       Re: POWEREXPAND

• To: mathgroup at smc.vnet.net
• Subject: [mg31918] Re: POWEREXPAND
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Wed, 12 Dec 2001 04:13:58 -0500 (EST)
• References: <9v49ui\$n28\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Peter,
With t>0

Sqrt[(-1/t)^3] =
Exp[1/2(Log[Abs[(-1/t)^3] + I Arg[(-1/t)^3])] =
Exp[1/2(Log[Abs[(-1/t)^3] + I Arg[(-1/t)^3])] =
Exp[1/2(Log[1/t^3] + I Pi)] =
I /t^(3/2)

Whereas

Sqrt[-1/t] = Exp[1/2(Log[Abs[-1/t]]+ Arg[-1/t])] =
Exp[1/2(Log[1/t]+ I Pi)] =
I/Sqrt[t]

So that

Sqrt[-1/t]^3  = ( I/Sqrt[t)]^3 = -I/t^(3/2)

Check:

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

I/t^(3/2)

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

-(I/t^(3/2))

--
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

"Blimbaum Jerry DLPC" <BlimbaumJE at ncsc.navy.mil> wrote in message
news:9v49ui\$n28\$1 at smc.vnet.net...
> Concerning the recent question about PowerExpand of (-1/t)^(3/2)
> Sqrt[-t]  note that the results depend on how you enter the formula....
>
> PowerExpand[Sqrt[(-1/t)^3] Sqrt[-t]]   gives the result   -1/t
>
> whereas
>
> PowerExpand[Sqrt[-1/t]^3 Sqrt[-t]   gives the result   1/t
>
>
> apparently   typing in (-1/t)^(3/2) Sqrt[-t]  in Mathematica
> corresponds to the second way...
>
>
> jerry blimbaum   NSWC Panama City, Fl
>

```

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