August 2011 - mathfun@mailman.xmission.com

[math-fun] Machine Go
by rcs＠xmission.com 11 Oct '11

11 Oct '11

Science Daily has a report of a recent success for a Go program. http://www.sciencedaily.com/releases/2009/05/090514083931.htm The program Huygens defeated two professional Go players. The program had handicaps of 6 or 7 stones. Question for the Go players among us: Does a handicap of 6 stones mean that the starting board has 6 stones added for the weaker player, or that 6 is added to the final score? Rich

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[math-fun] The Solvable Octic x^8-x^7+29x^2+29 = 0
by Bill Gosper 18 Sep '11

18 Sep '11

Really From: tpiezas(a)gmail.com Hello all, The octic, x^8-x^7+29x^2+29 = 0 (eq.1) (by Igor Schein) is solvable, but not as easy as merely factoring over a square root extension. Rather, this can be solvable by the 29th root of unity. My solution to (eq.1) is, [tweaked (with permission) by rwg on the basis of numerical evidence: {8 x} = {1-a-b+c+d-e-f-g, 1+a+b+c-d-e-f+g, 1-a+b-c-d+e-f-g, 1+a-b-c+d+e-f+g, 1-a-b-c-d-e+f+g, 1+a+b-c+d-e+f-g, 1+a-b+c-d+e+f-g, 1-a+b+c+d+e+f+g} ] where each {a,b,c,d,e,f,g} = Sqrt[4v_i+1] and the v_i are the 7 roots of the septic, 8903+47647v+39672v^2+7192v^3-522v^4-174v^5+v^7 = 0 (eq.2) The solution of which was given by Peter Montgomery as, v_i = 2(w^11+w^13+w^16+w^18)-2(w+w^12+w^17+w^28)-(w^2+w^5+w^24+w^27)+ (w^3+w^7+w^22+w^26)+(w^4+w^10+w^19+w^25)-(w^8+w^9+w^20+w^21) and one can set w_i = {t, t^7, t^23, t^25, t^16, t^20, t^24}, and t = exp(2Pi*I/29). P.S. Similarly, a solvable 32-deg equation in x can be solved by a 31-deg Lagrange resolvent in z in the form, x = z1^(1/2) +/- z2^(1/2) +/- ... +/- z31^(1/2) though, unfortunately, no explicit examples are yet known. - Tito Piezas III http://sites.google.com/site/tpiezas/ (Check this out! --rwg) [rwg: the septic roots are all real: v->-2 (2 cos((2 π)/29)+2 cos((3 π)/29)+cos((4 π)/29)-2 cos((5 π)/29)-cos((6 π)/29)+2 cos((7 π)/29)-sin(π/58)-sin((3 π)/58)-sin((7 π)/58)+sin((9 π)/58)+sin((11 π)/58)-sin((13 π)/58)) v->2 (cos(π/29)+cos((2 π)/29)-cos((4 π)/29)-cos((5 π)/29)-2 cos((6 π)/29)-2 sin(π/58)-sin((3 π)/58)-sin((5 π)/58)-sin((7 π)/58)-sin((9 π)/58)-2 sin((11 π)/58)+2 sin((13 π)/58)) v->2 (2 cos(π/29)-cos((2 π)/29)-cos((3 π)/29)+cos((4 π)/29)+cos((5 π)/29)-cos((7 π)/29)-2 sin((3 π)/58)-2 sin((5 π)/58)-2 sin((7 π)/58)+sin((9 π)/58)+sin((11 π)/58)-sin((13 π)/58)) v->-2 (2 cos(π/29)-cos((2 π)/29)+cos((3 π)/29)+cos((5 π)/29)+cos((6 π)/29)+cos((7 π)/29)+sin(π/58)-sin((3 π)/58)-2 sin((5 π)/58)-sin((7 π)/58)-2 sin((11 π)/58)+2 sin((13 π)/58)) v->-2 (cos(π/29)+cos((2 π)/29)-2 cos((3 π)/29)-2 cos((4 π)/29)-cos((5 π)/29)+cos((6 π)/29)-2 cos((7 π)/29)+sin(π/58)-sin((5 π)/58)-2 sin((9 π)/58)+sin((11 π)/58)-sin((13 π)/58)) v->2 (cos(π/29)+2 cos((2 π)/29)+cos((3 π)/29)+cos((4 π)/29)-2 cos((5 π)/29)+cos((6 π)/29)+cos((7 π)/29)+sin(π/58)+2 sin((3 π)/58)-sin((5 π)/58)+2 sin((7 π)/58)+sin((9 π)/58)) v->-2 (cos(π/29)-cos((3 π)/29)+2 cos((4 π)/29)-2 cos((6 π)/29)-cos((7 π)/29)-2 sin(π/58)+sin((3 π)/58)-sin((5 π)/58)+sin((7 π)/58)+2 sin((9 π)/58)-sin((11 π)/58)+sin((13 π)/58))]

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[math-fun] does anyone recognize ... ?
by rcs＠xmission.com 11 Sep '11

11 Sep '11

Does anyone recognize the sum (for n>=1) of log(n+1) -------- n! The numerical value is roughly 1.5586704364253894. My attempts to use ISC found no matches, but I'm no expert on looking for non-obvious matches. Rich

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[math-fun] quintic trivia
by Bill Gosper 05 Sep '11

05 Sep '11

"Exhaustive" search (81 cases) sez there are no solvable irreducible quintics with coeffs restricted to {-1,0,1}. After x^5-2 (modulo root negation and reciprocation), the smallest, in some sense, seems to be -1 - 2 x - x^2 + x^3 + x^4 + x^5, whose five solutions are -1/5 +w*u_00+w^2*u_01+w^3*u_10+w^4*u_11, where w = e^(2 pi i n/5), n an integer mod 5, and u_jk := j j k (-1) 1371 - 145 (-1) (sqrt 5) + 65 (-1) sqrt(470 (1 - ------)) sqrt 5 R(-----------------------------------------------------------) 12500 where R is the *real* fifth root of a real: R(#):=#/(#^4)^(1/5). If you use the default complex 5th root, the negative ones will need a correction factor of e^(4 i pi/5). Beware: This is the kind of 5th root they mean in the Wikipedia article<http://en.wikipedia.org/wiki/Quintic_function> and in the examples in the (glitchy) Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf> . (InputForm in case Gmail murdered the above: (1371 - 145*(-1)^j*Sqrt[5] + 65*(-1)^k*Sqrt[470]*Sqrt[1 - (-1)^j/Sqrt[5]])/12500 ). The u_jk^5 are the (real) roots of the quartic 1 + 325793750 u + 4331113281250 u^2 - 41839599609375 u^3 + 95367431640625 u^4 (Recall that the quintic only had 2s and 1s.) So a convenient way to exhibit the solution to an irreducible quintic ax^5+bx^4+... is the constant -b/(5a) + one of these quartics. Recall that quartic solutions in general have quadratic surds inside cubic surds inside quartic surds, so for the quintic we appear doomed to suffer quadratic surds inside cubic surds inside ... inside quintic surds. But no! For some magic Galwological reason, these quartics seem always to solve down to (real!) quadratics in quadratics, at worst. Sometimes, as with -3 + 3 x + 3 x^2 + 3 x^3 + 3 x^4 + x^5, the quartic even factors: (59049 - 12543750 u + 9765625 u^2) (-7776 + 1050000 u + 9765625 u^2). But usually it doesn't even decompose, so your quartic solver needs to be smart. --rwg Is there a name for these nice quartics? They're not usually biquadratic. How about "nice"?-) PS, for some reason, Preview saturates my Mac CPU while trying to scroll the later portions of the Intelligencer article<http://www.math.carleton.ca/%7Ewilliams/papers/pdf/244.pdf>.

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Re: [math-fun] q-trigs with sin^2 + cos^2 == 1
by Bill Gosper 02 Sep '11

02 Sep '11

>rwg> Eqn 6 in the paper is wrong. That's not "Jackson's derivative", unless >> the author means Jesse's or Michael's. > joerg> http://en.wikipedia.org/wiki/Q-derivative says > "The q-derivative is also known as the Jackson derivative." > > (I never heard that name before, though). Look closer at the eqn! --rwg

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Re: [math-fun] q-trigs with sin^2 + cos^2 == 1
by Bill Gosper 01 Sep '11

01 Sep '11

> Joerg> One can regularize to get that one >> (via abs( qexp(I*x) ) == 1 for some nice qexp()). >> The paper >> Jan L.\ Cie\'{s}li\'{n}ski: >> {Improved $q$-exponential and $q$-trigonometric functions}, >> arXiv:1006.5652v1 [math.CA], (29-June-2010). >> URL: \url{http://arxiv.org/abs/1006.5652}.} >> does it for the version where limit_{q-->1} gives >> the usual functions (exp(), cos(), sin()). >> rwg> Caution to the casual: In this paper, superscripts of e, E and Script-E > are *not* exponents. In fact, I don't even see a relation of the form > Script-Exp_q1(x)^2 ~ > Script-Exp_q2(2x), so I don't see much motivation, beyond nice > q-difference and > nice Pythagoras. Can we have q, Pythagoras, *and* identities? Ohh, sort of. Using Cie ́slin ́ski's regularized q-exponential rexpq[q_, x_] -> QPochhammer[-1/2 (1 - q) x, q]/QPochhammer[1/2 (1 - q) x, q] and your bisected product observation (below), i.e., QPochhammer[q x, q^2]^2/QPochhammer[-q x, q^2]^2 == ( QPochhammer[-x, -q] QPochhammer[x, q])/(QPochhammer[-x, q] QPochhammer[x, -q]) we have rexpq[q^2, q x]^2 == rexpq[-q, (-1 + q) x] rexpq[q, (1 + q) x] which, as q->1, becomes (e^x)^2 = ?(-1) * e^(2x), where ?(-1) must be some flavor of 1. (Cie ́slin ́ski doesn't explain negative q.) This could take some getting used to. >[...] >Joerg> The equation in question is >> E(+q^2, -q*x)^2 = E(+q, -x) * E(-q, +x) >> >> I stopped investigating, however, after looking >> into Whittaker/Watson chapter.21 and suspecting >> my finding is really a specialization of stuff >> given there. It's just a bisected product but it might be a key to finding rexpq and regularized q-trig identities . All we do is replace some occurrences of 2 with 1+q, and some occurrences of 0 with 1-q !-) --rwg

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Re: [math-fun] NYTimes: How to Fix Our Math Education
by Dan Asimov 30 Aug '11

30 Aug '11

I don't know. The idea of a bonus for getting all the answers right seems fine, but why not just award that to anyone who gets all the answers right? I once took an epidemiology course from Olli Miettinen at Harvard School of Public Health, where tests were not only true-false +-1, but each question also had a space where you were supposed to enter how sure you are that your answer is right -- a number between 0 and 1. The +-1 score was then multiplied by this factor for each question before the numbers were summed. --Dan Gareth wrote: << On Monday 29 August 2011 17:46:33 Michael Kleber wrote: << Why even bother, with a true/false exam? Only two choices makes it clear that "all wrong" and "all right" are equivalent. >> The 200% thing provides students with a way to say "I am very confident that all my answers are correct". Having that level of confidence and being right presumably correlates with knowing the material very well indeed -- better than can be identified merely by having got everything right, on a true/false exam -- so the exam gains a little bit of extra dynamic range. >> Sometimes the brain has a mind of its own.

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Re: [math-fun] q-trigs with sin^2 + cos^2 == 1
by Bill Gosper 29 Aug '11

29 Aug '11

> * Bill Gosper <billgosper(a)gmail.com> [Aug 29. 2011 16:54]: >> [...] > >> Unfortunately sin^2+cos^2 /=1 in q-land. > Joerg> One can regularize to get that one > (via abs( qexp(I*x) ) == 1 for some nice qexp()). > The paper > Jan L.\ Cie\'{s}li\'{n}ski: > {Improved $q$-exponential and $q$-trigonometric functions}, > arXiv:1006.5652v1 [math.CA], (29-June-2010). > URL: \url{http://arxiv.org/abs/1006.5652}.} > does it for the version where limit_{q-->1} gives > the usual functions (exp(), cos(), sin()). > Caution to the casual: In this paper, superscripts of e, E and Script-E are *not* exponents. In fact, I don't even see a relation of the form Script-Exp_q1(x)^2 ~ Script-Exp_q2(2x), so I don't see much motivation, beyond nice q-difference and nice Pythagoras. Can we have q, Pythagoras, *and* identities? The paper claims application to quantum mechanics and "quantum calculus", so if there is an application it would be another case of "q" being a fortuitous name. (Recall that q-Vandermonde sum for the expected rank of a random bitmatrix with bit probablities p and q.) > (Implementations are given in > http://www.jjj.de/pari/qtrig2r.gpi > the nonregularized ones are in > http://www.jjj.de/pari/qtrig2.gpi ) > > > I recently did essentially the same with the products > prod(n>=0, 1 - x*q^n) and > prod(n>=0, 1 + x*q^n) > > (See http://www.jjj.de/pari/qtrig1r.gpi > the nonregularized ones are in > http://www.jjj.de/pari/qtrig1.gpi ) > [the majority of lines in both files are > relations given as comments; no need to > understand pari syntax] > > Are the latter ones easily related to your q-trigs? Not that I can see--I've never found a convincing q-exponential underpinning. (But I wouldn't swear there isn't one.) Likewise, I haven't found a convincing tan, cot, etc. I should check if qsin(i x) is anything like a sinh. Then we might have qexp:=qcosh+qsinh. I'm unsanguine. --rwg I was hoping to test qsin(7 pi/22) for exact expressibility as a quintic solution, but Julian's result says even lowly q-cos pi/5 is quintic for q>1, only quadratic for q=1. > > For those in qtrig2r.gpi one can happily rediscover > all kinds of modular equations. I was slightly excited > to find a two-parameter version of > Theta4(q^2)^2 = Theta4(q) * Theta3(q) > which, expressed in my q-exp() is > E(+q^2, -q^2)^2 = E(+q, -q) * E(-q, +q) > The equation in question is > E(+q^2, -q*x)^2 = E(+q, -x) * E(-q, +x) > > I stopped investigating, however, after looking > into Whittaker/Watson chapter.21 and suspecting > my finding is really a specialization of stuff > given there. > > > Btw. your paper stops at the prime p=5. > I could, with some effort, get one modular equation > for p=7 (indeed Somos' q56_18_240a) but none for p>7. > > The _only_ thing I rely on is that my program > expressing things as eta-products is working > (and it's hard to imagine things going wrong there), > the rest comes from (the analogs of) most simple > trigonometric identities such as > tan() * cot() == 1. > (That is why I still much like my q-trigs...) > > > cheers, jj > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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Re: [math-fun] NYTimes: How to Fix Our Math Education
by Henry Baker 29 Aug '11

29 Aug '11

Wasn't there a math course at MIT in which you could score 200% if you got _all_ the answers wrong? Needless to say, you had to be pretty sure about your answers before turning in a test booklet like this! At 02:19 PM 8/28/2011, Eugene Salamin wrote: >Here's a news article [ http://news.yahoo.com/ind-vouchers-prompt-thousands-change-schools-17021644… ] with headline "Indiana vouchers prompt thousands to change schools". It looks like, given the freedom to choose and take their money with them, people will desert the public schools. > >Not all families have this freedom. Sometimes, it is reserved for economically or scholastically poor students. Here's an idea. A gifted student will normally strive for 100% on these performance exams that have become so important in recent times. Public schools like smart students since the high grades help their rating score and bring increased funding. But suppose instead these students conspire to achieve 0%. Say there are fifty 5-choice questions, 0.8^50 = 1.4e-5, so there's a subtle message in that zero score. > > -- Gene

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[math-fun] q-trigs with sin^2 + cos^2 == 1
by Joerg Arndt 29 Aug '11

29 Aug '11

* Bill Gosper <billgosper(a)gmail.com> [Aug 29. 2011 16:54]: > [...] > Unfortunately sin^2+cos^2 /=1 in q-land. One can regularize to get that one (via abs( qexp(I*x) ) == 1 for some nice qexp()). The paper Jan L.\ Cie\'{s}li\'{n}ski: {Improved $q$-exponential and $q$-trigonometric functions}, arXiv:1006.5652v1 [math.CA], (29-June-2010). URL: \url{http://arxiv.org/abs/1006.5652}.} does it for the version where limit_{q-->1} gives the usual functions (exp(), cos(), sin()). (Implementations are given in http://www.jjj.de/pari/qtrig2r.gpi the nonregularized ones are in http://www.jjj.de/pari/qtrig2.gpi ) I recently did essentially the same with the products prod(n>=0, 1 - x*q^n) and prod(n>=0, 1 + x*q^n) (See http://www.jjj.de/pari/qtrig1r.gpi the nonregularized ones are in http://www.jjj.de/pari/qtrig1.gpi ) [the majority of lines in both files are relations given as comments; no need to understand pari syntax] Are the latter ones easily related to your q-trigs? For those in qtrig2r.gpi one can happily rediscover all kinds of modular equations. I was slightly excited to find a two-parameter version of Theta4(q^2)^2 = Theta4(q) * Theta3(q) which, expressed in my q-exp() is E(+q^2, -q^2)^2 = E(+q, -q) * E(-q, +q) The equation in question is E(+q^2, -q*x)^2 = E(+q, -x) * E(-q, +x) I stopped investigating, however, after looking into Whittaker/Watson chapter.21 and suspecting my finding is really a specialization of stuff given there. Btw. your paper stops at the prime p=5. I could, with some effort, get one modular equation for p=7 (indeed Somos' q56_18_240a) but none for p>7. The _only_ thing I rely on is that my program expressing things as eta-products is working (and it's hard to imagine things going wrong there), the rest comes from (the analogs of) most simple trigonometric identities such as tan() * cot() == 1. (That is why I still much like my q-trigs...) cheers, jj

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