Re: difficult/unconventional series expansion

*To*: mathgroup at smc.vnet.net*Subject*: [mg106289] Re: [mg106259] difficult/unconventional series expansion*From*: "David Park" <djmpark at comcast.net>*Date*: Thu, 7 Jan 2010 02:32:18 -0500 (EST)*References*: <32317537.1262776392395.JavaMail.root@n11>

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] 0 If you approach from the other direction you obtain an imaginary value. David Park djmpark at comcast.net http://home.comcast.net/~djmpark/ From: Roman [mailto:rschmied at gmail.com] 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 Series[S[x],{x,Infinity,1}] 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 (x+1)E^(-x+(2x+1)/(2(x+1))E^(-x)) 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. Cheers! Roman.