November 2015 - mathfun@mailman.xmission.com

[math-fun] Chinese Remainders for Audio & Signal Processing ?
by Henry Baker 07 Nov '15

07 Nov '15

[Totally different thread from family planning!!] I sometimes play with MIDI synthesizers to construct & modify music. The basic operation of a MIDI synthesizer is to *add* sounds to one another to produce chords, etc. In order for the human ear to later pull apart these synthesized sounds, it is essential for this *addition* to obey *superposition* -- i.e., *linearity*: f(a+b)=f(a)+f(b) and f(c*a)=c*f(a). The problem with digital sound is that it typically uses too few bits to be able to preserve superposition/linearity; e.g., some systems use 12-16 bits, while others may use fixed precision up to 24 bits. Newer signal processing systems employ *floating point* arithmetic, which is sometimes a bit better, but still incapable of preserving superposition/linearity 100%. Has anyone (else) considered a digital sound/signal processing system based upon Chinese Remainders? The idea is to represent the digital samples (mod 2), (mod 3), (mod 5), etc., (mod p), until we have enough remainders to completely capture the full *dynamic range* of the music or signals. The wonderful thing about arithmetic (mod p) is that it forms a field, so we can do all of our digital filtering convolutions for each of the primes *separately*, and then reconstruct the final signal for playback using the Chinese Remainder Theorem. Superposition/linearity is faithfully preserved. Such a signal processing system would be useful in certain types of signal processing where the signal is extremely faint in comparison with another signal (or noise). Such systems include "spread spectrum" techniques, where the signal may be 10-20 dB below the noise floor. Linear system techniques *without roundoff errors* would be essential to performing the necessary computations. Video is also escaping from the 8-bit world; new cameras can get to 10-bit, 12-bit and even 14-bit dynamic ranges, so Chinese Remainder techniques might be useful for video, as well. An interesting question is in the final sound/signal *reconstruction*. It may be possible to reconstruct the signal from the Chinese Remainders *out of thin air* -- i.e., by letting the human ear (or eye) put everything back together again. But I'm not sure how to build a speaker or display which produces (mod p) sounds/brightnesses !.

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[math-fun] Chinese male excess
by Warren D Smith 07 Nov '15

07 Nov '15

This whole "war" thing is nonsense -- i.e. I see no evidence for it. And it is hard to produce evidence for it, because I think there has never before in human history been a country with a big male excess. They are exploring terra incognita. [Well, I think there are some cases caused by differential migration, e.g. Alaska was very male for a while, maybe still is. Also there were cases with a big male deficit because of, e.g. world war I killing off so many males there then was a big male deficit.] As far as I can tell, the Chinese 1-child policy worked excellently. It gave them the biggest economy in the world. (Despite all the total bull one sometimes hears about how cutting baby production will hurt the economy.) It has been a tremendous success. And further, one might wish that the rest of the world would adopt the same policy, in the sense that I believe the whole world is operating well above what renewable resources permit, i.e. a population crash is inevitable. That inevitable population decline could be done the relatively pleasant way, like the Chinese did it, or the doom and war way, like the rest of the world is aiming for right now with all their mighty stupidity and inability to plan a damn thing. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Chinese family planning
by Henry Baker 07 Nov '15

07 Nov '15

[Caution: thar be biases & politics ahead!] I realize that I've hijacked this thread way off topic, but here are some papers: http://belfercenter.ksg.harvard.edu/files/hudson_and_den_boer_spring_2002.p… A Surplus of Men, A Deficit of Peace Valerie M. Hudson and Andrea Den Boer Security and Sex Ratios in AsiaÂs Largest States "One overlooked wellspring of insecurity, we argue, is exaggerated gender inequality." "We first quantify the scale on which sex ratios are being altered in Asia, then estimate the number of resulting surplus young adult males currently present in AsiaÂs two largest states, China and India, as well as projected to the year 2020. Next, we discuss behavioral syndromes associated with surplus young adult male groups, and investigate the role of such groups in instability and violence within and between societies in several historical cases." "Given our focus on security, we are most interested in bare branches [surplus males] who are in the age group 15Â34/35, as this is the male age group that commits the preponderance of violence within a society. How many young adult bare branches are there now, and how many are there expected to be in the next two decades in the two most populous nations on earth, China and India? Generally speaking we will be seeing 29Â33 million young surplus males in China in twenty years and 28Â32 million young surplus males in India at that same time." "In sum, these young surplus males may be considered, relatively speaking, losers in societal competition." "The behavior of young surplus males also follows a broadly predictable pattern. Theory suggests that compared with other males in society, bare branches will be prone to seek satisfaction through vice and violence, and will seek to capture resources that will allow them to compete on a more equal footing with others." "Cross-culturally, an overwhelming percentage of violent crime is perpetrated by young, unmarried, low-status males." "The reproductive strategy chosenÂthat of high sex ratioÂcoupled with primogeniture ensured, in BooneÂs words, Âpolitical instability, warfare, and territorial expansion.Â" "Nevertheless, we understand how their analysis of 448 populations could result in the conclusion that Âwe are most likely to find unbalanced sex ratios when warfare is present.Â See Divale and Harris, ÂPopulation, Warfare, and the Male Supremacist Complex,Â American Anthropologist, Vol. 78, No. 3 (September 1976), pp. 531, 528." "India and Pakistan, both high sex-ratio societies, are unlikely to settle the conflict peacefully" "Nevertheless, it should never be forgotten by policymakers on either side of the PaciÂªc that the worst-case scenario implies that China may have close to 40 million young adult bare branches to spare in twenty years, and that the government may at that point ardently wish to see them give their lives in pursuit of a national interest. The alternative is to allow them to remain a threat to national interest, which may increasingly be seen as an untenable policy position by the government." "We suggest that societies with young adult sex ratios of approximately 120 and above are inherently unstable, and both China and India are, at this writing, nearing that level and will probably surpass that level in the next two decades." http://www.japss.org/upload/12.ross.pdf An Âarmy of bachelorsÂ? ChinaÂs Male Population as a World Threat http://anon-ftp.iza.org/dp3214.pdf Sex Ratios and Crime: Evidence from ChinaÂs One-Child Policy http://deepblue.lib.umich.edu/bitstream/handle/2027.42/97286/j.1538-165X.20… Bare Branches: The Security Implications of AsiaÂs Surplus Male Population by Valerie M. Hudson and Andrea M. den Boer. Cambridge,MA, MIT Press, 2004. 400 pp. $35.00. "Males are more violent than females, in part because of higher levels of testosterone. Men without women are far more violent than are men with women. Testosterone levels drop significantly and quickly when men marry." "The authors use historical cases to show high instability in countries with Âexaggerated sex ratios.Â Typically, high-sex-ratio countries have authoritarian governments" At 01:43 PM 11/6/2015, Fred Lunnon wrote: Is the proof constructive, and tell us who they're going to fight? More generally, does it offer an effective algorithm to discourage the human race from constantly turning every scientific advance into an agent of misery? WFL On 11/6/15, Henry Baker <hbaker1(a)pipeline.com> wrote: > The excess male Chinese are called "Chinese Remainders", and the fact that > this leads to war is called the "Chinese Remainder Theorem". ;-) > > At 12:56 PM 11/6/2015, Henry Baker wrote: > >>"It's estimated that by 2020, China will have 24 million more men than >> women of marriageable age on the mainland" >> >>http://english.cri.cn/7146/2013/02/26/2702s750680.htm >> >>Of course, artificially high sex ratios like this in human history are >> always resolved the same way: war. The young men kill one other until the >> problem has resolved itself.

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Re: [math-fun] Chinese family planning
by Henry Baker 07 Nov '15

07 Nov '15

Yes, but what happens the next & succeeding generations ? (Assuming 1:1 "traditional" marriages.) At 11:42 AM 11/6/2015, James Propp wrote: >Has anyone in the pop math biz tackled the mathematical side of the news >story about China's (now abandoned) one-child-per-family policy? > >Specifically, many families adopted the family planning algorithm "Have >kids till you have a son, then stop", which (under idealized assumptions) >gives rise to families of average size exactly 2. > >A non-mathematician friend of mine asked me at dinner last night why the >expected size of a family that stops when the first son is born is 2. I >began to give him an intuitive argument that doesn't involve calculation, >but he ended up preferring the argument that shows that 1/2 + 2/4 + 3/8 + >... = 2 by way of summing the formulas >1/2 + 1/4 + 1/8 + ... = 1 > 1/4 + 1/8 + ... = 1/2 > 1/8 + ... = 1/4 > ... >to obtain 1/2 + 2/4 + 3/8 + ... = 1 + 1/2 + 1/4 + ... = 2 > >Is there a place where this is explained, and explained well? > >Jim Propp

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[math-fun] Asimov's correlation problem
by Warren D Smith 06 Nov '15

06 Nov '15

The optimal solution was unit vectors that are the N+1 vertices of a regular simplex in N-dimensional space centered at the origin. (N=3 for Asimov's problem, regular tetrahedron.) These are anticorrelated somewhat, dot product of any two is -1/N minimizing correlation by making it maximally negative. My solution with 4 mutually orthogonal 8-vectors was optimal if we want zero correlation, which minimizes correlation if you meant the absolute value of the correlation...

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[math-fun] What comes after tetration?
by Bill Gosper 05 Nov '15

05 Nov '15

Ak[_,1]=2;Ak[0,x_]:=x+2;Ak[n_,x_]:=Ak[n - 1, Ak[n, x - 1]] In[3]:= TableForm[Table[Ak[n,x],{n,0,3},{x,1,6-n}]] Out[3]//TableForm= 2 4 5 6 7 8 2 4 6 8 10 2 4 8 16 2 4 16 (Rows: n = 0,1,2,3; Columns: x = 1,2,3,4) Let's try bumping the row length. In[4]:= TableForm[Table[Ak[n,x],{n,0,3},{x,1,7-n}]] During evaluation of In[4]:= $RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of 1-1. >> One minus one? It's trying for 65536 by adding up 2s recursively. Let's give it a hint: In[2]:= Ak[1,x_]:=2x In[3]:= TableForm[Table[Ak[n,x],{n,0,3},{x,1,7-n}]] Out[3]//TableForm= 2 4 5 6 7 8 9 2 4 6 8 10 12 2 4 8 16 32 2 4 16 65536 Trying for one more x, In[5]:= TableForm[Table[Short[Ak[n,x]],{n,0,3},{x,1,8-n}]] During evaluation of In[5]:= $RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of 2-1. >> Out[5]//TableForm= 2 4 5 6 7 8 9 10 2 4 6 8 10 12 14 2 4 8 16 32 64 2 4 16 65536 2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 (2 MaxFormatDepthExceeded MaxFormatDepthExceeded MaxFormatDepthExceeded)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))) Time for another hint: In[7]:= Ak[2,x_]:=2^x In[8]:= TableForm[Table[Short[Ak[n,x]],{n,0,3},{x,1,8-n}]] Out[8]//TableForm= 2 4 5 6 7 8 9 10 2 4 6 8 10 12 14 2 4 8 16 32 64 2 4 16 65536 20035299304068464649790723<<19677>>72339445587895905719156736 (Insanely) trying for another row, In[9]:= TableForm[Table[Short[Ak[n, x]], {n, 0, 4}, {x, 1, 8 - n}]] $RecursionLimit::reclim2: Recursion depth of 1024 exceeded during evaluation of 3-1. >> (See nice try <http://gosper.org/ack.png>. Hey, where's the diagonal scroll bar?) --rwg

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Re: [math-fun] Random variable puzzle
by Seb Perez-D 05 Nov '15

05 Nov '15

R=-1/sqrt(3) My proof is rather pedestrian, and involves proving that the variance-covariance matrix, with 1 on the diagonal and R off, is positive semi-definite. X can be constructed as U1+R*(U2+U3+U4) (and the others by permutation), where the Ui are independent, zero-mean random variables of equal variance. Cheers, Sébastien On Nov 3, 2015 22:02, "Dan Asimov" <asimov(a)msri.org> wrote: Suppose X, Y, Z, W are real random variables with a joint distribution such that each one has a finite mean and standard deviation. Suppose that all pairs of these random variables have the same correlation coefficient:* R = rho(X,Y) = rho(X,Z) = rho(X,W) = rho(Y,Z) = rho(Y,W) = rho(Z,W) . Find the minimum possible value of R. —Dan ____________________________________ * The definition of the correlation coefficient rho(U,V) of two random variables U and V is the expected product of their standardizations: rho(U,V) = E( ((U-mu_U)/sigma_U) * ((V-mu_V)/sigma_V)) ) where E is expectation, mu denotes mean and sigma denotes standard deviation. _______________________________________________ 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] math-fun Digest, Vol 153, Issue 4
by Warren D Smith 04 Nov '15

04 Nov '15

> From: Dan Asimov <asimov(a)msri.org> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: [math-fun] Random variable puzzle > Message-ID: <3C0B9325-43FC-4DDA-A42E-69EC9AF69FFF(a)msri.org> > Content-Type: text/plain; charset=utf-8 > > Suppose X, Y, Z, W are real random variables with a joint distribution such > that each one has a finite mean and standard deviation. > > Suppose that all pairs of these random variables have the same correlation > coefficient:* > > R = rho(X,Y) = rho(X,Z) = rho(X,W) = rho(Y,Z) = rho(Y,W) = > rho(Z,W) > . > > Find the minimum possible value of R. For the following 4x4 Hadamard matrix (entries +1 and -1) -+++ +-++ ++-+ +++- each row has mean=1/2, variance=1, and all rows are orthogonal. If we instead take these four rows of an 8x8 Hadamard +-+-+-+- ++--++-- +--++--+ ++++---- then each row has mean=0, variance=1, and all rows orthogonal.

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[math-fun] I just figured out what PseudoInverse is for
by Bill Gosper 02 Nov '15

02 Nov '15

While seeking to accelerate the zeta(2) series, I specialized a very general path-invariant system of 3x3 matrices to a very special case of Dixon's theorem. Out[75]= {{k, {{(1/2 + k)^2/((k + 1/2 (1 - 2 n)) (k + 1/2 (1 + 2 n))), 0, -((n^2 (-1 + 2 n))/( 4 k (k + 1/2 (1 - 2 n))))}, {0, (1/2 + k)^2/((k + 1/2 (1 - 2 n)) (k + 1/2 (1 + 2 n))), -(n^2/( 2 (1/2 + k) (k + 1/2 (1 - 2 n))))}, {0, 0, 1}}}, {n, {{( n^2 (1/2 (1 - 2 k) + n))/((1 + n)^2 (1/2 (1 + 2 k) + n)), 0, n}, {0, (n^2 (1/2 (1 - 2 k) + n))/((1 + n)^2 (1/2 (1 + 2 k) + n)), 1}, {0, 0, 1}}}} (Strictly speaking, Dixon's 3F2 doesn't cover zeta(2), but unlike pFq, matrix products are not required to sum from n=0.) When your 3x3 matrices all become triangular, as above, you can separate the system into two systems of 2x2s by deleting one or the other of the top two rows, and the corresponding column from each matrix in the system. Corey's path-invariant matrix package contains the indispensabie function sidestep, which takes a matrix and "conjugates" it with all the matrices in the current system in such a way as to preserve path invariance. Merely by replacing Inverse with PseudoInverse, sidestep accepts nonsquare matrices and performs the dimension reduction! To delete the top rows and left columns, In[76]:= sidestep[{{0, 0}, {1, 0}, {0, 1}}] Out[76]= {{k, {{(1 + 2 k)^2/( 1 + 4 k + 4 k^2 - 4 n^2), -(( 2 n^2)/((1 + 2 k) (1 + 2 k - 2 n)))}, {0, 1}}}, {n, {{(n^2 (1 - 2 k + 2 n))/((1 + n)^2 (1 + 2 k + 2 n)), 1}, {0, 1}}}} Path invariance check: In[77]:= picheck[%] Out[77]= True A triangular contour in this 2x2 system leads immediately to Out[350]= 10 HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, -(1/4)] In[351]:= N[%/\[Pi]^2] Out[351]= 1. (Recall the "Chinese proverb" pi^2 =10.) Congratulations to Mma for knowing how to evaluate %350, and not doing so automatically: In[352]:= FunctionExpand[%%] Out[352]= 10 (\[Pi]^2/6 + 2 (-(\[Pi]^2/10) - ArcCsch[2]^2) - 2 (-(\[Pi]^2/10) + ArcCsch[2]^2 + 1/2 (\[Pi]^2/15 - ArcCsch[2]^2)) - Log[1/2 + Sqrt[5]/2]^2 + 2 Log[1/2 + Sqrt[5]/2] Log[3/2 + Sqrt[5]/2]) In[353]:= FullSimplify[%] Out[353]= \[Pi]^2 --rwg

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[math-fun] George Boole's 200th BDay Nov 2nd
by Henry Baker 01 Nov '15

01 Nov '15

Celebrate Boole's Birthday with "(I Can't Get No) Satisfaction",* which conjectured that 3SAT is NP-Complete. ;-) http://georgeboole.com/ * (in polynomial time) Curiously, Boole's donation website won't take Bitcoin: https://community.ucc.ie/donation-euro-page---george-boole-200 Are there any digital currencies that solve satisfiability problems for their proof-of-work? (Boole-coin ?)

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