August 2020 - mathfun@mailman.xmission.com

[math-fun] Rolling polyhedra
by James Propp 26 Dec '20

26 Dec '20

I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees. For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is). Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)? Jim Propp

4 10

[math-fun] Chess variant
by David Wilson 13 Nov '20

13 Nov '20

In Paul Morphy's "Opera House" chess game, Morphy never moves a piece backward. I propose Opera House Chess, with initial position and moves as in chess, except pieces cannot move backward. Is it playable? With victory conditions modified?

4 4

[math-fun] One parameter quintic solution
by Bill Gosper 01 Nov '20

01 Nov '20

This is probably ho-hum, but while trying to simplify a Chowla-surdberg, I found that 0 = a + 80 x + 20 x^3 + x^5 has solutions {((-a - Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5), ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), (-a + Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^( 1/5), (-a - Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^(1/5)} Is it from a two-parameter family? C.f. https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics --rwg

1 3

[math-fun] Interesting limit question
by Dan Asimov 05 Sep '20

05 Sep '20

I found this interesting question in an undisclosed location: Let f : Z+ —> R be a function such that a) f(1) > 0 and b) f(n+1) = (1 + 1/f(n))^n for all n >= 1. Is it possible that f(n) —> oo as n —> oo ??? —Dan

6 6

[math-fun] Mersenne conjecture solved?
by Neil Sloane 01 Sep '20

01 Sep '20

There is a claim that there are infinitely many Mersenne primes: Fengsui Liu, <a href=" https://research.journalspress.com/index.php/science/article/view/567">There are Infinitely Many Mersenne Primes</a>, London Journal of Research in Science: Natural and Formal (2020) Vol. 20, Issue 3, Compilation 1.0. The link has been added to sequence A000668, but if the proof is invalid I should add a comment saying so. Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com

4 4

Re: [math-fun] Nested radicals considered harmful
by Dan Asimov 29 Aug '20

29 Aug '20

Of course, for degree ≥ 5, not all algebraic numbers are expressible in terms of iterated +/-/*/÷ and rational powers of integers. This discussion mentions some interesting related issues, in particular that matrices can be used to express algebraic numbers: <https://math.stackexchange.com/questions/657168/understanding-non-solvable-…>. —Dan ----- Maybe it's time to look for some other representation of algebraic numbers, instead of nested radicals? Unnested radicals work reasonably well, with near-canonical forms and tolerable computing efforts to manipulate -- I'm thinking +-*/, although / tends to make messes. And strict canonicalization fails because of a factoring problem: if your expression includes sqrt(499999), you won't know it's also 127sqrt(31) without a hint. But if your expression is sqrt(499999) + sqrt(52111), a gcd will provide the hint, without needing to do any factoring. The usefulness of nested radicals for calculating roots of polynomial equations is questionable: We have the cubic and quartic formulas, but using them to get good numerical results requires care: Knuth points out that even the quadratic formula can have numerical problems. Are we better off with a solid numerical routine? The other point of representations, besides numerics, is the benefit of canonical expressions, providing an at-sight equality check, and helping with simplification by allowing visibly equal terms to be combined, and occasionally cancelled. If you plug the quadratic formula back into the polynomial, you can simplify the result down to 0=0. But if you do the same thing with the cubic or quartic formulas, you need to know some special manipulations to get to 0=0. I don't have a good alternative to propose. My only candidate is things like "solve(x^3 = x+1, near 1.3)", which is about as hard to work with as the cubic formula, and offers no obvious scheme for canonicalization. But the present situation is not good. If you need to invoke Gosper or Ramanujan as part of routine simplification of expressions, you've already lost. -----

3 2

Re: [math-fun] Nested radicals considered harmful
by Henry Baker 29 Aug '20

29 Aug '20

Hi Rich, et al: Do any of these radical formulae work for characteristic not zero? At 11:46 PM 8/28/2020, rcs(a)xmission.com wrote: >Maybe it's time to look for some other representation of >algebraic numbers, instead of nested radicals? > >Unnested radicals work reasonably well, with near-canonical >forms and tolerable computing efforts to manipulate -- I'm >thinking +-*/, although / tends to make messes. And >strict canonicalization fails because of a factoring problem: >if your expression includes sqrt(499999), you won't know it's >also 127sqrt(31) without a hint. But if your expression is >sqrt(499999) + sqrt(52111), a gcd will provide the hint, >without needing to do any factoring. > >The usefulness of nested radicals for calculating roots of >polynomial equations is questionable: We have the cubic and >quartic formulas, but using them to get good numerical results >requires care: Knuth points out that even the quadratic >formula can have numerical problems. Are we better off with >a solid numerical routine? > >The other point of representations, besides numerics, is >the benefit of canonical expressions, providing an at-sight >equality check, and helping with simplification by allowing >visibly equal terms to be combined, and occasionally cancelled. >If you plug the quadratic formula back into the polynomial, >you can simplify the result down to 0=0. But if you do the >same thing with the cubic or quartic formulas, you need >to know some special manipulations to get to 0=0. > >I don't have a good alternative to propose. My only candidate >is things like "solve(x^3 = x+1, near 1.3)", which is about as hard >to work with as the cubic formula, and offers no obvious scheme >for canonicalization. But the present situation is not good. > >If you need to invoke Gosper or Ramanujan as part of routine >simplification of expressions, you've already lost. > >Rich > >---- >Quoting Bill Gosper <billgosper(a)gmail.com>: >>https://en.wikipedia.org/wiki/Tetracontagon >><https://en.wikipedia.org/wiki/Tetracontagon#:~:text=In%20geometry%2C%20a%20….> >>expresses >>cot(?/40) in radicals. >>Hilariously. They reexpress the triply nested radical >>Sqrt[1 + (1 + ?5 + Sqrt[5 + 2 ?5])^2] four different ways, all longer(!), >>and all >>triply nested. In fact, it denests to >> ??5 ?(1 + ?5) + (3 + ?5)/?2. >>?rwg >>https://community.wolfram.com/groups/-/m/t/980264

2 1

[math-fun] Yet another Heighway Dragon theorem. And some Terdragon bits.
by Bill Gosper 29 Aug '20

29 Aug '20

A couple of months ago, Julian found the first half of this identity for the (self avoiding) "median curve" <http://gosper.org/dragmed.png> of the usual dyadic rational sweep over 2^k+1 points. 1/2 (dragun[2^-k n] + dragun[2^-k (1 + n)]) == dragun[2^-k (1/2 - 1/4 (-1)^(1 + n) + n)] == dragun[2^(-3 - k) (-4 + 8 (1 + n) + Sqrt[2] Cos[1/4 (1 + n) π] - 2 Cos[1/2 (1 + n) π] - Sqrt[2] Cos[3/4 (1 + n)π] - (2 + Sqrt[2]) Sin[1/4 (1 + n) π] + 2 Sin[1/2 (1 + n) π] - (-2 + Sqrt[2]) Sin[3/4 (1 + n) π])] I just "conjectured" this 2nd half. The Heighway function never sees the trigs and radicals, which are artifacts of a period 8 perturbation of the sweep parameter. Mandelbrot seemed slightly embarrassed by his kludgy averaging|smoothing recipe for self avoidance. I'll bet he would have been pleased to see it fall out of these simple formulas. Though not nearly as pleased as just being able to exactly evaluate continuous fractal curves anywhere on the continuum. It turns out that the (self-avoidng) median curve of the Terdragon <http://gosper.org/Terdmed.png> is identical to merely phase shifting it by half a step. It's possible that this is a new theorem, since it's not obvious how to take half a step on a triadic fractal. But it's hard to find a nice sampling pattern for the median <http://gosper.org/Terdshftmed.png> curve of the median curve <http://gosper.org/Terdshftmed.png>, because all its points are double! Leaving you with an exponential search (over the two-valued inverses) for a simple pattern. —rwg

2 3

[math-fun] The breadth and depth of radical-denesting ignorance
by Bill Gosper 29 Aug '20

29 Aug '20

https://en.wikipedia.org/wiki/Tetracontagon <https://en.wikipedia.org/wiki/Tetracontagon#:~:text=In%20geometry%2C%20a%20….> expresses cot(π/40) in radicals. Hilariously. They reexpress the triply nested radical Sqrt[1 + (1 + √5 + Sqrt[5 + 2 √5])^2] four different ways, all longer(!), and all triply nested. In fact, it denests to √√5 √(1 + √5) + (3 + √5)/√2. —rwg https://community.wolfram.com/groups/-/m/t/980264

2 1

Re: [math-fun] The breadth and depth of radical-denesting ignorance
by Dan Asimov 29 Aug '20

29 Aug '20

Well, whaddya expect of people who think that 2_C_1 (2 choose 1) is clearer than 2. —Dan Bill Gosper wrote: ----- https://en.wikipedia.org/wiki/Tetracontagon <https://en.wikipedia.org/wiki/Tetracontagon#:~:text=In%20geometry%2C%20a%20….> expresses cot(π/40) in radicals. Hilariously. They reexpress the triply nested radical Sqrt[1 + (1 + √5 + Sqrt[5 + 2 √5])^2] four different ways, all longer(!), and all triply nested. In fact, it denests to √√5 √(1 + √5) + (3 + √5)/√2. —rwg https://community.wolfram.com/groups/-/m/t/980264 -----

1 0