|
[Date Index]
[Thread Index]
[Author Index]
Re: Reduce/Solve
- To: mathgroup at smc.vnet.net
- Subject: [mg49916] Re: [mg49912] Reduce/Solve
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Fri, 6 Aug 2004 03:09:32 -0400 (EDT)
- References: <200408051322.JAA05969@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On 5 Aug 2004, at 15:22, Dr A.H. Harker wrote:
>
> Dear All,
> I tend to think of Reduce as a more powerful tool than Solve,
> yet with
>
> eq = J == J0 (1 + r/d) Exp[-r/d]
> ss = Solve[eq, r]
> tt = Reduce[eq, r]
>
> I get useful output from Solve, but Reduce returns the expression
> unevaluated. What am I missing? Mathematica 5.0.0.0 under Windows XP.
>
> Tony
>
> Dr A.H. Harker
> Director of Postgraduate Studies
> Deputy Head, Condensed Matter and Materials Physics Group
> Department of Physics and Astronomy
> University College London
> Gower Street
> LONDON
> WC1E 6BT
> (44)(0)207 679 3404
> a.harker at ucl.ac.uk
>
>
>
>
Reduce is not "more power full than Solve". If this were true what need
would there be for Solve? Reduce attempts to give a complete solution
and when that is impossible it returns no answer. In your particular
case Solve produces a warning about using Inverse functions and tells
you that you may not have a complete solution. That tells you exactly
the reason why Reduce returns the original expression back to you.
Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/
Prev by Date:
RE: Problem with eval. of neg. cube root of neg. #
Next by Date:
Re: populate a list with random numbers from normal distribution?
Previous by thread:
Reduce/Solve
Next by thread:
Re: Reduce/Solve
|