February 2016 - mathfun@mailman.xmission.com

[math-fun] Flux of the gravitational wave event (was Re: Maps from earthshine?)
by Keith F. Lynch 23 Feb '16

23 Feb '16

Eugene Salamin <gene_salamin(a)yahoo.com> wrote: > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >> (One final aside on moonlight and gravitational waves: Has anyone >> else noticed that the peak flux of the gravitational wave event, >> GW150914, on Earth was seven times the flux of the light of the >> full Moon? If it was visible light, you could not only have seen >> it, but could have read by its light. LIGO isn't really very >> sensitive!) > http://www.aei.mpg.de/~schutz/download/lectures/AzoresCosmology/Schutz.Azor… > the flux (power per unit area) carried by a gravitational wave of frequency f and strain h is > F = (pi/4) (c^3/G) f^2 h^2 The equation I found has an 8 instead of the 4. I'll try to find it again. Except for the pi/8 or pi/4 constant factor, the equation is easy to deduce from dimensional analysis. > ?? = (3.2 mW/m^2) (h / 1e-22)^2 (f / 1 kHz)^2. > At the peak of the recently detected wave, h = 1e-21, f = 100 Hz, I had heard that the (Earth-received) frequency at the peak flux was about 250 Hz. So I got amout 10 mw/m^2. As a sanity check, I divided the reported peak power output of the event, 3.6E+49 watts, i.e. 200 solar masses per second annihilated, by the area of a sphere 1.3 billion light years in radius. I get about 20 milliwatts per square meter. What accounts for the factor of two discrepancy? Probably polarization. LIGO, if I understand correctly, is sensitive to only one of the two polarizations. Also, red shift. 200 solar masses per second at the event equals about 180 solar masses per second on Earth, as the event is receding from us at about a tenth of the speed of light. Also, the received frequency at the time of peak power, about 250 Hz, was originally about 270 Hz. That's an impressive orbital period (1/270 second) for two 30-solar-mass objects. ("Only" 3 solar masses were annihilated in total, because the event lasted less than a second.)

2 1

Re: [math-fun] Maps from earthshine?
by Keith F. Lynch 23 Feb '16

23 Feb '16

Mike Stay <metaweta(a)gmail.com> wrote: > Sounds a lot like http://graphics.stanford.edu/papers/dual_photography/ Interesting, especially that that dates only to 2005. I had basically the same idea maybe 40 years ago. My idea was to project a blank (solid white) raster from a TV onto whatever I wished to photograph. I realized that a photocell would pick up an NTSC video feed suitable for showing on a TV (complete with vertical and horizontal blanking intervals), and that the scene would look as if the camera was where the TV was, the light source was where the photocell was, and that if there was any other light source the image would look foggy. My proposed application was a poor man's black and white video camera, since in those days TVs and photocells were plentiful and video cameras weren't. Note that no computers were involved. In retrospect it wouldn't have worked very well because CRT TVs had a slow phosphor. But I'm sure it would have worked if they had had fast phosphors. And that, in turn, reminds me of a 2002 demonstration that it's possible to reconstruct an image from a CRT if you aim a photocell at any surface illuminated by that CRT and throw lots of computer power at it to compensate for the slow phosphor: http://csdl2.computer.org/persagen/DLAbsToc.jsp?resourcePath=/dl/proceeding… However, I don't see the application to constructing an Earth map from earthshine, given that the Moon is illuminated by all sun-illuminated parts of the Earth in its view at once, rather than one pixel at a time.

1 0

[math-fun] measures of nonEuclidean simplices
by Warren D Smith 23 Feb '16

23 Feb '16

A review of this topic is in appendix C of http://rangevoting.org/BestVrange.html for the 3-dimensional formulas see appendix C3...

1 0

[math-fun] Maps from earthshine?
by Keith F. Lynch 23 Feb '16

23 Feb '16

I'm always interested in how much can be figured out from how little. The recent gravitational wave observation, and the reconstruction of its source, is a perfect example. Earthshine is the illumination of the dark part of the Moon by the Earth. How bright it is depends on the phase of the Earth as seen from the Moon, how much of the illuminated part of the Earth that's visible from the Moon is cloud-covered, how much is ice-covered, how much is cloud-free ice-free land, and how much is cloud-free ice-free ocean. Also, its color depends on how what's doing the reflecting (clouds, ice, water, desert land, vegetation, etc.). My question is how good a map of the Earth it would be possible to construct from careful measurements of earthshine from one location on Earth. Assume its brightness can be accurately measured even in the daytime (or explain why it can't be). Note that the Moon's latitude as well as its longitude varies considerably, so it doesn't always have an equatorial view. Decades of observations can be combined so as to average out the clouds and concentrate on long-term features. I'm envisioning an alternate history in which photometry, advanced math, and computers were developed before world travel. (And in fact, the nature of earthshine was understood long before Columbus.) Also, a hypothetical skeptic who wonders if other continents really exist as depicted on maps, and only trusts his own observations, given that computers, telescopes, and photometers are cheaper than world travel. (One final aside on moonlight and gravitational waves: Has anyone else noticed that the peak flux of the gravitational wave event, GW150914, on Earth was seven times the flux of the light of the full Moon? If it was visible light, you could not only have seen it, but could have read by its light. LIGO isn't really very sensitive!)

5 8

[math-fun] area of spherical triangle
by Warren D Smith 22 Feb '16

22 Feb '16

Folke Eriksson: On the measure of solid angles, Mathematics Magazine 63,3 (June 1990) 184-187. http://www.maa.org/sites/default/files/Eriksson14108673.pdf Eriksson's formula can be re-written as follows. Let the coordinates of the vertices of the triangle (drawn on unit sphere) be the rows of 3x3 matrix X. Then NonEuclideanSignedArea = 2 * arctan( det(X) / (1+M12+M23+M13) ) where M=X Xtranspose and for spherical geometry K=+1. This formula also works to give the signed area of a hyperbolic nonEuclidean triangle where the first two coordinates of each unit-length 3-vector now are imaginary, and we change arctan to argtanh. And of course in 1 dimension lower (X now a 2x2 matrix) the corresponding formula is Measure = arccos( M12 ) or for hyperbolic use argcosh. But I do not believe there is any formula anywhere near as simple as these if we go to one dimension higher. Specifically mere arctrig will not suffice. Dilogarithms also are needed. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

2 1

[math-fun] terse transcendence
by Bill Gosper 21 Feb '16

21 Feb '16

with just double, triple, and small angle formulas: In[631]:= sin[x_] := x /; Abs[x - x^2] == Abs@x; sin[x_] := #*(3 - 4*#^2) &@sin[x/3] In[634]:= sin[Pi/3`33]^2 Out[634]= 0.75000000000000000000000000000000 In[633]:= sin[Pi/2`33 - I] - sin[I*1`33] I // tim During evaluation of In[633]:= 0.003148,0 Out[633]= 2.71828182845904523536028747135265 + 0.*10^-33 I In[682]:= Log@%633 Out[682]= 0.999999999999999999999999999999997 + 0.*10^-33 I In[673]:= asin[x_] := 2 asin[x/(Sqrt[1 - x] + Sqrt[1 + x])]; asin[x_] := x /; Abs[x + x^2] == Abs[x] In[674]:= 2 asin@1`33 Out[674]= 3.141592653589793 (We lose half our digits near turning points.) In[683]:= 6 asin[.5`33] // tim During evaluation of In[683]:= 0.002256,0 Out[683]= 3.14159265358979323846264338327950 versin[x_] := x^2/2 /; Abs[x] == Abs[x + x^2]; versin[x_] := 2 (2 - #) # &@versin[x/2] versin[Pi/5`33] 0.190983005625052575897706582817181 (3 - 4 %)^2 5.00000000000000000000000000000000 In[663]:= tan[x_] := x /; Abs[x + x^2] == Abs@x; tan[x_] := 2 #/(1 - #^2) &@tan[x/2] In[665]:= tan[Pi/4`33] Out[665]= 1.00000000000000000000000000000000 In[668]:= (# + I)/(I - #) &@tan[I/2`33] Out[668]= 2.71828182845904523536028747135284 In[669]:= Log@% Out[669]= 1.00000000000000000000000000000007 Unfortunately, due to excessive significance pessimism, this terse weierstrass p only works on machine floats: In[326]:=weierp[z_, g2_, g3_] := z^-2 /; Abs[z] == Abs[z + z^2] In[327]:= weierp[z_, g2_, g3_] := (g2/4 - 3 #^2)^2/((4 #^2 - g2) # - g3) - 2 # &@ weierp[z/2, g2, g3] In[328]:= weierp[#, 2, 0] & /@ WeierstrassHalfPeriods[{2., 0}] Out[328]= {0.707106781186547 + 0.*I, -0.707106781186547 + 0.*I} Overriding the pessimism defeats the terseness: weierp[z_, g2_, g3_] := (g2^2 + 32 g3 #1 + 8 g2 #1^2 + 16 #1^4)/( 16 (-g3 - g2 #1 + 4 #1^3)) &@SetPrecision[weierp[z/2, g2, g3], 44] In[357]:= SetPrecision[ weierp[#, 0, 3] & /@ WeierstrassHalfPeriods[{0, 3.}], 22] Out[357]= {0.9085602964160698294456, -0.4542801482080349147228 + 0.7868362975662361391456 I} In[358]:= %^3 Out[358]= {0.750000000000000000000, 0.750000000000000000000 + 0.*10^-22 I} Unlike with arcsin, I have yet to figure out how to extend this technique to InverseWeierstrassP (which we need for NeilB's rational squares in arithmetic progression formula). --rwg

1 0

[math-fun] Approximate disk-dissection
by James Propp 20 Feb '16

20 Feb '16

Does anyone know of work that's been done on approximate dissection and recomposition of disks? For instance, dividing a disk of radius 2 into pieces and then rearranging the pieces to form an approximation to two discs of radius 1? Jim Propp

4 7

[math-fun] simple in/ex-radii & altitude relation
by Fred Lunnon 20 Feb '16

20 Feb '16

A neat identity relates inradius r_IN , ex-radius r_EX , altitude h , of a simplex with distinguished base and apex --- *** 1/r_EX + 1/r_IN = 2/h *** --- the altitude equals the harmonic mean of the radii ! I have not come across this elsewhere --- reference, anyone? A fellow by name of Ilya Bogdanov on Math Overflow (carefully rinsing mouth with disinfectant) seems well versed in this stuff. There turn out to be a good few Ilya Bogdanovs in Russia ... does anybody have an email address for this one? WFL

3 7

[math-fun] π puzzle
by Bill Gosper 20 Feb '16

20 Feb '16

On 2016-02-18 23:27, Simon Plouffe wrote: > Hello, > I think this is Vieta formula for Pi in disguise, > > 99 terms of it and then multiply by 2, > > and it would correspond to about 100 terms of it, > http://keisan.casio.com/exec/system/1354861725 > > Best regards, > Simon Plouffe Interesting! But I think not. (And therefore am not?) Listing just a few terms reveals, e.g. the 6th to be In[842]:= 64 Sqrt[2 - Nest[Sqrt[2 + #] &, 0, 5]] Out[842]= 64 Sqrt[2 - Sqrt[2 + Sqrt[2 + Sqrt[2 + Sqrt[2 + Sqrt[2]]]]]] In[843]:= N@% Out[843]= 3.14128 This would have made a much better puzzle, had I noticed it. More rapidly convergent: In[848]:= 96*Sqrt[2 - Nest[Sqrt[2 + #] &, 1, 5]] Out[848]= 96 Sqrt[2 - Sqrt[2 + Sqrt[2 + Sqrt[2 + Sqrt[2 + Sqrt[3]]]]]] In[849]:= N@% Out[849]= 3.14145 This corresponds to In[804]:= 3 Nest[#/(Sqrt[1 - #] + Sqrt[1 + #]) &, .5`33, 52] 2^53 // tim During evaluation of In[804]:= 0.000608 Out[804]= 3.14159265358979323846264338327950 In[805]:= Sin@% Out[805]= 0.*10^-33 This shames my original puzzle, In[787]:= Nest[#/(Sqrt[1 - #] + Sqrt[1 + #]) &, 1`33, 98] 2^99 Out[787]= 3.141592653589793 by not losing significance (which I now almost understand), and, mysteriously requiring only half of the 33/.30103 iterations I'd expect. --rwg

2 1

Re: [math-fun] Perigal's Pythagorean proof
by Bill Gosper 19 Feb '16

19 Feb '16

On 2016-02-13 15:11, Gareth McCaughan wrote: > On 13/02/2016 18:38, Bill Gosper wrote: > >> gosper.org/Perigal.gif --rwg > > Cute picture, but is proving that it actually proves > Pythagoras any easier than just proving Pythagoras[?] Well, no. It's just a sketch of a pure-dissection proof that is a bit harder than the skewy ones sketched by gosper.org/Pythanim.gif, but more visually convincing. --rwg

17 21