$x^3-x-t=0$ has three roots that can be expressed in hypergeometric form, for example
$$x_1=-1-\frac{t}{2}{_2F_1\left(\frac{1}{3},\frac{2}{3};\frac{3}{2};\frac{27t^2}{4} \right)}+\frac{3t^2}{8} {_3F_2\left(\frac{5}{6},\frac{7}{6},1;\frac{3}{2},2;\frac{27t^2}{4} \right)}$$
See here for more information.
How can I get the solution of the general equation $x^n-x-t=0$ in hypergeometric form with Mathematica?
Since Root objects can be symbolically differentiated, we can find the closed form for its Taylor series (for an explicit n).
root = Root[#1^4 - #1 - t &, 1];
coeff = Refine[FunctionExpand[SeriesCoefficient[root, {t, 0, k}]], k >= 0];
Sum[coeff t^k, {k, 0, ∞}] // FullSimplify
-t HypergeometricPFQ[{1/4, 1/2, 3/4}, {2/3, 4/3}, -256t^3/27]
We can pack this into a function, but we need an alternative to FullSimplify because it becomes too slow.
PostProcess[f_] := Collect[f, _HypergeometricPFQ] /. {a_ F_HypergeometricPFQ :>
With[{r = Rationalize[Chop[N[a]]]},
r F /; Precision[r] === ∞
]
}
RootToHypergeometric[n_] := Module[{coeff, k},
coeff = Refine[FunctionExpand[
SeriesCoefficient[Root[#1^n - #1 - t &, 1], {t, 0, k}]], k >= 0];
PostProcess[Sum[coeff t^k, {k, 0, ∞}]]
]
Table[n -> RootToHypergeometric[n], {n, 4, 6}]
{ 4 -> -t HypergeometricPFQ[{1/4, 1/2, 3/4}, {2/3, 4/3}, -256 t^3/27], 5 -> -HypergeometricPFQ[{-1/20, 3/20, 7/20, 11/20}, {1/4, 1/2, 3/4}, 3125 t^4/256] + 1/4 t HypergeometricPFQ[{1/5, 2/5, 3/5, 4/5}, {1/2, 3/4, 5/4}, 3125 t^4/256] + 5/32 t^2 HypergeometricPFQ[{9/20, 13/20, 17/20, 21/20}, {3/4, 5/4, 3/2}, 3125 t^4/256] + 5/32 t^3 HypergeometricPFQ[{7/10, 9/10, 11/10, 13/10}, {5/4, 3/2, 7/4}, 3125 t^4/256], 6 -> -t HypergeometricPFQ[{1/6, 1/3, 1/2, 2/3, 5/6}, {2/5, 3/5, 4/5, 6/5}, -46656 t^5/3125] }
I don't really see a pattern for a solution in terms of n.
Edit
It looks like the closed form is in the link you provided.
It's been a while; I guess I can post my take on the question now.
The following routine for solving the trinomial equation $x^n-x+t=0$ (where $t$ is complex) is of course based on the results of this paper. It returns all $n$ roots, as opposed to just one root. The wrinkle in my implementation is that I elected to represent the (series of) hypergeometric functions presented in the paper in terms of the Meijer $G$-function. (Actually, the Fox $H$-function would have been the more appropriate function to use for the general solution, but since it's not yet built into Mathematica, I had to make do with MeijerG[] through repeated use of the Gauss multiplication formula for the gamma function.)
Here's the routine:
trinomRoots[n_Integer /; n > 1, t_] :=
Append[Table[Exp[-(2 π I j)/(n - 1)] - t/(2 π)^(n - 3/2) Sqrt[n/(n - 1)^3]
MeijerG[{Append[Range[n - 1]/n, (n - 2)/(n - 1)], {}},
{Range[0, n - 2]/(n - 1), {-1/(n - 1)}},
-(n^(n/(n - 1)) t Exp[(2 π I j)/(n - 1)])/(n - 1), 1/(n - 1)],
{j, 0, n - 2}],
Sqrt[n/(n - 1)^3] (t/Sqrt[2 π])
MeijerG[{Range[0, n - 1]/n, {}}, {{0}, Range[-1, n - 3]/(n - 1)},
-n^n (t/(n - 1))^(n - 1)]]
Here are a few examples:
Sort[trinomRoots[5, 1], Re[#1] <= Re[#2] &] // Quiet
{-1 - (Sqrt[5/2] MeijerG[{{1/5, 2/5, 3/5, 3/4, 4/5}, {}},
{{0, 1/4, 1/2, 3/4}, {-1/4}}, 5 5^(1/4)/4, 1/4])/(64 π^(7/2)),
-I - (Sqrt[5/2] MeijerG[{{1/5, 2/5, 3/5, 3/4, 4/5}, {}},
{{0, 1/4, 1/2, 3/4}, {-1/4}}, -5 I 5^(1/4)/4, 1/4])/(64 π^(7/2)),
I - (Sqrt[5/2] MeijerG[{{1/5, 2/5, 3/5, 3/4, 4/5}, {}},
{{0, 1/4, 1/2, 3/4}, {-1/4}}, 5 I 5^(1/4)/4, 1/4])/(64 π^(7/2)),
1 - (Sqrt[5/2] MeijerG[{{1/5, 2/5, 3/5, 3/4, 4/5}, {}},
{{0, 1/4, 1/2, 3/4}, {-1/4}}, -5 5^(1/4)/4, 1/4])/(64 π^(7/2)),
Sqrt[5/(2 π)]/8 MeijerG[{{1/5, 2/5, 3/5, 4/5}, {}},
{{0}, {-1/4, 1/4, 1/2}}, -3125/256]}
FunctionExpand[%[[1]]] // FullSimplify (* show explicit hypergeometric expression *)
-1 - (261 π^2)/(140 5^(3/4) Gamma[-(7/20)] Gamma[17/20] Gamma[21/20]
Gamma[49/20]) -
HypergeometricPFQ[{-1/20, 1/5, 9/20, 7/10}, {2/5, 3/5, 4/5}, 256/3125] -
1/5 HypergeometricPFQ[{3/20, 2/5, 13/20, 9/10}, {3/5, 4/5, 6/5}, 256/3125] +
1/25 HypergeometricPFQ[{7/20, 3/5, 17/20, 11/10}, {4/5, 6/5, 7/5}, 256/3125] -
1/125 HypergeometricPFQ[{11/20, 4/5, 21/20, 13/10}, {6/5, 7/5, 8/5}, 256/3125]
{N[%], N[Root[1 - #1 + #1^5 &, 1]]} (* check *)
{-1.167303978261419, -1.1673039782614187}
Sort[N[trinomRoots[9, -3 + 4 I], 20]]
{-1.15737300065622390026 - 0.25869616182022624031 I,
-1.04152925102213535542 + 0.52458846764227132511 I,
-0.73118138143615817856 - 0.96088154253626318503 I,
-0.48212736972957767713 + 1.05763503540895842021 I,
0.07216502145935959748 - 1.21942175005944057635 I,
0.29590581961582700752 + 1.14008946378455944936 I,
0.85499258660164957177 - 0.87742956898864238100 I,
0.97603686085833181128 + 0.69847391536413082742 I,
1.21311071430892712332 - 0.10435785879534763941 I}
% - Sort[x /. NSolve[x^9 - x + (-3 + 4 I), x, 20]] // Chop // Union
{0}
As shown above, use FunctionExpand[] on the output of trinomRoots[] to convert the MeijerG[] expressions to their corresponding hypergeometric forms.