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
>
>
>

```

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