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

**References**:**Re: Simplification question***From:*Curtis Osterhoudt <gardyloo@mail.wsu.edu>