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You can get very close to the solution in three iterations of Newton's method: f[x_, a_] := x/Sin[Pi/2 x]^a fInvNewton[y_?NumericQ, a_?NumericQ, iter_: 3] := Nest[# - (f[#, a] - y)/Derivative[1, 0][f][#, a] &, If[y > 2/Pi, y, ((Pi/2)^a y)^(1/(1 - a))], iter] Example: Plot[Evaluate@{fInvNewton[y, 0.5, 3], InverseFunction[f, 1, 2][y, 0.5]}, ...


4

If all you're interested in is the inverse power series, then don't calculate the InverseFunction; instead, use the InverseSeries function: fseries[α_] := Series[x (1/Sin[Pi x/2])^α, {x, 0, 5}] Table[InverseSeries[fseries[α], y], {α, -1, 1, 1/2}] Note that $\alpha$ has to be a rational number for this to work. In particular, if you replace the 1/2 ...


5

Let us define the equation: Clear[eq]; eq[m_, f_] := 1/(x - 1) - (m + 1)/(x^(m + 1) - 1) == f; where x stays for Exp[f/(k t)]. If one applies the function Solve to it, Mma clearly answers that it cannot provide exact solution of this equation. It should not be expected, therefore, that one can find any other analytical solution. Numerically, it is ...


1

Something like this: f[x_, m_] := 1/(x - 1) - (1 + m)/(x^(1 + m) - 1) g[y_, m_] := x /. NSolve[f[x, m] == y, x]



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