, {14, 1, 4}, {1, 14, 13}, {1, 13, 15, 3}, {15, 13, 2}, {7, 15, 2, 8, 16}, {3, 15, 7}, {9, 3, 7}, {9, 7, 16}, {17, 16, 8}, {9, 16, 17, 12}, {6, 8, 2}, {6, 11, 18, 17, 8}, {18, 12, 17}, {18, 10, 12}, {10, 18, 11}, {5, 4, 10, 11}, {5, 11, 6}, {10, 4, 1, 3, 9, 12}}]] In[776]:= PolyhedronData["TriangularHebesphenorotunda", "Volume"] Out[776]= 5.10875 Bleahh, 5/2 + (7 Sqrt[5])/6 In[458]:= PolyhedronData[%453[[-1]], "Centroid"] Out[458]= {0., 0., 0.} Bleahh. {0, 0, 0}. --rwg >So [ElongatedPentagonalRotunda] just barely won't stand on a square face. --rwg >You could use one as a carpenter's level. It must tip over, and >again when rotated 180º. But then why not just keep in your pocket a scrap of copper or PVC pipe?

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[math-fun] mult. group of GF(p) uniformly distributed?
by Henry Baker 14 Oct '16

14 Oct '16

Is this known or conjectured with a conjecture name? If X is a uniform random distribution over [0,p-1], p prime, is g^X a uniform random distribution, for a generator g ?

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[math-fun] Number of semiprimes
by Zak Seidov 13 Oct '16

13 Oct '16

What is the maximal number of semiprimes among 100 consecutive numbers? 38? -- Zak Seidov

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[math-fun] The Law of Sines
by Bill Gosper 11 Oct '16

11 Oct '16

Anish: In Holt McDougal Mathematics Grade 7, Common Core Edition Chapter 4 Proportional Relationships, Chapter Test Tell whether the figures are similar. Problem 16. We concluded "No, because the angles disagreed and the sides weren't proportional." But a really neat theorem from Trigonometry says that if a triangle has angles A, B, C, and opposite sides a, b, c, then sin(A)/a = sin(B)/b = Sin(C)/c. The first triangle has for these three ratios In[532]:= N@{Sin[99 º]/11, Sin[27 º]/5, Sin[54 º]/9} Out[532]= {0.08978984914501252, 0.09079809994790936, 0.08989077715277194} (You can type º by Esc deg Esc.) In[533]:= Sin[99 º]/11 == Sin[27 º]/5 == Sin[54 º]/9 Out[533]= False so it is geometrically impossible! Doubly so, because all three ratios disagree. The second triangle has ratios In[534]:= N@{Sin[102 º]/22, Sin[29 º]/11, Sin[49 º]/17} Out[534]= {0.04446125457880935, 0.0440736018405761, 0.04439468118957483} In[535]:= Sin[102 º]/22 == Sin[29 º]/11 == Sin[49 º]/17 Out[535]= False So the two triangles *are* similar--they're both impossible! Furthermore, since you can conclude absolutely anything from a false premise, they are also *not* similar, although they're both vegetarian Klingons performing karaoke. Problem 17. True. Unquestionably. The angles agree. The sides are proportional. And the figures are possible because quadrilaterals aren't rigid. ∆WYZ ~ ∆MNO in each pair. Find the unknown measures. 18. <Boggle!> The first triangle is again (doubly) impossible. But by (alleged) similarity, we can substitute angles into the second triangle, where the Law of Sines gives us two equations and two unknowns, which we can solve! Out[536]= Sin[37 º]/n == Sin[a]/10 == 1/5 Sin[27 º] In[537]:= Solve@% Out[537]= {{n -> 5 Csc[27 º] Sin[37 º], a -> ConditionalExpression[π - ArcSin[2 Sin[27 º]] + 2 π C[1], C[1] ∈ Integers]}, {n -> 5 Csc[27 º] Sin[37 º], a -> ConditionalExpression[ArcSin[2 Sin[27 º]] + 2 π C[1], C[1] ∈ Integers]}} Don't be scared by this. It's the best way to write the completely correct answer. (Csc[27º] = 1/Sin[27º]. ArcSin[2 Sin[27 º]] is the angle whose sin is twice the sin of 27º.) So n -> 5 Csc[27 º] Sin[37 º] becomes n -> 6.628057453815753, contradicting 5×4/6 = 6.66666... that we got from proportions in the impossible first triangle. The ConditionalExpression is Mathematica's slightly awkward notation for two infinite sets of solutions, where C[1] stands for an arbitrary constant and C[1] ∈ Integers means it's an integer. By trying C[1] = -1, 0, 1 we would discover that our only plausible solution is a ~ 115º, the first case of Out[537] when C[1] = 0. Instead of extracting this, we can now use, instead of Solve, the fancier solver Reduce, which handles both equations and inequalities. Just tell it a is obtuse: In[540]:= Reduce[{90 º < a < 180 º, %536}, {a, n}] Out[540]= a ==π- ArcSin[2 Sin[27º]] && n == 5 Csc[27º] Sin[37 º] The π and ArcSin are in radians. In terms of degrees, In[541]:= % /. π-> 180 º/. Csc@x_ -> 1/Inactive[Sin]@x Out[541]= a == 180 º - ArcSin[2 Sin[27 º]] && n == (5 Sin[37º])/Sin[27º] Numerically, In[544]:= N@Activate@% Out[544]= a == 2.003152519583483 && n == 6.628057453815753 Again, a is in radians. In degrees, In[546]:= MapAt[180 Inactive@º#/π &, %%, {1, 2}] Out[546]= a == 114.7721850931306 º && n == 6.628057453815753 instead of the book's a = 116º and n = 20/3. Nowhere that I can see does the book admit that these angles and sides are only approximate, nor are you told how to manipulate approximate quantities. 19. wants you to solve for x and c in the first triangle, which is "similar" to the (impossible) second one. By the Law of Sines, In[524]:= Solve[Sin@c/x == Sin[74 º]/11 == Sin[44 º]/8 == Sin[62 º]/10] (Debug) Out[524]= {} Mathematica rejects this as an inconsistent system with no solutions, but knowing that the second "triangle" is not a triangle, a valid solution is x = The Square Rutabaga of the Apocalypse, and c = The Heartbreak of Psoriasis. --Bill

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[math-fun] Folding a tie in thirds
by James Propp 11 Oct '16

11 Oct '16

James Tanton's video https://m.youtube.com/watch?v=G8cD5NMOMcs is fun but I don't plan to adopt his solution; when I have to fold a tie into thirds, I use several fingers plus a little physics to equalize the lengths. It's hard to describe; a little bit like pulleys. (I guess I could make a video clip if any of you are unsure of what I mean.) But the open-ended ending of Tanton's video makes me wonder how my physics-based approach could be generalized to other folding fractions. Surely someone has written about this? Also: my method ignores the fact that fingers have positive radius. I wonder if some variant of my method compensates for this. Jim Propp

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[math-fun] Rotundancy
by Bill Gosper 10 Oct '16

10 Oct '16

Mathematica likes to tabulate polyhedra with their centroid at the origin, and unit edges. An interesting exception is the Johnson solid In[459]:= ToRadicals[PolyhedronData["ElongatedPentagonalRotunda", "Centroid"]] Out[459]= {0, 0, (-3735 + 1689 Sqrt[5] - Sqrt[2 (1449845 - 3240379/Sqrt[5])])/13928} (Elongated = glued to a prism with square faces.) Even more interesting is how close this is to the origin: In[460]:= N@% Out[460]= {0., 0., 0.0003022667749277455} So it just barely won't stand on a square face. --rwg You could use one as a carpenter's level. It must tip over, and again when rotated 180º.

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[math-fun] Same Diagonal Dissection
by Ed Pegg Jr 08 Oct '16

08 Oct '16

I've been doing some strange dissections lately. One is a nice puzzle -- divide a 6x6x6 cuboid into different integer-sided cuboids with volumes from 12 to 24. The solution is unique. Divide a rectangle into smaller rectangles with different sizes, but all having diagonals with the same length. The following link shows a dissection with 12 rectangles. http://math.stackexchange.com/questions/1958738/same-diagonal-dissection Use 30 or less different rectangles to bound a 3D space. No two neighboring rectangles can be in the same plane. http://community.wolfram.com/groups/-/m/t/928487 Blanche dissections, and the Mondrian puzzle http://community.wolfram.com/groups/-/m/t/903043 Zak's triangle http://community.wolfram.com/groups/-/m/t/851275 One puzzle I haven't solved -- can a cuboid be dissected into differently sized cuboids with identical volume, in a way that isn't equivalent to a 2D or stack of 2D solutions? Can different cuboids with volume 720 fill a 20x24x24 cuboid?

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[math-fun] trivial example of a discontinuous dimension limit
by Bill Gosper 08 Oct '16

08 Oct '16

The centroid of a triangle as one side vanishes. It approaches 2/3 of the way toward the vanishing side, instead of the midpoint of a line segment. --rwg

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[math-fun] Nine every time
by James Propp 08 Oct '16

08 Oct '16

Art Benjamin recently gave a talk in which he performed the following trick: An audience member secretly picks a positive integer N whose digits are distinct and increasing. The audience member uses a calculator to compute 9N. The magician is able to divine the sum of the digits of 9N without knowing N. That's because that sum is always 9. Art tells me he learned this fact from an article in Mathematics Magazine (Vol 87, No 3, June 2014) called "Surprises, Surprises, Surprises", written by Felix Lazebnik. Lazebnik's statement is slightly more general: "Consider any positive integer N whose (decimal) digits read from left to right are in non-decreasing order, but the last two digits (tens and ones) are in increasing order. Prove that the sum of digits of 9N is always exactly 9." Lazebnik says that he heard about the problem from applied mathematician named Valery Kanevsky, who tells me he does not know the problem's provenance. Do any of you know anything about this? Jim Propp

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