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Re: Re: Simplification question
*To*: mathgroup at smc.vnet.net
*Subject*: [mg59121] Re: [mg59057] Re: [mg59032] Simplification question
*From*: Murray Eisenberg <murray at math.umass.edu>
*Date*: Fri, 29 Jul 2005 00:42:05 -0400 (EDT)
*Organization*: Mathematics & Statistics, Univ. of Mass./Amherst
*References*: <200507280626.CAA12220@smc.vnet.net>
*Reply-to*: murray at math.umass.edu
*Sender*: owner-wri-mathgroup at wolfram.com
One can use TrigExpand first:
Simplify[#, Assumptions->{m â?? Integers}] & /@ TrigExpand[Cos[3 m Pi]]
(-1)^(3*m)
But Mathematica doesn't want to simpify (-1)^(3m) to (-1)^m:
Simplify[(-1)^(3m), Assumptions -> {m â?? Integers}]
(-1)^(3*m)
Amusingly:
FullSimplify[(-1)^(3m), Assumptions -> {m â?? Integers}]
Cos[3*m*Pi]
Curtis Osterhoudt wrote:
> My only guess is that Cos can return either a positive or negative 1
> depending on the integer. It might be nice if Mathematica could return a
> piecewise answer depending on what m was, but ...
>
> snoofly wrote:
>
>
>>Can someone explain this to me please.
>>
>>Clear[m]
>>
>>Simplify[Sin[3 m Pi], Assumptions -> m \[Element] Integers]
>>0
>>
>>Simplify[Cos[3 m Pi], Assumptions -> m \[Element] Integers]
>>
>>Cos[3 m \[Pi]]
>>
>>I'm not sure why Mathematica cannot deduce that the second simplification
>>should be (1)^m.
>>
>>
>>
>>
>
>
>
>
--
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
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