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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.




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