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Re: Re: How to evaluate Exp[I Pi(1+x)]?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg52833] Re: [mg52770] Re: How to evaluate Exp[I Pi(1+x)]?
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Tue, 14 Dec 2004 05:59:31 -0500 (EST)
*References*: <200412100123.UAA18952@smc.vnet.net> <cpehuc$6hr$1@smc.vnet.net> <200412130922.EAA23340@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
This works but for a different reason, which is that the result holds
even if you do not assume that x is real (see my latest posting).
(If it were necessary that x is real your approach would not work since
ComplexExpand would not pass this information to FullSimplify.)
On 13 Dec 2004, at 18:22, Dr. Wolfgang Hintze wrote:
> I remember that ComplexExpand[expr] assumes parameters in expr to be
> real. Hence the following was the shortest succesful expansion I found:
>
> In[4]:=
> FullSimplify[ComplexExpand[E^(I*Pi*(1 + x))]]
>
> Out[4]=
> -E^(I*Pi*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.
>
> Regards,
> Wolfgang
>
> 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.
>>>
>>>
>>>
>>>
>>>
>>
>> Hello?
>>
>> FullSimplify[ComplexExpand[Exp[I*Pi*(1 + x)]],
>> x $B":(B Reals]
>>
>>
>> -E^(I*Pi*x)
>>
>>
>>
>> Andrzej Kozlowski
>> Chiba, Japan
>> http://www.akikoz.net/~andrzej/
>> http://www.mimuw.edu.pl/~akoz/
>>
>>
>
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