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: ReplaceAll / sequential replacements versus joint replacement

  • To: mathgroup at smc.vnet.net
  • Subject: [mg55448] Re: [mg55429] ReplaceAll / sequential replacements versus joint replacement
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 24 Mar 2005 03:41:48 -0500 (EST)
  • References: <200503231034.FAA03987@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 23 Mar 2005, at 11:34, Ronald Wendner wrote:

> Suppose there is an expression e1 which depends, among others, on 
> variables
> a and b.
> Replace both a and b by zero:
>
> e1/.{a->0,b->0} Mathematica gives, say, result1.
>
> Next, consider
>
> e2 = e1/.{a->0};
> e2/.{b->0} Mathematica gives, say, result2.
>
> How can result 1 differ from result 2?
>
> For my specific problem, I also attach a short Mathematica notebook.
> [Contact the author to obtain the notebook - moderator]
>
> I'd appreciate any help.
>
> Many thanks,
> Ron
>
>
They certianly can differ, for example:


(a - 2*b)/(a - b) /. {b -> 0, a -> 0}

Indeterminate

vs

(a - 2*b)/(a - b) /. a -> 0 /. b -> 0

2


Andrzej Kozlowski
Chiba, Japan
http://www.akikoz.net/andrzej/index.html
http://www.mimuw.edu.pl/~akoz/


  • Prev by Date: Re: ReplaceAll / sequential replacements versus joint replacement
  • Next by Date: Hypergeometric integral looks wrong ?
  • Previous by thread: Re: ReplaceAll / sequential replacements versus joint replacement
  • Next by thread: Re: ReplaceAll / sequential replacements versus joint replacement