November 2015 - mathfun@mailman.xmission.com

Re: [math-fun] "OneSixOne" palindrome approx to phi
by Henry Baker 24 Nov '15

24 Nov '15

Note the patterns: all even, but not divisible by 8. Notice the pattern in the other primes? 2 b= 4 = 2 100ln= 139 b= 10 = 2 5 100ln= 230 b= 26 = 2 13 100ln= 326 2 b= 68 = 2 17 100ln= 422 b= 178 = 2 89 100ln= 518 b= 466 = 2 233 100ln= 614 2 b= 1220 = 2 5 61 100ln= 711 b= 3194 = 2 1597 100ln= 807 b= 8362 = 2 37 113 100ln= 903 2 b= 21892 = 2 13 421 100ln= 999 b= 57314 = 2 28657 100ln= 1096 At 09:14 PM 11/23/2015, rcs(a)xmission.com wrote: >The limiting ratio is a tad less than e; maybe phi^2? --Rich > >----------- >Quoting Henry Baker <hbaker1(a)pipeline.com>: > >>At 01:38 PM 11/23/2015, Henry Baker wrote: >>>But why stop with 161; cf(phi) is also a palindrome. >>> >>>Harmony and balance will cost you 2400 euros. >>> >>>Smaller amounts of harmony and balance are denoted by "Tau" and >>>cost you only 900 euros. >>> >>>So, for which bases is phi a palindrome for the first 3 digits?

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[math-fun] "OneSixOne" palindrome approx to phi
by Henry Baker 24 Nov '15

24 Nov '15

But why stop with 161; cf(phi) is also a palindrome. Harmony and balance will cost you 2400 euros. Smaller amounts of harmony and balance are denoted by "Tau" and cost you only 900 euros. So, for which bases is phi a palindrome for the first 3 digits? "ONESIXONE are the three first figures of the number that governs beauty, the golden ratio. The golden ratio attributes an aesthetic character to objects. Some even believe it has a mystical importance. Along the history it has been included in the design of various works of architecture and art. The golden ratio is synonym with balance, harmony and beauty because of its sound, its emotional charge, its visual and graphic balance. The sense of harmony that transmits being a palindrome number and representativeness of this number in the philosophy of the brand, as well as its their relationship with nature and architecture." http://onesixone.es/the-brand/

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Re: [math-fun] "Abba-Ima" and Thue-Morse
by Bill Gosper 23 Nov '15

23 Nov '15

On 2015-11-23 08:19, James Propp wrote: > I don't have time to record it and post the recording, but here's how it > goes, IIRC: > > Leader: A-bba! > Group: I-ma! > > Leader: I-ma! > Group: A-bba! > > Leader: Abba Ima! > Group: Ima Abba! > > Leader: Ima Abba! > Group: Abba Ima! > > Leader: Abba Ima Ima Abba! > Group: Ima Abba Abba Ima! > > Leader: Abba Ima Ima Abba Ima Abba Abba Ima! > Group: Ima Abba Abba Ima Abba Ima Ima Abba! > > (Actually, I'm not sure the last call-and-response pair is included outside > of gatherings of especially nerdy Jews; most song-leaders probably fear > over-taxing the short-term memories of the singers.) > > Jim Propp > > On Mon, Nov 23, 2015 at 10:13 AM, James Propp <jamespropp(a)gmail.com> wrote: > >> Does anyone know the background of the incursion of the Thue-Morse >> sequence into 20th century Jewish popular folk music? >> >> There's a style of call-and-response mini-song often tucked into the >> middle of "David Melech Yisrael" dating back at least to the 1970s, in >> which two opposed Hebrew words or phrases, like mother/father and yes/no, >> are alternated in Thue-Morse style (see >> https://en.m.wikipedia.org/wiki/Thue–Morse_sequence) >> >> Was there some particular musician who was responsible for this? >> >> (Also, can anyone find a relevant clip on YouTube?) >> >> Jim Propp >> https://en.m.wikipedia.org/wiki/Thue–Morse_sequence is surprisingly good, but doesn't quite subsume http://www.inwap.com/pdp10/hbaker/hakmem/series.html Item 122 nor http://www.inwap.com/pdp10/hbaker/hakmem/topology.html Item 115 In particular, the non-regular continued fraction for t(1/2), In[284]:= 1/(3 - 1/(2 + ContinuedFractionK[1 - 2^2^k, 2^2^(k + 1), {k, 0, n}])) In[287]:= Table[%284, {n, 0, 4}] Out[287]= {7/17, 106/257, 27031/65537, 1771476586/4294967297, 7608434000728254871/18446744073709551617} is about twice as efficient as interpreting partial strings of T in binary: In[290]:= RealDigits[106/257, 2] Out[290]= {{{1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0}}, -1} (Remember to prefix a leading 0.) (The former is alternately mediant(x,x,1/1) and mediant(x,1/1), where x = the latter.) I got excited when the article mentioned fractals, but instead of Hilbert, named von Koch! In[302]:= hilbert /@ %287 Out[302]= {{2/5 + I}, {7/17 + I}, {106/257 + I}, {27031/65537 + I}, {1771476586/4294967297 + I}} I.e., the exact Hilbert function backs up one term, and adds i ! Likewise to the 1's complement: In[303]:= 1 - %287 Out[303]= {10/17, 151/257, 38506/65537, 2523490711/4294967297, 10838310072981296746/18446744073709551617} In[304]:= hilbert /@ % Out[304]= {{3/5 + I}, {10/17 + I}, {151/257 + I}, {38506/65537 + I}, {2523490711/4294967297 + I}} Thus bolstering the claim that x = t(1/2) and x = 1-t(1/2) are the only solutions to Hilbert(x) = x+i . I wonder if the wikisnobs will accept HAKMEM as unoriginal research. --rwg

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[math-fun] "Abba-Ima" and Thue-Morse
by James Propp 23 Nov '15

23 Nov '15

Does anyone know the background of the incursion of the Thue-Morse sequence into 20th century Jewish popular folk music? There's a style of call-and-response mini-song often tucked into the middle of "David Melech Yisrael" dating back at least to the 1970s, in which two opposed Hebrew words or phrases, like mother/father and yes/no, are alternated in Thue-Morse style (see https://en.m.wikipedia.org/wiki/Thue–Morse_sequence) Was there some particular musician who was responsible for this? (Also, can anyone find a relevant clip on YouTube?) Jim Propp

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Re: [math-fun] EllipticE valuation [Was: EllipticK valuation [Was: Elliptic K π approximation]]
by Bill Gosper 23 Nov '15

23 Nov '15

Appended below is "I'm having no luck finding a table of special values of EllipticE. This may be due to a simpler formula than above to produce them from EllipticK. It looks like they're something K + something/K." Borwein & Borwein, (2.3.17) E = K - pi/K Theta4Dot/Theta4 where Theta4Dot:= d/ds Theta(q), s:= -log(q)/pi Two years after my logderivetas remark below, I don't understand it. Here is how to get DedekindEta'[x] in radicals from DedekindEta[x] in radicals. Empirically, for rational x^2: Out[131]= \[Alpha] -> 1/(4 x) + Derivative[1][DedekindEta][x]/ DedekindEta[x])/(\[Pi] DedekindEta[x]^4 is algebraic. E.g., for x = 2 Sqrt[-3] In[181]:= %131 /. x -> Sqrt[-3]*2 Out[181]= \[Alpha] -> (-(I/(8 Sqrt[3])) + Derivative[1][DedekindEta][2 I Sqrt[3]]/ DedekindEta[2 I Sqrt[3]])/(\[Pi] DedekindEta[2 I Sqrt[3]]^4) In[182]:= RootApproximant[%[[2]]] Out[182]= Root[-4 + 5 #1^2 - 10 #1^4 - 2 #1^6 - 6 #1^8 + 8 #1^10 - 5 #1^12 + 9 #1^14 + 2 #1^16 &, 10] Insufficient precision. In[184]:= RootApproximant[N[%181[[2]], 669]] Out[184]= Root[1 + 143822848 #1^6 + 1048576 #1^12 &, 8] In[185]:= ToRadicals[%] Out[185]= ((-1)^(1/3) (-70226 - 40545 Sqrt[3])^(1/6))/(2 2^(2/3)) In[186]:= Strad[%] During evaluation of In[186]:= PossibleZeroQ::ztest1: Unable to decide whether numeric quantity Sqrt[26+15 Sqrt[3]]-(70226+40545 Sqrt[3])^(1/6) is equal to zero. Assuming it is. >> Out[186]= (I Sqrt[26 + 15 Sqrt[3]])/(2 2^(2/3)) In[187]:= Strad[%] Out[187]= (I (5 + 3 Sqrt[3]))/(4 2^(1/6)) Since there are various means to find DedekindEta[2 I Sqrt[3]] -> ((3 - Sqrt[3])^(1/4) Gamma[1/3]^(3/2))/(2 2^(7/8) Pi) , we have Derivative[1][DedekindEta][2 I Sqrt[3]] == -(( I (3 - Sqrt[3])^(1/4) Gamma[1/3]^( 3/2) (-1 - (Sqrt[3] (3 - Sqrt[3]) (5 + 3 Sqrt[3]) Gamma[1/3]^6)/( 64 2^(2/3) Pi^3)))/(16 2^(7/8) Sqrt[3] Pi)) and finally 2 EllipticE[(Sqrt[3] - 1)^2/8] == pi[(Sqrt@3 - 1)/Sqrt@8] == 2 ((2^(1/3) Pi^2)/( 3^(3/4) Gamma[1/3]^3) + ((3 + Sqrt[3]) Gamma[1/3]^3)/( 8 2^(1/3) 3^(3/4) Pi)) where pi[e]:= circumference of ellipse of unit diameter and eccentricity e: In[238]:= pi[e_] := 2 EllipticE[e^2] In[239]:= pi[0] Out[239]= Pi In[240]:= pi[1] Out[240]= 2 It would now be almost routine to produce tables of special values of EllipticK and EllipticE from my table of DedekindEta[complexsurd]. The obstacles are knowing if you've used enough precision in RootApproximant, and taming indigestible surdbergs. --rwg On Wed, Nov 7, 2012 at 3:05 PM, Bill Gosper <billgosper(a)gmail.com> wrote: The ("complete") first kind elliptic integral K(m) and parameter m are both expressible in 𝜗 constants, and those three+ 𝜗 constants are all expressible in terms of Dedekind η. So Out[840]= EllipticK[(16 η[q]^8 η[q^4]^16)/η[q^2]^24] == (π η[q^2]^10)/(2 η[q]^4 η[q^4]^4) Crude but effective: d/dq and eliminate EllipticK to get EllipticE[(16 \[Eta][q^4]^8)/(\[Eta][q]^8 + 16 \[Eta][q^4]^8)] == -( 1/(\[Eta][q]^8 + 16 \[Eta][q^4]^8))(-((\[Pi] \[Eta][q]^4 \[Eta][q^2]^10)/( 2 \[Eta][ q^4]^4)) - (\[Pi] \[Eta][ q^2]^9 (6 Sqrt[3] q^(3/2) logderiveta[q] \[Eta][q] \[Eta][q^2] \[Eta][q^4] - 30 Sqrt[3] q^(5/2) logderiveta[q^2] \[Eta][q] \[Eta][q^2] \[Eta][q^4] + 24 Sqrt[3] q^(9/2) logderiveta[q^4] \[Eta][q] \[Eta][q^2] \[Eta][ q^4]) (\[Eta][q]^8 + 16 \[Eta][q^4]^8))/(2 q^( 1/3) \[Eta][q]^4 \[Eta][ q^4]^4 (6 Sqrt[3] q^(7/6) logderiveta[q] \[Eta][q] \[Eta][q^4] - 24 Sqrt[3] q^(25/6) logderiveta[q^4] \[Eta][q] \[Eta][q^4])) The remaining problem is those logderivetas, but lo, ((1/(4*x) - ((logderiveta[(1/(E^x))]))/(E^x))/((DedekindEta[((x*I)/(2*Pi))])^4)) is algebraic for (x/π)^2 rational. Unfortunately, (x/π)^2= (i/2-1/√12)^2 (as in the q for k=(-1)^(1/3)) isn't rational, so EllipticE[(-1)^(1/3)] == E^(-(I π/12)) ((2 2^(1/3) π^2)/(3^(3/4) Gamma[1/3]^3) + Gamma[1/3]^3/(4 2^(1/3) 3^(1/4) π)) required Legendre's relation plus numerical empiricism. This, with the K((-1)^(1/3)) result gives closed forms for all 2F1[a+1/2,b+1/2,c;(-1)^(1/3)], a,b,c integers. Likewise for every k for which we find both K(k) and E(k), which are contiguous 2F1[k]. BtW, if you accept MatProd[mat[k],{k,2,a}] as closed form, then I can write 2F1[a+1/2,b+1/2,c;k] in closed form. --rwg Added 2012-11-11 13:22 And for such x, we have many DedekindEta[((x*I)/(2*Pi)) in our collection. In fact, people.math.carleton.ca/~*williams*/papers/*pdf*/*299*.*pdf claims to give a complete solution. Early on, it appears to omit half the odd denominators, but I assume this is remedied later in the derivation. Anyway, as a fairly strenuous example: EllipticE[1/(1 + (-1 + GoldenRatio^(3/2))^8/(16 GoldenRatio^4))] == ( 2 5^(3/4) GoldenRatio^4 π^(3/2))/( Sqrt[1 + Sqrt[5]] (1525 + 682 Sqrt[5])^(1/4) Gamma[1/20] Gamma[9/ 20]) + ((-7 - 3 Sqrt[5] + Sqrt[190 + 85 Sqrt[5]]) Gamma[1/ 20] Gamma[9/20])/( 8 (1525 + 682 Sqrt[5])^(1/4) Sqrt[5 (1 + Sqrt[5]) π]) I'm having no luck finding a table of special values of EllipticE. This may be due to a simpler formula than above to produce them from EllipticK. It looks like they're something K + something/K. At the other extreme, has anyone seen *EllipticK[(-1)^(1/3)] == ((-1)^(1/12) 3^(1/4) Gamma[1/3]^3)/( 4 2^(1/3) π) ? --rwg

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[math-fun] function question
by David Wilson 21 Nov '15

21 Nov '15

Is 1. Is phi(n) + sigma(n) - 2n >= 0 for all n > 1? 2. Is phi(n) + sigma(n) - 2n = 5 for any n?

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Re: [math-fun] Turning a fraction into an integer
by Bill Gosper 20 Nov '15

20 Nov '15

See https://cp4space.wordpress.com/2013/10/24/enumerating-the-rationals/ Here's a fast implementation of this forward and backward Wilf-Calkin map that NeilB did in Mathematica. In[345]:= newman2pos[71426/7623] // tim During evaluation of In[345]:= 0.001112,0 Out[345]= 999999999 In[347]:= pos2newman[67107847] // tim During evaluation of In[347]:= 0.000729,0 Out[347]= 355/113 calkinwilf[num_, denom_] := Block[{lis = {}, n = num, d = denom}, (While[ n > 1 || d > 1, (lis = Prepend[lis, Boole[n > d]]; If[n > d, n -= d, d -= n])]; lis)]; wilfcalkin[bitlis_List] := Block[{n = 1, d = 1}, (Do[ If[bitlis[[i]] == 0, d += n, n += d], {i, 1, Length[bitlis]}]; Rational[n, d])] newman2pos[num_, denom_] := FromDigits[Prepend[calkinwilf[num, denom], 1], 2]; newman2pos[frac_] := newman2pos[Numerator[frac], Denominator[frac]] pos2newman[0] = 0; pos2newman[pos_] := Sign[pos]*wilfcalkin[IntegerDigits[pos, 2][[2 ;; -1]]] --rwg

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[math-fun] Turning a fraction into an integer
by Eric Angelini 20 Nov '15

20 Nov '15

Hello Math-Funsters, I was looking for a way to represent any fraction with one single integer. Do you know if there is a common way to do that or a paper about it? I came up with this (heavy?) idea: concatenate N and D and encode the place of the slash "/". So, the fraction we want to represent is N/D. We build the integer NDBA with four concatenated strings of digits. Starting from the right: A is a single digit B is a string that has length A (if A=0 then A=10) D is the denominator of the fraction N/D N is the numerator of the fraction N/D To reconstruct the fraction N/D from NDBA we: - read A - compute the length of B - read B - insert the symbol "/" after B digits, starting the count from the leftmost digit of NDBA - erase A and B Examples: The fraction 1/2 --> 1211 [or 12.1.1 --> 1/2] The fraction 113/355 --> 11335531 [113/355.3.1] The fraction 2015/1234567890 --> 2015123456789041 [2015/1234567890.4.1] The fraction 1234567890/888 --> 1234567890888102 [1234567890888.10.2] etc. This system works for N having less than 10 billion digits, of course -- which seems ok in our everyday life (we would have an integer like ND.9999999999.0 with ten 9's) Best, É. P.-S. Many integers will not represent any fraction -- but this is another story (10211 would produce the fraction 1/02, for instance).

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Re: [math-fun] laser cool liquids?
by Henry Baker 20 Nov '15

20 Nov '15

Nevertheless, the ability to cool a particular neuron or cell (or cell portion -- e.g., nucleus or mitochondrion) might be hugely valuable in unraveling how these things work. At 07:29 AM 11/20/2015, Warren D Smith wrote: >Because you need like 1 blue photon per molecule, and if even a >tiny fraction of those photons, like 1%, get wasted/lost, that'd be hugely >inefficient. > >And.. sure enough, looking at the paper they are using megawatt per >cm^2 irradiance levels for their "refrigerator." My god.

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[math-fun] Turning a fraction into an integer and reverse: three efficient algorithms
by Warren D Smith 20 Nov '15

20 Nov '15

What we want is three algorithms. Algorithm 1 inputs integer N and outputs rational p/q. Algorithm 2 ("ordinal") inputs p/q and outputs integer N (inverse of alg 1). Algorithm 3 inputs rational p/q, and effectively converts it to integer, increments integer, and converts it back to rational which it outputs ("successor"), except it hopefully does it faster than that. The Stern sequence doubling recurrences in A002487 show how alg 1 can be implemented in a number of arithmetic ops proportional to the number of bits in the integer. As indeed Dijkstra pointed out in both https://www.cs.utexas.edu/users/EWD/transcriptions/EWD05xx/EWD570.html and https://www.cs.utexas.edu/users/EWD/transcriptions/EWD05xx/EWD578.html To redo Dijkstra my way, given integer N we may compute the rational p/q where p=fusc(N) and q=fusc(N+1) by letting (p,q)=Alg1(N) where, recursively, Alg1(N){ if(N>0){ if(N is even){ (p,q)=Alg1(N/2); return( p, p+q ); }else{ (p,q)=Alg1((N-1)/2); return( p+q, q ); } } return( 0,1 ); } This can be rewritten as an iterative nonrecursive algorithm: Alg1(N){ p=0; q=1; for( the bits b of N from most-signif to least-signif do ){ if(b){ p = p+q; }else{ q = p+q; } } return( p,q ); } That in turn tells us how to write the inverse algorithm Alg2(p,q){ N=0; for(j=1; p>0; j=2*j){ if(q>p){ q = q-p; }else{ p = p-q; N += j; } } return(N); } which note also runs in a number of arithmetic ops proportional to the number of bits in N. A002487 says Moshe Newman pointed out how to implement Alg3 in a constant number of arithmetic ops: the successor of x=p/q is s=1/(2*floor(x) + 1 - x) = 1/(floor(x) + 1 - frac(x)). Does this really count as a constant number of ops -- how hard is the GCD you need to reduce the new fraction? Well, letting p=q*t+u where 0<=u<q so that p/q = t + u/q, evidently GCD(u,q)=1 since p/q was already in lowest terms, so then we see s=q/(t*q+q-u)=q/(p+q-2*u) is already in lowest terms unless u=0. If u=0 (which only happens when q=1) then s=1/(t+1) is lowest terms. Hence, no GCDing ever needs to be done. This yields the very simple rational-successor algorithm Alg3(p,q){ u = (p mod q); return( q, p+q-2*u ); } --Warren D. Smith http://RangeVoting.org

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