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Re: Re: Simplification question


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
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University of Massachusetts                413 545-2859 (W)
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Amherst, MA 01003-9305


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