Re: convert an expression to an infinite series
- To: mathgroup at smc.vnet.net
- Subject: [mg29625] Re: convert an expression to an infinite series
- From: "Orestis Vantzos" <atelesforos at hotmail.com>
- Date: Fri, 29 Jun 2001 01:35:58 -0400 (EDT)
- Organization: National Technical University of Athens, Greece
- References: <9heu94$6ne$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Consider your expression a function of one of its variables, say m. A very realistic candidate for the series would then be the Taylor series of your expression as a function of m, around say 0. As the Taylor series involves the derivatives of your expression, what you need is a general formula for them. I don't think it is very easy, but with smart use of Solve and Eliminate you could probably deduce a recursive formula for the derivatives which you could attempt to solve with RSolve. Orestis PS. If it's all Greek to you, it's probably because I am Greek ;-) "Toshiyuki (Toshi) Meshii" <meshii at mech.fukui-u.ac.jp> wrote in message news:9heu94$6ne$1 at smc.vnet.net... > Hello, > > Does anyone know a way to convert an expression to an infinite series? > Let me show the idea by a concrete problem. > > In[29]:= > Sum[E^(-2*m*n*L), {n, 0, Infinity}] > > Out[29]= > E^(2*L*m)/(-1 + E^(2*L*m)) > > Yes, Mathematica 4.1 knows that the infinite sum converges to a specific > value. > Then, my question is that how can I expand the expression > E^(2*L*m)/(-1 + E^(2*L*m)) > in an infinite series on n, in this case > Sum[E^(-2*m*n*L), {n, 0, Infinity}] > > Is there any way or is it an one way path? > > -Toshi > > >