March 2013 - mathfun@mailman.xmission.com

[math-fun] This is how you solve a Rubik's Cube
by Robert Baillie 04 Mar '13

04 Mar '13

http://www.wimp.com/solvecube/

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[math-fun] Mathematica help: formula manipulation
by Bill Gosper 04 Mar '13

04 Mar '13

Despite its bulk, OEIS's Fibonacci article (A000045) could use another formula or two. How about after FORMULA G.f.: x / (1 - x - x^2). F(n) = ((1+sqrt(5))^n-(1-sqrt(5))^n)/(2^n*sqrt(5)). Alternatively, F(n) = ((1/2+sqrt(5)/2)^n-(1/2-sqrt(5)/2)^n)/sqrt(5). F(n) = F(n-1) + F(n-2) = -(-1)^n F(-n). F(n) = round(phi^n/sqrt(5)). ... we add the first order recurrence F(n+1) = F(n)/2 + √((-1)^n + 5 F(n)^2/4), n>=0 ? And for laughs, the World's shortest fibula, F(n+1) = U_n(i/2)/i^n , (U:= Chebyshef 2nd kind) Deriving the former, I got perfectly frustrated by Mathematica. It knows In[267]:= Simplify[(-1)^(2*n), n \[Element] Integers] Out[267]= 1 >From In[232]:= 1 + (-1)^n Fibonacci[n]^2 + (-1)^n Fibonacci[n] Fibonacci[1 + n] = = (-1)^n Fibonacci[1 + n]^2 all I wanted was to In[268]:= Assuming[n \[Element] Integers, Solve[%232, Fibonacci[n + 1]]] Out[268]= {{Fibonacci[ 1 + n] -> -(1/2) (-1)^-n (-(-1)^n Fibonacci[n] - I^n Sqrt[4 + 5 (-1)^n Fibonacci[n]^2])}, {Fibonacci[ 1 + n] -> -(1/2) (-1)^-n (-(-1)^n Fibonacci[n] + I^n Sqrt[4 + 5 (-1)^n Fibonacci[n]^2])}} Who ordered i^n? Cosmetics aside, which is the correct root? It depends on n mod 8 ! (n>0). Um, returning to cosmetics, how to punt i^n? In[273]:= Assuming[ n \[Element] Integers, %268 /. x : I^n*Sqrt[_] :> Sqrt[Simplify[Expand[x^2]]]] Out[273]= {{Fibonacci[ 1 + n] -> -(1/2) (-1)^-n (-(-1)^n Fibonacci[n] - Sqrt[ I^(2 n) (4 + 5 (-1)^n Fibonacci[n]^2)])}, {Fibonacci[ 1 + n] -> -(1/2) (-1)^-n (-(-1)^n Fibonacci[n] + Sqrt[ I^(2 n) (4 + 5 (-1)^n Fibonacci[n]^2)])}} OK, the big gun: Assuming[n \[Element] Integers, %268 /. x : I^n*Sqrt[_] :> Sqrt[FullSimplify[Expand[x^2]]]] {{Fibonacci[1 + n] -> -(1/2) (-1)^-n (-(-1)^n Fibonacci[n] - Sqrt[ LucasL[n]^2])}, {Fibonacci[1 + n] -> -(1/ 2) (-1)^-n (-(-1)^n Fibonacci[n] + Sqrt[LucasL[n]^2])}} Wise ass. I got fairly close with In[288]:= %268 /. x : I^n*Sqrt[_] :> Sqrt[Assuming[n \[Element] Integers, Simplify[ComplexExpand[x^2]]]] Out[288]= {{Fibonacci[ 1 + n] -> -(1/ 2) (-1)^-n (-(-1)^n Fibonacci[n] - Sqrt[(-1)^n (4 + 5 (-1)^n Fibonacci[n]^2)])}, {Fibonacci[ 1 + n] -> -(1/ 2) (-1)^-n (-(-1)^n Fibonacci[n] + Sqrt[(-1)^n (4 + 5 (-1)^n Fibonacci[n]^2)])}} but never got that radicand to simplify, except by manually forcing (-1)^(2n)->1. --rwg

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Re: [math-fun] Maxima help: polynomial manipulation
by Bill Gosper 04 Mar '13

04 Mar '13

MRob>I think a few people here are familiar with Maxima. I'd also be perfectly happy with a solution in PARI or in Sage, if that turns out to be a better way of doing this. I've built up a sum like this: 1.2345/(x+1) + 2.3456/(x+2) + 3.4567/(x+3); (I won't show the output, since the alignment is funny, and if there's a way to turn off Maxima's ASCII pretty-printing please tell me) Then I want to "ratsimp" it, to get a ratio of polynomials. If I do this, Maxima gives me: (35184 x^2 + 129625 x + 106786) / (5000 x^3 + 30000 x^2 + 55000 x + 30000) The problem with this is that the imprecise numbers have all been made into integers, and the coefficients in the denominator are huge. What I really want is this: (7.0368 x^2 + 25.925 x + 21.3572) / (x^3 + 6 x^2 + 11 x + 6) I'm trying to prevent the automatic conversion from floats to rational fractions, so as to keep the coefficients in the denominator small. The "ratepsilon" and "bftorat" flags don't seem to help. ----- In Macsyma, block([fancy_display:false],playback(10))$ 3.4567 2.3456 1.2345 (d26) ------ + ------ + ------ x + 3 x + 2 x + 1 (c27) XTHRU(%); (d27) (x + 3) (1.2345 (x + 2) + 2.3456 (x + 1)) + 3.4567 (x + 1) (x + 2) ------------------------------------------------------------------ (x + 1) (x + 2) (x + 3) (c28) MAP(EXPAND,%); 2 7.0368 x + 25.925 x + 21.3572 (d28) ------------------------------ 3 2 x + 6 x + 11 x + 6 (c29) COMBINE(EXPAND(D27)); 2 7.0368 x + 25.925 x + 21.3572 (d29) ------------------------------ 3 2 x + 6 x + 11 x + 6 (c31) BLOCK([KEEPFLOAT : TRUE],RATSIMP(D26)); 2 7.0368 x + 25.925 x + 21.3572 (d31) ------------------------------ 3 2 x + 6 x + 11 x + 6 In 2.4, you can paste a displayed expression into an input section and it will become inputform, modulo a few bugs. Also, string(d26) gives "3.4567/(X+3)+2.3456/(X+2)+1.2345/(X+1)" --rwg PS, you undermoused: 77451915729367 should be 777451915729367 in MRob> This reminds me of something Neil Sloane mentions in his OEIS talks, namely that the function f(n) = floor(2n/log(2)) has the same value as the function g(n) = ceiling(2/(2^(1/n)-1)) (Sloane's sequence A78608) for all integer n from 1 to 77451915729367, but differs at n=77451915729368 (and at some other larger values, seeoeis.org/A129935)

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[math-fun] Maxima help: polynomial manipulation
by Robert Munafo 03 Mar '13

03 Mar '13

I think a few people here are familiar with Maxima. I'd also be perfectly happy with a solution in PARI or in Sage, if that turns out to be a better way of doing this. I've built up a sum like this: 1.2345/(x+1) + 2.3456/(x+2) + 3.4567/(x+3); (I won't show the output, since the alignment is funny, and if there's a way to turn off Maxima's ASCII pretty-printing please tell me) Then I want to "ratsimp" it, to get a ratio of polynomials. If I do this, Maxima gives me: (35184 x^2 + 129625 x + 106786) / (5000 x^3 + 30000 x^2 + 55000 x + 30000) The problem with this is that the imprecise numbers have all been made into integers, and the coefficients in the denominator are huge. What I really want is this: (7.0368 x^2 + 25.925 x + 21.3572) / (x^3 + 6 x^2 + 11 x + 6) I'm trying to prevent the automatic conversion from floats to rational fractions, so as to keep the coefficients in the denominator small. The "ratepsilon" and "bftorat" flags don't seem to help. -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

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Re: [math-fun] max negative charge on an ion -- chemists?
by Warren Smith 03 Mar '13

03 Mar '13

Thanks to E.Salamin for the tip; I then found a few papers A.I. Boldyrev & J.Simons: Theoretical Search for Small Linear Doubly Charged Anions, J. Chem. Phys. 98 (1993) 4745-??. M.K. Scheller, R.N. Compton, L.S. Cederbaum: Gas Phase Multiply Charged Anions, Science 270 (1995) 1160-1166. ALEXANDER I. BOLDYREV, MACIEJ GUTOWSKI, AND JACK SIMONS Small Multiply Charged Anions as Building Blocks in Chemistry, Acc. Chem. Res. 29 (1996) 497-502 http://simons.hec.utah.edu/papers/paperso/201.pdf the conclusion seems to be (based on both experiments and computation) that apparently no N-atom doubly-negatively-charged ions exist in isolation for N=1,2,3, despite fact many such ions exist inside crystals and sometimes in solutions too. The 4-atom (LiF3) 2- is electronically and locally geometrically stable, but not chemically thermodynamically stable. Both 2 and 3 charged negative ion species are known to exist but usually with a lot bigger formulas than you'd think from chemistry textbook negative ion formulas (most of which cannot exist in isolation).

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[math-fun] max negative charge on an ion -- chemists?
by Warren Smith 03 Mar '13

03 Mar '13

Plenty of atoms can be ionized to -1 charge. (I'm not sure if this is true of all of them. E.g. perhaps Helium and other noble gases cannot be; also perhaps Zinc, Magnesium, Beryllium, Nitrogen, Manganese are immune.) But no atom can be ionized to -2 charge -- i.e. no such ion is stable in isolation. There is, in fact, a naive approximation, called Thomas-Fermi-von Weizsacker, to quantum mechanics in which it is a theorem (by Benguria & Lieb) that the max possible negative charge you can put on an N-atom molecule, N>0, is at most N extra electrons. I would conjecture that asymptotically, the most negative charge you can put on an N-atom molecule is going to be constant*sqrt(N). This would be achieved for a hollow metal sphere, regarded as a giant "molecule," with thickness of order 1 atoms, radius r where N is of order r^2; charging to order sqrt(N) would give it a constant electric field at surface. But this conjecture is completely open. QUESTION: what is the most negative charge f(N) that can be put (stably) on an N-atom molecule for small values of N? E.g. sulfite (SO3) or carbonate (CO3) each with -2 charge both show f(4)>=2, assuming this ion actually is stable in isolation in vacuum. Arsenite AsO3 with charge -3 shows f(4)>=3 assuming same. The AsS (charge -3) and allylenide [C=C=C] with charge -4 ions each would be enough to refute that Thomas-Fermi-von Weisacker theorem for actual quantum mechanics, if these ions can exist stably in isolation (which I doubt, but they exist inside some ionic crystals). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Fixed points of Google translate ?
by Bill Gosper 02 Mar '13

02 Mar '13

What prompted me was the implausible Google Translate of the fourth "sentence" of Lorem ipsum,Cum sociis natoque penatibus et magnis from "Latin" into "This helps us prevent automated submissions". Discovering this in a bogus Yahoo News story, http://news.yahoo.com/blogs/the-upbeat/cum-sociis-natoque-penatibus-et-magn… made me think Yahoo had hacked Google Translate into believing this highly idiomatic interpretation. But G-Ting that English back into "Latin" nearly recovers the Lorem ipsum, which then loops. And the first "sentence", "Lorem ipsum dolor sit amet, consectetur adipisicing elit", G-Ts to "This page is half the battle WIN!" Shame on me for imagining Yahoo pulling one on Google. --rwg Mitchell Riley>Here is a site that does English <-> Japanese until a fixed point is found. http://www.translationparty.com/ When we found the site a few years ago, my friend discovered that "Naruto" expanded forever. The step from English to Japanese would give the word both in English and Japanese, so the total length doubled each round trip. The translator isn't fooled by that any more. It looks like it's very nearly fooled by "Naruto Naruto", after a few doublings it is ended by the strange appearance of some other word. On 3 March 2013 00:29, Henry Baker <hbaker1(a)pipeline.com> wrote: A offline comment by Bill G. prompted me to think about the following problem: Let gt(x) be "Google translate" of some corpus x from some language D into some language R. Let gt^-1(y) be the "Google translate" of y in the language R back to the language D. Let rt(x) by the "round trip" translate of x in D to R and back to D. What are the fixed points of rt(x) ? They obviously imply fixed points of gt(rt(x)). It would be interesting to build a simple process to grab some random text from the web & see if rt^n(x) converges to a fixed point for some n. What types of phrases might cause rt^n(x) to "explode" without bound? Are there any "implosions", where rt^n(x) becomes empty? Are there cycles, such that rt^n(x) never converges, but rt^(n+m)(x)=rt^n(x) for some m and for all n>some p ? For a given language, and a word w within that language, there must exist at least one comprehensible sentence containing that word w. Determine the fixed points for each of these sentences.

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[math-fun] Fixed points of Google translate ?
by Henry Baker 02 Mar '13

02 Mar '13

A offline comment by Bill G. prompted me to think about the following problem: Let gt(x) be "Google translate" of some corpus x from some language D into some language R. Let gt^-1(y) be the "Google translate" of y in the language R back to the language D. Let rt(x) by the "round trip" translate of x in D to R and back to D. What are the fixed points of rt(x) ? They obviously imply fixed points of gt(rt(x)). It would be interesting to build a simple process to grab some random text from the web & see if rt^n(x) converges to a fixed point for some n. What types of phrases might cause rt^n(x) to "explode" without bound? Are there any "implosions", where rt^n(x) becomes empty? Are there cycles, such that rt^n(x) never converges, but rt^(n+m)(x)=rt^n(x) for some m and for all n>some p ? For a given language, and a word w within that language, there must exist at least one comprehensible sentence containing that word w. Determine the fixed points for each of these sentences.

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Re: [math-fun] Fixed points of Google translate ?
by Henry Baker 02 Mar '13

02 Mar '13

In a village in which no man shaves himself, who shaves the barber? http://en.wikipedia.org/wiki/Barber_paradox Perhaps I should ask Wolfram Alpha? Alpha Answer: "Wolfram|Alpha doesn't understand your query" At 08:16 AM 3/2/2013, Fred lunnon wrote: > I conjecture that the fixed point is independent of the input, and constitutes >the Question of Life, the Universe and Everything. > > [The answer was earlier conjectured by Adams to be 42, until I pointed out >in a post prompted by Conway that it is actually 24.] WFL > >On 3/2/13, Henry Baker <hbaker1(a)pipeline.com> wrote: >> A offline comment by Bill G. prompted me to think about the following problem: >> >> Let gt(x) be "Google translate" of some corpus x from some language D into some language R. >> >> Let gt^-1(y) be the "Google translate" of y in the language R back to the language D. >> >> Let rt(x) by the "round trip" translate of x in D to R and back to D. >> >> What are the fixed points of rt(x) ? >> >> They obviously imply fixed points of gt(rt(x)). >> >> It would be interesting to build a simple process to grab some random text from the >> web & see if rt^n(x) converges to a fixed point for some n. >> >> What types of phrases might cause rt^n(x) to "explode" without bound? >> >> Are there any "implosions", where rt^n(x) becomes empty? >> >> Are there cycles, such that rt^n(x) never converges, but rt^(n+m)(x)=rt^n(x) for some m and for all n>some p ? >> >> For a given language, and a word w within that language, there must exist at least one >> comprehensible sentence containing that word w. Determine the fixed points for each >> of these sentences.

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[math-fun] telescoping triggish products
by Bill Gosper 02 Mar '13

02 Mar '13

Decades ago in an MIT talk, I gave a method to enumerate a family of telescoping products akin to Product[(1/2)*(1 + x^2^(-n)), {n, Infinity}] == (-1 + x)/Log[x] . (Which telescopes because you can shift the index by changing the x variable.) Recently, Mourad Ismail sent a paper reminding me to carry out the enumeration, on which Neil and I started yesterday. Neil found Product[(1/2)*(z^(-2)^(-k) + 1), {k, 1, Infinity}] == (z - 1)/(z^(2/3)*Log[z]) and fears it to be centuries old, after getting severely scooped on a continued fraction identity last week. I found Product[-(-1)^(2/3) + (-1)^(1/3)*z^(-2)^(-k), {k, Infinity}] == (-1 + (-1)^(2/3)*z)/((-1 + (-1)^(2/3))*z^(2/3)) which looks oscillatory and nonconvergent, but is really just equivalent to Product[2*Sin[Pi/6 + (-1/2)^k*t], {k, Infinity}]==(2*Cos[Pi/6 + t])/Sqrt[3] from http://www.tweedledum.com/rwg/idents.htm . (Nice(?) exercise.) Two of the above identities come from the scheme (a*x^4 + b*x^2 + c)/(c*x^2 + b*x + a) reducing to a polynomial, which Reduce promises contains three more. And there are other schemes... --rwg

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