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You could define a function that generates the explicit components for you: ClearAll[p] p[i_] := Array[HoldForm[p][i], 4] so you can use p[2] for your $p_2$ vector: p[2] (* Out: {p[2][1], p[2][2], p[2][3], p[2][4]} *) With that in hand we can try your simple equality; perhaps I am misunderstanding your notation here, but I am not sure how your equality ...


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ClearAll[f] f[x0_, y0_] := D[{Sqrt[x^2 + y^2], ArcTan[x, y]}, {{x, y}}] /. {x -> x0, y -> y0} f[0.3, 0.5] // MatrixForm I would caution you against using MatrixForm in the definition, as that would leave you with results that cannot be easily used in further computation.


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You can write something like f[{r_, θ_}] := Module[{M = D[{Sqrt[x^2 + y^2], ArcTan[x, y]}, {{x, y}}]}, Block[{x = r Cos[θ], y = r Sin[θ]}, M]] I wouldn't be surprised if this has a slightly simpler formulation. The symbolic result agrees with what I expect: Assuming[r > 0, Simplify[f[{r, θ}]]] (* {{Cos[θ], Sin[θ]}, {-(Sin[θ]/r), Cos[θ]/r}} *) ...


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Since this is a simple homework problem, I don't think it would fair to just hand you the whole answer, but here are some hints. To compute an exact expression for the mean curvature, use ArcCurvature andIntegrate. When you have the exact expression, let us call it avgCrv, then you can get an approximate numerical value with N[avgCrv] 0.561 You can ...


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centers = Import["C:\\Users\\W10\\Downloads\\cluster2_centres.csv", "Data"]; orientations = Import["C:\\Users\\W10\\Downloads\\cluster2_orientations.csv", "Data"]; angles = MapThread[VectorAngle, {centers, orientations}]; You can use Scale to create an oblate spheroid from Sphere: ClearAll[oblate, translaterotate] oblate[a_, c_] := Scale[Sphere[], {a, a, ...


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You can manually generate the table from the recurrence equation. Clear["Global`*"] A = {{2, 0}, {0, 4}}; b = {0, 0}; x0 = {2, 1}; x[0] = x0; x[k_Integer?Positive] := x[k] = x[k - 1] - Norm[A.x[k - 1] + b]^2/((A.x[k - 1] + b).A.(A.x[k - 1] + b))*(A.x[k - 1] + b) (table = x /@ Range[0, 20]) // Column The closed-form solution is sol[...


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