November 2017 - mathfun@mailman.xmission.com

[math-fun] muffin problem -- clusters and bounds
by Huddleston, Scott 01 Dec '17

01 Dec '17

I enclose examples of some of the bounds in my 28 Feb. post, and a derivation of the most involved example. S(12/7) = 3/7 = r/4 = (3-r)/3 Muffins: 1/7 * { [3,4]*12 } Persons: 1/7 * { [4*3]*4, [3*4]*3 } S(7/4) = 5/12 = (3-r)/3 = (2r-1)/6 Muffins: 1/12 * { [5,7]*6, [6*2] } Persons: 1/12 * { [5*3,6]*2, [7*3]*2 } S(40/23) = 19/46 = (2r-1)/6 = (12-5r)/8 Muffins: 1/46 * { [19,27]*24, [20,26]*12, [23*2]*4 } Persons: 1/46 * { [19*3,23]*8, [20*4]*3, [27*2,26]*12 } S(228/131) = 54/131 = (12-5r)/8 = (7r-6)/15 Muffins: 1/131 * { [54,77]*120, [55,76]*24, [57,74]*60, [65,66]*24 } Persons: 1/131 * { [54*3,66]*24, [54*2,55,65]*24, [57*4]*15, [77*2,74]*60, [76*3]*8 } S(449/258) = 319/774 = (7r-6)/15 = (73-32r)/42 Muffins: 1/774 * { [319,455]*210, [320,454]*30, [325,449]*42, [336,438]*30, [337,437]*90, [384,398]*42, [387*2]*5 } Persons: 1/774 * { [319*3,390]*42, [319*2,325,384]*42, [320*3,387]*10, [336,337*3]*30, [455*2,437]*90, [455,454,438]*30, [449*3]*14 } Here's a derivation of 449/258 example. 449 muffins -> 898 pieces, used in 124 4-parts with 496 pieces and 134 3-parts with 402 pieces. We split muffins between a 3-part and a 4-part when possible, but 4-parts contain 94 more pieces than 3-parts. We divide these 94 pieces (47 muffins) as evenly as possible among the 124 4-parts. 47 pairs of 4-parts contain both halves of a muffin, which I'll call an A^1 cluster and write A^1 = [a1*3,m]*2 where m denotes the two muffin halves and the other pieces are denoted a1, indicating a set of pieces of possibly different sizes that occure in A^1 clusters. 34 4-parts contain only muffin pieces whose mates appear in 3-parts. These I call A^0 clusters and write A^0 = [a0*4] The 449/258 example uses 30 A^0 clusters ([a0*4])*30 (120 a0 pieces, 30 4-parts) 47 A^1 clusters ([a1*3,m]*2)*47 (282 a1 pieces, 94 4-parts) Note that the A^1 cluster by itself sets an upper bound on S -- the cluster is 2 persons (size 2r), it cntains 1 complete muffin (size 1, pieces m*2), and it has 6 unmatched muffin pieces (a1*6), giving S(449/258) <= avg(a1) = (2r-1)/6. A^0 clusters also bound S by S(449/258) <= avg(a0) = r/4, but the A^1 cluster bound is tighter. Note that this (2r-1)/6 bound holds for examples 7/4 and 40/23. The muffin mates of a0 pieces (resp. a1 pieces) are z0 pieces (resp. z1 pieces). We distribute z0 and z1 pieces as evenly as possible across the 134 3-parts (402 pieces). With 282 z1 pieces, at least 14 3-parts must consist of only z1 pieces. These are Z^0 clusters, Z^0 = [z1*3] = [z10*3] where I refine piece labels so that z1 pieces appearing in Z^0 clusters are labelled z10. I create general Z^k clusters by finding minimal sets of persons that contain complete a0/z0 muffins. Maximally even distribution of the 120 z0 pieces places one per 3-part. Then a Z^1 cluster, closed under a0/z0 muffins, is Z^1 = [a0*4],[z0,z1*2]*4 = A^0,[z0,z11*2]*4 where z1 pieces appearing in Z^1 clusters are relabelled z11. Z^0 clusters bound S by S(449/258) <= avg(a10) = 1-avg(z10) = 1-r/3 = (3-r)/3 and Z^1 clusters provide an even tighter bound S(449/258) <= avg(a11) = 1-avg(z11) = 1-(5r-4)/8 = (12-5r)/8. The 449/258 example uses 14 Z^0 clusters = ([z10*3])*14 ( 42 z10 pieces, 14 3-parts) 30 Z^1 clusters = (A^0,[z0,z11*2]*4)*30 (240 z11 pieces, 120 3-parts) As before, we refine the labelling of a1 pieces into a10 and a11 pieces, according to z10 or z11 complements. The next level of clustering is constructed by taking closure under a10/z10 muffins. The 282 a1 pieces in the 47 A^1 clusters are divided into 240 a11 pieces and 42 a10 pieces. Thus at least 5 A^1 clusters must contain only a11 pieces, and with maximally even distribution we have 42 A^1 clusters with 1 a10 piece and 5 a11 pieces. Continuing to refine pieces according to their appearance in A^10 or A^11 clusters, we have A^10 = A^1 = {a110*6, ...} (6 a110 pieces and some other pieces) A^11 = Z^0, (A^1)*3 = (a111*15, ...) with 449/258 consisting of 5 A^10 clusters ( 5 A^1 clusters, 30 a110 pieces) 14 A^11 clusters (42 A^1 clusters, 252 a111 pieces) A^11 clusters provide bound S(449/258) <= avg(a111) = (7r-6)/15 The 449/258 example lies on the intersection of this bound and the next level cluster bound. The next cluster level takes a110/z110 muffin closure and induces clusters Z^10 = Z^1 Z^11 = A^10, (Z^1)*6 with 449/258 consisting of 0 Z^10 clusters 5 Z^11 clusters (i.e., only Z^11 clusters) and bound S(449/258) <= avg(a1111) = 1-avg(z1111) = (73-32r)/42. I conclude with some computer generated additional information on these bounds. If anyone sort of understands this post and would like to use my program as is (i.e., no manual and documentation is mostly my posts), drop me a note and I'll send the C++ source (300 lines). B<4>(r) = (1r-0)/4; S'(r) <= B<4>(r) on (3/2,2/1] tight bound on (3/2,8/5] and at r = (4+8k)/(2+5k) (3/2,8/5] contains mediant(3/2,8/5) = 11/7 S'( (3/2,8/5] ) = (3/8,2/5]; S'(11/7) = 11/28 B<>(r) = (1-0r)/2 B<4>(r) = B<>(r) at r = 2/1 = 2.0000000 S'(2/1) = 1/2 = 0.50000000 B<4,1>(r) = (3-1r)/3; S'(r) <= B<4,1>(r) on [12/7,2/1) tight bound on [9/5,2/1) and at r = (12+9k)/(7+5k) [9/5,2/1) contains mediant(9/5,2/1) = 11/6 S'( [9/5,2/1) ) = [2/5,1/3); S'(11/6) = 7/18 B<4>(r) = (1r-0)/4 B<4,1>(r) = B<4>(r) at r = 12/7 = 1.7142857 S'(12/7) = 3/7 = 0.42857143 B<4,1,1>(r) = (2r-1)/6; S'(r) <= B<4,1,1>(r) on (12/7,7/4] tight bound on (12/7,59/34] and at r = (21+59k)/(12+34k) (12/7,59/34] contains mediant(12/7,7/4) = 19/11 S'( (12/7,59/34] ) = (17/42,7/17]; S'(19/11) = 9/22 B<4,1>(r) = (3-1r)/3 B<4,1,1>(r) = B<4,1>(r) at r = 7/4 = 1.7500000 S'(7/4) = 5/12 = 0.41666667 B<4,1,1,1>(r) = (12-5r)/8; S'(r) <= B<4,1,1,1>(r) on [40/23,7/4) tight bound on [148/85,7/4) and at r = (80+148k)/(46+85k) [148/85,7/4) contains mediant(47/27,7/4) = 54/31 S'( [148/85,7/4) ) = [7/17,13/32); S'(54/31) = 51/124 B<4,1,1>(r) = (2r-1)/6 B<4,1,1,1>(r) = B<4,1,1>(r) at r = 40/23 = 1.7391304 S'(40/23) = 19/46 = 0.41304348 B<4,1,1,1,1>(r) = (7r-6)/15; S'(r) <= B<4,1,1,1,1>(r) on (40/23,228/131] tight bound on (40/23,1119/643] and at r = (228+1119k)/(131+643k) (40/23,1119/643] contains mediant(40/23,47/27) = 87/50 S'( (40/23,1119/643] ) = (142/345,265/643]; S'(87/50) = 103/250 B<4,1,1,1>(r) = (12-5r)/8 B<4,1,1,1,1>(r) = B<4,1,1,1>(r) at r = 228/131 = 1.7404580 S'(228/131) = 54/131 = 0.41221374 B<4,1,1,1,1,1>(r) = (73-32r)/42; S'(r) <= B<4,1,1,1,1,1>(r) on [449/258,228/131) tight bound on [9349/5372,228/131) and at r = (1347+9349k)/(774+5372k) [9349/5372,228/131) contains mediant(134/77,47/27) = 181/104 S'( [9349/5372,228/131) ) = [1107/2686,2267/5502); S'(181/104) = 75/182 B<4,1,1,1,1>(r) = (7r-6)/15 B<4,1,1,1,1,1>(r) = B<4,1,1,1,1>(r) at r = 449/258 = 1.7403101 S'(449/258) = 319/774 = 0.41214470 B<4,1,1,1,1,1,1>(r) = (61r-60)/112; S'(r) <= B<4,1,1,1,1,1,1>(r) on (449/258,764/439] tight bound on (449/258,149516/85913] and at r = (10696+149516k)/(6146+85913k) (449/258,149516/85913] contains mediant(449/258,764/439) = 1213/697 S'( (449/258,149516/85913] ) = (11909/28896,35408/85913]; S'(1213/697) = 32173/78064 B<4,1,1,1,1,1>(r) = (73-32r)/42 B<4,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1>(r) at r = 764/439 = 1.7403189 S'(764/439) = 2533/6146 = 0.41213798 B<4,1,1,1,1,1,1,1>(r) = (1101-487r)/615; S'(r) <= B<4,1,1,1,1,1,1,1>(r) on [160212/92059,764/439) tight bound on [1313469/754729,764/439) and at r = (160212+6567345k)/(92059+3773645k) [1313469/754729,764/439) contains mediant(19549/11233,764/439) = 20313/11672 S'( [1313469/754729,764/439) ) = [1555262/3773645,111271/269985); S'(20313/11672) = 986147/2392760 B<4,1,1,1,1,1,1>(r) = (61r-60)/112 B<4,1,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1,1>(r) at r = 160212/92059 = 1.7403187 S'(160212/92059) = 37941/92059 = 0.41213787

2 1

Re: [math-fun] Parenthesizing
by Allan Wechsler 01 Dec '17

01 Dec '17

Rich politely shielded me from the jeers of the rest of the list in pointing out an "algebraic" identity that applies to 3 as it does to any number, namely: (a^a)^(a^a) = (a^(a^a))^a In general, if K and L are two different exponentiation-towers, then (a^K)^L = a^(KL) = (a^L)^K. This inspires two new questions: (1) Is this "KL law" the only algebraic law that exponentiation-towers obey in general, or are there other such general relations not inferrable from the KL law? (2) In particular for a=3, are there any equivalences that aren't true for general a? (Probably a careful reading of Guy and Selfridge [1973], cited by Neil earlier, would answer both questions.) On Wed, Nov 29, 2017 at 2:30 PM, <rcs(a)xmission.com> wrote: > (3^3)^(3^3) = (3^(3^3))^3 = 3^(3^4) = 3^81 . > The 3 and 3^3 in the exponent are multiplied, which commutes. > > Rich > > ----------- > > Quoting Allan Wechsler <acwacw(a)gmail.com>: > >> I got my head in a little twist over this whole thread. Reassure me: do >> all >> the different parethesizations of 3^3^3^...^3, for n 3's, give different >> values? The original problem (using principle values of a^b) is >> interesting >> because i is special, right? >> >> >> On Wed, Nov 29, 2017 at 9:46 AM, James Propp <jamespropp(a)gmail.com> >> wrote: >> >> Apropos of parenthesizing exponential towers, here's an easier question >>> (probably not new): Look at all the (Catalan-many) ways to parenthesize >>> a_1 >>> - a_2 - ... - a_n. Not all are distinct as linear functions of a_1, a_2, >>> ..., a_n; e.g., >>> >>> (a - (b - c)) - d = a - (b - (c - d)). >>> >>> How many different functions can be obtained? I don't know the answer. Is >>> it 2^(n-2)? >>> >>> >>> Jim Propp >>> _______________________________________________ >>> 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 >> >> > >

4 6

[math-fun] Complexity of detecting whether the intersection of two regular languages is empty?
by Mike Stay 30 Nov '17

30 Nov '17

Given two regular expressions, how hard is it to tell whether they unify, i.e. whether the intersection of the languages they denote is empty? -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

4 7

[math-fun] Lights Out online
by James Propp 30 Nov '17

30 Nov '17

Anyone know of a good website that lets you play Lights Out? I found some websites that offer playing for free, but they aren't the sort of sites I'd recommend to others, because they have offensive ad-content or require downloads or just seem sketchy in one way or another. I'd especially like to know of a site that lets you play a mod 3 version, where clicking on a cell increments it and its neighbors by 1 mod 3. (The usual version is mod 2.) Thanks, Jim Propp

2 2

[math-fun] Parenthesizing
by James Propp 29 Nov '17

29 Nov '17

Apropos of parenthesizing exponential towers, here's an easier question (probably not new): Look at all the (Catalan-many) ways to parenthesize a_1 - a_2 - ... - a_n. Not all are distinct as linear functions of a_1, a_2, ..., a_n; e.g., (a - (b - c)) - d = a - (b - (c - d)). How many different functions can be obtained? I don't know the answer. Is it 2^(n-2)? Jim Propp

3 2

[math-fun] Complex exponentiation
by Dan Asimov 29 Nov '17

29 Nov '17

When z is a nonzero complex number z = r exp(i*theta) and alpha is any complex number, then classically: z^alpha is defined to be the set of complex values given by (*) {exp(alpha * log(z)) where log(z) takes exactly all the values of the form {c + 2K*pi*i | K in Z} for some complex number c with exp(c) = z, e.g., c = ln(r) + i*theta . Now (*) exp(alpha * (ln(r) + i*theta). Let alpha = a + bi for a, b real. Then we have z^alpha = {exp((a + bi)*(ln(r) + i*theta)) = {exp(a*ln(r) - b*theta) + i(a*theta + b*ln(r))) = {exp(a*ln(r) - b*theta_0) i(a*theta_0 + b*ln(r))) * exp(2pi*(-b + a*i))^K where (say) -pi < theta_0 <= pi defines theta_0 uniquely, so the last equation reads: z^alpha = {u * v^K | K in Z} for complex numbers u, v with u = exp(a*ln(r) - b*theta_0) i(a*theta_0 + b*ln(r))), and v = exp(2pi*(-b + a*i)). Now, when people talk about i^i^i^i^.... they usually mean by taking the parentheses from the top down — and using the principal logarithm: i^z = exp((pi*i/2)*z) and iterating: i^...^z = i^(i^(...^(i^z)...)) = exp((pi*i/2)*(exp((pi*i/2)*...*exp((pi*i/2)z)...)) But what if we *didn't* restrict to using the principal logarithm, but instead considered *all* the determinations of i^previous stuff each time we exponentiated. Then at each stage there will be countably many determinations. Now (as usual) let the *number of exponentiations* approach oo. QUESTION: What are the limit points of the determinations: -------- Those points z of C that are *limits* of any sequence of points of the form z_n = something born at exactly n exponentiations, i.e., such that for every epsilon > 0 there exists N such that there is a point z_n that is a result of n > N exponentiations with |z - z_n| < epsilon. —Dan

2 1

[math-fun] some biggish complex numbers
by Neil Sloane 29 Nov '17

29 Nov '17

There was a discussion here recently about a non-integral sequence where the 43d term was falsely claimed to be "too big to compute exactly" Here is a different question. Sequence https://oeis.org/A198683 is defined to be the "Number of distinct values taken by i^i^...^i (with n i's and parentheses inserted in all possible ways) where i = sqrt(-1) and ^ denotes the principal value of the exponential function." The first 11 terms are 1, 1, 2, 3, 7, 15, 34, 77, 187, 462, 1152 and there seem to be two different opinions about the 12th term, which may be either 2919 or 2926. The number of ways to insert the parens is of course a Catalan number, but the tricky part is testing for equality when the numbers get large (or small). Jon Schoenfield has been trying to settle the 12th term, and his comments (see also the "History" tab in that sequence) are very interesting. He points out that one of the values for n=11 is about 4.1007...*10^41232950809707420597749203381002924. Maybe one of experts on this list could help?

1 0

Re: [math-fun] Miscounting in the Cheddar Gorge ...
by Dan Asimov 29 Nov '17

29 Nov '17

It seems that 98.6 is an overestimate for average human body temp. In 1992 they got a figure of 98.2: https://www.ncbi.nlm.nih.gov/pubmed/1302471 —Dan ----- From: Andy Latto <andy.latto(a)pobox.com> On Sat, Nov 18, 2017 at 11:11 AM, Leo Broukhis <leobro(a)gmail.com> wrote: >> This was converted from 37.0 Celsius. Do you have a source for the value being given to three significant figures in Celsius? I looked online, and found "Normal body temperature is considered to be 37°C (98.6°F); however, a wide variation is seen. Among normal individuals, mean daily temperature can differ by 0.5°C (0.9°F), and daily variations can be as much as 0.25 to 0.5°C" on the NIH site at https://www.ncbi.nlm.nih.gov/books/NBK331/, quoted from "Clinical Methods: The History, Physical, and Laboratory Examinations. 3rd edition.". If it varies among individuals by .5 decrees C, and daily variations can be as much as .5 degrees C, 37 seems a better description than 37.0 -----

5 4

[math-fun] weird looking EllipticK valuation
by Bill Gosper 28 Nov '17

28 Nov '17

Stumbling through old mail I was startled by Hypergeometric2F1[1/3, 2/3, 1, 1 - GoldenRatio] == 2 Sqrt[GoldenRatio] π/(5^(7/12) Gamma[1/3] Gamma[11/15] Gamma[14/15]) I have no idea where I got it, but a fairly immediate consequence is EllipticK[1/64 (47 + 17 I Sqrt[3] - 21 Sqrt[5] - 7 I Sqrt[15])] == ((1 + Sqrt[5]) (17 + 17 I Sqrt[3] + 21 Sqrt[5] - 7 I Sqrt[15])^(1/4) π^2)/( 2 Sqrt[2] 3^(3/4) 5^(7/12) Gamma[1/3] Gamma[11/15] Gamma[14/15]) I don't have any η(i√15)s in my identity collection, but I strongly suspect Gamma[1/3] Gamma[11/15] Gamma[14/15] will be Chowla-Selberg's prediction for the transcendental part. The algebraic part was mostly E^(I ArcCot[1/8 (1 + Sqrt[3])^4 (Sqrt[3] + Sqrt[5])^2]) . (Huh? That's algebraic? Obviously!) --rwg Many of these 2F1s come out as algebraic*ElliptickK[algebraic] π^rational. MROB: Let me know if you ever add K to ries. --Bill

1 1

[math-fun] A completely unrelated question triggered by the word "convex"
by Dan Asimov 28 Nov '17

28 Nov '17

Suppose you have a finite set of points in the plane. The process known as "convex peeling" arose as one way of defining the "median" of a finite set of 2-dimensional data, analogous to the usual one for 1-dimensional data: Let the set of points be X. Define X_0 = X and X_(n+1) = X_n - {p in X_n | p is on the boundary of the convex hull of X_n} for n >= 0. This sequential "filtration" of X is known as convex peeling. About 37 year ago someone showed me that if you start with a large number of points uniformly distributed in a square and then proceed to remove them in stages via "convex peeling" — the boundaries of the last groups of points remaining are exceedingly circular. So that there appears to be a true statement, something like: ----- With probability = 1, the shape of the normalized result of convex peeling [toward the end of the process] approaches [in some sense] a circle, as the number of points approaches oo. ----- But I've never seen a proof of this. Any ideas? —Dan

3 3