Re: transform
- To: mathgroup at smc.vnet.net
- Subject: [mg41442] Re: [mg41406] transform
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Tue, 20 May 2003 03:23:15 -0400 (EDT)
- References: <200305180902.FAA05910@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
flemming wrote:
>
> How do I transform a function to a sumform f.eks:
> ArcTanh[x]-->Sum[x^(2*n+1)/(2*n+1),{n,0,Infinity}],
> in mathematca 3.0?
> Thanks in advance
You can get the structure of terms as below. This is using version 4.2
of Mathematica but 3.0 gives something similar.
In[2]:= <<DiscreteMath`RSolve`
In[4]:= InputForm[SeriesTerm[ArcTanh[x], {x,0,n}]]
Out[4]//InputForm= (KroneckerDelta[Mod[-1 + n, 2]]*UnitStep[-1 + n])/n
In version 4.2 you cannot actually use your Sum as an unevaluated
infinite series form of ArcTanh without use of Unevaluated or Hold or
some such. The reason is that the Sum will evaluate to ArcTanh
(regardless of whether you use the form for terms that you gave, or that
given by SeriesTerm).
In[4]:= Sum[SeriesTerm[ArcTanh[x], {x,0,n}] * x^n, {n,0,Infinity}]
Out[4]= ArcTanh[x]
In[5]:= Sum[x^(2*n+1)/(2*n+1),{n,0,Infinity}]
Out[5]= ArcTanh[x]
Daniel Lichtblau
Wolfram Research
- References:
- transform
- From: "flemming" <flemmingjuul@hansen.tdcadsl.dk>
- transform