April 2013 - mathfun@mailman.xmission.com

[math-fun] "critical exponents" for columns resisting buckling
by Warren D Smith 26 Apr '13

26 Apr '13

Assume a 3D universe. Cut a cylindrical column out of a block of solid material. Apply force F pushing the two ends of the column toward each other. According to Euler, the column will become unstable against "buckling" if F > (pi/L)^2 * Y * I where Y is Young's modulus of elasticity, I is the areal moment of inertia of column cross section, and L is column length. (Euler's analysis is valid for "slender" columns and concerns stability against infinitesimal "sideways" perturbations.) If other convex shapes are permitted besides circular cylinders, the same thing happens except that the constant pi^2 changes. Specifically, Keller claimed to have computed the optimum convex shape for a 3D column (we always assume ball joints at both ends; Keller also examined having "clamped" ends on one or both sides but his analyses of those cases were attacked) Joseph B.Keller: The shape of the strongest column, Arch. Rat. Mechanics Anal. 5,1 (1960) 275-285. The volume of the optimum convex column is 0.866 times the volume of the uniform column of the same strength. It turns out that Keller's result had already been found by Thomas Clausen in 1851. I prefer to re-express the Euler/Clausen/Keller results another equivalent way as: the NUMBER OF ATOMS N of column-material needed to prevent Euler's buckling-instability (as a function of F and L) is N > constant * F^(2/3) * L^(7/3) for a convex column, valid in the "slender" limit where L is large with F growing at most sublinearly as a function of L. [Derivation: multiply both sides of original Euler ineq by (L^3)/R where R=column radius to get F*L*L*L/R > const * R*R*L = const*N, now solve original Euler for R to find R^3 = const*F*L^2, substitute that in to get N<F^(2/3)*L^(7/3) for instability.] Note the interesting exponent 7/3. Now consider the following "Sierpinski fractal" nonconvex column. Start with the union of a regular tetrahedron of edge length L, with balls centered at the 4 vertices, each of radius R1. Now "remove the middle octahedron" from this tetrahedron, leaving 4 half-edgelength regular tetrahedra each containing one of the 4 corners of the original tet, and union the result with balls (centered at the edge midpoints of the original tet) of radii R2 each. Now on each of these 4 remaining tets, do the same "remove middle octahedron" operation, unioning with balls of radii R3 at the joints... and so on. The recursion stops once we reach the atomic length scale, i.e. after lg(L/atomlength) recursive levels where lg(x)=log(x)/log(2). Each recursion (ignoring the balls) removes exactly half the material, so that the amount left after all levels of recursing is a fraction of order atomlength/L. This constitutes a number of atoms of order L^2, not (which it originally was) L^3. We proclaim the two "endpoints" of this "column" to be any two among the 4 vertices of the original tetrahedron and the handy R1-radius balls easily allow considering "ball joints" at them. If the two end-balls are pushed toward each other by epsilon*L, then this would be compatible with their edge-midpoint moving sideways by sqrt(epsilon)*L (ignoring higher order terms since epsilon assumed tiny). This in turn shortens the 4 length=L/2 edges at the next recursive level by an amount of order sqrt(epsilon)*L each, which can be accomodated by sideways motion of their midpoints by amounts of order L*epsilon^(1/4), and so on. The motion by epsilon*L causes an energy-gain of F*epsilon*L. But there are elastic energy expenditures caused by those distortions (i.e. "sideways" motions). We get instability if the gain exceeds the expenditures. Considering the elastic energies for distorting those balls, the expenditures ought to be of order R2^3*epsilon + 4*R3^3*epsilon^(1/2) + 16*R4^3*epsilon^(1/4) + 64*R5^3*epsilon^(1/8) +... independently of L and F (depends only on epsilon and the Rk's), which is upper bounded by R2^3 + 4*R3^3 + 16*R4^3 + 64*R5^3 +... [we assume the Rk are chosen so that this series converges, for example Rk^3 proportional to F*L*5^(-k)] and is lower bounded by epsilon times that. This suggests we ought to get stability if the number N of atoms obeys N > constant * [ F*L + L^2 ] for this kind of "Sierpinski fractal column" when L is large. =============================================================== We can also do the same kind of analysis but in a 2D universe. In that case a 2D Euler would have argued that for a convex column of length L, stability requires N > constant * F^(1/2) * L^2 atoms, valid in the "slender" limit where L is large but with F shrinking toward 0 as a function of L. [Keller's analysis also works in 2D by the way...] In constrast if we use a "Sierpinski triangle fractal" based on starting with equilateral triangle of side length L, remove middle triangle each recursion then adjoin balls of radius Rk at edge-midpoint joints (at recursion level k)... where the Rk shrink geometrically as a function of k and R1^2 is proportional to L*F, then * number of recursion levels is lg(L/atomlength). * number of atoms (ignoring balls) at deepest recursive level is L^2 * (3/4)^#levels which is of order L^1.584962501 where lg(3)=1.584962501. This causes stability (presumably) if N > constant * [ F*L + L^1.584962501 ]. ================================================================ This way of doing things focuses attention on the "critical exponents" which in 3D and 2D (we have shown nonrigorously) are <=2 and <=1.585 respectively; improvements versus Euler/Clausen/Keller's 7/3=2.3333 and 2 respectively. Can these exponents be shrunk by using better fractals? Can they be justified rigorously? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] 19 by 23 stable reflector with 43-tick repeat time
by Adam P. Goucher 26 Apr '13

26 Apr '13

Discovered by MikeP (Playle?) of the conwaylife.com forums a few minutes ago: http://cp4space.wordpress.com/2013/04/26/recent-discoveries-in-conways-life/ Oscillators with periods 43 and 53 now exist, leaving just four unsolved periods... Sincerely, Adam P. Goucher

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[math-fun] A grid with MAX
by Eric Angelini 25 Apr '13

25 Apr '13

Hello MathFunsters, I've seen this on "Le Monde" webpage. http://bit.ly/12RaSEi Draw a 3x3 square and write "1" in the upper left corner-square. Fill now one by one the other squares according to these rules: - select an empty square - write in it the sum of its neighboring squares (a "corner-square" has 3 neighbours, an "edge-square" 5 and the "center-square" 8) - when all squares are marked, record the highest written value. Example: +-----+-----+-----+ | 1 | | | | | | | +-----+-----+-----+ | | | | | | | | +-----+-----+-----+ | | | | | | | | +-----+-----+-----+ +-----+-----+-----+ | 1 | | | | | | | +-----+-----+-----+ | | | | | | | | +-----+-----+-----+ | | | 0 | | | | | +-----+-----+-----+ +-----+-----+-----+ | 1 | 1 | | | | | | +-----+-----+-----+ | | | | | | | | +-----+-----+-----+ | | | 0 | | | | | +-----+-----+-----+ +-----+-----+-----+ | 1 | 1 | | | | | | +-----+-----+-----+ | | 2 | | | | | | +-----+-----+-----+ | | | 0 | | | | | +-----+-----+-----+ +-----+-----+-----+ | 1 | 1 | 3 | | | | | +-----+-----+-----+ | | 2 | | | | | | +-----+-----+-----+ | | | 0 | | | | | +-----+-----+-----+ +-----+-----+-----+ | 1 | 1 | 3 | | | | | +-----+-----+-----+ | | 2 | 6 | | | | | +-----+-----+-----+ | | | 0 | | | | | +-----+-----+-----+ +-----+-----+-----+ | 1 | 1 | 3 | | | | | +-----+-----+-----+ | | 2 | 6 | | | | | +-----+-----+-----+ | | 8 | 0 | | | | | +-----+-----+-----+ +-----+-----+-----+ | 1 | 1 | 3 | | | | | +-----+-----+-----+ | | 2 | 6 | | | | | +-----+-----+-----+ | 10 | 8 | 0 | | | | | +-----+-----+-----+ +-----+-----+-----+ | 1 | 1 | 3 | | | | | +-----+-----+-----+ | 22 | 2 | 6 | <--- MAX value = 22 | | | | +-----+-----+-----+ | 10 | 8 | 0 | | | | | +-----+-----+-----+ My best MAX is 40: is it the highest possible MAX in a 3x3 square? What are, if we consider all nxn squares, the MAX values for the first n? The sequence of such MAX starts with S(max) = 1,4,... then what? This is perhaps old hat, sorry -- but there are too many seqs in the OEIS that start with 1,4,... Best, E.

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[math-fun] Fractal/'supersparse' matrix theory ??
by Henry Baker 25 Apr '13

25 Apr '13

I'm trying to understand how certain fractal structures behave -- e.g., medieval Gothic churches as approximations to fractals, or aerogel-type structures, etc. Typical finite element analysis of bulk structures is 'dense', in that it produces dense matrices. Fractal structures, on the other hand, produce extremely sparse matrices that are barely connected. Is there any branch of matrix analysis that deals with such 'supersparse' matrices (I'll call them that until I find that someone else has already given them a better name) ?? I'm also trying to find out if there are any existing codes for analyzing such structures; I'd be willing to bet that existing codes aimed at dense structures would perform poorly relative to a code that is optimized for fractal structures. Obviously, finite matrix theory can only model truly fractal structures down to some finite level, but perhaps some scheme for recursive block matrices might be the right way to deal with these things.

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[math-fun] cast iron snowflake
by Bill Gosper 24 Apr '13

24 Apr '13

I found an old Macsyma notebook equating the area enclosed by Koch's Snowflake (easy geometric sum) with the area enclosed by my(?) Fourier series for it, using A = π Sum n |c_n|^2. (Which, apparently, not everybody knows: http://www.mathworks.com/matlabcentral/answers/52325-2d-integration-over-an… ) I probably sent this before: Sum[Product[1 - Tan[((k + 1/2)*π)/(-2)^n], {n,∞}]^2/(k + 1/2)^3, {k, -∞,∞}] ==π^3/2 But I don't remember the enormous difficulty in numerically testing it that I now encounter. After a couple of million terms, most are negligible, but those with k of the form, e.g., (-2 - (-2)^j)/3, j>0, are not. But adding these in for 22<j<99 still gives me < two significant figures! Presumably, there are other families of non-negligible k. This is annoying. I have the expansion for "Snowflakes" of any dimension ≤ 2, and was hoping to watch what happens to the enclosed area as the dimension approaches 2 and the order of approximation approaches ∞, and the enclosed area is suddenly all boundary. --rwg

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[math-fun] Modulus Nine
by Hans Havermann 24 Apr '13

24 Apr '13

Start with 0,0,...,0,1 (n-1 0's and a single 1); thereafter the next term is the sum of the previous n terms, modulus 9. The sequence will repeat; for example, n=2: 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 0, 1, … (period 24); and n=3: 0, 0, 1, 1, 2, 4, 7, 4, 6, 8, 0, 5, 4, 0, 0, 4, 4, 8, 7, 1, 7, 6, 5, 0, 2, 7, 0, 0, 7, 7, 5, 1, 4, 1, 6, 2, 0, 8, 1, 0, 0, 1, … (period 39). Periods can get quite large; for instance, n=24 has period 421900912158 but, empirically, periods for n= one-less-than-powers-of-three & powers-of-three appear to be not only more manageable but surprisingly regular: {2,3} -> {24,39} {8,9} -> {240,273} {26,27} -> {2184,2271} {80,81} -> {19680,19929} {242,243} -> {177144,177879} {728,729} -> {1594320,1596513} {2186,2187} -> {14348904,14355471} {6560,6561} -> {129140160,129159849} {19682,19683} -> {1162261464,1162320519} Can anyone come up with an exact formulation for these?

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[math-fun] "Ponder This"
by Keith F. Lynch 23 Apr '13

23 Apr '13

There's a monthly math puzzle at http://researchweb.watson.ibm.com/ponder/ Many of them are quite good. But I found this month's puzzle rather lame. I treated it as a speed programming challenge. It took me 17 minutes to solve it and email them the solution. One can replace the letters in the expression I/M = B/A+C/K with different digits from 1 to 6 to form a valid equation. For example: 5/6 = 2/4+1/3. Substituting back letters to digits, using the same solution ACK.IBM is 413.526. The problem: Replace the letters with different digits such that D/O+N/T = P/E+R/I+S/H and send us PONDER.THIS translated back to a number. This is unsatisfying for three reasons. For one thing, there are multiple valid solutions. Hundreds of them. For another, each letter only occurs once. They might as well have just had ten blanks and said to use each digit once and only once. Finally, rearranging the digits at the end, and sticking a decimal point in, is gratuitous. (I'm also annoyed that they put it online several days early, so I wasn't the first to solve it. They list solvers in order of when they submitted their solutions.) A decent cryptarithm or alphametic has a unique solution and duplicate digits. A classic is SEND + MORE = MONEY. That one dates to before the computer age; you're supposed to use cleverness instead of raw CPU power. (The first step is to realize that when a sum of two numbers is longer than either summand, the leftmost digit must be 1. Hence so is the other M, which then constrains the S, etc.) I decided to come up with a problem along the same lines as theirs, but more elegant, to submit to them for use during a future month. I'd start with PONDER=THIS, and find operations I can stick in such that it will have a unique solution. One example I quickly found was P*ON*DER=T*HIS. Or, if you prefer (as I do), _*__*___=_*___. Then I figured, anything they can do, I can do meta. So I decided to submit, not that problem, but the problem of finding such a problem. Unfortunately, that suffers from the same flaw. With multiplication as the sole operation, there are 12 solutions, of which the above is just one. They all use the same 5 numbers (two single-digit, one two-digit, and two three-digit) in different arrangements. This happens because multiplication is commutative. Actually, that's two operations, since concatenation (gluing two or more digits together to create a multi-digit number) counts. I'm currently looping over all combinations of addition, subtraction multiplication, division, concatenation, and decimal point. An example of this is _*___*_+_=_/_._-_ which unfortunately has multiple solutions. 0*367*8+4=9/1.5-2 is one of the solutions. The precedence is concatenation (including the decimal point) first, then multiplication and division together left to right, then addition and subtraction together left to right. So 3-2+1 means 2, not 0, and 8/4/2 means 1, not 4. I believe this is standard. Looping through operations is more challenging than looping through numbers. I had to write my own parser. To keep it fast and simple, I always alternate between digit and operations. No parentheses, no prefix plus or minus, and I use a character for concatenation rather than just having the digits next to each other. (The comma is suppressed during printing, of course.) The parser rejects invalid arrangements such as division by zero, a multi-digit number with a leading zero not followed by a decimal point, and a number with more than one decimal point. Amusingly, the ASCII characters * + , - . / are consecutive. So if I use the comma for concatenation, I can just loop over the numeric values of those ASCII characters, 42 through 47. I'm not using floating point. I keep track of numerators and denominators. And when determining whether the resulting two fractions are equal, I don't bother with common denominators, I simply multiply the first one's numerator by the second one's denominator and compare that with the product of the second one's numerator and the first one's denominator. If those products are equal, the fractions are equal. Neither do I ever reduce fractions (so 1.2*3.4*5/6 is stored as 2040/600, not 17/5). Why waste the CPU time? The numbers never get big enough to get an overflow error on a 32-bit machine. That's an interesting problem itself: What's the largest numbers I can get in my fractions? Or when cross-multiplying the final two fractions to see if they're equal? What if my equal sign was elsewhere in the equation, e.g. __=________? (Actually, even if I got an overflow error in the final step, does it really hurt anything if it's the *same* overflow error for both cross-multiplies? No, I don't plan to program that riskily, but it's a thought.) I'm not doing exponentiation, and not because the usual exponential symbol, ^, is elsewhere in the ASCII code. Trying to figure out how to represent the value of something like 4^5^6^7^8^9 gives me a (metaphorical) headache. Especially since with exponentiation, precedence is right to left, so that would be a rather large number. Also, if I put in exponentiation, then by symmetry I ought to also put in its opposite. But unlike addition and multiplication it has two opposites, roots and logs. And both of them would require me to deal with irrational numbers. My hope is that I'll find some set of operations that has a single solution. Something like, "Arrange ten blank spaces, an equal sign, three multiplication signs, two decimal points, a plus sign, and a minus sign such that there's only one way to fill in the blanks with unique digits to make the equation correct." The downside is that this will take about a month to run. And that's just for the six-four (______=____) arrangement. If the equal sign can go anywhere (except on the ends, of course), it's more like half a year. Has this already been done? In conclusion, their problem wasn't so lame after all, since it inspired the above. (Unless you consider the above to be lame.)

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Re: [math-fun] "Ponder This"
by Keith F. Lynch 23 Apr '13

23 Apr '13

Allan Wechsler <acwacw(a)gmail.com> wrote: > Dan Asimov <dasimov(a)earthlink.net> wrote: >> ELEVEN + TWO = TWELVE + ONE > Dan, this is just an anagram, right? Unless I'm missing something > truly devilish, it's arithmetically impossible. I've confirmed that there is no solution. Since there are only seven distinct letters, I also tried base 7. Then all bases through 40. Still no solution. (Larger bases are very slow to test.) Would one expect there to be a (possibly non-unique) solution in some base, eventually, just by chance? If so, I propose the opposite challenge -- finding an "anti-alphametic," an arrangement that has no solution in any base. Getting back to my original puzzle under this subject line, I'm about halfway through searching the six-four case, ______=____ with interspersed * + - . / So far I've found about 430,000 arrangements with non-unique solutions and 220 with unique solutions. Of those 220, just 2 have negative values: __*_._-_._=_/__-_ and _*_._/_-_/_=_-__._ Of those 220, 13 contain all of * + - . / and 4 contain all of those once and only once. Those 4 are closely related so I'll only give one of them: __*_.___=_/_-_+_ Again, you're to replace the underscores with each of the ten digits once and only once, and there is a unique solution. I'll re-ask my question: Given that I'm never simplifying fractions, and that I cross-multiply them at the end to check for equality, what are the largest numbers that could appear? Am I at risk for integer overflow? For instance _*_.____=__-_/_ has the unique solution 4*2.8750=13-9/6, which yeilds the fractions 115000/10000 and 69/6. Cross-multiplying those fractions to check if they're equal gives 115000*6 = 10000*69 = 690000.

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Re: [math-fun] math-fun Digest, Vol 122, Issue 28
by Stuart Anderson 22 Apr '13

22 Apr '13

When I tried to add the one letter words 'I' and 'a' to the answer, the wikianswers would not allow me to post, and returned this error message; We could not post your answer because it contained potentially obscene words. > ---------------------------------------------------------------------- > > Message: 1 > Date: Mon, 22 Apr 2013 03:57:00 +1000 > From: Stuart Anderson <stuart.errol.anderson(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] How many different words are there in the word > mathematics? > Message-ID: > < > CA+3-r9NJm6WL6WqJk-WajbkvTuMiSj1787+pwWap31WNobBTJA(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > spoiler below; > > > > > 229 according to > > http://wiki.answers.com/Q/How_many_different_words_are_there_in_the_word_ma… > > > > > > >

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[math-fun] How many different words are there in the word mathematics?
by Stuart Anderson 21 Apr '13

21 Apr '13

spoiler below; 229 according to http://wiki.answers.com/Q/How_many_different_words_are_there_in_the_word_ma…

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