March 2009 - mathfun@mailman.xmission.com

[math-fun] 4 interesting constants and 3 new sequences.
by Simon Plouffe 23 Mar '09

23 Mar '09

Hello everybody, Here are 4 strange constants (and sequences). I was studying the numbers of 5^n, 2^n and n! written backward and realized that powers of 5 appear in a stange manner (as well as many others!). Consider 5^3 = 125, so the backward value is 0.521, 5^10 = 9765625, so backward value is 0.5265679, see A071583. It appears that there is an upper limit to this process, the constant is 0.5265679578796997657885576975995789586775656... the lower limit being 0.521302330431131124210313300023141021034302... The interesting fact is the following, after the 2nd term (or second digit), only the numbers in {5,6,7,8,9} appears. The constant does NOT correspond to anything I know. The lower limit has the same property, only the numbers in {0,1,2,3,4} appears after the first term. The phenomena appears also with powers of 2. The upper limit for 2^n is 0.8889899898988998 the lower limit is 0.21101001010110010... The number 0.8889899898988998 appear in A023415 of David Wilson. again, only the numbers in {8,9} appear in the first constant and only digits in {0,1} in the second (after the first digit). The most surprising fact is that the sum : 8.8898998989889987 + 2.1101001010110010 = 11 EXACTLY. These are experimental results as you may guess. INTERESTING NOTE : if we take backward values of n! then the 2 constants are the same and the sum is still 11 exactly. Can someone explain this strange phenomena ? Simon Plouffe ############################################################################################## %I A158624 %S A158624 5,2,6,5,6,7,9,5,7,8,7,9,6,9,9,7,6,5,7,8,8,5,5,7,6,9,7,5,9,9,5,7,8,9,5, %T A158624 8,6,7,7,5,6,5,6 %N A158624 Upper limit of backward value of 5^n. %C A158624 Digits are all in {5,6,7,8,9} after 2nd term. %F A158624 No known formula. %C A158624 The lower limit is 0.521302330431131124210313300023141021034302... %e A158624 5^3 = 125 so the backward value is 0.521, 5^10 = 9765625, so backward value is 0.5265679. The upper limit of all values is a constant, constant appears to be 0.5265679578796997657885576975995789586775656... %K A158624 cons,nonn %O A158624 1,1 %A A158624 Simon Plouffe (simon.plouffe(AT)gmail.com) Mar 23 2009. %R A158624 see A158625, A071583. ################################################################################################ %S A158625 5,2,1,3,0,2,3,3,0,4,3,1,1,3,1,1,2,4,2,1,0,3,1,3,3,0,0,0,2,3,1,4,1,0,2,1,0,3,4,3,0,2 %N A158625 Lower limit of backward value of 5^n. %C A158625 Digits are all in {0,1,2,3,4} after first term. %F A158625 No known formula. %C A158625 The upper limit is 0.5265679578796997657885576975995789586775656... %e A158625 5^3 = 125 so the backward value is 0.521, 5^10 = 9765625, so backward value is 0.5265679. The lower limit of all values is a constant, constant appears to be 0.521302330431131124210313300023141021034302... %K A158625 cons,nonn %O A158625 1,1 %R A158625 see A158624, A004094. %A A158625 Simon Plouffe (simon.plouffe(AT)gmail.com) Mar 23 2009. ################################################################################################# These are additional comments and values to sequence A023415 %N A023415 Upper limit of backward value of 2^n and n!. %C A023415 The sum of constants in A158626 and A158627 is conjectured to be 11 exactly. %A A023415 Simon Plouffe (simon.plouffe(AT)gmail.com) Mar 23 2009. ################################################################################################# %S A158627 2,1,1,0,1,0,0,1,0,1,0,1,1,0,0,1,0 %N A158627 Lower limit of backward value of 2^n and n!. %C A158727 The sum of constants in A158626 and A158627 is conjectured to be 11 exactly. %A A158627 Simon Plouffe (simon.plouffe(AT)gmail.com) Mar 23 2009. important comment : the value of A158627 is deduced from sequence A023415 if conjecture about 11 is true then the sequence A158627 would be : [2, 1, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1 , 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1]: There are no apparent pattern in this sequence.

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Re: [math-fun] If We Have Free Will, Then So Do Electrons
by Dan Asimov 22 Mar '09

22 Mar '09

I don't know what I think of the existence of free will per se, but just on heuristic grounds, this *implication* is not surprising. I.e., if determinism applied to all sufficiently small regions of the universe, then it would seem to necessarily apply to larger regions. So if the latter fails, the former must, too. --Dan Ray wrote: << http://science.slashdot.org/article.pl?sid=09/03/20/1229233 snahgle writes "Mathematicians John Conway (inventor of the Game of Life) and Simon Kochen of Princeton University have proven that if human experimenters demonstrate 'free will' in choosing what measurements to take on a particle, then the axioms of quantum mechanics require that <http://arxiv.org/abs/quant-ph/0604079>the free will property be available to the particles measured, or to the universe as a whole. . . . >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] If We Have Free Will, Then So Do Electrons
by Dan Asimov 22 Mar '09

22 Mar '09

What is the definition of a measurement at A being "influenced" by what happened at B ? --Dan Gene wrote: << [A] source S produces a pair of particles that are detected at A and B at space-like separation, so that A and B cannot influence each other (if you believe causality and relativity). It can happen that the measurements at A and B are correlated, e.g. if the z components of spins are measured, A has spin up if and only if B has spin up. These correlations become evident only when the recorded measurements are brought together and compared. The cause and effect relation responsible for such correlations is the common origin of the particles at S. There is no possible measurement at A that can be influenced by what happened at B. >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] prime ladders
by rcs＠xmission.com 22 Mar '09

22 Mar '09

The Prime Ladder puzzle is to change one prime into another, by changing a digit at a time. All the intermediate numbers must also be prime. Example: 101 into 997: 101 -> 107 -> 197 -> 997 Challenge: Change 1009 into 9973. Rich

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Re: [math-fun] error function: series development at infinity
by Axel Vogt 20 Mar '09

20 Mar '09

> Message: 6 > Date: Thu, 19 Mar 2009 23:22:54 +0100 > From: "wouter meeussen" <wouter.meeussen(a)pandora.be> > To: "math-fun" <math-fun(a)mailman.xmission.com> > Subject: [math-fun] error function: series development at infinity > Message-ID: <000e01c9a8e1$3fb9f7f0$c887c554@pcwou> > Content-Type: text/plain; charset="iso-8859-1" > > dear math funners, > > does anyone know of a series development of erf(x) round x-> infinity, or equivalently > series of erf (1/x) around x=0? > Alternatively, > how to get approximations to log(1-erf(x)) for x around, say, 50 to 1000? > Googled for it in vain, > > Wouter. One can try Mills ratio R(x) = cN(x)/D(N)(x) where cN(x) = 1 - N(x) is the complementary cumulative normal distribution and D(N) is the derivative. Then R(1/x) is a power series in x (!) with alternating coefficients a(n) = (-1)^n * (2*n)!/n!/(2^n) = (-1)^n * 2^n/Pi^(1/2)*GAMMA(n+1/2) (I have no proof at hand for this). They can be computed by recursion or approximately for large n and since they are alternating one has an error estimate for finite sums. May be that is just a re-formulation of the formula in Abramowitz.

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[math-fun] If We Have Free Will, Then So Do Electrons
by Ray Tayek 20 Mar '09

20 Mar '09

http://science.slashdot.org/article.pl?sid=09/03/20/1229233 snahgle writes "Mathematicians John Conway (inventor of the Game of Life) and Simon Kochen of Princeton University have proven that if human experimenters demonstrate 'free will' in choosing what measurements to take on a particle, then the axioms of quantum mechanics require that <http://arxiv.org/abs/quant-ph/0604079>the free will property be available to the particles measured, or to the universe as a whole. Conway is giving <http://www.princeton.edu/main/news/archive/S23/69/84A24/index.xml?section=a…>a series of lectures on the 'Free Will Theorem' and its ramifications over the next month at Princeton. A followup <http://www.ams.org/notices/200902/rtx090200226p.pdf>article strengthening the theory (PDF) was published last month in Notices of the AMS." --- vice-chair http://ocjug.org/

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[math-fun] error function: series development at infinity
by wouter meeussen 20 Mar '09

20 Mar '09

dear math funners, does anyone know of a series development of erf(x) round x-> infinity, or equivalently series of erf (1/x) around x=0? Alternatively, how to get approximations to log(1-erf(x)) for x around, say, 50 to 1000? Googled for it in vain, Wouter.

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[math-fun] tight muffin bound proofs for several intervals
by Huddleston, Scott 19 Mar '09

19 Mar '09

In this note I prove that the muffin bound S(r) is continuous on several intervals, including [4/3,8/5]-{3/2} and [9/5,15/7]-{2}. I first prove this for optimal "binary" solutions, which are those in which every muffin splits in exactly 2 pieces, then address the general optimum. Let S'(r) denote the optimum binary solution value. Consider example r = 11/6 in interval [9/5,2). The first level decomposition of a solution to 11/6 is: r = m/p = 11/6, 22 muffin pieces [3] x 2 Z parts, 6 pieces [4] x 4 A parts, 16 pieces 5 excess muffins (10 pieces) in A There are fewer Z pieces (6) than A pieces (16), so if a solution with all Z pieces the same size (11/18) and no piece smaller than a Z piece complement (7/18) is feasible, it will be optimal. The 6 Z piece complements must be A pieces. Pack them in the 6 A parts as evenly as possible, accounting for 6 of the 11 muffins. The remaining 5 muffins (10 pieces) all pack in A parts. Call these "m" pieces. This gives a tentative solution [a,z]*6, [m*2]*5; [z*3]*2; [a*2,m*2]*2, [a,m*3]*2 The "fill set" is this decomposition is the portions of the A parts that contain m pieces. This fill set has 2 2-parts, the [m*2] subparts of the [a*2,m*2] 4-parts, and 2 3-parts. The key idea of the proof is that we can reduce problems r in [9/5,2) to smaller muffin problems whose first level structure is r's fill set. Then an affine transform of the reduced problem solution realizes a solution for the fill set: fill set for 11/6 has 2 2-parts and 2 3-parts fill set reduces to muffin problem 5/4 r = m/p = 5/4, 10 muffin pieces [2] x 2 Z parts, 4 pieces [3] x 2 A parts, 6 pieces 1 excess muffins (2 pieces) in A with lexical optimum [z*2]*2; [a*2,m]*2 = [5/8*2]*2; [3/8*2,1/2]*2 The 2-parts and 3-parts in the reduced problem have average piece sizes (5/4)/2 = 15/24 and (5/4)/3 = 10/24 respectively. The 2-parts in the 11/6's fill set have average piece size (11/6-7/18*2)/2 = = 57/108. The fill set 3-parts have average piece size (11/6-7/18)/3 = 52/108. The affine map from the 5/4 solution to an 11/6 fill set solution thus maps 15/24 |-> 57/108 and 10/24 |-> 52/108, which shifts piece sizes closer to 1/2 by a factor of 9/2. The solved 11/6 A-parts from this map are: 1/108 * { [42*2,57*2]*2, [42,52*3]*2 } Theorem: S(r) = 1-r/3 on [9/5,2). Proof: S(r) <= 1-r/3 is a long known lower bound. Here we realize solutions for all r in [9/5,2), proving the bound is tight. There is a function Q which reduces each instance r in [9/5,2) to a smaller problem Q(r) matching r's fill set, constructed as in the above example for r = 11/6. Q turns out to be a Mobius transform. It is uniquely determined by the image of 3 of its points, e.g, Q(11/6) = 5/4 Q(9/5) = 1 Q(15/8) = 3/2 and we have Q(r) = (7r-12)/(2r-3). Q maps [9/5,2) to [1,2). For reduced problem t = Q(r), we have numerator(t) < numerator(r) and t's total piece count divides r's fill set piece count, i.e., 2k * numerator(t) = r's fill set piece count for some k>0. The affine map g from solution(t) to solution(fill_set(r)) shifts each piece size x toward 1/2 with some scale factor g_s: g(x) = (x-1/2)*g_s + 1/2 A straightforward calculation finds g_s = g_s(r) = 2r/3-1. Let S_2(r) = g(S'(Q(r))), the size of the smallest piece in r's fill set. Here's a direct check that S_2(r) > S(r) on [9/5,2): For r in [9/5,15/8], 1 <= t = Q(r) <= 3/2, S'(t) > 1/4, and 1/5 <= g_s(r) <= 1/4. Thus S_2(r) = g(S'(t)) > g(1/4) >= 7/16 > 2/5 >= S(r). For r in [15/8,2), 3/2 <= t = Q(r) < 2, S'(t) > 1/3, and 1/4 <= g_s(r) < 1/3. Thus S_2(r) = g(S'(t)) >= g(1/3) >= 4/9 > 3/8 >= S(r). Since S_2(r) is the smallest fill piece size and S_2(r) > S(r), we've just proved this reduction finds the optimal second largest distinct piece size for r in [9/5,2). Note finally that g defines a bijection between binary solutions to Q(r) and binary solutions to r's fill set which preserves lexical optimality. QED. We have S'(r) = S(r) for all nonintegral r >= 3/2. The S'/S bounds and reduction functions for intervals [4/3,8/5] and [9/5,15/7] are: S'(r) = 1-r/2 on [4/3,3/2), Q<3,1>(r) = (5r-6)/(2r-2) Q: [4/3,3/2) |-> [1,3/2) S(r) = r/4 on (3/2,8/5], Q<4>(r) = (23-14r)/(26-16r) Q: [8/5,3/2) |-> [1,3/2) S(r) = 1-r/3 on [9/5,2), Q<4,1>(r) = (7r-12)/(2r-3) Q: [9/5,2) |-> [1,2) S(r) = r/5 on (2,15/7], Q<5>(r) = (20-5r)/(9-2r) Q: [15/7,2) |-> [1,2) I haven't yet calculated the affine scale factor function for 3 of these cases, though it's easy to find for any individual instance. Note that 1/4 < S'(r) <= 1/3 on [4/3,3/2) for the binary solution optimum, but the general optimum is S(r) = 1/3. This general optimum feasibility is also provable by a Mobius reduction to binary solutions of smaller instances. S(r) = 1/3 for r = m/p in [4/3,3/2) is achieved by splitting muffins into either 2 pieces or exact thirds so there are exactly 3p pieces and all persons get 3 pieces. The "fill sets" in this range are the person pieces left after removing all exact 1/3 pieces, a set of 2-parts and 3-parts. r = 4/3 has all 2-parts for its fill set, and r approaching 3/2 approaches a fill set of all 3-parts. The reduction which maps these fill sets to smaller problems is Q_3(r) = 3r-3 For example, consider r = 22/15 in [4/3,3/2). We achieve S(r) = 1/3 with 1 muffin [1/3*3] and 21 2 piece muffins [m*2]*21. The solution for persons has the form [1/3,m*2]*3, [m*3]*12 Removing the [1/3] pieces leaves a fill set of 3 2-parts [m*2] and 12 3-parts [m*3]. Q_3(r) = 3r-3 reduces this to Q_3(22/15) = 7/5 with binary (not general) muffin decomposition r = m/p = 7/5, 14 muffin pieces [2] x 1 Z parts, 2 pieces [3] x 4 A parts, 12 pieces 5 excess muffins (10 pieces) in A and optimal binary solution [z*2]; [a,m*2]*2, [m*3]*2; 7/5's fill set reduces (by Q_2(r) = Q(r) = (5r-6)/(2r-2)) to Q(7/5) = 5/4 with previously seen optimal solution [z*2]*2; [a*2,m]*2 = [5/8*2]*2; [3/8*2,1/2]*2 This maps affinely back to optimal binary solution 1/20 * { [14*2]; [6,11*2]*2, [9*2,10]*2 } for 7/5. Replicated 3 times, this maps affinely into the fill set for 22/15, giving the optimal person solution 1/60 * { [20,34*2]*3; [26,31*2]*6, [29*2,30]*6 } for 22/15. Note that "ternary" optimal solutions don't map affinely into fill sets in these reductions. They are a top level exception, but only binary solutions (in this note) are targets in the recursive reduction to smaller problems. There are two exceptions S'(r) < S(r) where the binary optimum is not the general optimum. One exception is ternary optima where S'(r) < 1/3 = S(r). I've just proved continuity and optimality of both the general and the binary optimum on [4/3,3/2), by Mobius reductions that prove fill set solution feasibility. The other exception is S'(r) = 1/2 < 1 = S(r), which only occurs at integral r. This case and other optimal solutions that use unsplit muffins are simply handled by rejoining any combinable pairs that a binary solution might have.

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Re: [math-fun] (exp(Pi)-ln(3))/ln(2)
by Dan Asimov 18 Mar '09

18 Mar '09

The discrepancy is only 4/5. Why must you be so fussy? --Dan Rich wrote: << 31.8 = 159/5 --------- Quoting Dan Asimov <dasimov(a)earthlink.net>: > Mike Stay wrote: > > > << > 31.80000000332526 = (exp(Pi)-ln(3))/ln(2) > . . . > >> > > Fascinating, especially because this says > > (exp(Pi)-ln(3))/ln(2) ~ 163/5, > . . . >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] (exp(Pi)-ln(3))/ln(2)
by Dan Asimov 17 Mar '09

17 Mar '09

Mike Stay wrote: << 31.80000000332526 = (exp(Pi)-ln(3))/ln(2) Does this run of zeros have something to do with the j function? >> Fascinating, especially because this says (exp(Pi)-ln(3))/ln(2) ~ 163/5, whereas (from the J function) the well-known approximation exp(Pi sqrt(163)) ~ 262537412640768743.9999999999992500725971, also involves 163 prominently. (163 is the largest number d such that the algebraic integers of the imaginary quadratic field Q(sqrt(-d)) form a unique factorization domain. I hope I said that correctly.) --Dan P.S. Anyone know the prime factorization of 262537412640768744 ??? _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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