Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2005
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2005

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Simplification question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59072] Re: Simplification question
  • From: Peter Pein <petsie at dordos.net>
  • Date: Thu, 28 Jul 2005 02:26:29 -0400 (EDT)
  • References: <dc76pp$jvg$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

snoofly schrieb:
> 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.
> 
see http://groups.google.com/groups?selm=dc16i2%246q5%241 at online.de&rnum=1

the behaviour of Mathematica is rather strange (in my eyes):

 FullSimplify[(-1)^(3*m), m \[Element] Integers]
yields
 Cos[3*m*Pi]

-- 
Peter Pein
Berlin
http://people.freenet.de/Peter_Berlin/


  • Prev by Date: Re: Simplification question
  • Next by Date: Re: How to simplify an expression in version 5
  • Previous by thread: Re: Simplification question
  • Next by thread: Re: Simplification question