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1

For a numerical approximation you may try something like: M = {{s, ab, ac}, {ab, s, bc}, {ac, bc, s}}; disc = Discriminant[CharacteristicPolynomial[M, x], x] // FullSimplify f[e_] := 2 Norm[e]^-3 (1 - 3 Sin[θ]^2 Cos[φ - ArcTan[e[[1]], e[[2]]]]^2) Cos[{x, y}.e]; ab = f[{1, 0}]; ac = f[{0, 1}]; bc = f[{1, 1}] + f[{-1, 1}]; s = f[{2, 0}] + f[{0, 2}]; ...


11

In the version 10.2, there is a builtin DensityPlot3D function, which can be used to visualize orbitals. a0=1; With[{ρ = 2 r/(n a0)}, Sqrt[(2/(n a0))^3 (n - l - 1)!/(2 n (n + l)!)] Exp[-ρ/2] ρ^ l LaguerreL[n - l - 1, 2 l + 1, ρ] SphericalHarmonicY[l, m, θ, ϕ]] DensityPlot3D[(Abs@ψ[{3, 2, 0}, {Sqrt[x^2 + y^2 + z^2], ArcTan[z, Sqrt[x^2 + ...


2

I believe that NeumannValue[-xload, x == span], as given in the question, is correct, as can be seen from the following simple test, for which the solution is obvious. test[Y_, ν_] := Inactive[Div][({{Y, 0}, {0, Y}}.Inactive[Grad][u[x, y], {x, y}]), {x, y}]; sol = NDSolveValue[{test[Ey, n] == NeumannValue[-xload, x == span], u[0, y] == 0}, u, {x, 0, ...


3

Starting with bbgodfrey's excellent suggestion to solve ab == ac == bc == 0, we can obtain a fairly compact list of all of the solutions. If we Reduce the equations with conditions on the variables we get a complicated result, so it's easier to Reduce first and apply conditions after: Reduce the equations and throw out some obviously inconsistent results: ...


4

Further edited to simplify results It is not difficult to show that the three eigenvalues, w, of M are equal if and only if M is diagonal; i.e., ab = ac = bc = 0. f[e_] := 2 Norm[e]^-3 (1 - 3 Sin[θ]^2 Cos[ϕ - ArcTan[e[[1]], e[[2]]]]^2) Cos[{x, y}.e] abeval = f[{1, 0}] (* 2 Cos[x] (1 - 3 Cos[ϕ]^2 Sin[θ]^2) *) aceval = f[{0, 1}] (* 2 Cos[y] (1 - 3 Sin[θ]^2 ...


3

This is an incomplete answer, but we will be able to show that there is no solution for most values of θ and ϕ. We will also be able to draw a plot of the regions of interest that you should check further to find solutions, should they exist. M = {{s - w, ab, ac}, {ab, s - w, bc}, {ac, bc, s - w}}; {d, c, b, a} = CoefficientList[Det[M], w]; disc = ...


4

Look at what you get when you do e.g. f[1,1,1] and Conjugate[f[1,1,1]]. As the docs for Conjugate state, "Conjugate does not always propagate into arguments", and here it does not propagate into Cos and Sin. A solution seems to be to do f[kx_, ky_, t_] := -t E^(-I kx a) (1 + 2 E^(I (3 kx a)/2)*Cos[Sqrt[3]/2 ky a]); BlockA[kx_, ky_, t_] := ...


1

One approach is as follows. First note that 0 <ArcTan[(600 - z), 155] < ArcTan[(100 - z), 155] < 2 Pi, which allows the Conditional expressions to be eliminated. Plot[{ArcTan[(600 - z), 155], ArcTan[(100 - z), 155]}, {z, 0, 750}] Then the Integrals can be performed as follows: Assuming[2 Pi > x2 > x1 > 0, With[{A = 1/137, B = 1, p = ...


6

Borrowing the analytically-fitted color matching functions cieX, cieY, cieZ and sRGBGamma from this answer, here is a function for generating the colors of the black body spectrum. The conversion being done here assumes a luminance (Y in the XYZ system) of 1: With[{planck = 1/((Exp[1.43877696*^7/(#1 #2)] - 1) #1^5) &, tab = Through[{cieX, cieY, ...



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