October 2013 - mathfun@mailman.xmission.com

[math-fun] Puzzling the 120-cell
by Joerg Arndt 14 Oct '13

14 Oct '13

The preprint Saul Schleimer, Henry Segerman, Puzzling the 120-cell, http://arxiv.org/abs/1310.3549 may be appealing to some people here (just a felling).

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[math-fun] Moore-Schulman base-3/2 conjecture
by Warren D Smith 14 Oct '13

14 Oct '13

> 1. Fractional bases (Cris Moore) > From: Cris Moore <moore(a)santafe.edu> > If we're talking about fractional bases, allow me to introduce you to a > conjecture of Leonard J. Schulman and me: > > Consider writing zero in base 3/2. That is, for i=0,1,2... let a_i be an > integer, and let > > 0 = \sum_{i=0}^\infty a_i (2/3)^i . > > Furthermore, assume that a_i is in {+1,0,-1} for each i, and that a_0 = 1. > Thus the first term is 1, and the remaining terms need to cancel it out. > > Conjecture: all initial subsequences have a nonzero density of nonzero > coefficients. That is, there is a constant C > 0 such that, for every n, at > least Cn of the coefficients a_1, a_2, ... a_n are nonzero. > > I am equally interested in any nearby conjecture: e.g. replace 3 with 5 or > 7, let a_i belong to {-2,-1,0,1,2}, and so on. > > If you can prove this, it will have nice consequences for a construction in > computer science called tree codes: see http://arxiv.org/abs/1308.6007 > > - Cris --That's an interesting conjecture! Seems like P-adics ought to be of some use: http://en.wikipedia.org/wiki/P-adic_number but I don't know anything about them and have not tried. Also, you were asking for the sum to equal 0. However, if it must equal x, then it seems to me it is easy to then prove your conjecture for almost all x, simply based on a volume argument. That is, the percentage of reals in any nonzero width interval, which are representable in base (3/2) using below any positive constant fraction of nonzero digits in {-1, 0, +1}, obviously is zero, so almost all x require at least a constant fraction of nonzero digits, to get represented. Indeed, let's do an entropy calculation. Let the fraction of nonzero digits be F. Then entropy per digit in nats then is E = (F-1)*ln(1-F) - F*ln(F/2). In order to represent a positive fraction of reals, we must have exp(E) >= 3/2. I find that F > 0.1038573417. So now the Moore-Schulman conjecture is proven... except not for x=0, just for "almost all x." Now let us consider the "special x" which ARE representable using less than, say, 1%, of the digits nonzero. We can prove properties these special x must have. For example, we can easily show that any special x must have an infinite number of rational approximations p/q, where q is a power of 3, such that |x-p/q| < q^(-1.01). (Which yet again proves these x have measure zero.) The set of special x is closed under multiplication by (2/3).

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[math-fun] Messages in pi (was Re: math novels?)
by Keith F. Lynch 14 Oct '13

14 Oct '13

Hans Havermann <gladhobo(a)teksavvy.com> wrote: > MathWorld's article on 'base' cautions that "the representation of a > given integer in an irrational base may be nonunique" I believe that's true of the representation of any number in any non-integer base (e.g. 2.5 in base 1.5). I was speaking of *any* representation, not necessarily the canonical greedy-algorithm representation. Sorry if that wasn't clear. This is unfortunate, since it means I can't test bases for integer termination by calculating the representation of integers in the obvious way. The rule for integer representations that terminate may not be obvious. In base sqrt(2), the rule is to leave odd powers blank. In base phi, the rule is to never have two consecutive "1"s. Also, the tiniest amount of precision loss wrecks the algorithm. What's the square of the square root of 2? 1.9999999999999999? That's less than 2, so 2 in base sqrt(2) can't start with 100. Sigh. > (but this is true also of integer bases: 1=.999? in base ten). Only of numbers that terminate. (In base 10, those are rationals whose denominator only contains 2 and 5 as prime factors.) I think that in non-integer bases, every number (integers, rationals, and irrationals) has multiple representations. The canonical form is to use the greedy algorithm, so 1.00000 is standard and 0.99999... is not. > More relevant perhaps is the issue of normalcy. I think you mean "normality." "Normalcy" is a neologism invented by President Harding. > My understanding of the concept is that it cannot be applied to > irrational bases. What then is the frequency distribution of the > digits (0, 1, 2, 3) of seventeen in base pi? 1/pi, 1/pi, 1/pi, and (pi-3)/pi respectively, assuming 17 is normal in base pi. I'd be very surprised if it isn't. Of course multiple representations are possible, and perhaps you could go out of your way to skew the statistics. However, you can't get away with simply never using a "3" -- I checked. Weirdness: Sagan spoke of pi in base eleven. I just looked at eleven in base pi, and it doesn't appear to contain any "3"s at all! Just how far in do you have to go to find a "3"? If there are no "3"s at all, does that count as a message? (This is with the usual greedy algorithm, of course.) Your ellipses, including one of mine that you quoted, are turning into question marks. I suspect some Microsoft product at your end is "helpfully" turning each instance of three consecutive dots into the single Microsoft-proprietary character that represents an ellipsis (octal 205, decimal 133, hexadecimal 85), and the list software is then "helpfully" replacing any character whose high bit is "1" with a question mark. Please beat your software into submission. Thanks. And, moderator, please tweak the list software to allow ISO-8859-1 accented letters through. It's okay to nuke Microsoft droppings and UTF-8 mojibake. Thanks. You get extra credit for turning Microsoft droppings back into what they were supposed to have been, e.g. the above-mentioned character back into three dots.

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[math-fun] Messages in pi (was Re: math novels?)
by Bill Gosper 13 Oct '13

13 Oct '13

Here are some 17 base π Mma timings: In[55]:= tim[RealDigits[17, \[Pi], 69]] During evaluation of In[55]:= 0.000157 Out[55]= {{1, 2, 0, 2, 2, 0, 0, 2, 1, 1, 0, 1, 0, 2, 0, 2, 3, 0, 0, 2, 0, 0, 0, 3, 0, 0, 1, 0, 0, 0, 1, 2, 1, 0, 0, 1, 2, 2, 2, 0, 0, 1, 1, 2, 2, 2, 2, 1, 2, 1, 0, 1, 2, 0, 0, 1, 2, 1, 1, 2, 1, 2, 0, 1, 2, 1, 1, 0, 1}, 3} In[56]:= tim[RealDigits[17, \[Pi], 512];] During evaluation of In[56]:= 0.000851 In[57]:= tim[RealDigits[17, \[Pi], 1024];] During evaluation of In[57]:= 0.002248 In[58]:= tim[RealDigits[17, \[Pi], 2^11];] During evaluation of In[58]:= 0.006922 In[60]:= Table[Timing[RealDigits[17, \[Pi], 2^n]][[1]], {n, 12, 18}] Out[60]= {0.020484, 0.106171, 0.556247, 2.749796, 14.463928, \ 79.579053, 367.510315} --rwg Dan Asimov <dasimov(a)earthlink.net> wrote: The thing I mainly recall about Contact is that a message is sent by somehow *altering* the digits of pi. Keith Lynch>You are misremembering. There's no implication that pi had ever had a different value. (At least not in the paperback. Does anyone have a first edition hardback they can check?) Dave Dyer <ddyer(a)real-me.net> wrote: On the other hand, since PI has an infinite number of nonrepeating digits, such finite patterns MUST exist. Keith Lynch>That doesn't follow. Last I heard, nobody had ever proven pi normal. Normality, despite its name, is very weird. It's known that "nearly all" numbers are normal, i.e. non-normal numbers are of measure zero, but, to the best of my knowledge, nobody has ever proven any particular number to be normal. (Some numbers contrived for the purpose have been proven normal in one base, but not in all positive integer bases at once.) If pi *is* normal, as it probably is, then it's true that pi in binary must contain all possible bit strings, including copies of every book and DVD -- and the complete archives of this list, including posts that haven't been written yet. But of course all of these are buried enormously deep. Finding anything "interesting" in the first trillion digits would be, well, interesting. Also seehttp://www.netfunny.com/rhf/jokes/01/Jun/pi.html which is apparently the best known thing I've ever written. If pi is normal, does it necessarily contain ASCII text of a proof that pi is normal? :-) James Propp <jamespropp(a)gmail.com> wrote: I wonder whether Sagan was falling prey to the confusion about whether pi is a mathematical constant or a physical constant. Keith Lynch>Possibly. But he was obviously aware that it could be calculated, not just measured. A dimensioned physical constant such as the speed of light could be made to contain any desired message by a choice of how our system of units is set up. A dimensionless constant such as the fine structure constant or the proton-to-electron mass ratio would be the same number in any system of units. But those are of course measured, not calculated. They're both only known to about nine significant digits, and it would be an enormous amount of work to learn each additional digit, so there's no possibility of storing lengthy messages in them that we could read even if someone had write access to them. Anyhow, a message in a physical constant would merely prove that the writer is cleverer or more powerful than us, not that he is infinitely clever or powerful. A message in pi would prove the latter. Except that we could never rule out the possibility that our computers which "found" the message had been hacked. And that's a logically preferred hypothesis, since it would require only a finite amount of cleverness and power. Similarly for *any* alleged proof of a being's omniscience or omnipotence. JPropp>Personally, I think we should look for a message from God in the digits of seventeen. I mean, everyone says that after the decimal point there's just a whole lot of 0's --- but how far out have they looked? Keith Lynch>No, the digits of seventeen *in base pi*, 120.220021101020230020003... This of course never repeats or terminates. And I'd bet that nobody has ever checked it for hidden messages. :-) As an aside, half the time this list seems to be more about Mathematica than about math. To how many digits can Mathematica calculate 17 in base pi? (I calculated the above digits using plain old C. It took me all of ten minutes to write the code.)

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[math-fun] Non-integer bases in which all integers terminate
by Keith F. Lynch 13 Oct '13

13 Oct '13

A question insipired by my previous post, in which I mentioned seventeen in base pi (120.2200211010202300200030...). In what non-integer real-number bases greater than 1 can the representations of all integers terminate? (The representations may need to be non-canonical, i.e. don't necessarily use a greedy algorithm.) Examples are base sqrt(2) and base phi. I believe all such bases are irrational but not transcendental. (Are they all quadratic?) Can anyone prove this? Can anyone list all such bases? I mention "greater than 1" since there's a symmetry around 1. If all integers terminate in base b, they also terminate in base 1/b. Just reverse the order of the digits and the radix point. For instance 2013 in base 1/10 is 0.3102. They also terminate in base -b and base -1/b. Also, what can be said about bases in which the representations of some integers repeat rather than terminate? I know that N can be non-canonically represented as 1.111111111111111... in base N/(N-1), but I don't think any other integers repeat in that same base except powers (and roots?) of N.

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Re: [math-fun] Riddle about repeated operations (SPOILER)
by James Propp 13 Oct '13

13 Oct '13

Q. What do you get when you differentiate sin x with respect to x a hundred times? A. You get cos x every time. :-) Jim Propp P.S. The chain of musings that led me to this riddle started with the literal inaccuracy of the way English speakers describe m times n as what you get when you "add m to itself n times". This is wrong in two ways: we actually add n-1 times, not n times, and what we add m to is not "itself" (except the first time) but rather the running total of all the m's previously added.

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Re: [math-fun] math novels?
by Bill Gosper 12 Oct '13

12 Oct '13

Keith Lynch> Carl Sagan's novel _Contact_ is mostly about communications with extraterrestrials. But a far more interesting subplot is about a very different sort of communication. The aliens point out to mankind that if you calculate pi in base eleven, near the beginning there's a square number of consecutive digits that are all 0 or 1, and when plotted as a square they give a picture of a circle. A little further into the number there are reams of meaningful binary data. Thus Marc LeBrun's suggestion that a monster π calculation be bound up and published as _The Saganic Verses_. --rwg

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[math-fun] Riddle about repeated operations
by James Propp 12 Oct '13

12 Oct '13

(This is original, though I suspect the basic idea is not new.) Q. What do you get when you differentiate sin x with respect to x a hundred times? No, the answer is not sin x. See the next post for the answer. Jim Propp

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[math-fun] Inverse Newman iteration
by Bill Gosper 12 Oct '13

12 Oct '13

NJAS>See one of the comments in http://oeis.org/A002487 for one possible answer. The reference is to D. E. Knuth, Problem 10906, American Mathematical Monthly; solution by Moshe *Newman*, American Mathematical Monthly, 110 (No. 7, 2003), 642-643. On Sat, Oct 12, 2013 at 12:15 PM, Warren D Smith <warren.wds(a)gmail.com>wrote: What is "Newman's rational counter"? Apparently, Moshe Newman discovered that iterating 1/(2 Floor[#] + 1 - #)& starting with 0 enumerates the nonnegative rationals. I was using the reverse sequence In[1106]:= NestList[2*Ceiling[1/#] - 1 - 1/# &, 3/8, 22] Out[1106]= {3/8, 7/3, 4/7, 5/4, 1/5, 4, 3/4, 5/3, 2/5, 5/2, 3/5, 4/3, 1/4, 3, 2/3, 3/2, 1/3, 2, 1/2, 1, 0, Indeterminate, Indeterminate} So 3/8 is the "20th rational". as the actual counter, which I should have called n(r). E.g., n((k-1)/(k+1))/2 - 1 = 2^(k/2-2) (3 + √2 + (-)^k (3 - √2)) goes 1, 3, 2, 6, 4, 12, 8, 24, 16, ... (alternately multiplying by 3 and 2/3, LinearRecurrence[{0, 2}, {1, 3},9], A164073), in case you ever need slower-than-fibonacci exponential positive integers. For monotone, 2^((k-5)/2) (4 + 3 Sqrt[2] - (-1)^k (4 - 3 Sqrt[2])), 2, 3, 4, 6, 8, 12, 16, 24, 32,... which is n(1/(8/(3 + (-1)^k) + (k-5)/2)). (You can go way slower with those Pisot-Vijayaraghavan numbers.) --rwg

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[math-fun] Newman's rational counter
by Warren D Smith 12 Oct '13

12 Oct '13

What is "Newman's rational counter"?

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