Re: How to evaluate Exp[I Pi(1+x)]?
- To: mathgroup at smc.vnet.net
- Subject: [mg52770] Re: How to evaluate Exp[I Pi(1+x)]?
- From: "Dr. Wolfgang Hintze" <weh at snafu.de>
- Date: Mon, 13 Dec 2004 04:22:09 -0500 (EST)
- References: <200412100123.UAA18952@smc.vnet.net> <email@example.com>
- Sender: owner-wri-mathgroup at wolfram.com
I remember that ComplexExpand[expr] assumes parameters in expr to be
real. Hence the following was the shortest succesful expansion I found:
FullSimplify[ComplexExpand[E^(I*Pi*(1 + x))]]
Simplify is not enough.
By the way, while doing much complex function work recently I devolped
the habit to always use ComplexExpand as the first opration.
Otherwise even the simplest operations Re, Im, Abs don't work.
Andrzej Kozlowski wrote:
> On 10 Dec 2004, at 10:23, hello wrote:
>>Is there a way to evaluate Exp[I Pi (1+x)]? I am expecting to see the
>>result to be:
>>-Exp[I Pi x]
>>because Exp[I Pi] can be reduced to shorter form.
> FullSimplify[ComplexExpand[Exp[I*Pi*(1 + x)]],
> x $B":(B Reals]
> Andrzej Kozlowski
> Chiba, Japan
Prev by Date:
Re: Suse 9.2 + mathematica 5.0 problem
Next by Date:
How to express ODE?
Previous by thread:
Re: How to evaluate Exp[I Pi(1+x)]? version=2.55
Next by thread:
Re: Re: How to evaluate Exp[I Pi(1+x)]?