MathGroup Archive 2010

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: difficult/unconventional series expansion

What about:

S[x_] := Log[E^x - 1]/(E^x - 1) + (1 + 1/(E^x - 1)) Log[
    1 + 1/(E^x - 1)]

S2[y_] = S[1/y];  
Series[S2[y], {y, y0, 1}] // Normal;  
Limit[%, y0 -> 0, Direction -> -1]


If you approach from the other direction you obtain an imaginary value.

David Park
djmpark at  

From: Roman [mailto:rschmied at] 

Dear group,

I am trying to do an unconventional series expansion, and find it
difficult with Mathematica. Given the function
   S[x_] = Log[E^x-1]/(E^x-1)+(1+1/(E^x-1))Log[1+1/(E^x-1)]
I am looking for the behavior for very large x.
[For the interested: S(x) is the entropy (in units of the Boltzmann
constant) of a harmonic oscillator, with x=\hbar\omega/(kT). So I'm
looking for the low-temperature behavior of the entropy.]

The problem is that simply doing
does nothing (since this is not a series expansion in the usual
sense). But I know what I would like to find: the lowest terms of the
"series expansion" are
Do you know how to find this expression automatically in Mathematica?
I am interested in a general technique, not just the results for this
particular function.


  • Prev by Date: Re: why an extra mark in legend?
  • Next by Date: Re: More /.{I->-1} craziness
  • Previous by thread: difficult/unconventional series expansion
  • Next by thread: Re: difficult/unconventional series expansion