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: [mg59067] Re: [mg59032] Simplification question
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Thu, 28 Jul 2005 02:26:14 -0400 (EDT)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

The rules that it applies don't come up with a simpler form

LeafCount[Cos[3*m*Pi]]

5

Simplify[Cos[3*m*Pi]//TrigToExp,Assumptions->Element[m,Integers]]

(-1)^(-3*m)

LeafCount[%]

5

Simplify[#,Assumptions->Element[m,Integers]]&/@(Cos[3*m*Pi]//
TrigExpand)

(-1)^(3*m)

LeafCount[%]

5


Bob Hanlon

> 
> From: "snoofly" <snoofly at snoofly.com>
To: mathgroup at smc.vnet.net
> Date: 2005/07/27 Wed AM 01:25:00 EDT
> Subject: [mg59067] [mg59032] Simplification question
> 
> 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.
> 
> 


  • Prev by Date: Re: Operating with binary numbers
  • Next by Date: Re: silly questions?
  • Previous by thread: Re: Re: Simplification question
  • Next by thread: Re: Simplification question