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Update 2: In Mathematica 9.0.1 this takes only 19 seconds on first evaluation and 10 seconds on subsequent evaluations. The results returned by M9 and M10 are equivalent but not given in identical form. Update: While I was writing this I tried Eigenvectors[m], which finished in 100 seconds on my machine. I'm leaving the NullSpace-based method below ...


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NullSpace works on both numerical and symbolic matrices. and implied by your link, under "Details" subsection, "Method" would sort of refer to the algorithm NullSpace uses. Possible settings for the Method option include "CofactorExpansion", "DivisionFreeRowReduction", and "OneStepRowReduction". The default setting of Automatic switches among these ...


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the result of Eigenvalues[m]: ( using g instead of \[gamma] ) {{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,g^2,0,0,0,0,0,-g,0,0,-g,0,0,0,0,0,1},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,g,0,0,0,0,0,-1,-g,0,0,0,0,0,1,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,g,0,0,0,0,-1,-g,0,0,0,0,1,0,0},{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,-1,0,0,-1,0,0,1,0,0,0}, ...


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By restriction I had in mind clipping: something like this. x1[t_] := Evaluate[Sin[Clip[phi1[t] /. sol, {-Pi/4, Pi/4}]]] y1[t_] := Evaluate[-Cos[Clip[phi1[t] /. sol, {-Pi/4, Pi/4}]]] x2[t_] := Evaluate[Sin[Clip[phi1[t] /. sol, {-Pi/4, Pi/4}]] + Sin[Clip[phi2[t] /. sol, {-Pi/4, Pi/4}]]] y2[t_] := ...


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χ /. First[NSolve[SetV == #/(Cpf (1 - χ)), χ]] & /@ demand or equivalently χ /. First@Solve[SetV == dem/(Cpf (1 -χ)), χ] (-dem + Cpf SetV)/(Cpf SetV) and now (-# + Cpf SetV)/(Cpf SetV)& /@ demand {(-1.92*10^6 + Cpf SetV)/(Cpf SetV), (-2.07*10^6 + Cpf SetV)/( Cpf SetV), (-2.37*10^6 + Cpf SetV)/( Cpf SetV), (-2.72*10^6 + Cpf SetV)/( ...


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Simple replace should do the trick χ /. {{χ -> (-1.92`*^6 + Cpf SetV)/(Cpf SetV)}} Yields {(-1.92*10^6 + Cpf SetV)/(Cpf SetV)}



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