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: symbolic approximation (formular manipulation)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55018] Re: symbolic approximation (formular manipulation)
  • From: dh <dh at metrohm.ch>
  • Date: Thu, 10 Mar 2005 05:24:18 -0500 (EST)
  • References: <d0mnef$71b$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hello Zhe,
Try to formulate clearly what you want to do and the answer is obvious.
Because R1<<R2 you want to replace R1+R2 by R2
Because R2<<R3 you want to replace R2+R3 by R3
This can be done by ReplaceALL, abbreviated by /.

(R1+R2)/(R1(R2+R3))  /. {(R1+R2)->R2, R2+R3->R3}

have fun, Daniel

Zhe Hu wrote:
> A formula in electrical engineering can be simplified after assuming,
> e.g. R1<<R2, within 10% error tolerance.
> 
> How this kind of "simplify" or approximation can be performed by Mathematica?
> 
> For example, R1<<R2, R2<<R3
> 
> R1+R2                      R2                                      R2
> ---------------  --->  ------------------  ---------> ------------------
> R1(R2+R3)               R1(R2+R3)                       R1 R3
> 
> 
> 
> I thought about replacing: R1->0.00001*R2, but that would eliminate R1
> completely from the formula (denumerator), which is not favorable in
> some cases.
> 


  • Prev by Date: Re: symbolic approximation (formular manipulation)
  • Next by Date: Re: String comparison
  • Previous by thread: Re: symbolic approximation (formular manipulation)
  • Next by thread: Re: symbolic approximation (formular manipulation)