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2

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: ...


2

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 ...


2

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, ...


2

Here's my take. You can use either newVisibleSpectrum[] or myVisibleSpectrum[] as the underlying ColorFunction; I'll use the latter. (* smooth step function *) smoothStep3 = Compile[{{a, _Real}, {b, _Real}, {x, _Real}}, With[{t = Min[Max[0, (x - a)/(b - a)], 1]}, t*t*(3 - 2 t)], ...


5

(with many thanks to halirutan and kirma for their kind assistance) Here's a different take. In this article, piecewise Gaussian functions that approximate the CIE color matching functions are presented. For this answer, instead of just taking the coefficients from the paper directly, I used their proposed model in FindFit[], using a 1 nm tabulation of the ...


3

As noted in the comment above, k1 etc must be called with its argument list. With this change the r1 etc equations become r1 = k1[r, ϕ] dens[r, ϕ]^2 nH[r] nOH[r]; r2 = k2[r, ϕ] dens[r, ϕ]^2 nH[r] nH2O[r]; r17 = k17[r, ϕ] dens[r, ϕ]^2 nH2[r] nO[r]; r18 = k18[r, ϕ] dens[r, ϕ]^2 nH2[r] nOH[r]; r62 = k62[r, ϕ] dens[r, ϕ]^2 nO[r] nOH[r]; r63 = k63[r, ϕ] ...


24

You can also construct the image from Graphics primitive, which ultimately may give you more control: spectrum[list_List] := Graphics[ {Thickness[0.005], ColorData["VisibleSpectrum"][#], Line[{{#, 0}, {#, 1}}]} & /@ list, PlotRange -> {{380, 750}, {0, 1}}, PlotRangePadding -> None, ImagePadding -> All, AspectRatio -> 1/5, ImageSize ...


12

I prefer ListDensityPlot here as it gives flexibility to plot a range of data points. First of all, I define a function which generates a narrow spectrum around our desired wavelength: spec[wavelength_, width_] := Flatten[Table[{{x, 0, x}, {x, 1, x}}, {x, wavelength - width, wavelength + width, 0.1}], 1]; where we can specify wavelength and width of ...


3

For isotherms, ContourPlot will work here a = 27; b = 1; R = 1; p = (R*T/(V - b)) - a*(1/V^2); ContourPlot[p, {T, 0, 12}, {V, b, 8}, PlotPoints -> 180, Frame -> True] You can also visit here. Update For change of axis, do this ContourPlot[p, {V, b, 8}, {T, 0, 12}, PlotPoints -> 180, Frame -> True, FrameLabel -> Automatic]


2

All done by Morphological Components clusteringb[config_] := Module[{output, cm, cindices, csizes}, output = MorphologicalComponents[Image@Abs@config, CornerNeighbors -> False]; cm = ComponentMeasurements[ output, {"Label", "Mask", "Count"}][[All, 2]]; {cindices, csizes} = Transpose[{{#1, #2["NonzeroPositions"]}, {#1, #3}} & @@@ cm]; ...


0

It is a physical phenomenon occurring in a very short time( milliseconds). A = 1; \[Phi] = 0.0257275275; \[Omega] = 151.6*10^3; Cj = 17*10^(-9); Cs = 35*10^(-9); Rt = 223.9; I0 = 5*10^-6; L = 15*10^-3; Cd[t_] := Piecewise[{{Cj/Sqrt[1 - A*Cos[\[Omega]*t]/\[Phi]], t < 0}, {Cs*Exp[A*Cos[\[Omega]*t]/\[Phi]], t >= 0}}]; qn = {V'[t] == (j[t] - ...


4

(I know this is an old question, but it just got nudged and I saw another way to do it.) You can also take advantage of the vector potential to do this: $$ \vec{A}(\vec{r}) = \frac{\mu_0 I}{4\pi} \oint \frac{d\vec{r}'}{|\vec{r} - \vec{r}'|} $$ Using much the same symmetry arguments as given in the answer above, we can first look at the vector potential in ...


1

After removing the superfluous θ[0] == 0 from bc, the key to the solution is a subtle change in the use of Piecewise in the definition of Cd: Cd[x_] := Piecewise[{{Cj/Sqrt[1 - x/ϕ], x < 0}}, Cs*Exp[x/ϕ]] eqn = {V'[t] == (j[t] - Id[V[t]])/Cd[V[t]], j'[t] == (V0[t] - Rt*j[t] - V[t])/L}; bc = {V[0] == A, j[0] == 0 (*, \[Theta][0] == 0*)}; pl = ...


5

March - this is not a complete answer, but instead addresses the gathering of indices since that seems to be important. Here's a comparison of some methods, tested only on my cigar-lounge netbook so caveat lector. OP - Virgil's original (position based), Virgil - Virgil's adaptation of my first comment, CMT. 2/3 - my second and third comments (gathered ...



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