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1

How funny - I was just doing this the other day (different equation, with a series in x^(1/2)). Here is what I came up with (actually, I'm following Isaac Newton's method, I believe): Clear[sc]; (* series coefficient of x in f[b, x] == 0 *) sc[f_, n_] := Nest[ Append[#, a /. First @ Solve[ SeriesCoefficient[ f[b, #.b^(Range @ Length @ # - ...

0

The (accepted) answer given by episanty gave me another idea I want to share. While SeriesCoefficient leaves the infinite sum untouched, Derivative does not, so we can get the coefficients manually. The issue episanty points out is that the lower limit of the sum errors out at $0^0$. We can prevent that behavior by modifying Sum (eek!) and everything works ...

2

You could always build a pattern matcher that will recover the coefficients you want: seriescoefficient[ Sum[coeff_ b_^n_, {n_, min_, max_}], {b_, b0_, m_}] := coeff /. {n -> m} will evaluate seriescoefficient[Sum[a[n] b^n, {n, 0, ∞}], {b, 0, 3}] to a[3]. This may or may not work, or require different amounts of care for conditions, depending on ...

2

Not sure if this is what you are after, but could implicitly write all vars as functions of t, then take derivatives and solve for first few in y to form a Taylor series. For the given example it is as below. exprs = {p1*x[t] + p2*x[t]^2 - y[t] Sin[x[t]] + p3*x[t]^3 - t}; dexprs = D[exprs, t]; d2exprs = D[dexprs, t]; augexprs = Join[exprs, dexprs, ...

1

I'm not sure how sound this solution is, but you can substitute x in f[x] with the inverse series of g[x] and then take the serie fS[y_, t_, x_, xo_, n_] := With[{s = InverseSeries[Series[t, {x, xo, n}]]}, Series[ y /. x -> s, {x, xo, n}]] fS[p1 x + p2 x^2, Sin[x] + p3 x^3, x, 0, 4]

3

You'd have to work out whether this is well behaved and convergent on your own so taking your own word on how sane the approach is, you can define your expression: ClearAll[expr]; expr = (γ Cosh[s L] + s δ Sinh[s L])^(-1); then replace -Log[q] for s and expand around 1 to some order and replace back q as E^(-s) (these are inverses of each other in the ...

1

Probably because it has a rather complicated branch point at z=1. It is actually very easy to derive the explicit expansion: singular[x_] = Csc[Sqrt[3] Pi] (-I Sqrt[3] Pi^2 + Pi (1 + 1/(Sqrt[3] (-1 + x))) - 2 Sqrt[3] Pi HarmonicNumber[Sqrt[3]] - Sqrt[3] Pi Log[-1 + x]); finite[x_] = (3 - 2 Sqrt[3]) HypergeometricPFQ[{1, 4, Sqrt[3]}, {2 + Sqrt[3], 4 + ...

5

You can do : expr = (-1)^(1/4) π (Cot[(-1)^(3/4) π / Sqrt[x]] + I Cot[(-1)^(1/4) π / Sqrt[x]])/(4 Sqrt[x]); FullSimplify[ComplexExpand[expr, TargetFunctions -> {Re, Im}], Assumptions -> {x > 0}] (* (π (Sin[(Sqrt[2] π)/Sqrt[x]] - Sinh[(Sqrt[2] π)/Sqrt[x]])) / (2 Sqrt[2] Sqrt[x] (Cos[(Sqrt[2] π)/Sqrt[x]] - Cosh[(Sqrt[2] π)/Sqrt[x]])) *) ...

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