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1

Apparently, nobody brought up this possibility: SeriesCoefficient[-72 + 9 (-2 + x)^2 + 4 (3 + y)^2, {x, 2, 0}, {y, -3, 0}] -72 where it is assumed that you know the terms subtracted from the corresponding variables.


6

A bit complicated, this one: With[{m = 5, r = 3}, CoefficientList[Sum[(-x)^n/n!, {n, 0, m}]^r, x] == Table[Sum[FactorialPower[r, k] BellY[n, k, Table[(-1)^i, {i, m}]], {k, 0, r}]/n!, {n, 0, m r}]] True Recall that the partial Bell polynomials are a way to express FaĆ  di Bruno's formula, which applies here since ...


5

define a function of two variables, f[x_, m_] = 2 m (Sqrt[4 x^2 (1 - m^2) + m^4 + 4 m^2 x]/(1 - m^2) + m^2/(4 (1 - m^2) Sqrt[m^2 - 1]) ArcSin[(2 (1 - m^2) x + 4 m^2)/(4 m^3)]); then tell InverseFunction to invert w.r.t. the first argument: inv = InverseFunction[f, 1, 2]; Show[{Plot[f[x, 2], {x, -10, 10}, PlotRange -> All], ...


4

The expressions for the coefficients of the Pochhammer symbol are in fact well-known (see e.g. Concrete Mathematics): $$\prod_{k=0}^{n-1}(x+k)=\sum_{k=0}^n \left[{n}\atop{k}\right]x^k$$ where $\left[{n}\atop{k}\right]$ is a Stirling cycle number. In Mathematica, this corresponds to (-1)^(n - k) StirlingS1[n, k]. Table[Product[x + k, {k, 0, n - 1}] == ...


5

expr = (-1 + (E^(w[1, 0] - 1.55432 w[1, 1]) - E^(-w[1, 0] + 1.55432 w[1, 1]))^2 / (E^(w[1, 0] - 1.55432 w[1, 1]) + E^(-w[1, 0] + 1.55432 w[1, 1]))^2) expr /. Power[E, Plus[x_, y__]] :> Inactive[Times][Power[E, x], Power[E, y]]/. {Power[E, Times[c_, w[i_, j_], ___]] :> Power[T[i, j], Sign[c]], Power[E, w[i_, j_]] ...



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