September 2011 - mathfun@mailman.xmission.com

[math-fun] sextics and septics
by Bill Gosper 12 Sep '11

12 Sep '11

Julian changed 5 to 6 in my quintic solver and now it solves (solvable) sextics, without even changing the internal Solve to be the quintic solver! So I changed 6 to 7 and solved (eq.2) from Tito's -------------------------------- 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). [...] ------------------------ But instead of Montgomery's elegant solution featuring 29th roots of unity, I got many pages of septic surds of sextic surds of quadratic surds, implying improbable-looking identities. I don't expect these Montgomery-style elegances to apply very often, so mountains of radicals may be the general rule. But when I bumped 7 to 8 and tried (eq.1), I couldn't find a septic resolvent, even using 999 digits of WorkingPrecision. ? Even if this gets fixed, it has no hope of recovering Tito's 16th degree resolvent, due to a small matter of 17! possibilities. Julian also showed me an elegant way (satfun) to condense all n solutions of the nth degree eqn into a single f(k mod n). Here's the sextic solver: satfun[L_List] := Block[{n = 1 + Length[L[[1]]]}, Do[If[GCD @@ (Arg[v1/v2]/\[Pi]) == 2/n, Return[v2*(v1/v2)^#]], {v1, L}, {v2, L}]] (sixto5th = Exp[2*I*\[Pi]* Flatten[Outer[List, Range[0, 5], Range[0, 5], Range[0, 5], Range[0, 5], Range[0, 5]], 4]/6]) // Short Ssext[q_] := Block[{rts = #[[1, 2]] & /@ NSolve[q == 0, WorkingPrecision -> 666], boa, foos, S4 = Permutations[Range[4]]}, foos = InverseFourier[ rts[[#]]/Sqrt[6]]^6 & /@ (Join[{}, #] & /@ (Permutations[ Range[6]])); foos = (Together[#[[1, 2]]]^(1/6) &) /@ Solve[0 == Rationalize[ Expand[Times @@ (# - Rest[foos[[Ordering[ Denominator[Rationalize[Plus @@ #, 9.^-666]] & /@foos, 1][[1]]]]])]]]; boa = Rationalize[Plus @@ rts/6]; Evaluate[boa + foos.satfun[ Select[sixto5th, MemberQ[Chop[#.foos + boa - rts], 0] &]]] &] E.g.; In[78]:= MinimalPolynomial[2^(1/3) + Sqrt[2], x] Out[78]= -4 - 24 x + 12 x^2 - 4 x^3 - 6 x^4 + x^6 In[79]:= Ssext[%] Out[79]= -(-1)^#1 Sqrt[2] + 2^(1/3) E^((2 I \[Pi])/3) (E^(-((2 I \[Pi])/3)))^#1 & In[80]:= % /@ Range[6] Out[80]= {2^(1/3) + Sqrt[2], -Sqrt[2] + 2^(1/3) E^(-((2 I \[Pi])/3)), Sqrt[2] + 2^(1/3) E^((2 I \[Pi])/3), 2^(1/3) - Sqrt[2], Sqrt[2] + 2^(1/3) E^(-((2 I \[Pi])/3)), -Sqrt[2] + 2^(1/3) E^((2 I \[Pi])/3)} In[81]:= MinimalPolynomial[Cos[\[Pi]/21]] Out[81]= 1 + 16 #1 + 32 #1^2 - 48 #1^3 - 96 #1^4 + 32 #1^5 + 64 #1^6 & In[82]:= Ssext[%@x] Out[82]= -(1/12) - 1/4 (-1)^#1 Sqrt[7/3] + ((7/2 (71 + 39 I Sqrt[3]))^(1/6) E^(( 2 I \[Pi])/3) (E^(-((I \[Pi])/3)))^#1)/( 4 Sqrt[3]) + ((7/2 (71 - 39 I Sqrt[3]))^(1/6) E^(-((2 I \[Pi])/3)) (E^((I \[Pi])/3))^#1)/(4 Sqrt[3]) + 1/12 7^(1/3) (-(1/2) I (-13 I + 3 Sqrt[3]))^(1/6) E^((2 I \[Pi])/ 3) (E^(-((2 I \[Pi])/3)))^#1 + 1/12 7^(1/3) (1/2 I (13 I + 3 Sqrt[3]))^(1/6) E^(-((2 I \[Pi])/3)) (E^((2 I \[Pi])/3))^#1 & In[83]:= Rationalize[N[ArcCos[% /@ Range[6]]/\[Pi], 69]] Out[83]= {1/21, 11/21, 5/21, 13/21, 17/21, 19/21} --rwg scaly anteater's transacetylase labor-intensive inventoriables

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Re: [math-fun] does anyone recognize; exercises
by N. J. A. Sloane 11 Sep '11

11 Sep '11

Rich said: > I suggest abandoning the distinction between student/teacher for purposes of access. I agree! The OEIS has always been completely open to everyone to consult - like a book in a library. And one of its explicit goals is to help people do IQ tests. I recommend that any index to problems follow the same policy. Neil

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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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Re: [math-fun] does anyone recognize; exercises
by Dan Asimov 11 Sep '11

11 Sep '11

This sounds like a great idea but there's a potential drawback: It would be shame if a lot of old chestnut problems that are great the first time you encounter them were listed with their answers easy to find, so that newcomers to these problems might be too tempted to just look up the answer instead of working on them. (Or does Google already make this the case?) --Dan Rich wrote: << While we're on the subject of math-related databases, I'll mention another collection we need: Worked Problems & Puzzles. Mathists generate and solve a lot of problems. Most of these are not profound enough to publish. It would still be worthwhile to collect them, to smooth the path for followers. Collecting is the easy part; the hard task is organization & making it searchable. For example, some of the geometry problems we've discussed over the years: How to make them searchable beyond adding keywords? Searching for "ellipse intersection" is too unspecific. >> Sometimes the brain has a mind of its own.

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[math-fun] Tilings of hyperbolic space
by Jason 10 Sep '11

10 Sep '11

http://bulatov.org/math/1101/Bulatov%20V.%20(2011)%20Tilings%20of%20the%20h…

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Re: [math-fun] Bealesque identities
by Adam P. Goucher 09 Sep '11

09 Sep '11

> Dear Adam Goucher, > Although my messages aren't reaching math-fun > (except by courtesy of Dan Asimov), they have generated quite a > bit of correspondence. I don't think that your proof is correct. I realised that several minutes later: I merely proved the existence of an inconic, rather than a circumconic. Sorry! How embarrassing... Anyhow, coupled with the original (unproven) theorem, this means that the hexagon formed by the incentres belongs to an infinite family of hexagons as a result of Poncelet's Porism. (You're the same Richard Guy who co-authored with Conway and Berlekamp, I presume?) Sincerely, Adam P. Goucher

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Re: [math-fun] Bealesque identities
by Dan Asimov 09 Sep '11

09 Sep '11

Richard Guy, who is having trouble posting to math-fun, sent this message: -------------------------------------------------------------------------- Apologies for butting in, but my messages to math-fun are bounced back by a spam filter, so I don't know how to send. What I wanted to say was: The following problem was given to me by Julian Salazar, a high school student, who said he got it second-hand from members of the US IMO team. There are several subsidiary problems. The medians of a triangle divide it into six smaller triangles. Show that the six incentres lie on an ellipse. 1. Who originated the problem? 2. Did he/she have a proof? 3. Experiment suggests that if the three ``cevians'' are drawn through any point, not only the centroid, then the result is still true, except that if the point is outside the triangle, then the conic is a hyperbola, and if the point is on an edge of the triangle, the conic is a pair of straight lines. 4. In the original problem, is the centre of the ellipse among Clark Kimberling's three-thousand-odd triangle centres? 5. Same question as 4., but taking the defining point to be other of the more familiar triangle centres, besides the centroid. 6. If we apply Conway's extraversion to the problem, are there any selections of six excentres which lie on a conic? I must be getting old, because any attempt at finding a proof bogs down in a morass between Euclidean and projective geometry. Or, just possibly, it isn't true?!? Happy new academic year to all. R. -------------------------------------------------------------------------- Sometimes the brain has a mind of its own.

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Re: [math-fun] about 10000 numbers, a certain document from M.I. Shamos.
by N. J. A. Sloane 08 Sep '11

08 Sep '11

I am on vacation so I am not reading emails in the order in which they were received! Concerning Professor Shamos's table of numbers, a few years ago, when I was visiting Pittsburgh, we agreed that it would be nice to include them in the OEIS. As for the question of decimal expansion versus continued fraction expansion, the OEIS policy (written down in one of the old help pages) is that we like to get both! the decimal expansion is often more convenient for human use, but the continued fraction is not base dependent so is more satisfactory to a mathematician. Neil

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Re: [math-fun] about 10000 numbers, a certain document from M.I. Shamos.
by Dan Asimov 08 Sep '11

08 Sep '11

I pretty much agree with Marc et al. that the decimal digits of an interesting number aren't usually a natural way to describe that number, so perhaps they do deserve their own database, of which Shamos's paper could be a start. By the way, does anyone know how to search a .pdf document for mathematical symbols? --Dan Sometimes the brain has a mind of its own.

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Re: [math-fun] about 10000 numbers, a certain document from M.I. Shamos.
by Bill Gosper 08 Sep '11

08 Sep '11

mlb> * And of course the formulas shouldn't be buggy--yet already folks have pointed out errors in this data set. Indeed this book seems to have a distressing density of errors. E.g., in just the equations for 0, the !! sum is wrong, and the following one doesn't even converge. Not wrong, but the 1st and 3rd sums are just special cases of 0=sum(a^k*(k-a)/k!,k>=0). And the sin^3 sum is a special case of the Fourier series for a step function that spends 1/3 of its time at 0. And Neil just noticed the dx is missing from the integral. And that typographical anomaly in the 2nd formula for 1 is not 2k^2-1. Etc. --rwg

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