February 2013 - mathfun@mailman.xmission.com

[math-fun] The Evolution of Genomes incl. j*nk DNA
by Henry Baker 19 Feb '13

19 Feb '13

FYI -- Interesting video lecture re mathematical perspective on genome evolution and "junk DNA": Mike Lynch, Indiana Univ. & KITP The Evolution of Genomes Jan. 11, 2010 http://online.kitp.ucsb.edu/bblunch/lynch/

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[math-fun] Richard Crandall
by rcs＠xmission.com 19 Feb '13

19 Feb '13

The recent Mersenne prime discovery prompted me to do some mousing around the web, and I discovered that Richard Crandall died during the Xmas holidays. He's known for (a) the Mersenne Prime posters with 1 point type; (b) the "irrational radix" variation of FFT multiply that GIMPS uses; (c) some serious work on computing zeta function values; and (d) the NEXT elliptic curve patent. http://en.wikipedia.org/wiki/Richard_Crandall http://www.perfscipress.com/ his press: posters & some interesting books http://blog.stephenwolfram.com/2012/12/remembering-richard-crandall-1947-20… (The picture with the Wolfram piece is the blog owner, not Crandall.) Rich

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[math-fun] planar graph of US states
by Warren Smith 19 Feb '13

19 Feb '13

you'll have an easier time drawing graph on paper thus verifying/denying planarity, than you will trying to code up planarity algorithm. It'll take <5 minutes. (Jeez.) [And the fraction of, e.g, N vertex planar graphs within N vertex graphs is asymptotically known... almost all graphs are nonplanar...]

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[math-fun] meteorite
by Warren Smith 18 Feb '13

18 Feb '13

A fragment 0.5 meter in size would not be falling at over bullet speed, and would not land explosively; instead it would be at terminal velocity in air, which is pretty low. Subsonic probably. It was ejected from the explosion 30-50 km up, then it fell, slowing down to terminal velocity as it did so. The impact with the ice could then have shattered it. Recall my earlier post about threshold size; much above that size meteors do not lose much speed due to atmosphere hence land explosively; if much below it they lose essentially all speed and end up at terminal velocity in air, which is low (and size dependent) and if small the meteorite will be pretty undamaged.

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[math-fun] Newtrinos
by Warren Smith 18 Feb '13

18 Feb '13

The neutrino experiments MiniBooNE http://en.wikipedia.org/wiki/MiniBooNE http://www-boone.fnal.gov/about_boone/interestingfacts.html and its predecessor LSND http://en.wikipedia.org/wiki/LSND find evidence conflicting with the usual view that there are 3 neutrino flavors. Instead, they think there must be an extra two "sterile" neutrino types and with comparatively large mass for at least one, namely over about 1eV. This would be an excellent detection of the long-sought "dark matter" except that 1eV is too large to be compatible with cosmological observations, i.e. there would be *too much* dark matter. Oops. Also, these detectors seem to think neutrino flavor oscillations behave much differently than for antineutrinos, which also would be a massive shock to physics ("CPT violation" claimed, which I find absolutely unbelievable, so they are probably wrong, at least about that) plus a large "matter/antimatter asymmetry." I got this from this Los Alamos magazine article http://www.lanl.gov/science/1663/january2012/story2full.shtml it would be nice to know more... -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Gamma product request, inclusion-exclusion & other residue tricks, and readability
by Bill Gosper 18 Feb '13

18 Feb '13

WDS> So is what Gosper now wants, this?: PRODUCT(over n & k with n>0, 0<k<n, gcd(k,n)=1)OF (z-k/n) or is it this?: PRODUCT(over k with 0<k<n, gcd(k,n)=1)OF (z-k/n)! Anyway, about inclusion exclusion and other games with residues mod n: A. I recently rediscovered, which was known to Gauss, the facts that the residues mod n which are relatively prime to n, (1) form a multiplicative abelian group, (2) which is cyclic if and only if n is odd-prime-power, 2 times odd-prime-power, or n=2 or n=4, (3) which is "bicyclic" if n>4 is a power of 2, that is, all such residues are powers of 5 times +1 or -1, and the powers of 5 times +1 constitute a cyclic subgroup of half the full size, where full size of course is EulerTotient(n). These facts often allow interesting results to be got for these "nice" n. B. Assume n is squarefree for simplicity. Let "weight(x)" be the number of primes in x's factorization. Use the fact that SUM(0<=k<=n) Binomial(n,k) * (-1)^k = (1-1)^n = 0. Conclude SUM(divisors d of n) (-1)^weight(d) = 0. Also SUM(divisors d of n with d>1) (-1)^weight(d) = -1. The application of this is SUM(k with 0<k<n and gcd(k,n)=1)OF something = SUM(divisors d of n) (-1)^weight(d) * SUM(k with 0<k<n and d divides k) something C. Products are sums if you use logs. That is, B and C allow you to do inclusion-exclusion where everything involved is a sum over an additive subgroup of residues mod n, i.e. nice. The function (-1)^weight(d) more generally can be replaced by MobiusFunction(d) if we allow n that are not squarefree -- that will work for many uses. <wds --------------------- We never actually understood this, and had to rediscover it ourselves: gamid[n_Integer] := Product[Gamma[z + k/n], {k, Select[Range[n], CoprimeQ[#, n] &]}] == Product[(k^(1/2 - k z) (2 \[Pi])^(1/2 (-1 + k)) Gamma[k z])^ MoebiusMu[n/k], {k, Divisors[n]}] Test: In[38]:= gamid/@{4,6,8,9,10,12,15,16} Out[38]= {Gamma[1/4+z] Gamma[3/4+z]==(2^(3/2-6 z) \[Pi] Gamma[4 z])/Gamma[2 z],Gamma[1/6+z] Gamma[5/6+z]==(2^(1-4 z) 3^(-3 z) \[Pi] Gamma[z] Gamma[6 z])/(Gamma[2 z] Gamma[3 z]),Gamma[1/8+z] Gamma[3/8+z] Gamma[5/8+z] Gamma[7/8+z]==(2^(5/2-16 z) \[Pi]^2 Gamma[8 z])/Gamma[4 z],Gamma[1/9+z] Gamma[2/9+z] Gamma[4/9+z] Gamma[5/9+z] Gamma[7/9+z] Gamma[8/9+z]==(8 3^(1/2-15 z) \[Pi]^3 Gamma[9 z])/Gamma[3 z],Gamma[1/10+z] Gamma[3/10+z] Gamma[7/10+z] Gamma[9/10+z]==(2^(2-8 z) 5^(-5 z) \[Pi]^2 Gamma[z] Gamma[10 z])/(Gamma[2 z] Gamma[5 z]),Gamma[1/12+z] Gamma[5/12+z] Gamma[7/12+z] Gamma[11/12+z]==(2^(2-12 z) 3^(-6 z) \[Pi]^2 Gamma[2 z] Gamma[12 z])/(Gamma[4 z] Gamma[6 z]),Gamma[1/15+z] Gamma[2/15+z] Gamma[4/15+z] Gamma[7/15+z] Gamma[8/15+z] Gamma[11/15+z] Gamma[13/15+z] Gamma[14/15+z]==(16 3^(-12 z) 5^(-10 z) \[Pi]^4 Gamma[z] Gamma[15 z])/(Gamma[3 z] Gamma[5 z]),Gamma[1/16+z] Gamma[3/16+z] Gamma[5/16+z] Gamma[7/16+z] Gamma[9/16+z] Gamma[11/16+z] Gamma[13/16+z] Gamma[15/16+z]==(2^(9/2-40 z) \[Pi]^4 Gamma[16 z])/Gamma[8 z]} With respectful thanks nevertheless, --rwg wds> Gosper has an annoying habit of using non-ASCII characters I cannot read, which come out as "?" on my screen. Or maybe they really are "?". Who knows? hmm, that might explain why I'm not getting any answers. Off list I've been merrily exchanging emails with Chinese and all sorts of exotica from the Special Characters menu. But this time the culprit is not math-fun--I can see the Uppercase Gamma in my Subject summary window, where there's a ? in Warren's reply. Who else is having this problem? Was anybody besides Warren pwned by my BLACK RIGHT-POINTING SMALL TRIANGLE (?), (which GMail has suddenly stopped inserting on double-clicking the font table)? Maybe Warren is alone on a network made of juice cans and string? Anyway, he's right. I should've stipulated 0<k<n under the product. But he might have guessed my intent from the n=12 example, which is plain ASCII, except for the pi. Haven't we been using ? quite a while now? It's just option p on the Mac keyboard. --rwg --Your ? here came out as ? on my reader. I have a mac. When I type "option-p" into this here window, I do not get pi. I get p. I just did it. Gosper's posts often are hard for me to read, they often have had wrong line-breaks and unreadable characters combined. The solution is to post using only ASCII characters and manually placing all your line breaks exactly where they should be, using fairly short lines only --not depending on tools to insert them for you (since they may place them wrong). If you do that, everybody can read it. If you do not, some people will have trouble reading it. Your choice. <WDS ω. ⊅ε®рιи₲τοи ∫⋔ιθ ϚϘ∪∃∀⋔∫ ⊙∬⎮₣α₲εϩļ

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[math-fun] meteorite lands near Chelyabinsk Russia, causes injuries
by Warren Smith 17 Feb '13

17 Feb '13

Allegedly over 900 sought medical help and a few hospitalized, but nobody killed. Supposedly Russian air defense systems shot it down (!). http://rt.com/news/meteorite-crash-urals-chelyabinsk-283/ http://www.usatoday.com/story/news/world/2013/02/15/russia-meteorite/192199…

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[math-fun] Reverse is harder?
by Warren Smith 16 Feb '13

16 Feb '13

True, proving P unequal NP would be a needed first step to prove reverse is superpolynomially harder like cryptologists want. So... let us lower our goals... Can we prove for some problem, that the forward computation is some polynomial, while the backward is some higher-degree polynomial (at least)? Or... can we even prove a logarithmic gap -- I already proved one, but only for parallel computing; I want now a gap for ordinary sequential compute time. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Sums of two cubes
by Allan Wechsler 16 Feb '13

16 Feb '13

I am sure I am very late to this party, but I have been looking at positive sums of two (not necessarily positive) cubes, catalogued on OEIS as A045980. The corresponding set for squares is easily characterized in terms of prime factorizations: the exponents of primes of the form 4k-1 must be even. (I hope I have that characterization right; if it's not, it's close.) The experience for sums of two squares inspires me to factorize the first few dozen entries in A045980. The result is very suggestive, about as suggestive as it's possible to be with actually, you know, *suggesting* anything. The most striking feature is a marked preference for primes of the form 6k+1. Of course if a number n is the sum of two cubes, then nm^3 is also, for all integers m. But the converse is not true. For example, 1^3 + 7^3 = 344 = 2^3*43, but 43 is not the sum of two cubes. The property is not multiplicative: for example, 2 and 7 are sums of two cubes, but 14 isn't. That having been said, it displays frustrating *hints* of multiplicativity. After staring at these numbers for a couple of days, I have come up with a hard question related to this spooky para-multiplicativity. For any k, let Ck be the set of integers n such that kn is the sum of two squares. C1, for instance, is A045980 itself. C2 is {1, 4, 8, 13, 14 ...}, which is not in OEIS. (In fact, none of the Ck's I looked at were in OEIS except C1.) Here is my question: do there exist *any* two integers j and k such that Cj = Ck?

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[math-fun] Reverse is harder?
by Warren Smith 15 Feb '13

15 Feb '13

Peter Ryan: If you have a list of randomly ordered strings we have the following provable asymmetry: 1) choosing an element at random 2) finding a given string in the set.

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