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Hi Sander,

Thanks for the compliment. You're also doing well with the recent Chaos game. Could be interesting to see a variation of your code for a mapping between the coherent logogram ring structure and diffuse pixel-dust.

But I am still worried that maybe the aliens have some extra-sensory perception for geometric powers of $r$, so I did some more surface calculation starting with the condition for minimal surfaces:

$$(1+h_x^2)h_{yy} - 2 h_x h_y h_{xy} + (1+h_y^2)h_{xx}=0,$$

another one of Lagrange's equations. I have never seen an expansion solution of this equation, but can produce one easily. Here we include only odd powers of $r$ and angular dependence $2\pi/3$ periodic. Since the constraint is in cartesian coordinates, we define the following variables:

r = Sqrt[x^2 + y^2]; P = y/r; Q = x/r;
Q3 = 4 Q^3 - 3 Q; P3 = 3 P - 4 P^3;

And construct the surface as a ring over these variables

indList[n_ /; OddQ[n]] := Flatten[Table[{n, i - j, j}, {i, 1, Floor[n/3], 2}, {j, 0, i}], 1]
indList[n_ /; EvenQ[n]] :=  Flatten[Table[{n, i - j, j}, {i, 0, Floor[n/3], 2}, {j, 0, i}], 1]
params = Flatten[indList[2 # + 1] & /@ Range[7], 1];
(*params = Flatten[indList[#]&/@Range[2,15],1];*)
AbsoluteTiming[ \[CapitalPsi] =  Expand[Simplify[     Expand[Total[
       Times[c @@ #, r^#[[1]], Q3^#[[2]], P3^#[[3]] ] & /@   params]]]]; ]


d\[CapitalPsi]dx = D[\[CapitalPsi], x];
d\[CapitalPsi]dy = D[\[CapitalPsi], y];
d2\[CapitalPsi]dx2 = D[\[CapitalPsi], {x, 2}];
d2\[CapitalPsi]dy2 = D[\[CapitalPsi], {y, 2}];
d2\[CapitalPsi]dxdy = D[D[\[CapitalPsi], x], y];

AbsoluteTiming[ testExp = Expand[((1 + d\[CapitalPsi]dx^2)*d2\[CapitalPsi]dy2 - 
      2 d\[CapitalPsi]dx d\[CapitalPsi]dy d2\[CapitalPsi]dxdy + (1 + 
         d\[CapitalPsi]dy^2)*d2\[CapitalPsi]dx2)]; ]

And solve,

Sols[n_] := Fold[Join[#1, Expand@Solve[#2 /. #1 ][[1]]] &, {}, Function[{a}, # == 
         0 & /@ (Coefficient[Coefficient[testExp, x, #], y, a - #] & /@
          Range[0, a])][#] & /@ (2 Range[n] + 1)]

AbsoluteTiming[ With[{Sol6 = Sols[6]}, SolList =    Cases[Sol6, 
HoldPattern@Rule[c[n_ /; n < Evaluate[2 #], _, _], _]] & /@ 
     Range[3, 8]; ]]

These values cause cancellation of the first few coefficients in the expansion of the mean curvature numerator

Expand[Function[{a}, ((Coefficient[Coefficient[testExp, x, #], y, 
            a - #] //. SolList[[-1]]) & /@ Range[0, a])][#] & /@ Range[0, 13]] // TableForm

The solution is a formal power series, not gauranteed to converge. However, if the coefficients decrease quickly as powers of $r$ increase, convergence should occur. We choose coefficients $1/n!$ for terms with $r^n$. A test of the error is given by change in amplitude with change in iteration parameter, relative to the total amplitude at radius $r=1$.

ErrorFunctions =   Evaluate[Partition[ \[CapitalPsi] /. 
        SolList /. {c[x_, _, 0] :> 1/x!, c[_, _, _] -> 0}, 2, 1] /.
     {x_, y_} :> 100 Subtract[y, x]/Mean[{y, x}] /. {x -> 
      Cos[\[Theta]], y -> Sin[\[Theta]]}];

LogPlot[ErrorFunctions[[{1, 3, 5}]], {\[Theta], 0, 2 Pi},
 AxesLabel -> {"Angle", "Percent
   Error"}, ImageSize -> 500, PlotRange -> Full]

As iteration of the recursive solution algorithm increases, the difference oscillates more rapidly and with a decreasing amplitude. By a sixth approximation, already converges to $0.05\%$. Good enough.

It's also possible to measure convergence of the mean curvature to zero as a function of the iteration parameter

MeanCurvatureConvergence =   N[(testExp //. SolList[[#]] /. {c[x_, _, 0] :> 1/x!, 
c[_, _, _] -> 0} /. x -> Cos[Pi] /. y -> Sin[Pi]) & /@ Range[6]];

ListLogPlot[MeanCurvatureConvergence, PlotRange -> All, 
AxesLabel -> {"Iteration", "Curvature
Numerator"}, Joined -> True, ImageSize -> 500]

So yes, the surface as written appears to converge toward minimal form,

\[CapitalPsi]6 =  Expand[\[CapitalPsi] /. SolList[[-1]] /. {c[x_, _, 0] :> 1/x!,  c[_, _, _] -> 0}]
Expand[TrigReduce[\[CapitalPsi]6 /. {x -> \[Rho] Cos[\[Phi]],     y -> \[Rho] Sin[\[Phi]]}]]

Finally, the logogram plot, "earth" + "walk",

This deformed logogram looks quite similar to those above. However the equations underlying this surface have no $r^5$ term. It seems that a $2\pi/3$ periodic minimal surface, which expands in whole powers of $r$, cannot have an $r^5$ term. Compare with previous expansions, which are not minimal and do have a term like $r^5$.

We have yet to find parameter values that lead to periodic surfaces, but already have obtained a surface geometry that could be tested in nature. The famous soap film experiment should yield surfaces as seen at math.hmc.edu and daviddarling.info .

Part 2 : Overlap & Fourier Analysis

Code available at: Github Branch .

• Fourier Notebook

Time-Independent Calculations

Taking each logogram as a two-dimensional wavefunction $$ | \Psi_n \rangle : \langle i,j | \Psi_n \rangle = \text{pixel data (normalized) }, $$ calculate overlap for each pair $M_{n,m} =M_{m,n} = \langle \Psi_n | \Psi_m \rangle $ :

By row take the largest value other than $M_{n,n}=1$, and plot the superposition

Create wavefunctions

$$| \cos( n \phi ) \rangle: \langle i,j | \cos( n \phi ) \rangle = \cos(n \arctan(i,j)),$$ $$| \sin( n \phi ) \rangle: \langle i,j | \cos( n \phi ) \rangle = \sin(n \arctan(i,j)),$$

and use these to calculate the Fourier Amplitudes

$$FC_{n,m}=\langle \cos( n \phi ) | \Psi_m \rangle^2, $$ $$FS_{n,m}=\langle \sin( n \phi ) | \Psi_m \rangle^2, $$

Sum to obtain the total squared amplitude

$F_{m,n} = FC_{m,n}+FS_{m,n} $

For each wavefunction index by $m$, list plot the $n$ sequence. Also plot the most symmetric graphs.

more fanfiction:

Scientist: One plus one plus one, equals three. Three spatial dimensions, represented by approximate cyclic symmetry.

Linguist: Time goes around the circle. Clockwise or counterclockwise, endlessly repeating.

Scientist: Minus $d\tau^2$ equals $ds^2$. Surely the Aliens have some conception of special relativity. Logogram number 22, I'm starting to think that "spacetime" is a better translation than "time".

Linguist: The three lobes bear some resemblance to the partials for "suffering", "liberation". This third one, I don't know, looks like "stillness", possibly "death".