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Re: PowerExpand in mathematica

  • To: mathgroup at
  • Subject: [mg31868] Re: PowerExpand in mathematica
  • From: "Allan Hayes" <hay at>
  • Date: Sat, 8 Dec 2001 05:51:44 -0500 (EST)
  • References: <9uq8cg$ada$>
  • Sender: owner-wri-mathgroup at

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

    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]


Check this against

    Simplify[(-1/t)^(3/2) Sqrt[-t],t>0]


    Simplify[(-1/t)^(3/2) Sqrt[-t],t<0]



Allan Hayes
Mathematica Training and Consulting
Leicester UK
hay at
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Peter" <ptrsn2001 at> wrote in message
news:9uq8cg$ada$1 at
> 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

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