Re: PowerExpand in mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg31868] Re: PowerExpand in mathematica
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Sat, 8 Dec 2001 05:51:44 -0500 (EST)
- References: <9uq8cg$ada$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Peter, The correct answer, assuming t>0, which is what PowerExpand assumes, is 1/t. We can trace this from the definition z^p = Exp[p(Log[Abs[z]] + Arg[z] I)] (notethat Log is here working on real nunbers and Arg gives the principal value, in (-Pi,Pi] ) as follows Exp[3/2( Log[Abs[-1/t]]+ I Arg[-1/t])] *Exp[1/2 (Log[Abs[-t]]+ I Arg[-t])] Exp[3/2( Log[1/t]+ I Pi)] Exp[1/2 (Log[t]+ I Pi)] Exp[3/2( -Log[t]+ I Pi)] Exp[1/2 (Log[t]+ I Pi)] Exp[3/2( -Log[t])]*Exp[3/2I Pi] *Exp[1/2 Log[t]]*Exp[ 1/2I Pi] Exp[3/2( -Log[t])+1/2 Log[t]]*Exp[3/2I Pi+1/2 I Pi] Exp[ -Log[t]]*Exp[2I Pi] 1/t*1 1/t Check this against Simplify[(-1/t)^(3/2) Sqrt[-t],t>0] 1/t Also Simplify[(-1/t)^(3/2) Sqrt[-t],t<0] -(1/t) -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Peter" <ptrsn2001 at yahoo.com> wrote in message news:9uq8cg$ada$1 at smc.vnet.net... > Hi, > > Is this a known bug, feature? Or I am doing something wrong? > ------------------------------- > Mathematica 4.0 for Linux > Copyright 1988-1999 Wolfram Research, Inc. > -- Motif graphics initialized -- > > In[1]:= PowerExpand[(-1/t)^(3/2) Sqrt[-t]] > > 1 > Out[1]= - > t > --------------------------------- > The correct answer is -1/t, of course. > > Thanks, > Peter >