July 2016 - mathfun@mailman.xmission.com

Re: [math-fun] 32-bit computers almost gone
by Henry Baker 06 Jul '16

06 Jul '16

Yes, but... 1. Memory is cheap. Even uP will have decent sized memories. Every delta in line widths produces at least O(delta^2) in memory size, if not O(delta^3). 2. SIMD-type processors are cheap and quite energy-efficient. There's really no reason for a traditional 16-bit processor anymore. 3. Most crypto codes are 1-1 functions, and therefore *reversible*. In theory, reversible functions can be computed with asymptotically zero energy dissipation. At 07:31 AM 7/6/2016, Richard Howard wrote: >BTW, there is an interesting mathy problem here--how do you do security >when you have only a few microjoules and a few kB of memory in a 16 bit >processor? > >RSA is not even remotely feasible.

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[math-fun] 32-bit computers almost gone
by Henry Baker 06 Jul '16

06 Jul '16

FYI -- Does this mean that even cheap microprocessors -- e.g., in the "Internet of Things" -- will be 64 bits? OpenWRT -- open source SW originally used in routers -- is still 32-bits; I wonder how long this will last. https://openwrt.org/ http://www.theregister.co.uk/2016/07/05/linux_letting_go_32bit_builds_on_th… Linux letting go: 32-bit builds on the way out Ubuntu joins calls for users to let go of i386 versions 5 Jul 2016 at 00:56, Richard Chirgwin Major Linux distributions are in agreement: it's time to stop developing new versions for 32-bit processors. Simply: it's a waste of time, both to create the 32-bit port, and to keep 32-bit hardware around to test it on. At the end of June, Ubuntu developer Dimitri Ledkov chipped into the debate with this mailing list post, saying bluntly that 32-bit ports are a waste of resources. ...

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[math-fun] computer word sizes of 2^2^n
by Henry Baker 06 Jul '16

06 Jul '16

Is it just me, or does anyone else find this property pretty cool: (%i1) x^256-1,factor; 2 4 8 16 32 64 128 (%o1) (x - 1) (x + 1) (x + 1) (x + 1) (x + 1) (x + 1) (x + 1) (x + 1) (x + 1) I.e., the factors of 2^2^n-1 are (2^n-1) and (2^2^k+1) for k<n. This property has to be useful for some recursions on computer word sizes, but I don't recall Knuth using it anywhere. Is anyone aware of *any* interesting use of this property?

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[math-fun] back to kindergarten
by Neil Sloane 06 Jul '16

06 Jul '16

Take a sheet of paper, fold it n times (in the naive way), then cut off the 4 corners. How many holes? Answer: https://oeis.org/A274230, surprisingly a new sequence There are some conjectures for a formula which should not be hard to prove? My question is, is there a 3-D analog? Start with a very large soft brick. Dan? Neil

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[math-fun] A001519?
by Zak Seidov 06 Jul '16

06 Jul '16

If a(n)=3*a(n-1)-a(n-2), then (a (n+1) - a (n))^2 = a (n)*(n + 1) - 1. How to prove/disprove it? Thx, Zak https://oeis.org/A001519

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[math-fun] Solve functional equation(?)
by Zak Seidov 04 Jul '16

04 Jul '16

Solve functional equation, for f(x), with any suitable initial values, f (a + b) = (f (a) + f (b))/(1 - f (a)*f (b)). -- Zak Seidov

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[math-fun] ₁Ψ₁ headzup
by Bill Gosper 03 Jul '16

03 Jul '16

http://mathworld.wolfram.com/RamanujanPsiSum.html gives Ramanujan's celebrated closed form with q/a/x and b/a/x printed as q/a x and b/a x . --rwg

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[math-fun] How I miss Mizan Rahman, or, what the heck is this?
by Bill Gosper 02 Jul '16

02 Jul '16

Out[1131]= k q Sum -------------------------------, 1 - 2 k 2 1 + 2 k 2 (1 - q ) (1 - q ) k,-∞,∞ = 1 2 4 (1 + q) QFactorial[-(-), q ] 2 ----------------------------- 2 4 (1 - q ) q-Pochhammers in the sum but not the summand? "q-Mittag Leffler"? It q-retracts to 16 Sum[-----------, {k,-∞,∞}] == 2 2 (1 - 4 k ) 2 2 π Mathematica is dumbstruck. For minutes on end. Are these actually common? Test for q=1/2: In[1132]:= N[Activate[%1131/.q->.5]] Out[1132]= True Term ratio: In[1180]:= DiscreteRatio[%1131[[1,1]],k] Out[1180]= (q (1-q^(1-2 k))^2 (1-q^(1+2 k))^2)/((1-q^(-1-2 k))^2 (1-q^(3+2 k))^2) Nine terms of the sum give 27 term agreement with the product, despite the lack of q^k²: In[1189]:= Series[%1131/.\[Infinity]->9/.-\[Infinity]->-9//Activate,{q,0,29}] Out[1189]= 1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5549 q^28+6114 q^29+O[q]^30==1+5 q+12 q^2+24 q^3+44 q^4+68 q^5+102 q^6+150 q^7+202 q^8+270 q^9+354 q^10+446 q^11+561 q^12+693 q^13+838 q^14+1002 q^15+1195 q^16+1407 q^17+1638 q^18+1906 q^19+2179 q^20+2491 q^21+2842 q^22+3202 q^23+3610 q^24+4042 q^25+4504 q^26+5008 q^27+5550 q^28+6114 q^29+O[q]^30 In[1190]:= Subtract@@% Out[1190]= -q^28+O[q]^30 I got it by mildly abusing the (bilateral) "q-Dixon's theorem" in http://math.sun.ac.za/~hproding/pdffiles/Triples-long.pdf which is not the (unilateral) q-Dixon's I get from path invariant matrices. --rwg

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[math-fun] ambiguous cylinder illusion
by Thane Plambeck 02 Jul '16

02 Jul '16

http://youtu.be/oWfFco7K9v8 if you search youtube for the phrase in the subject line, you'll find some other videos that at least give a hint at what is going on -- Thane Plambeck tplambeck(a)gmail.com http://counterwave.com/

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[math-fun] Puzzle about non-empty sets
by James Propp 01 Jul '16

01 Jul '16

The sum of the elements of the empty set is 0; the product of the elements of the empty set is 1. What is the smallest non-empty set of real numbers that shares this property with the empty set? (What if we require all the numbers to be rational? I haven't figured this part out yet.) Jim Propp

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