Polylog equations

[Valeri Astanoff]

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

[da...@wolfram.com]

Could iterate Solve for this.

In[8]:= Solve[PolyLog[5/2,x]==z /. Solve[PolyLog[3/2,x]==y,x]]

Solve::ifun: Inverse functions are being used by Solve, so some
solutions may
not be found; use Reduce for complete solution information.

Solve::svars: Equations may not give solutions for all "solve"
variables.

5 3
Out[8]= {{z -> PolyLog[-, InverseFunction[PolyLog, 2, 2][-, y]]}}
2 2

Daniel Lichtblau
Wolfram Research

[sashap]

On Jan 21, 6:04 am, Valeri Astanoff <astan...@gmail.com> wrote:

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

[David W.Cantrell]

I hope this is not the best way, but it's all that came to mind:

z(y) is given by

PolyLog[5/2, x /. FindRoot[PolyLog[3/2, x] == y, {x, 1/2}]]

David

[David W.Cantrell]

I haven't seen my previous reply or any replies from others appear in the
newsgroup yet. But Valeri and I have had some correspondence by private
email. From that, here's some information that may interest others.

According to Valeri, Albert Einstein gave an approximation for z(y) in
writing about quantum theory of ideal gases: z(y) is approximately

y - 0.1768 y^2 - 0.0034 y^3 - 0.0005 y^4

But those appear to be merely the first four terms of a Maclaurin series,
which I found easily with Mathematica:

In[7]:= Normal[
PolyLog[5/2, InverseSeries[Series[PolyLog[3/2, x], {x, 0, 4}], y]]]

Out[7]= y - y^2/(4*Sqrt[2]) + (1/8 - 2/(9*Sqrt[3]))*y^3 +
(-(3/32) - 5/(32*Sqrt[2]) + 1/(2*Sqrt[6]))*y^4

In[8]:= N[%]

Out[8]= y - 0.176777 y^2 - 0.00330006 y^3 - 0.000111289 y^4

and so it would seem that, if Einstein had intended to give the first four
terms of that Maclaurin series, either his numerical work was not quite
right or I (or Mathematica) made a mistake.

Valeri also noted that z(Zeta[3/2]) is Zeta[5/2], which is about 1.34148...
For comparison, using Einstein's approximation, one gets 1.3219, while
using four terms of the actual Maclaurin series, one gets 1.34194...

David W. Cantrell

[Valeri Astanoff]

On 26 jan, 02:25, "David W.Cantrell" <DWCantr...@sigmaxi.net> wrote:

> Valeri Astanoff <astan...@gmail.com> wrote:
> > Good day,
>
> > Given these polylog equations:
>
> > In[1]:= Assuming[0 < x < 1,

> > =A0 =A0 =A0 =A0 Solve[PolyLog[3/2,x]==y && PolyLog[5/2,x]==z,z,x=

]]
>
> > Out[1]= {{}}
>
> > what is the best way to get z(y)?
>
> I haven't seen my previous reply or any replies from others appear in the
> newsgroup yet. But Valeri and I have had some correspondence by private
> email. From that, here's some information that may interest others.
>
> According to Valeri, Albert Einstein gave an approximation for z(y) in
> writing about quantum theory of ideal gases: z(y) is approximately
>
> y - 0.1768 y^2 - 0.0034 y^3 - 0.0005 y^4
>
> But those appear to be merely the first four terms of a Maclaurin series,
> which I found easily with Mathematica:
>
> In[7]:= Normal[

> =A0PolyLog[5/2, InverseSeries[Series[PolyLog[3/2, x], {x, 0, 4}], y]]]

>
> Out[7]= y - y^2/(4*Sqrt[2]) + (1/8 - 2/(9*Sqrt[3]))*y^3 +

> =A0 =A0(-(3/32) - 5/(32*Sqrt[2]) + 1/(2*Sqrt[6]))*y^4

>
> In[8]:= N[%]
>
> Out[8]= y - 0.176777 y^2 - 0.00330006 y^3 - 0.000111289 y^4
>

> and so it would seem that, if Einstein had intended to give the first four=

> terms of that Maclaurin series, either his numerical work was not quite
> right or I (or Mathematica) made a mistake.
>

> Valeri also noted that z(Zeta[3/2]) is Zeta[5/2], which is about 1.34148..=

.
> For comparison, using Einstein's approximation, one gets 1.3219, while
> using four terms of the actual Maclaurin series, one gets 1.34194...
>
> David W. Cantrell

Good day,

[David, thanks again for your help]

For those interested, and since it seems to have disappeared
from my previous posts, I give again the reference to Einstein's
manuscript containing the polynomial approximation where the y^4
term should have been -0.0001*y^4 instead of -0.0005*y^4
[exactly (-18-15*Sqrt[2]+16*Sqrt[6])/192 as computed with
InverseSeries]

Preussische Akademie der Wissenschaft
(1924, SB Phys.-math., S. 261-267)
Quantentheorie des einatomigen idealen Gases.

http://www.lorentz.leidenuniv.nl/history/Einstein_archive/Einstein_1925_manu=
script/Pages/Einstein_1925_15a.html

V.Astanoff