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Re: Polylog equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg84990] Re: Polylog equations
  • From: sashap <pavlyk at gmail.com>
  • Date: Tue, 22 Jan 2008 01:52:29 -0500 (EST)
  • References: <fn21oa$in5$1@smc.vnet.net>

On Jan 21, 6:04 am, Valeri Astanoff <astan... at gmail.com> wrote:
> Good day,
>
> Given these polylog equations:
>
> In[1]:= Assuming[0 < x < 1,
>         Solve[PolyLog[3/2,x]==y && PolyLog[5/2,x]==z,z,x]]
>
> Out[1]= {{}}
>
> what is the best way to get z(y)?
>
> Thanks,
>
> V.Astanoff

One way to get an approximation is to use ComposeSeries and
InverseSeries;

f[n_Integer?Positive, x_] :=
 Module[{y}, (ComposeSeries[
       InverseSeries[Series[PolyLog[3/2, y], {y, 0, n}]],
       Series[PolyLog[5/2, y], {y, 0, n}]] // Simplify // Normal) /.
   y -> x]

In[49]:= f[2, PolyLog[3/2, x]] - PolyLog[5/2, x] /. x -> x + O[x]^4

Out[49]= SeriesData[x, 0, {-1/8 + 2/(9*Sqrt[3])}, 3, 4, 1]

The series obtained converges for 0<y<1:

Show[ParametricPlot[{PolyLog[3/2, x], PolyLog[5/2, x]}, {x, 0, 1},
  PlotStyle -> Directive[Thick, Red]],
 Plot[Evaluate[Table[f[n, y], {n, 2, 6}]], {y, 0, PolyLog[3/2, 1]}]]

Oleksandr Pavlyk




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