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taylor = (vars - point).# &; init := D[f[vars], {vars, j}] /. Thread[vars -> point]; taylorPoly[m_] := Sum[1/j! Nest[taylor, init, j], {j, 0, m}] Example vars = {x, y, z}; point = {0, 0, 0}; f[vars_] = Sin[y - x ] + Exp[x - y + 2 z]; taylorPoly[2] // FullSimplify 1/2 (2 + 4 z + (x - y + 2 z)^2) taylorPoly[3] // FullSimplify 1/6 (6 + (3 + 2 x - 2 ...


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n = 0; a = 1; L = 0; b = 1/100; m = 1/2; g = (2 m a r)/(n + L + 1); f[r_] = AiryAi[(2 m b)^(1/3) r]; r0 = r /. FindMaximum[r^(L + 1) f[r], r, WorkingPrecision -> 20][[2]] // Rationalize[#, 0] &; r0 // N // InputForm (* 4.103398736759 *) r1inv = Series[1/r, {r, r0, 4}] // Normal // Simplify; r0 == r /. Solve[r1inv - 1/r == 0, r, ...


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The code at http://oeis.org/A006784 works for me: EngelExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Ceiling[1/Expand[#[[1]] #[[2]] - 1]], Expand[#[[1]] #[[2]] - 1]} &, {Ceiling[1/(A - Floor[A])], A - Floor[A]}, n - 1]] res = EngelExp[N[E/Pi, 500000], 27] returns: {2, 2, 3, 3, 7, 23, 43, 58, 30503, 32703, ...


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Maybe with the general Taylor formula you can calculate the Taylor remainder easier. This applies to any vars. taylor = (vars - point).# &; init := D[f[vars], {vars, j}] /. Thread[vars -> point]; taylorPolynom[m_] := Sum[1/j! Nest[taylor, init, j], {j, 0, m}] remInit := D[f[vars], {vars, j}] /. Thread[vars -> point + t (vars - point)] ...



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