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[math-fun] Another chess variant
by Dan Asimov 24 Nov '18

24 Nov '18

I just saw this a couple of days ago on YouTube: The idea is to pit K of one piece against L of another in the context of avoiding checkmate. Two different pieces, one for each player, are chosen, like K white queens and L black knights. The W and B kings are in their usual starting positions; the rest of the squares on the board are filled up with (say) the K W queens and L B knights. Like (lowercase = B, uppercase = W): n n n n k n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n n Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q Q K Q Q Q (Of course the numbers K of W and L of B pieces need not be of form 8J + 7.) It might be interesting to determine which ratios of which pieces lead to a fairly even game. —Dan

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[math-fun] Ford circles
by Henry Baker 24 Nov '18

24 Nov '18

I just discovered this paper on "Ford circles" (1938) for fractions and continued fractions. Very very cool & elementary. Suitable for high school students. Either https://trungtuan.files.wordpress.com/2017/06/ford1938.pdf or https://www.maths.ed.ac.uk/~v1ranick/papers/ford.pdf

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[math-fun] Hopefully simple question re matrices
by Henry Baker 23 Nov '18

23 Nov '18

Euclidean algorithm / continued fraction on *square matrices* ?? 3 issues: left/right variants; how to stop; how to get "smaller" remainder. Left/right variants: I accept that non-commutativity will produce 2 variants. How to stop: det(remainder)=0 What is the definition of "smaller" in: remainder = dividend - quotient*divisor, such that norm(remainder)<norm(divisor) I.e., what should be the definition of norm(), or are there multiple definitions that work?

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Re: [math-fun] The Case Against Quantum Computing
by Dan Asimov 22 Nov '18

22 Nov '18

I am a complete ignoramus about quantum computing, so this article roughly doubled my knowledge of it. I will say this: That article by Mikhail Dyakonov is *extremely* clear and well-written. —Dan Hans Havermann schrieb: ----- In the IEEE Spectrum last week, Mikhail Dyakonov presented his overview of the field: https://spectrum.ieee.org/computing/hardware/the-case-against-quantum-compu… -----

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[math-fun] Network on sphere puzzle
by Dan Asimov 19 Nov '18

19 Nov '18

Let a network N on the sphere S^2 = {p in R^3 | ||p|| = 1} be defined as a finite set of K >= 2 points in S^2 with some pairs of these points connected by a shortest (or tied for shortest) geodesic. We say a network N is "minimal" if for some epsilon > 0, all networks obtained from N by varying the position of each point by less than epsilon (and connecting each pair of corresponding points by a geodesic) cannot decrease the total length of N (the sum of the lengths of all geodesics). Puzzle: ------- What are the minimal networks on S^2 ? Extra credit:* ------------- What are the minimal networks on the square torus T^2 = R^2 / Z^2 ? —Dan ————— * I don't know the answer to this one yet. But I suspect Neil Sloane does.

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Re: [math-fun] Candela
by Keith F. Lynch 18 Nov '18

18 Nov '18

Simon Plouffe <simon.plouffe(a)gmail.com> wrote: > This is why they decided to make the kilogram definition based > on the Planck constant , there was also a project to define the > kilogram using a specific portion of 1 element based on the mole : > 6.022 x 10^23 atoms, they made 1 prototype of a <perfect> sphere > for it but that idea was abandoned. > Personaly, I prefered the second definition (the sphere), mainly > because it uses Pi. ;-). It does. In the new system, the kilogram depends on Planck's reduced constant, which is defined as being exactly 6.62607015*10^-34/(2 pi) joule seconds. :-)

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[math-fun] Fwd: Rational-coefficient polynomial, applied to integers
by Bill Gosper 17 Nov '18

17 Nov '18

If I were your doctor I would prescribe Mozart. ---------- Forwarded message --------- From: Julian Ziegler Hunts <julianj.zh(a)gmail.com> Date: Sat, Nov 17, 2018 at 8:47 PM Subject: Re: [math-fun] Rational-coefficient polynomial, applied to integers To: Bill Gosper <billgosper(a)gmail.com> Two more obvious errors in (v): that should be a product, not a sum, and there's a fencepost error. The actual formula is product(p^(a(j)(d-jp+1)), j=1,…,l-1)*p^((m-h)(d-lp+1)), which can probably be rearranged to something nicer. Apart from that, I think everything is correct. The faster method I mentioned is the following: if m≤p, then a polynomial which is identically zero mod p^m is a sum of multiples of (x^p-x)^m, p*(x^p-x)^(m-1), …, p^(m-1)*(x^p-x), and this condition can, I think, be checked faster than plugging in (it definitely can for m very small, e.g. m=3). Julian On Sat, Nov 17, 2018 at 1:01 AM Bill Gosper <billgosper(a)gmail.com> wrote: > Man, before falling asleep (like now) I can't even stay focused enough > to make your edits, let alone find bugs. Instead I wake up to find > characters repeated thousands of times. —Bill > > On Sat, Nov 17, 2018 at 12:14 AM Julian Ziegler Hunts < > julianj.zh(a)gmail.com> wrote: > >> Naturally I found an error while struggling to fall asleep: the sum in >> (v) is the number of such polynomials of degree at most d. >> >> I think the (somewhat) simpler/faster methods in (iv) apply up to p^p, >> but fail for p^(p+1), though they may be able to be modified. >> >> Julian >> >> On Fri, Nov 16, 2018 at 11:47 PM Julian Ziegler Hunts < >> julianj.zh(a)gmail.com> wrote: >> >>> You can forward it if you want to. It's not like everything posted to >>> math-fun is correct, and if someone else finds an error or refinement >>> before I do, that's fine with me. >>> >>> Julian >>> >>> On Fri, Nov 16, 2018 at 11:40 PM Bill Gosper <billgosper(a)gmail.com> >>> wrote: >>> >>>> Do you want me to hold off forwarding this to math-fun? —Bill >>>> >>>

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Re: [math-fun] Candela
by Keith F. Lynch 17 Nov '18

17 Nov '18

"Adam P. Goucher" <apgoucher(a)gmx.com> wrote: > There seems to be a lot of recent excitement about the redefinition > of the kilogram, but I'm more worried about the candela. Kilograms are more fundamental. I'm surprised they even keep the candela in SI. Lots of old metric-system units aren't in SI, for instance the calorie. I'm interested in the philosophical implications of the several changes. They aren't just changing the values of the various units, they're subtly changing what they're units *of*. For instance an earlier change in SI had the side effect of defining the vacuum speed of light as a specific constant. Before, people could debate whether that speed was constant, and could measure it. Today, it's constant by definition and any attempt to measure the speed is actually measuring the length of the meter. Since the fine structure constant is dimensionless, SI can't, and didn't attempt to, define its value. Its value depends on the speed of light, Planck's constant, the charge of the electron, and the permeability of space. Suppose that it's discovered that the fine structure constant isn't constant. Is it meaningful to say which of the four things it depends on has changed, or is it completely arbitrary? Previously, the speed of light and the permeability of space had defined values, so we'd have the choice of saying that Planck's constant or the charge of the electron had changed. With this week's change in SI, Planck's constant and the charge of the electron have been given defined values, and the permeability of space has lost its defined value, so we'd be required to say that if the fine structure changes, the permeability of space had changed. Getting back to the kilogram, there are several definitions of mass: Energy divided by the square of the speed of light. Resistance to acceleration (which can be subdivided into resistance in each of the X, Y, and Z directions). Tendency to attract other masses (which again might vary with direction). Tendency to be attracted by other masses (which again might vary with direction). Resistance to torque (which might vary with direction of the axis). Quantity of atoms of a specified type or mixture of types. Formerly the last definition was used, now the first one is being used. General relativity implies that those definitions are all equivalent. SI now implicitly assumes that general relativity is correct. (Nitpick: This week's changes won't actually take effect until next May.) Now, back to the candela: > On the face of it, it seems to have a relatively simple definition: > "The candela is the luminous intensity, in a given direction, of a > source that emits monochromatic radiation of frequency 540*10^12 Hz > and that has a radiant intensity in that direction of 1/683 watt per > steradian." > where luminous intensity is defined as: > I_v = 683 V(gamma) I_e > where I_e is the radiant intensity (in watts per steradian) and > V(gamma) is the 'standard luminosity function' (which must be equal > to exactly 1 when gamma = 540*10^12 Hz, for these definitions to be > consistent). > But this raises the question: what is V(gamma) in general? I've > found a tabulation of values for integer * 10^-9 metre wavelengths > in the visible interval: > http://donklipstein.com/photopic.html > but no indication as to how to compute V(gamma) for any of > the 2^(aleph_null) other wavelengths beside the 401 provided. > Any ideas? It can't be computed, only measured. Unlike other SI units, it's subjective. That table came from asking volunteers which spots of colored light, of various wavelengths and luminous fluxes, appeared brighter. It varies between individuals. The table even mentions that younger people generally see better in the blue. (That's because eye lenses turn yellowish with age, much as many clear plastics do.) It varies strongly with luminous flux (Purkinje effect). And it would be unethical to measure it outside the range of that table, since a light bright enough to be seen might be bright enough to damage the eye. Of course anyone can come up with a heuristic formula based on that table. Or based on another table made by testing a different set of people. At first I thought that that table isn't consistent with SI, since SI implies that V(gamma) is 1 for 540 THz, and the table states that it's 1 for 555 nm, which is not the same. 540 THz is about 555.171 nm, which has a lower value in the table, and 555 nm is about 540.167 THz. But then I realized that I was assuming a vacuum. Obviously, these tests were done in air (again, for ethical reasons). To make those numbers consistent would require an index of refraction of about 1.0003, which is indeed reasonable for typical comfortable temperatures and pressures of air.

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Re: [math-fun] Candela
by Dan Asimov 17 Nov '18

17 Nov '18

I recall that in college ('68-'72) a standard joke when someone asked "What's new?" was to answer "E/h" (since E = h nu describes Planck's constant h (I think) ... where nu is the frequency of a photon. If so, I'd guess that for an arbitrary (constant) frequency nu (OK, gamma), V(gamma) just needs to be multiplied by the ratio gamma / (540*10^12 Hz), or in other words for any frequencies gamma_1, gamma_2: V(gamma_1) / V(gamma_2) = gamma_1 / gamma_2. (If this is dumb, sorry about that.) —Dan Adam Goucher wrote: ----- ... "The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540*10^12 Hz and that has a radiant intensity in that direction of 1/683 watt per steradian." where luminous intensity is defined as: I_v = 683 V(gamma) I_e where I_e is the radiant intensity (in watts per steradian) and V(gamma) is the 'standard luminosity function' (which must be equal to exactly 1 when gamma = 540*10^12 Hz, for these definitions to be consistent). But this raises the question: what is V(gamma) in general? I've found a tabulation of values for integer * 10^-9 metre wavelengths in the visible interval: http://donklipstein.com/photopic.html but no indication as to how to compute V(gamma) for any of the 2^(aleph_null) other wavelengths beside the 401 provided. Any ideas? -----

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[math-fun] Fwd: Rational-coefficient polynomial, applied to integers
by Bill Gosper 17 Nov '18

17 Nov '18

(some trig trivia at end —rwg) ---------- Forwarded message --------- From: Julian Ziegler Hunts <julianj.zh(a)gmail.com> Date: Fri, Nov 16, 2018 at 11:33 PM Subject: Re: [math-fun] Rational-coefficient polynomial, applied to integers To: Bill Gosper <billgosper(a)gmail.com> I'm sure this has already been answered by now, at least the first part, but: (i) Easy proof that if there's an irrational coefficient the polynomial takes irrational values: find a basis over Q, including 1, for the vector space spanned by the coefficients (and 1), then write the polynomial as a linear combination of rational polynomials with coefficients in the basis. An irrational basis element has a nonzero polynomial coefficient, and the value of the polynomial can only be rational when this component is zero, which occurs only finitely many times. (ii) You don't need to test 0 through K-1, just up to the largest prime-power factor of K (minus one), or better yet make it symmetric around zero. (iii) After clearing denominators, you need to check that the polynomial is always zero mod p^m for any prime power p^m dividing K. There are probably general results on how to do this, but there are some obvious tricks to get started: since (x^p-x)^m is always zero, you can mod out by it; the polynomial needs to be a multiple of x(x-1)…(x-p+1) mod p^m, a multiple of x(x-1)…(x-2p+1) mod p^(m-1), …, a multiple of x(x-1)…(x-p^2+1) mod x^(m-p+1), a multiple of x(x-1)…(x-p^2-p+1) mod x^(m-p-1), … . (iv) Further consideration of the above shows that it is necessary and sufficient to test any r consecutive integers, for r large enough that r! is divisible by K. This is, of course, usually much smaller than K. And when it isn't, it's because K has large prime factors with small powers; we can consider those primes separately and use simpler methods which apply to small powers (I think you might be able to make these methods work for larger powers; ask me tomorrow). (v) The above method can give an explicit description of all polynomials of degree at most d which are zero mod p^m; for example, it allows us to count that there are sum(p^(a(j)(d-jp)), j=1,…,l-1)+p^((m-h)(d-lp)), where a(j) is the power of p dividing pj, l=r/p is minimal such that (lp)! is divisible by p^m, and h is the power of p dividing ((l-1)p)!. (I'm not entirely confident about (iv) and (v), and I'd like to double-check them, but I think they should be right.) I think the (somewhat) simpler/faster methods in (iv) apply up to p^p, but fail for p^(p+1), though they may be able to be modified. Julian On Fri, Nov 16, 2018 at 10:06 AM Bill Gosper <billgosper(a)gmail.com> wrote: > On 2018-11-15 15:06, Allan Wechsler wrote: > > For a polynomial P with coefficients in Z, it's trivial that P(n) is > > integer if n is. > > > > The converse is not true. There are lots of polynomials that always give > > integer answers to integer questions, but whose coefficients are not > > integers. For example, n(n+1)/2 = (1/2)n + (1/2)n^2 always takes integer > > values for integer n, even though the coefficients aren't integers. > > > > Is there a way to quickly eyeball a polynomial in general to see if it > is Z > > -> Z? > > > > If the coefficients are rational, one can find K = the LCM of the > > denominators, multiply through by K, and test it for all the integers > from > > 0 to K-1 to see if the result is always divisible by K. But I am hoping > > there is a simpler way. > > > > If any of the coefficients are irrational, my intuition is that the > > polynomial is never Z->Z, but I haven't been able to think of an easy > proof. > > > That's how Macsyma did it, in an internal three-valued Lisp function named > divby1: > (c59) COS(ODD^2*%PI/8); > > 2 > odd > ---- + 7/8 > %pi 8 > (d59) - cos(---) (- 1) > 8 > > (c60) ?LSP("(divby1 '((MEXPT SIMP) $ODD 2)))"); > > (d60) 8 1 true > —rwg > > (c80) SIN(%PI*BERNPOLY(INTEGER,6)/6); 6 5 4 2 2 integer - 6 integer + 5 integer - integer ----------------------------------------------- %pi 4 (d80) sin(---) (- 1) 252 Mathematica partly gets it: In[43]:= FullSimplify[Sin[\[Pi] BernoulliB[6, n]/6], n \[Element] Integers] Out[43]= Sin[1/252 (\[Pi] + 21 (-1 + n)^2 n^2 (-1 + 2 (-1 + n) n) \[Pi])] In[42]:= Table[Sin[\[Pi] BernoulliB[6, n]/6], {n, 9}] Out[42]= {Sin[\[Pi]/252], -Sin[\[Pi]/252], -Sin[\[Pi]/252], Sin[\[Pi]/252], Sin[\[Pi]/252], -Sin[\[Pi]/252], -Sin[\[Pi]/252], Sin[\[Pi]/252], Sin[\[Pi]/252]} Sin[π/7] and Sin[π/9] aren't constructible and thus won't come out in real radicals, but at least you can (c88) HALFANGLES(HALFANGLES(d80)); %pi sqrt(cos(---) + 1) 63 (d88) sqrt(1 - ------------------) sqrt(2) 6 5 4 2 2 integer - 6 integer + 5 integer - integer ----------------------------------------------- 4 (- 1) /sqrt(2) I don't know if Mathematica can partially TrigToRadicals. —rwg (Amazing: OS-X or GMail flags "constructible" and accepted "constructable" for many seconds before flagging that too!)

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