November 2012 - mathfun@mailman.xmission.com

Re: [math-fun] when and why was it agreed that 0.999...=1?
by Henry Baker 14 Nov '12

14 Nov '12

Anyone who's dealt with 1's complement arithmetic (used briefly in some computers in the 1960's) has already faced this issue: 11111.... = 00000... = 0 Perhaps we should bring 1's complement back for pedagogical purposes? There's lots of "redundant" number systems, with 2 or more names of the same value. In fact, deep in computer arithmetic units numbers are "recoded" all the time -- e.g., from digits {0,1}, to digits {0,1,-1}. I don't think that the Romans ever standardized on "IV" versus "IIII". Perhaps there isn't any way of disallowing 0.999..., without screwing everything else up even more. At 08:33 PM 11/13/2012, Gary Antonick wrote: >Hi all, > >I'm wondering if anyone knows when and why it was agreed that 0.999...=1.

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[math-fun] EllipticE valuation [Was: EllipticK valuation [Was: Elliptic K π approximation]]
by Bill Gosper 14 Nov '12

14 Nov '12

On Wed, Nov 7, 2012 at 3:05 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > The ("complete") first kind elliptic integral K(m) and parameter m are > both expressible in > 𝜗 constants, and those three+ 𝜗 constants are all expressible in terms > of Dedekind η. So > > Out[840]= EllipticK[(16 η[q]^8 η[q^4]^16)/η[q^2]^24] == > (π η[q^2]^10)/(2 η[q]^4 η[q^4]^4) > Crude but effective: d/dq and eliminate EllipticK to get EllipticE[(16 \[Eta][q^4]^8)/(\[Eta][q]^8 + 16 \[Eta][q^4]^8)] == -( 1/(\[Eta][q]^8 + 16 \[Eta][q^4]^8))(-((\[Pi] \[Eta][q]^4 \[Eta][q^2]^10)/( 2 \[Eta][ q^4]^4)) - (\[Pi] \[Eta][ q^2]^9 (6 Sqrt[3] q^(3/2) logderiveta[q] \[Eta][q] \[Eta][q^2] \[Eta][q^4] - 30 Sqrt[3] q^(5/2) logderiveta[q^2] \[Eta][q] \[Eta][q^2] \[Eta][q^4] + 24 Sqrt[3] q^(9/2) logderiveta[q^4] \[Eta][q] \[Eta][q^2] \[Eta][ q^4]) (\[Eta][q]^8 + 16 \[Eta][q^4]^8))/(2 q^( 1/3) \[Eta][q]^4 \[Eta][ q^4]^4 (6 Sqrt[3] q^(7/6) logderiveta[q] \[Eta][q] \[Eta][q^4] - 24 Sqrt[3] q^(25/6) logderiveta[q^4] \[Eta][q] \[Eta][q^4])) The remaining problem is those logderivetas, but lo, ((1/(4*x) - ((logderiveta[(1/(E^x))]))/(E^x))/((DedekindEta[((x*I)/(2*Pi))])^4)) is algebraic for (x/π)^2 rational. And for such x, we have many DedekindEta[((x*I)/(2*Pi)) in our collection. In fact, people.math.carleton.ca/~*williams*/papers/*pdf*/*299*.*pdf claims to give a complete solution. Early on, it appears to omit half the odd denominators, but I assume this is remedied later in the derivation. Anyway, as a fairly strenuous example: EllipticE[1/(1 + (-1 + GoldenRatio^(3/2))^8/(16 GoldenRatio^4))] == ( 2 5^(3/4) GoldenRatio^4 π^(3/2))/( Sqrt[1 + Sqrt[5]] (1525 + 682 Sqrt[5])^(1/4) Gamma[1/20] Gamma[9/ 20]) + ((-7 - 3 Sqrt[5] + Sqrt[190 + 85 Sqrt[5]]) Gamma[1/ 20] Gamma[9/20])/( 8 (1525 + 682 Sqrt[5])^(1/4) Sqrt[5 (1 + Sqrt[5]) π]) I'm having no luck finding a table of special values of EllipticE. This may be due to a simpler formula than above to produce them from EllipticK. It looks like they're something K + something/K. At the other extreme, has anyone seen *EllipticK[(-1)^(1/3)] == ((-1)^(1/12) 3^(1/4) Gamma[1/3]^3)/( 4 2^(1/3) π) ? --rwg

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[math-fun] SUSY, hence string theory, experimentally refuted?
by Warren Smith 13 Nov '12

13 Nov '12

The following http://www.cam.ac.uk/research/news/supersymmetry-squeezed-as-lhc-spots-ultr… http://arxiv.org/abs/1211.2674 http://www.guardian.co.uk/science/life-and-physics/2012/nov/13/supersymmetr… http://www.sciencedaily.com/releases/2012/11/121113095018.htm all claim a new particle decay has been detected which, while consistent with Standard Model, is very hard to explain with SUSY models since they would seem to predict it must be very rare. "Therefore" supersymmetry is wrong, hence string theory is dead meat. I do not either endorse or dispute this, I'm just pointing out it is there. Presumably running the experiment longer will clarify things. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Puzzle from 'The Tile Packing Problem''
by Stuart Anderson 13 Nov '12

13 Nov '12

Inspired by this paper http://people.cs.kuleuven.be/~ares.lagae/publications/LD07TPP/LD07TPP.pdf I made a jigsaw puzzle composed of 25-ominos from the recursive tile packing of the complete set of corner tiles over 4 colours (figure 8). This is the postscript; %!PS %%Creator:postscript & polyomino design; Stuart Anderson, stuart.errol.anderson(a)gmail.com %%Title: 256 tiles, 4 coloured, corner tiling: =>256 tile 25-omino tiling %%BoundingBox: 0 0 1100 1100 % Complete set of unit square, corner-to-corner, 4 coloured tiles, % 4^4 = 256 tiles, tiled as 16x16 toroid % from 'The Tile Packing Problem' by Ares Lagae & Philip Dutre % squares with coloured corners replaced by 25-ominos % /TileSet4Colours [ % corner colour codes for 256 tiles using edge-to-edge 4 colour matching 056 108 057 028 180 062 156 182 046 248 239 027 228 059 252 047 206 179 222 119 237 251 095 245 143 242 159 118 157 246 223 183 043 232 123 221 151 214 103 249 015 240 079 241 111 217 087 229 186 174 250 127 125 109 185 254 031 244 063 236 187 110 121 173 198 163 218 231 219 167 234 203 115 253 255 191 238 139 226 155 045 184 126 189 078 177 158 054 220 247 207 243 175 058 172 122 131 210 199 195 035 200 099 233 075 209 055 204 147 230 171 202 036 104 041 040 168 010 160 170 026 116 205 051 092 181 190 030 165 138 162 154 166 042 152 134 082 213 039 216 071 225 235 091 149 022 164 090 133 146 086 037 088 117 141 114 061 140 178 094 089 101 153 102 025 100 073 161 106 201 003 208 215 007 224 107 066 145 070 129 098 137 034 136 130 050 044 120 093 053 188 142 004 080 021 020 132 018 148 038 024 212 135 194 083 197 211 023 017 084 085 069 033 072 097 169 074 113 029 052 076 049 124 077 064 065 081 005 128 002 144 150 006 192 067 193 019 196 227 011 000 016 068 001 032 008 096 105 009 048 060 012 112 013 176 014] def /tmpstr 30 string def % temporary used to convert number to text /unit {12 mul} def % Convert units->points (1/12 inch) /white{ 1.0 1.0 1.0 setrgbcolor} def /black{ 0.0 dup dup setrgbcolor} def /grey{0.5 setgray} def /red {1 0 0 setrgbcolor} def /Courier-Bold findfont 14 scalefont setfont 2.5 setlinewidth 1 setlinejoin % rounded linejoin 1 setlinecap % rounded linecap /cp {closepath} def /st {stroke} def /gs {gsave} def /gr {grestore} def /l {90 rotate }def % left turn /r {-90 rotate }def % right turn /fh {0.5 unit 0 rlineto}def % 1/2 unit forward /f {1 unit 0 rlineto}def % 1 unit forward /f2 {2 unit 0 rlineto}def % 2 units forward /f3 {3 unit 0 rlineto}def % 3 units forward /Cornertile { % draw cornertile based on colour code 0, 1, 2 or 3 /Colour exch def % function accepts colour code as argument to a switch statement coded using ifelse statements Colour 0 eq {fh l f r f r f l f r f r f l f l f r fh} {Colour 1 eq {fh l f2 r f r f2 l f r f r f2 l f l f2 r fh} {Colour 2 eq {fh r f l f l f r f r f l f r f r f l fh} {Colour 3 eq {fh r f2 l f l f2 r f r f l f2 r f r f2 l fh} if } ifelse } ifelse } ifelse } def black 0 0 1400 1400 rectfill % fill blackground 32 72 translate % move pen off origin (screen bottom right) 1 1 16 { % for loop=1 to 16; ++ /y exch def % y is equal to outer loop variable 1 1 16 { % for loop=1 to 16; ++ /x exch def % x is equal to inner loop variable /number {TileSet4Colours y 1 sub 16 mul x 1 sub add get} def % extract tile numbers from array, read rows, columns y 2 mod x 2 mod eq {white}{black}ifelse % use checkerboard black/white alternating fill for tiles gs % save gfx state newpath % start new path x unit 5 mul 17 y sub unit 5 mul moveto % moveto grid point % taking integer divisors and remainders from powers of 4 gives corner colours, 255 given as an example number 64 idiv Cornertile % eg 255 64 idiv -> 3 topleft corner colour /Remainder1 { number 64 mod } def % eg 255 64 mod -> 63 %6 Remainder1 16 idiv Cornertile % eg 63 16 idiv -> 3 bottomleft corner colour /Remainder2 { Remainder1 16 mod } def % 63 16 mod -> 15 %8 Remainder2 4 idiv Cornertile % 15 4 idiv -> 3 bottomright corner colour /Remainder3 { Remainder2 4 mod } def % 15 4 mod -> 3 %10 Remainder3 Cornertile % 3 1 idiv -> 3 topright corner colour gs cp fill gr % closepath fill tile interior gs grey stroke gr % stroke the tile edge path y 2 mod x 2 mod eq {black}{white} ifelse % text fill contrast to tile fill -1 unit -2.9 unit rmoveto % position text number tmpstr cvs show % show number of tile } for } for showpage % %%EOF I have a much simpler construction I'm working on. Stuart

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Re: [math-fun] [EXTERNAL] Optimal Dome
by Henry Baker 12 Nov '12

12 Nov '12

The modern (Googlable) name for the "erg" function seems to be the "erfi" function: erfi(z) = erf(iz)/i http://mathworld.wolfram.com/Erfi.html At 08:45 AM 11/12/2012, Schroeppel, Richard wrote: > What's "erg"? --Rich

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Re: [math-fun] Optimal Dome
by Henry Baker 12 Nov '12

12 Nov '12

After a bit more thought, I think that the following would be feasible. An erf/erg dome cut off at some point with radius r, r fairly large, so when you're close to the outer edge of the dome, it looks like a flat wall at a very slight angle towards the dome. We now place a "skirt" around the outside of this dome which "starts" perhaps halfway up (down?) the dome side and gracefully (e.g., exp(-r) or exp(-r^2)) expands out from the dome in all directions. Note that the volume between the ground plane and underneath the skirt outside the dome is _filled in_ with material. The vertical weight of the erf/erg dome is carried by the lip of this dome itself; the only purpose of the skirt & filled in material is to provide the horizontal force towards the origin. If the dome is quite high, the angle of the wall at the lip is very nearly vertical, so the horizontal force required is relatively weak. The purpose of the skirt is to spread the horizontal force over as large a perimeter as possible. One way to analyze the circular "buttress" is to look at a section of it and turn it 90 degrees, so that the infinite lip is pointing down. The almost-horizontal top (after rotation) is now 1/2 of a standard arch bridge, except that in our case, the force coming down from the top is the original horizontal spreading force of the dome, and most importantly, the material of the bridge doesn't have to hold itself up (because gravity is pulling essentially orthogonally to the spreading force of the dome). At 01:08 PM 11/12/2012, Warren Smith wrote: >On 11/12/12, Henry Baker <hbaker1(a)pipeline.com> wrote: >> You've got a valid point, Warren. >> >> 1. I think it might be possible to get a pure compressive dome, but the base >> might have to be infinite. > >--1. The area of dome-cap goes like radius^2 or greater. >But its perimeter is 2*pi*radius. >Hence the compressive stress would be at least proportional to radius for a >uniform-thickness dome. >Hence growing the radius to infinity is impossible with uniform thickness. > >So, suppose the thickness were q(r), nonuniform. Roughly how would q grow? >Integral q(r) * r * dr is proportional to a lower bound on >weight for a radius-r cap, and this for uniform compressive stress >would be proportional to r*q (perimeter*thickness). > >Hence q' * r + q = C * r * q >with solution q(r) = K*exp(C*r)/r. > >So therefore, an infinite-width dome with uniform compressive stress would >perhaps be possible provided its thickness grew as above, i.e. roughly >exponentially with r. > >> 2. I think that there may be a solution that looks something roughly like a >> Gaussian revolved around the y axis. >> >> Basically, the "skirt" of the Gaussian pushes out on an ever increasing >> radius. > >--2. I don't think so because a nonconvex object like that will get >tension at points of negative curvature.

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[math-fun] moRe: EllipticE valuation [Was: EllipticK valuation [Was: Elliptic K π approximation]]
by Bill Gosper 12 Nov '12

12 Nov '12

Jörg>* Bill Gosper <billgosper(a)gmail.com> [Nov 12. 2012 16:42]: On Wed, Nov 7, 2012 at 3:05 PM, Bill Gosper <billgosper(a)gmail.com> wrote: [...] In http://oeis.org/A194094 Michael Somos gives: (2/Pi)*elliptic_E(k) = theta_3(q)^2 - 2 * (theta_4(q) / theta_3(q))^2 * Dq ( theta_4(q)^-2 ) = theta_3(q)^2 + 4 Dq (theta_4(q)) / (theta_4(q) * theta_3(q)^2) where Dq (f) := q * df/dq -------- This is presumably d/dq of EllipticK[EllipticTheta[2, 0, q]^4/EllipticTheta[3, 0, q]^4] == 1/2 \[Pi] EllipticTheta[3, 0, q]^2 but it can't be right--there can't be a naked q. I get EllipticE[EllipticTheta[2, 0, q]^4/EllipticTheta[3, 0, q]^4] == (Pi*(-EllipticTheta[2, 0, q]^4 + EllipticTheta[3, 0, q]^4)* Derivative[0, 0, 1][EllipticTheta][2, 0, q])/ (2*EllipticTheta[3, 0, q]*(EllipticTheta[3, 0, q]* Derivative[0, 0, 1][EllipticTheta][2, 0, q] - EllipticTheta[2, 0, q]*Derivative[0, 0, 1][EllipticTheta][3, 0, q])) which is a lot smaller than what I got by converting the 𝜗 to η before differentiating, but has the problem of needing special values for the two 𝜗 and 𝜗'. Can it really be that no one has tabulated EllipticE[E^-(surd π)]? --rwg

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Re: [math-fun] Optimal Dome
by Henry Baker 12 Nov '12

12 Nov '12

You've got a valid point, Warren. I think it might be possible to get a pure compressive dome, but the base might have to be infinite. I think that there may be a solution that looks something roughly like a Gaussian revolved around the y axis. Basically, the "skirt" of the Gaussian pushes out on an ever increasing radius. You could cut off the Gaussian at some point beyond the "circle of inflection" and put a ring around it, but the tension in the ring falls to zero as the radius of the ring increases. At 08:19 AM 11/12/2012, Warren Smith wrote: >--well, now you've blown it. Your stated goal was to get pure >compression. But as you just admitted, there is horizontal tension. >But in your defense, I claim this is impossible to avoid, at least for >a convex dome that eventually has vertical tangent; you unavoidably >get compression along vertical "meridians" but tension along "lines of >latitude."

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Re: [math-fun] Optimal Dome
by Warren Smith 12 Nov '12

12 Nov '12

Incidentally Heyman's "erg" function is called by Abramowitz+Stegun "Dawson's integral." I suppose it just arises from erf but with imaginary argument. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Optimal support points for uniform beam
by Warren Smith 12 Nov '12

12 Nov '12

Here is another "optimal shape / elasticity" problem of considerable interest. Consider a horizontal beam supported at N points: ========================== ^ ^ ^ Question: for each fixed N=1,2,3,..., what is the best choice of the N support points? (Assume thin uniform 1D beam with small-deflection approximation.) I presume the optimal points have even-symmetry and we can assume wlog the beam goes from x=-1 to x=+1. The beam curve shape y(x) will minimize gravitational+elastic energy: integral ( E*y''(x)^2 + g*y ) dx = minimum subject to constraint that y(x) = 0 whenever x is one of the N support points x=x1, x=x2, x=x3,..., x=xN. Here g is a gravitational-related constant and E an elastic-related constant. The Euler-Lagrange equation shows y(x) will satisfy the differential equation g + E*y''''(x) = 0 with solution y(x) = -x^4 * g/(E*24) + AnyCubicPolynomial(x). Thus y(x) + x^4 * g/(E*24) will simply be a "cubic spline" fit at the N support points (y, y', and y'' all continuous). The question then is where to place the points to minimize the maximum value of |y''(x)|, i.e. minimize the max-stress. It would be good to tabulate the points numerically for N=1,2,...,10. After all, civil engineers have been building these things for centuries, might as well figure out for them how to do it right. :) ---------- When N=1, obviously the answer is 1 midpoint at x=0. When N=2, the answer is not so obvious...

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