August 2014 - mathfun@mailman.xmission.com

[math-fun] Alan Adler, of Aerobie fame,
by Bill Gosper 14 Sep '14

14 Sep '14

finds Euler's product over primes for zeta(s) somewhat magical, and wonders if we funsters know other really neat sums or products involving primes. Hence yesterday's "Minor twist". Otherwise, all I could remember was our Valentine's Day 2011 discussion of "Euler's crazy pi product". --rwg

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[math-fun] Stupid question about geometrical transformations
by Fred Lunnon 04 Sep '14

04 Sep '14

What is generated by the union of projective and conformal (Moebius) groups? Since these two intersect in similarities, the super-group in n-space has dimension at least (n^2-1) + (n+2)(n+1)/2 - (n+1)n/2 - 1 = n^2 + n - 1 ; just how big is it? How should such transformations be represented for computational purposes? Why don't I know the answers to these apparently obvious questions? [Uh, maybe don't answer that one right now ...] Physicists have previously devoted some thought to this matter: in particular, a promising paper by Wolfgang Bertram (2001) at http://www.emis.de/journals/AG/2-4/2_329.pdf launches into discussing "Jordan functors", which will however surely cost this innocent much gruesome effort to decode. [Pascual Jordan certainly seems put himself about, despite which I don't recall ever having encountered him before this week.] Fred Lunnon

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Re: [math-fun] dragon gif
by Bill Gosper 01 Sep '14

01 Sep '14

Jörg>Telepathy? If have been thinking about _exactly_ this for some time, but did not code it up because my home-grown software does not really make it easy... Extra points if you do that gif-ery for the curve http://jjj.de/tmp-ryde/t-shirt-R13-15-hires-cropped.png (L-system at bottom, turns are by 120 degrees). Total curve should be light gray, and the moving part should be 1/13 of the total (so the same shape will appear 13 times, as indicated in the image). ... plus you get a T-shirt! (If you use the image to, well, print one. I got one and it looks quite OK). Best, jj Holy cr@p, http://gosper.org/semizerp.htm and http://gosper.org/semizerp1.htm on steroids! Another 13fold recursion, replete with inhomogeneous scaling: Mandelbrot's recursive Snowflake. Can such be likewise "teardropped"? He also had a 7fold recursion, I think. --rwg * Bill Gosper <billgosper(a)gmail.com> [Aug 31. 2014 11:27]: The Dragon *function* (not curve, dammit) continuously mapping [0,1] into the box[-1/3-i/3, 7/6+2i/3] with the image of a circulating interval of length 1/10 highlighted, pausing at various multiples of 1/10 to show anatomical measurements. gosper.org/dragondrift.gif --rwg

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Re: [math-fun] Another experimental multiple choice
by Bill Gosper 31 Aug '14

31 Aug '14

Allan>Gosper, I was already familiar with dragons, so I knew that in the tail chains each connected piece has half the area of the next (except for one weird joint in the middle). So I was able to get the right answer (assuming I did).<ACW But isn't that a fairly immediate consequence of the two components being congruent and perpdendicular and similar to their union? Join them perpendicularly, doubling the area. Downscale by rt2 to recover the original. But the 45deg rotation and downscale moves each noncentral blob down to the next smaller one, so areas go by 2^n. But thank you. That's exactly the sort of answer I was soliciting. And you may have been the only solver! Too hard for Math-fun? More likely too boring. --rwg (I'm surprised nobody said: Duh, you spoiled your own problem with the dragon gif post!) On Sun, Aug 31, 2014 at 4:26 PM, Bill Gosper <billgosper(a)gmail.com> wrote: On Sat, Aug 30, 2014 at 2:10 PM, Bill Gosper <billgosper(a)gmail.com> wrote: I tweaked it. gosper.org/Screen Shot < http://gosper.org/Screen%20Shot%20Dragon.png> Dragon < http://gosper.org/Screen%20Shot%20Dragon.png>.png < http://gosper.org/Screen%20Shot%20Dragon.png> . --rwg Did anyone (besides a helpful private respondent, who mentions that the figure remains vertically squished despite my struggles with (hypersucky) OpenOffice) find this too easy or too hard? If you solved it, did you use any prior or outside knowledge about Dragons? --rwg On Sat, Aug 30, 2014 at 12:19 PM, Bill Gosper <billgosper(a)gmail.com> wrote: Trying to minimize math jargon. gosper.org/Dragon Screen Shot.png < http://gosper.org/Screen%20Shot%20Dragon.png> Fair question? --rwg PS, pure evil: https://mail.google.com/mail/?ui=2&ik=97b2a014c7&view=att&th=1481fbf0be9db9… (Problem (4).) Subject: [math-fun] fib(1/2) Binet's and another formula give Sqrt[1/5 - (2 I)/5] and ((1 - I) (Cosh[ArcCsch[2]/2] + I Sinh[ArcCsch[2]/2]))/Sqrt[5] This is that "other formula". Axel Vogt raised a good point: In[730]:= MinimalPolynomial[((1 - I) (Cosh[ArcCsch[2]/2] + I Sinh[ArcCsch[2]/2]))/Sqrt[5], z] Out[730]= 1 - 2 z^2 + 5 z^4 I don't know why FullSimplify didn't try this. FullSimplify[TrigToExp[2*I^(n - 1)/Sqrt[5] Sin[n*ArcSec[2*I]]],n\[Element] Integers] comes within hailing distance of Binet.

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[math-fun] Another experimental multiple choice
by Bill Gosper 31 Aug '14

31 Aug '14

Trying to minimize math jargon. gosper.org/Dragon Screen Shot.png Fair question? --rwg

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Re: [math-fun] fib(1/2)
by Axel Vogt 31 Aug '14

31 Aug '14

> Subject: [math-fun] fib(1/2) > > Binet's and another formula give Sqrt[1/5 - (2 I)/5] and > ((1 - I) (Cosh[ArcCsch[2]/2] + I Sinh[ArcCsch[2]/2]))/Sqrt[5] > but FullSimplify can't get the former from the latter. > But it agrees they're equal. --rwg Using Maple: converting to exp gives an expression in radicals which 'radnormal' at least simplifies to -1/10*(2*I*5^(1/2)-5^(1/2)-5)/(1/2+1/2*5^(1/2))^(1/2); The package 'gfun' has a command 'algfuntoalgeq' which gives an equation for that, 8000-16000*I-40000*y^2 and solving it gives 1/5*(5-10*I)^(1/2) after picking that root.

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[math-fun] dragon gif
by Bill Gosper 31 Aug '14

31 Aug '14

The Dragon *function* (not curve, dammit) continuously mapping [0,1] into the box[-1/3-i/3, 7/6+2i/3] with the image of a circulating interval of length 1/10 highlighted, pausing at various multiples of 1/10 to show anatomical measurements. gosper.org/dragondrift.gif --rwg

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[math-fun] fib(1/2)
by Bill Gosper 31 Aug '14

31 Aug '14

Binet's and another formula give Sqrt[1/5 - (2 I)/5] and ((1 - I) (Cosh[ArcCsch[2]/2] + I Sinh[ArcCsch[2]/2]))/Sqrt[5] but FullSimplify can't get the former from the latter. But it agrees they're equal. --rwg

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[math-fun] Sampling from 2 different Gaussians
by Henry Baker 31 Aug '14

31 Aug '14

I'm pretty rusty with my probabilities. Suppose that I have 2 different Gaussian distributions, each with its own mean & variance. I want to take one sample from each of the distributions & compare them. Suppose that we call the first distribution A and the second distribution B; let a be the sample from A and b be the sample from B. Is there a relatively simple closed form expression for p(a>b) in terms of Amean, Avar, Bmean, Bvar ? What about a simple closed form for asymptotics -- e.g., a,b are sampled from a tail of A,B respectively -- e.g., a,b are each in the top decile of their respective distribution? top 1% ? I'm particularly interested in the cases where (Amean=Bmean,Avar>Bvar) and (Avar=Bvar,Amean>Bmean).

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[math-fun] Fundamental Theorem of Algebra: the video game
by James Propp 30 Aug '14

30 Aug '14

Viewing David Eisenbud's Numberphile video http://www.numberphile.com/videos/fundamental_theorem_of_algebra.html makes me wonder if anyone's created a better animation of this proof (note that the animation in the video treats the image of a circle centered at 0 under a polynomial map of degree n as just a circle, rather than a funky curve with winding number n). For that matter, it'd be neat to have an app that shows two copies of the complex plane: you get to move z around in one plane, which causes f(z) to move around appropriately in the other plane, and your goal is to move f(z) toward 0 by moving z as needed. Does such an app exist? Jim Propp

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