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| 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... |
| Ok, I understand now how to decribe a space-group by (4,4)matrices. But what exactly is meant by generators? In my opinion one has to determine the rotation (or whatever) group in the upper left half. This can be done by nxtgens = ... |
| Perhaps like this ( I think you should avoid complcated varable-names) NIntegrate[(Integrate[ lam2[s,u] Vbar[u],{u,0,s}])^2,{s,0,T}] |
| Hello Jean Michel, @Michael: an elegant and most impressive approach using deep functionalities of Mathematica. Here is a simple if not crude and basic approach, showing that there are, depending on the choice of a0, lots of solutions. For the... |
| The (second) method works on ellipses as well : el1 = fb1/fa1 Sqrt[fa1^2 - x^2]; el1n = - fb1/fa1 Sqrt[fa1^2 - x^2]; el2 = fb2/fa2 Sqrt[fa2^2 - (x - d)^2]; el2n = -fb2/fa2 Sqrt[fa2^2 - (x - d)^2]; aa = -fb1/fa1... |
| I think there is a way to get Epilogs from older plots, See GetEpilog in the notebook attached - near the end. |
| What is your question? I don't know what you mean. Something like this perhaps? (There is some additional work to be done for the Feder, and make sure your other code has run) a = 5; Animate[{Feder = ParametricPlot[{tt +... |
| Or more elementary FromCharacterCode /@ Range[66, 90, 2] FromCharacterCode /@ Range[98, 122, 2] |
| The stone-age-old version V7 gives the following answer: 1/2 (-2 x - ((-1 - Sqrt[1 - x] Sqrt[1 + x^2] + x (1 - 2 Sqrt[1 - x] Sqrt[1 + x^2] + x (-1 + x + Sqrt[1 - x] Sqrt[1 + x^2]))) ((-1 + I) Sqrt[2] Sqrt[(-1 + I) (-1 + x)] (1... |
| Take care of the notation. The Matrix of the eigenvectors is a list of vectors, which must be transposed to get the known results: eva = Eigenvectors[a] // Transpose Inverse[eva].a.eva // FullSimplify // MatrixForm K = {{4, -5},... |