March 2014 - mathfun@mailman.xmission.com

[math-fun] "The Bookie of Mensa"
by James Propp 21 Mar '14

21 Mar '14

Anyone remember a short comic piece from the 1980s called "The Bookie of Mensa"? It imagined a professional oddsmaker specializing in unsolved scientific and mathematical issues. (I think he declined to offer odds on the Riemann Hypothesis: "too rich for my blood".) I just did a Google search, but couldn't find it. (The title is riffing on Woody Allen's "The Whore of Mensa".) Jim Propp

2 1

[math-fun] Two circles intersecting
by Warren D Smith 20 Mar '14

20 Mar '14

>Looks like Warren was right. >(I'm not re-opening the MH370 case, >only admiring what logic and geometry >can produce) --actually, not really. If the Australian crash site claims (1500 miles West of Perth) thing is true, that means I was about "right" in the sense the plane went thru roughly the 2nd intersection point, BUT my axiom had been that those controlling the plane were not idiots, and hence would have flown to the other intersection point, which would be within reach of land (in fact on land). So: my axiom would be falsified by the Australian claims. So I would (or rather my axiom would) predict the Australian crash claims are a delusion and will not pan out. By the way, there were satellite pings each hour, though I only had the data for the last at 8:11am Malay time. If I did have that further data, then there would be more circles and under axioms like mine one could make additional deductions and basically trace the entire path taken by the plane. It is just absurd that the press et al have not done this. -- Warren D. Smith http://RangeVoting.org

1 0

[math-fun] Asimov's 9s question
by Warren D Smith 20 Mar '14

20 Mar '14

D.Asimov's question: Given an infinite string of random digits, what is the probability digits N to 2N are all 9s, for at least one value of N>0? I claim one can exactly solve this. The key is to set up the argument to count mutually-exclusive events. Consider a finitely-long string, L long. There are 10^L such. 10^(L-2) begin 99. 9*10^(L-4) begin x999 where x is not 9. 9*10^(L-5) begin xy9999 where y not 9. et cetera. So it seems to me all problems of this ilk can be solved by expressing as sums which then can be done in closed form.

1 0

[math-fun] patterns in algebraic numbers of degree 4
by Simon Plouffe 20 Mar '14

20 Mar '14

Hello, I made smaller versions here http://www.plouffe.fr/f_8192_base2.tif and http://www.plouffe.fr/f_4095_base2.tif Best regards, Simon Plouffe

1 1

[math-fun] patterns in algebraic numbers of degree 4
by Simon Plouffe 20 Mar '14

20 Mar '14

Hello, I made smaller versions here http://www.plouffe.fr/f_8192_base2.tif and http://www.plouffe.fr/f_4095_base2.tif Best regards, Simon Plouffe

1 0

[math-fun] Two circles intersecting
by Eric Angelini 20 Mar '14

20 Mar '14

Looks like Warren was right. (I'm not re-opening the MH370 case, only admiring what logic and geometry can produce) Best, É.

1 0

[math-fun] a large image of the binary value of a 4'th degree algebraic numbers, 3.2 billion bits.
by Warren D Smith 19 Mar '14

19 Mar '14

>That image prooves the point, it represent 3.2 billion bits >of that algebraic number. > f(N) = 1+1/4*(2*4^N+2*(16^N+1)^(1/2))^(1/2)/(2^N)^2 > where N=32767 --I don't think my computer is powerful enough to handle that image. (I have no doubt yours is better than mine...) But I had no trouble reading Gosper's file http://gosper.org/plouffe.png which makes a nice change from his usual unreadability :) Anyhow, if by "proves the point" S.P. meant that the generating function explanation I gave cannot explain this, then I think you are wrong, rather it proves the gen.fn. DOES explain this. I think you may have fallen into the following trap. You correctly observed my Catalan-number example (which by the way was not originally due to me) stops seeming nonrandom quite soon, and the reason is the exponential growth of the Cat.numbers causing an example based upon the Cat.# generating function F(x) to "start looking random" after only about |log(x)| series coefficients, which is about the |log(x)|^2 bit in the binary number F(x). Note the SQUARING, which you may have forgotten to do. In your case, x is about 4^(-32767) or maybe 1/8 times this or something (I forget right now the exact definition I had to make for x, but only the order of magnitude matters so to hell with the 1/8 for the moment). The point is, log(x) = 65536 or so. Therefore, remembering the squaring, we expect nonrandomness out to about the 65536^2 bit, which is the 4 billionth bit. There is no precise point at which the nonrandomness ends, at least as far as I am concerned, but this is the neighborhood where it should fade out. So I repeat, your findings are compatible with my fairly simple gen.fn. explanation. I also repeat, that I described how to try to construct considerably-more-impressively-long nonrandom example numbers (and I see J.Propp mentioned a similar idea). You should try some of those suggestions with your number-compute+visualize software & see what you find. Hopefully it will confirm my predictions but I have not tried this computation myself and I am sure you have both better software & hardware than I; and using the ideas I outlined it should be possible using a more systematic exploration (which I have not done, and might not be completely trivial either) to optimize them. -- Warren D. Smith http://RangeVoting.org

1 0

[math-fun] a large image of the binary value of a 4'th degree algebraic numbers, 3.2 billion bits.
by Bill Gosper 19 Mar '14

19 Mar '14

Simon>Hello, here is 1 image of 65535 x 50000 of f(N) where N=32767 the image is here :http://www.plouffe.fr/f_32767_base_2.tif the format is TIF , a black and white image, ordinary bitmap. to see the image , preferably use Photoshop and a fairly recent version. The image will be there in about 30 min. Recall : f(N) = 1+1/4*(2*4^N+2*(16^N+1)^(1/2))^(1/2)/(2^N)^2 by zooming on the image you will see what I meant and why I used false colors. That image prooves the point, it represent 3.2 billion bits of that algebraic number. Best regards, Simon Plouffe ps : the file is about 220 megs. ------- My Quicktime is gagging on this. I hope you have doubled the raster width, as in gosper.org/plouffe.png --rwg

1 0

[math-fun] a large image of the binary value of a 4'th degree algebraic numbers, 3.2 billion bits.
by Simon Plouffe 19 Mar '14

19 Mar '14

Hello, here is 1 image of 65535 x 50000 of f(N) where N=32767 the image is here : http://www.plouffe.fr/f_32767_base_2.tif the format is TIF , a black and white image, ordinary bitmap. to see the image , preferably use Photoshop and a fairly recent version. The image will be there in about 30 min. Recall : f(N) = 1+1/4*(2*4^N+2*(16^N+1)^(1/2))^(1/2)/(2^N)^2 by zooming on the image you will see what I meant and why I used false colors. That image prooves the point, it represent 3.2 billion bits of that algebraic number. Best regards, Simon Plouffe ps : the file is about 220 megs.

1 0

[math-fun] a strange class of algebraic numbers
by Bill Gosper 18 Mar '14

18 Mar '14

Nice. But your plots are line-wrapped. Out[698]= 1 + 2^(-1 - k) Sqrt[2^k + 2 Sqrt[1 + 2^(2 (-1 + k))]] You have, e.g. In[700]:= ArrayPlot[ Partition[RealDigits[%698 /. k -> 63, 2, 63^2][[1]], 63]] Try In[705]:= ArrayPlot[ Partition[ Join[ConstantArray[0, 32], RealDigits[%698 /. k -> 63, 2, 2*63^2][[1]]], 2*63]] --rwg Violating two embargoes at once: Climate change makes planes disappear.

1 0