May 2018 - mathfun@mailman.xmission.com

Re: [math-fun] radix notation, |r|=1, e.g., r=exp(i)
by Henry Baker 07 May '18

07 May '18

Ok, so here's a complete algorithm for representing *any* complex number z=x+yi=L*e^(i*theta) using digits "0" and "1" in radix exp(i)=e^i, to any precision you please. We first pick integer k, s.t. e^(i*k) ~ e^(i*theta) to the desired degree of precision. We then approximate L ~ sum(2*cos(j)), for j suitably chosen, as described below. Then z = L*e^(i*theta) = sum(2*cos(j),j in suitable set)*e^(i*k). I don't believe that there are any "collisions", whereby the product produces any bit positions which require a digit other than "0" or "1", but if there were, one could always choose a different set of j's in a way to avoid such collisions. As an example, let's approximate -pi = pi*exp(i*pi). We first approximate exp(i*pi) to within 1 angular degree. k=22 works, so exp(22*i) ~ exp(pi*i). We now approximate pi using a cosine series. pi = sum(2*cos(j), suitable j), so pi ~ 1 + 2*cos(6) + 2*cos(11) + 2*cos(14) + 2*cos(77) ~ 3.14. Finally, we have pi*e^(pi*i) ~ (r^0+r^6+r^-6+r^11+r^-11+r^14+r^-14+r^77+r^-77)*r^22 = r^-55+r^8+r^11+r^16+r^22+r^28+r^33+r^36+r^99 Check using Maxima: (%i1) r^-55+r^8+r^11+r^16+r^22+r^28+r^33+r^36+r^99; 99 36 33 28 22 16 11 8 1 (%o1) r + r + r + r + r + r + r + r + --- 55 r (%i2) %,r=exp(%i),rectform; (%o2) %i (sin(99) - sin(55) + sin(36) + sin(33) + sin(28) + sin(22) + sin(16) + sin(11) + sin(8)) + cos(99) + cos(55) + cos(36) + cos(33) + cos(28) + cos(22) + cos(16) + cos(11) + cos(8) (%i3) bfloat(%); (%o3) - 2.77994517385043b-2 %i - 3.140593309047521b0 (%i4) log(%); (%o4) 1.144450908369626b0 - 3.132741228718346b0 %i (%i5) exp(realpart(%)); (%o5) 3.140716342230068b0 (%i6) At 08:17 AM 5/5/2018, Henry Baker wrote: >I've been looking at expressing numbers in radix-r notation, >for |r|=1, e.g., r=exp(i)=e^i. > >Real r, |r|=1 is too boring, so I looked at complex r, |r|=1. > >r=exp(i)=e^i is "representative" of this class, and is easy >to deal with in a computer algebra system. > >We first notice that "most significant bit" and "least >significant bit" have no meaning, because |r^n|=1 for >all real n. > >We're obviously not in Kansas, anymore, Toto! > >Zero = "0", as usual. > >One = "1", as usual. > >Since r^k+r^(-k) = 2*cos(k), we can represent real numbers >in the range [-2,2] to any degree of accuracy we wish with >only 2 "1" bits using an appropriate k. > >Every other real number outside this range can be expressed >(very inefficiently) using a "greedy" algorithm that subtracts >2*cos(k) for various k's. "Inefficient" here means that >expressing n takes O(n) 1 bits, and most likely O(n) bits in >total. > >I also plotted all of the complex numbers I could construct >using m bits on both sides of 1 -- i.e., binary numbers from >2^(-m) up to 2^m. These form pretty & mostly symmetrical >patterns with blotches that eventually merge as m grows in >size. (I don't have any way to show you these plots.)

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[math-fun] Geometry puzzle
by Dan Asimov 06 May '18

06 May '18

Define an open disk on the unit sphere S^2 centered at the point c in S^2 to be a set of the form D = D(c,p,r) = {p in S^2 | dist(p, c) < r} where 0 < r <= pi/2 and dist is distance measured along the sphere. Every such D is contained in some hemisphere. Puzzle: ------- Suppose we have a collection of disjoint open disks on the unit sphere. Is is possible that for every point p of S^2, either p or its antipodal point -p lies inside some disk of the collection? —Dan

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[math-fun] semi-OT: Lord's "The Reciprocal Fibonacci Function"
by Joerg Arndt 05 May '18

05 May '18

Could somebody with access to Michael Lord: The Reciprocal Fibonacci Function, Mathematics Magazine 91(2):134-137, (March-2018}). https://doi.org/10.1080/0025570X.2018.1426946 email the pdf to me? Muchos thanksias in advancias!

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[math-fun] radix notation, |r|=1, e.g., r=exp(i)
by Henry Baker 05 May '18

05 May '18

I've been looking at expressing numbers in radix-r notation, for |r|=1, e.g., r=exp(i)=e^i. Real r, |r|=1 is too boring, so I looked at complex r, |r|=1. r=exp(i)=e^i is "representative" of this class, and is easy to deal with in a computer algebra system. We first notice that "most significant bit" and "least significant bit" have no meaning, because |r^n|=1 for all real n. We're obviously not in Kansas, anymore, Toto! Zero = "0", as usual. One = "1", as usual. Since r^k+r^(-k) = 2*cos(k), we can represent real numbers in the range [-2,2] to any degree of accuracy we wish with only 2 "1" bits using an appropriate k. Every other real number outside this range can be expressed (very inefficiently) using a "greedy" algorithm that subtracts 2*cos(k) for various k's. "Inefficient" here means that expressing n takes O(n) 1 bits, and most likely O(n) bits in total. I also plotted all of the complex numbers I could construct using m bits on both sides of 1 -- i.e., binary numbers from 2^(-m) up to 2^m. These form pretty & mostly symmetrical patterns with blotches that eventually merge as m grows in size. (I don't have any way to show you these plots.)

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[math-fun] Normal subgroups
by James Propp 05 May '18

05 May '18

Has anyone proved that normally subgroups aren't normal? (That is, for large n, if you pick a group G of order n uniformly at random, and then pick a subgroup H of G uniformly at random, the probability that H is a normal subgroup of G goes to 0 as n goes to infinity.) Jim

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Re: [math-fun] Normal subgroups
by Dan Asimov 05 May '18

05 May '18

The problem was not what I wrote way below: Obviously with a finite number of groups of order n, one can easily pick one "at random". But how about the subgroups? Are we picking one from *all* subgroups, including all those conjugate to one subgroup? That method will tend to decrease the probability of a normal subgroup, since every *other* subgroup will get counted once for each of its conjugates but a normal subgroup has only one conjugate. On the other hand, picking just one subgroup from each conjugacy class will increase the probability of a normal subgroup. So: What is the probability measure being considered? —Dan ----- > (That is, for large n, if you pick a group G of order n uniformly at > random, and then pick a subgroup H of G uniformly at random, the > probability that H is a normal subgroup of G goes to 0 as n goes to > infinity.) ----- >Jim Propp wrote: >----- >for large n, if you pick a group G of order n uniformly at random >----- > >I wonder how to define a probability on groups of order n. > >And in particular: If there are several ways to do this, >do they result in the same probabilities? > >—Dan > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Normal subgroups, and the eat-quality of cheese
by James Propp 05 May '18

05 May '18

I envisioned giving equal probability to each of the subgroups of cheese considered up to eat quality ... Let me try that again: I envisioned giving equal probability to each of the subgroups of G considered up to equality (as opposed to conjugacy). (I was speaking into my smart phone, and it changed my mathematical sentence into a curious culinary one.) Jim On Friday, May 4, 2018, Dan Asimov <dasimov(a)earthlink.net> wrote: > The problem was not what I wrote way below: Obviously with a finite > number of groups of order n, one can easily pick one "at random". > > But how about the subgroups? Are we picking one from *all* > subgroups, including all those conjugate to one subgroup? > > That method will tend to decrease the probability of a normal > subgroup, since every *other* subgroup will get counted once > for each of its conjugates but a normal subgroup has only > one conjugate. > > On the other hand, picking just one subgroup from each conjugacy > class will increase the probability of a normal subgroup. > > So: What is the probability measure being considered? > > —Dan > > > ----- > > (That is, for large n, if you pick a group G of order n uniformly at > > random, and then pick a subgroup H of G uniformly at random, the > > probability that H is a normal subgroup of G goes to 0 as n goes to > > infinity.) > ----- > > > >Jim Propp wrote: > >----- > >for large n, if you pick a group G of order n uniformly at random > >----- > > > >I wonder how to define a probability on groups of order n. > > > >And in particular: If there are several ways to do this, > >do they result in the same probabilities? > > > >—Dan > > > >_______________________________________________ > >math-fun mailing list > >math-fun(a)mailman.xmission.com > >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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[math-fun] Stupid Bell 103 modem question
by Henry Baker 05 May '18

05 May '18

Did anyone ever get this modem to digitally communicate via tin cans and a wire? It might be a fun high school science project, now that the 103 is now just open source software on typical Windows/Mac/Linux laptops.

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Re: [math-fun] Normal subgroups
by Dan Asimov 04 May '18

04 May '18

Jim Propp wrote: ----- for large n, if you pick a group G of order n uniformly at random ----- I wonder how to define a probability on groups of order n. And in particular: If there are several ways to do this, do they result in the same probabilities? —Dan

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Re: [math-fun] 60-Year-Old Maths Problem Partly Solved By Amateu
by Dan Asimov 04 May '18

04 May '18

This is beyond a doubt the same result discussed here recently: ----- The chromatic number of the plane is at least 5. ----- In fact there's now a polymath project, led in part by Terry Tao, to try to simplify the proof or improve this bound. —Dan P.S. No offense, but this is a rather poor description of the problem. It doesn't even mention the plane! But then again, math is not The Grauniad's strong point: ----- The problem is as follows. Imagine a collection of dots connected by > lines. The dots can be arranged any way at all, the only rule is that > all the connecting lines must be of equal length. For instance, in a > square the diagonal would not be joined up, but the outer edges would > be. Now, colour in all the dots so that no two connected points have the > same colour. How many colours are required. For a square, the answer > would be two. But the Hadwiger-Nelson problem asks what the minimum > would be for any configuration -- even one that extends across a plane > of infinite size./ -----

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