4

If you use BoundaryDiscretizeRegion, you can control the resolution of the boundary: BoundaryDiscretizeRegion[r2, MaxCellMeasure -> "Length" -> 0.005]


3

It may be possible using RiccatiSolve which can solve an equation of the form $$a^{T }.x+x.a-x.b.r^{-1}.b^{T }.x+q=0$$ If we assume $x=x^T$ and then choose values $a=0$, $b=I$, $r=A^{-1}$, and $q=B$, the above equation becomes $$0^{T }.x+x.0-x^T.I.(A^{-1})^{-1}.I^{T }.x+B=0$$ $$ x^T.A.x=B$$ The assumption $x=x^T$ specifies conditions that the matrices $...


3

a1 = data[[All, 1]]^# & /@ {2, 1, 0} ; a2 = Table[data[[All, 1]]^(3 - i), {i, 3}] a1 == a2 == A True


1

A={data[[All, 1]]^2, data[[All, 1]], Table[1, Length@ddata]}; B=data[[All, 2]]


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