October 2019 - mathfun@mailman.xmission.com

Re: [math-fun] lituus?
by Bill Gosper 20 Oct '19

20 Oct '19

Thanks Gareth! lituus <http://gosper.org/lituus.gif> Now what's a convincing portrayal of the sector area constancy? Hey Eric, gifs in MathWorld? —rwg On Sun, Oct 20, 2019 at 4:10 PM rwg <rwg(a)ma.sdf.org> wrote: > > > -------- Original Message -------- > Subject: Re: [math-fun] lituus? > Date: 2019-10-17 19:06 > From: Gareth McCaughan <gareth.mccaughan(a)pobox.com> > To: math-fun(a)mailman.xmission.com > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > On 14/10/2019 12:42, Bill Gosper wrote: > > > MathWorld: > > "The lituus is the locus of the point [image: P] moving such that the > > area > > <http://mathworld.wolfram.com/Area.html> of a circular sector > > <http://mathworld.wolfram.com/CircularSector.html> remains constant." > > What circle? What sector? Pic? —rwg > > It's r^2 theta = const. > > So for a point P at radius r and angle theta, take a circular > sector spanning angles 0..theta and radius r. As the spiral > winds inward around the origin, the radius decreases just enough > to keep that area the same. For "almost all" of the curve we > have theta>2pi and of course we need to count the area with > multiplicity. > > So e.g. if we take the constant to be 1, then at angle 2pi > (i.e., the first time the curve crosses the positive x-axis) > we have r = sqrt(1/2pi) and that circle is covered just once; > at angle 4pi where it crosses again we have r = sqrt(1/4pi) > so the circle is sqrt(2) times smaller in radius but is covered > twice. >

1 0

[math-fun] A Quercus point with Geo-location
by Brad Klee 20 Oct '19

20 Oct '19

According to science news articles, in order to accomplish their yearly dismissals, deciduous trees have specially-evolved abscission cells. The release of used-up leafs is not a passive firing, rather it is an active process involving climate-dependent chemical signals. Naively we could hypothesize that the spatial distribution of active abscission cells is uniform random on the entire canopy volume at every instant of time; however, a walk around the block, and I could easily find many counterexamples--especially oaks, including the Pin Oak (Quercus Palustris). Here is a picture taken today, of tree T1438 (with a day moon in the background, and now with GPS metadata, cool!): https://0x0.st/z3Ma.jpg https://0x0.st/z3Mm.png Most of the tree remains green, but already a few branches have gone entirely bare. Then there are a few working groups of mostly bronze color, which seem likely to fall soon. This data suggests that distribution function for chemical signalling is highly correlated on the small scale of a single leaf, but remains at least partially random on the large scale of the entire tree. It seems impossible to derive chemical equations of motion and the ensuing statistics, but still I am curious, is there a better available empirical explanation as to what the signals are, where they come from, and how they flow through the tree's network of branches? And how much do these effects depend on species? --Brad [1] https://www.npr.org/templates/story/story.php?storyId=114288700 [2] https://www.esf.edu/pubprog/brochure/leaves/leaves.htm

1 0

[math-fun] Fwd: Re: "LHS Sometimes returns precision errors? What the bleep is it doing numerics for an identity map?"
by Bill Gosper 20 Oct '19

20 Oct '19

Date: 2019-10-19 09:23 From: R B <bossipm(a)outlook.com> Some googling gave me https://nanopdf.com/download/sines-and-cosines-for-fractions-of_pdf from which Sin[π/34] can be derived simply. Arie Bos "So, sin π/n and cos π/n can be found for n = 2k ×3p ×5q ×17r ×257s ×65537t where k is a positive or zero integer and the indices p, q, r, s, t are all either zero or 1." RubbEESSHimo! In[90]:= #@Sin[π/7] & /@ {ToRadicals, Developer`TrigToRadicals} During evaluation of In[90]:= Developer`TrigToRadicals::obs: Developer`TrigToRadicals has been superseded by ToRadicals, [I believe this to be completely false.] and is now obsolete. It will not be included in future versions of the Wolfram Language. [And I pray this will not happen.] Out[90]= {-(1/2) (-1)^(5/14) (-1 + (-1)^(2/7)), -(1/2) I (1/3 (1/ 2 (1 - I Sqrt[7]) - ((-1 + I Sqrt[3]) (1/2 (-1 + I Sqrt[7]) + 1/2 (-1 - I Sqrt[7]) (1/2 (-1 + I Sqrt[3]) + 1/4 (-1 + I Sqrt[3])^2)))/( 2 (6 + 3/4 (-1 + I Sqrt[3]) (-1 + I Sqrt[7]) + 1/2 (-1 - I Sqrt[7]) (1 + 3/4 (-1 + I Sqrt[3])^2))^(1/3)) - 1/4 (-1 + I Sqrt[3])^2 (6 + 3/4 (-1 + I Sqrt[3]) (-1 + I Sqrt[7]) + 1/2 (-1 - I Sqrt[7]) (1 + 3/4 (-1 + I Sqrt[3])^2))^(1/3)) + 1/3 (1/2 (-1 - I Sqrt[7]) + ((-1 + I Sqrt[3])^2 (1/2 (-1 - I Sqrt[7]) + 1/2 (-1 + I Sqrt[7]) (1/2 (-1 + I Sqrt[3]) + 1/4 (-1 + I Sqrt[3])^2)))/( 4 (6 + 3/4 (-1 + I Sqrt[3]) (-1 - I Sqrt[7]) + 1/2 (-1 + I Sqrt[7]) (1 + 3/4 (-1 + I Sqrt[3])^2))^(1/3)) + 1/2 (-1 + I Sqrt[3]) (6 + 3/4 (-1 + I Sqrt[3]) (-1 - I Sqrt[7]) + 1/2 (-1 + I Sqrt[7]) (1 + 3/4 (-1 + I Sqrt[3])^2))^(1/3)))} In[325]:= MinimalPolynomial@% Out[325]= {-7 + 56 #1^2 - 112 #1^4 + 64 #1^6 &, -7 + 56 #1^2 - 112 #1^4 + 64 #1^6 &} In[328]:= %%%[[1]]@Sin[π/7] // FullSimplify Out[328]= 0 How can that bullbleep page remain up? —rwg > -----Oorspronkelijk bericht----- > Van: math-fun <math-fun-bounces(a)mailman.xmission.com> Namens Bill > Gosper > Verzonden: zaterdag 19 oktober 2019 07:46 > Aan: math-fun(a)mailman.xmission.com > CC: Wolfram Technical Support <support(a)wolfram.com> > Onderwerp: [math-fun] "LHS Sometimes returns precision errors? What the > bleep is it doing numerics for an identity map?" > Urgentie: Hoog > > Brad, there's a really interesting answer. E.g., In[68]:= Sin[π/34] // > FunctionExpand > > Out[68]= 1/(4 Sqrt[2/( > 15 + Sqrt[17] - Sqrt[2 (17 - Sqrt[17])] - Sqrt[ > 2 (34 + 6 Sqrt[17] + Sqrt[2 (17 - Sqrt[17])] - Sqrt[ > 34 (17 - Sqrt[17])] + 8 Sqrt[2 (17 + Sqrt[17])])])]) > > (Rather cumbrous, Now undo it.) > > In[69]:= ArcSin@% // FullSimplify // Timing > > Out[69]= {0.312125, π/34} > > How could it take .3 sec to look that up? Alternatively, how could it > get π/34 > so quickly? Or at all? > > Here Mathematica recognizes Macsyma's nicer rendition: > In[70]:= ArcSin[(-\[Sqrt](\[Sqrt]17 + 3) \[Sqrt](4 \[Sqrt]17 - > 2 \[Sqrt](34 - 2 \[Sqrt]17)) + \[Sqrt](34 - > 2 \[Sqrt]17) + \[Sqrt]17 - 1)/16] // FullSimplify // Timing > > Out[70]= {0.948516, π/34} > > even though it wasn't able to simplify its version to Macsyma's. So > where did > it get the notion of π/34? It guessed it numerically! It is then > fairly routine to > show that the two surds are identical, or that Macsyma's satisfies > In[72]:= > MinimalPolynomial[Sin[π/34], x] // Timing > > Out[72]= {0.002874, > 1 - 8 x - 40 x^2 + 80 x^3 + 240 x^4 - 192 x^5 - 448 x^6 + 128 x^7 + > 256 x^8} > > At least this is how Macsyma did it. Without even requiring a ratsimp > or > radcan. > —rwg > _______________________________________________ > math-fun mailing list

2 1

[math-fun] Zulu Bungy was tragically put out of business (collapse side panel)
by Bill Gosper 19 Oct '19

19 Oct '19

https://www.google.com/maps/place/37%C2%B056'05.0%22N+22%C2%B059'02.0%22E/@37.9217471,22.9981443,557a,35y,38.17t/data=!3m1!1e3!4m5!3m4!1s0x0:0x0!8m2!3d37.934722!4d22.983889?hl=en when the state-owned cement company substituted spanakopita flour during a shortage. Seriously, can anyone explain this effect? —rwg

3 2

[math-fun] FullSimplify[ArcTan[Cos[x], Sin[x]]] == ArcTan[Cos[x], Sin[x]] ?
by Brad Klee 18 Oct '19

18 Oct '19

Ouch! I think it's just a modulus function, that's what the graph looks like at least: Show[ Plot[ArcTan[Cos[x], Sin[x]], {x, -10 Pi, 10 Pi}, PlotStyle -> Thick], Plot[Mod[x + Pi, 2 Pi] - Pi, {x, -10 Pi, 10 Pi}, PlotStyle -> Directive[Dashing[.05], Thick, Red]], PlotRange -> All, ImageSize -> 500 ] Or is it supposed to be: Show[ Plot[ArcTan[Cos[x], Sin[x]], {x, -10 Pi, 10 Pi}, PlotStyle -> Thick], Plot[Pi - Mod[Pi - x, 2 Pi], {x, -10 Pi, 10 Pi}, PlotStyle -> Directive[Dashing[.05], Thick, Red]], PlotRange -> All, ImageSize -> 500] Gotta watch those boundary values, and don't forget all of the broken evaluation over complexes: ArcTan[Cos[x],Sin[x]]==(Pi-Mod[Pi-Re[x],2*Pi]+I*Im[x]) LHS Sometimes returns precision errors? What the bleep is it doing numerics for an identity map? Try: ArcTan[Cos[I #], Sin[I #]] & /@ Range[100] ArcTan[Sin[I #]/Cos[I #]] & /@ Range[100] Refine[ArcTanh[Tanh[x]], {x \[Element] Reals}] Surprisingly, the second alternative does quite a bit better, managing to get up to n=75 rather than n=18, but still dies at 76 and thereafter. This is okay as long as we only need a thin strip around the real axis, but in general, no, "Indeterminate" is not a good return. --Brad

3 2

[math-fun] lituus?
by Bill Gosper 18 Oct '19

18 Oct '19

MathWorld: "The lituus is the locus of the point [image: P] moving such that the area <http://mathworld.wolfram.com/Area.html> of a circular sector <http://mathworld.wolfram.com/CircularSector.html> remains constant." What circle? What sector? Pic? —rwg

2 1

[math-fun] A faster algorithm for π(x)
by Varun Rajkumar 15 Oct '19

15 Oct '19

Earlier, I had the doc π(n) and the Sieve of Eratosthenes <https://docs.google.com/document/d/1yzj1QZZzmqVG4ow6ScTR5bX5mDOKxNVkQGX7G-H…>, but I have a better algorithm: This is the same method as in the doc (read the doc and edit it) but we don't add 1 because then it will be hard to compute. What do you think of it? It works, I tried it. -Varun

2 1

[math-fun] A Permutation Cipher Text
by Brad Klee 13 Oct '19

13 Oct '19

https://0x0.st/zwRU.txt Easy to brute force if you get the gist of the permutation scheme but looks difficult to reverse engineer otherwise. Codebreaking algorithms? --Brad

3 4

Re: [math-fun] A Permutation Cipher Text
by Keith F. Lynch 12 Oct '19

12 Oct '19

Mike Stay <metaweta(a)gmail.com> wrote: > Keith F. Lynch <kfl(a)keithlynch.net> wrote: >> Some years ago it occurred to me that exclusive-oring a text with >> a subset of the square roots, in binary, of the first N primes, >> followed by the above scrambling algorithm (in ASCII order), >> followed by the same exclusive-oring again, would be a pretty >> good encryption algorithm. Does anyone disagree? > The scrambling is irrelevant if the ciphertext has length N and > the square roots of those N primes are GF(2)-linearly independent. I'm assuming that the square roots of the primes, expressed in binary, are all normal ("patternless") and uncorrelated with each other (i.e. the exclusive-or of any subset of them is also normal), though as far as I know neither has actually been proven. (I'm speaking of the digits after the radix point only.) > Just apply Gaussian elimination and you've reduced the cipher to a > one-time pad. > If N is less than the length of the ciphertext, then there will > likely be exploitable statistical correlations. N *is* less. It's just large enough that it's impractical to try all 2^N possible subsets of the first N primes, and will remain so for the foreseeable future. The main reason for the scrambling step is because I use the *same* subset of the primes for multiple files. (Well, almost the same. I append the low 24 bits of the file creation time (Unix time, seconds since 1970 began) to the subset.) Specifically, I hash, scramble, hash again, reverse-scramble, then hash again. This has the advantage that encryption and decryption are identical. Also, since it's an odd number of hashes (with the same subset), the number of 1-bits is not preserved, as it would be with an even number of hashes regardless of scrambling. "Hash" means exclusive-or the k-bit text with the first k bits of a subset of the first n primes exclusive-ored together. "Scramble" means reverse the text between each pair of identical characters, wrapping around so that the first and last characters aren't unmoved. I do this once for every character, in ASCII order. "Reverse scramble" is the same as scramble, but in reverse ASCII order. If what I'm proposing is sound, then scrambling would also allow for re-use of one time pads on multiple files.

2 1

[math-fun] Efficient way to find solutions?
by Mike Stay 12 Oct '19

12 Oct '19

I'd like to efficiently find solutions to constraints of the form lower1 <= ((x * 0x000001) mod 2^24) < upper1 && lower2 <= ((x * 0x088405) mod 2^24) < upper2 && lower3 <= ((x * 0x652819) mod 2^24) < upper3 && ... where the constants are consecutive powers of 0x088405. Brute-forcing the space isn't much work to do for a single set of upper and lower bounds, but when I've got 2^40 such sets, it's daunting. Any ideas? -- Mike Stay - metaweta(a)gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com

4 6