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Hans Dolhaine
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Hello Dan. Strictly spoken I sum not over generators, but over the whole group. That is the trick that Conjugate[T[a]] . H . T[a] equals H. The details are outlined in the text cited. Correction: it must be of course ...
Hello Bryan. Do that, and I appreciate if you would tell me what they say.
What about this? sol = Solve[{(x/a)^2 + (y/b)^2 == 1, y - (k x + m) == 0}, {y, x}]; xx1 = {x, y} /. sol[[1]] xx2 = {x, y} /. sol[[2]] dd = xx1 - xx2 // FullSimplify chordlength = Sqrt[dd.dd] // FullSimplify
In case you want more (here 15) contours and you know minf and maxf, the minimum and maximmum value for your contours you could for example use n = 15; minf = -4; maxf = 4; cons = Table[minf + (j - 1) (maxf - minf)/(n - 1), {j,...
I must admit that I don't understand what you mean. I think you should use Conjugate at the appropriate places.
Hello Zoeyy, you don't say anything about the background (physics) of your data, so I have no idea of a reasonable fit. What do you think of this? fff = FindFit[data1, a (1 - b Erf[x - c]), {{a, 2200}, {b, 3}, {c, 12}}, x] h = a (1 - b...
Hi sinval, if ( if ! ) you could do your integration, you would get a function of r and theta. I wondered if it were possible to get an impression of it. So I tried a numerical integration. But there were lots of error messages ( I didn't...
Hello all, having heateq = D[fT[x, t], t] - a D[fT[x, t], x, x]; fLT = LaplaceTransform[heateq, t, s] (*-fT[x,0]+s LaplaceTransform[fT[x,t],t,s]-a LaplaceTransform[(fT^(2,0))[x,t],t,s]*) is there a more simple and concise way to...
I think this could be done like this heateq = D[fT[x, t], t] - a D[fT[x, t], x, x] Do a Laplace-Transformation, giving an ODE for the PDE fLT = LaplaceTransform[heateq, t, s] Solve the equivalent ODE sol = DSolve[-T0 + s...
Hello Zhao, gg is already unitary, means as there are no complex numbers it is hermitian. Map[MatrixForm, Transpose[#].# & /@ gg, 1] You can create another subgroup of gg from the matrices with gg[[3,3]] == 1 (I did not (yet) test...