July 2020 - mathfun@mailman.xmission.com

Re: [math-fun] A COMPLETELY different kind of "squared square"
by Dan Asimov 03 Jul '20

03 Jul '20

Here's a periodic tiling with 3 different-size squares: <https://www.pngkit.com/png/full/462-4628967_a-plane-composed-of-a-series-of…>. —Dan Allan Wechsler wrote: ----- I take back my intemperate guess that a square toroidal dissection into _any_ number of distinct squares is possible. In fact I don't know what the next smallest number of parts is, after 1 and 2. -----

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[math-fun] A COMPLETELY different kind of "squared square"
by Dan Asimov 03 Jul '20

03 Jul '20

A squared square <https://en.wikipedia.org/wiki/Squaring_the_square> is an NxN square that's tiled with smaller KxK squares, where no K occurs twice. I've long wondered about squared *tori*, (using the NxN square torus), which may be easier to find than squared squares. (If you don't require the K to be different, there is an amusing example of squared tori — where the tiles are all 1x1 squares — for any Pythagorean hypotenuse, i.e., any N such that N^2 = A^2 + B^2 for integers A, B.) Amusing because the sides of the tiles are not parallel to those of the NxN square. Can there be a smaller NxN squared torus than the smallest squared square? But how about the famous expression 1^2 + 2^2 + 3^2 + ... + L^2 = N^2 for L = 24 and N = 70, the unique nontrivial such equation. QUESTION: --------- Can a 70x70 square torus be tiled with one KxK square tile for each K in the range 1 ≤ K ≤ 24 ??? Possibly in a non-parallel fashion? —Dan

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[math-fun] A different kind of "square square"
by James Propp 03 Jul '20

03 Jul '20

The title of Eric's post gave me an idea for a different game to play with numbers. The prime p=101030101 could be called a "square prime" (in base ten) because the number of decimal digits of p is a perfect square AND if you write the digits of p in the form of a perfect square you get a pattern with all the symmetries of a square: 1 0 1 0 3 0 1 0 1 There are 44 square primes with 9 digits: 101030101, 101060101, 111181111, 111191111, 121272121, 121282121, 131333131, 141484141, 141494141, 151545151, 151585151, 161636161, 171767171, 181888181, 191939191, 191969191, 303050303, 313111313, 323222323, 323232323, 363646363, 363676363, 363686363, 373717373, 383828383, 383838383, 383888383, 727212727, 727272727, 727282727, 757545757, 757585757, 777767777, 797939797, 797949797, 909070909, 919141919, 919171919, 919191919, 929292929, 959555959, 979797979, 989868989, 989898989 On the other hand, there are only two 9-digit "square squares": 121242121 = 11011^2 and 404090404 = 20102^2. There are no 16-digit square squares (probably for congruential reasons but I don't see what they are; do you?). The are six 25-digit square squares: 1020100000204020000010201 = 1010000000101^2, 1020102020205020202010201 = 1010001000101^2, 1020104040208020404010201 = 1010002000101^2, 1232120202329232020212321 = 1110009100111^2, 1232122422348432242212321 = 1110010100111^2, and 1232124642369632464212321 = 1110011100111^2. (Note the one whose square root is not a palindrome.) Are there infinitely many square squares? Jim Propp

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[math-fun] Better than additive persistence
by Michael Kleber 02 Jul '20

02 Jul '20

I just heard a delightful claim from a work colleague; so far I have no idea how to check or prove it. The "additive persistence" of a number is how many times you need to apply the operation "n -> sum of the digits when you write n in base 10" before you get to a single digit. For example 199 has additive persistence 3, since 199 -> 1+9+9 = 19 -> 1+0 = 10 -> 1+0 = 1. Sometimes you can get to a single digit more quickly if you're allowed to only insert + signs between *some* of the digits in the base-10 writing, rather than all of them. For example, if you're allowed to choose 199 -> 1+99 = 100, then you can get to a single digit in only 2 steps instead of 3. Here's the remarkable claim: By appropriate choice of where to insert + signs, you can always reach a single digit in at most 4 steps! It is extremely not obvious to me that any finite number of steps suffices, much less that it's 4. Right now I believe it only because the person I heard the problem from has a good record of relaying puzzles with correct solutions :-) --Michael -- Forewarned is worth an octopus in the bush.

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Re: [math-fun] Solar panel calculations
by Henry Baker 01 Jul '20

01 Jul '20

Yes, I've been to that 'sunroof' web site, but it doesn't give me any details, so I haven't the slightest idea what calculations it is actually performing. Furthermore, it doesn't know the pitch of my roof, unless it is doing some clever image-matching from different satellite angles. At 08:16 AM 6/29/2020, Thomas Colthurst wrote: >I did this! Well, I and a small team. google.com/sunroof > >It even does the integral over every pixel of your roof, so >as to take into account shading from nearby trees and houses (as >well as non-flat roofs, horizon effects, chimneys and other roof >obstructions, ...). > >You might also like NREL's PVWatts calculator <https://pvwatts.nrel.gov/>, >which is less >accurate (because it doesn't do the full ray-tracing over the roof >thing) but provides more intermediate information. > >-Thomas C > >On Mon, Jun 29, 2020 at 10:52 AM Henry Baker <hbaker1(a)pipeline.com> wrote: >> I'm considering putting a solar PV system on my roof, and I >> thought that I should be able to calculate most of the numbers >> myself from first principles. I have searched for so-called >> online 'calculators', but found that most of them are aimed at >> idiots. >> >> Here's what I expected to find: >> >> A formula giving the exact sun position throughout the year, >> given latitude and longitude, as well as a factor which >> handles the distance from the sun w.r.t. time. >> >> Now this only provides the *input* energy. >> >> I now need to overlay this with average *weather* data >> on a per-hour basis to get the expected net energy. >> >> Finally, I need to *integrate* these data over an >> entire year to get the values for my own house. >> >> There may be such an online calculator, but I haven't >> been able to find it yet. >> >> It may be possible to factor the weather info out >> of the integration, so that I could 'correct' my >> location sunlight & weather data with my own lat/lon >> azimuth (angle of my house) and roof angle, but I'd >> need to see the equations to find out if this factoring >> is possible. >> >> I presume that someone on this list has already done >> these calculations for their own home?

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[math-fun] Volume of a spherical tetrahedron
by Dan Asimov 01 Jul '20

01 Jul '20

I recently became curious to know how to calculate the volume of a regular spherical tetrahedron — that is, a subset of the unit 3-sphere S^3 bounded by 4 spherical triangles, such that all 6 dihedral angles are equal. The 2D case is just a "regular spherical triangle" with its 3 angles equal. It's well known that if the common angle is t, then its area is A(t) = 3t - π. I've perused a few papers on the volume of a spherical tetrahedron, and this is astonishingly more complicated than the 2D case. We can, however use the limiting cases and the regular 4-polytopes to assert a few of its values. Let V(t) denote the volume for dihedral angle = t, and let a = arccos(1/3). Note that 2π^2 is the volume of the entire 3-sphere S^3. Then: V(a) = 0 V(2π/5) = π^2/300 V(π/2) = π^2/8 V(2π/3) = 2π^2 V(π) = π^2 V(2π-a) = 2π^2 If we only knew the *form* of V(t), this might be (more than) enough to determine it for all t. But I don't. —Dan

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[math-fun] A "numerical coincidence"
by Bill Gosper 01 Jul '20

01 Jul '20

ries: Sum[(-4)^n/Binomial[2*n, n], {n, 0, Infinity}] == 1/2 + Log[-1 + Sqrt[2]]/(2*Sqrt[2]) Can Maple do this? Craziness: Plouffe's old Inverter turns it into a much easier arcsinh sum, that Mathematica can do. But Maple seemingly could't when Plouffe tabulated it. —rwg

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Re: [math-fun] A "numerical coincidence"
by Axel Vogt 01 Jul '20

01 Jul '20

> Subject: [math-fun] A "numerical coincidence" > Message-ID: > <CAA-4O0GVEC2VoDb5eGBBJxoYywkO5XX8Bv3bWc2-xOAe35wpFQ(a)mail.gmail.com> > Content-Type: text/plain; charset="UTF-8" > > ries: > Sum[(-4)^n/Binomial[2*n, n], {n, 0, Infinity}] == 1/2 + Log[-1 + > Sqrt[2]]/(2*Sqrt[2]) > > Can Maple do this? sum((-4)^n/binomial(2*n,n) *x^n, n = 0 .. infinity, formal) gives it as sqrt and arcsin, limit x = 1 turns it to almost your result.

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Re: [math-fun] Number of ways to pick 4 points from an m X n grid?
by Keith F. Lynch 01 Jul '20

01 Jul '20

Dan Asimov <dasimov(a)earthlink.net> wrote: > One method I did not see listed in that linked MathWorld article > is to choose the points from a standard normal distribution in the > plane, i.e., with density given by > p(x,y) = (2?)^(-1) exp(-(x^2 + y^2)/2). > This is of course circularly symmetric, but it's not at all clear to > me that four points picked from this distribution will have the same > chance of their convex hull being a quadrilateral as four points > picked from the uniform distribution on a disk. It seems to be different. I've only started, but the first 3 billion give me 0.64905, as contrasted with 0.704480 for a circle and 0.694444 for a square. I'll let it run for a couple days. To make sure I'm generating a standard normal distribution in the plane correctly, please tell me what the means of the r values (distances from the origin) should be. I'm measuring the arithmetic, quadratic ("RMS"), and cubic means. In researching this, I found https://www.jstor.org/stable/pdf/30037457.pdf which says Show that the expected number of vertices of the convex hull of n points sampled from the standard normal distribution in the plane is asymptotically 2 sqrt(2 pi ln n). That doesn't answer your question, but it does answer one of mine (albeit for a different distribution). I guessed, in a post here on Sunday, that it might be proportional to either the square root of n or to the natural log of n. Apparently it's a combination of both. > Affine transformations preserve ratios between areas. Right. And also preserve straight lines. What I didn't realize was that all triangles are affine transformations of each other. I noticed this for rectangles when I realized that changing the aspect ratio of a rectangle in my program would mean multiplying by a number and then immediately dividing by the same number. So the answer has to be the same for all rectangles. Also on Sunday, I posted: For the circular five-point case I calculated 0.3561808623 and posted 0.356181. That page doesn't give the solution. Assuming the solution is in the same form as the circular four-point case, 1 - (i / (j pi^2)), where i and j are integers, (35 and 12 in the 4-point case), if I try 1 - (305 / (48 pi^2)), I get 0.356188312... which differs from my answer by 7 in the sixth digit. Does anyone know if that's the correct formula? That was with about 9 billion sets. With 59 billion (and counting) I get 0.35618796, so it's looking like it really is 1 - (305 / (48 pi^2)), as it differs from that formula by only 3 in the *7th* digit. That's a hell of a way to find a formula.

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