Re: Solve with assumptions
- To: mathgroup at smc.vnet.net
- Subject: [mg58293] Re: Solve with assumptions
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Sun, 26 Jun 2005 01:33:55 -0400 (EDT)
- Organization: The Open University, Milton Keynes, England
- References: <d86937$c7n$1@smc.vnet.net><d89261$s7f$1@smc.vnet.net> <d9isr2$cpk$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Mukhtar Bekkali wrote:
> Great.
>
> How do you pull the solution out from Reduce? That is, with Solve I can
> have x=x/.Solve[f[x]==0,x] and get x and use it later on. The output
> of Reduce is identity, x==number.
>
Hi Mukhtar,
*ToRules* is your friend in this case. For example,
In[1]:=
Reduce[x^2 - 1 == 0, x]
Out[1]=
x == -1 || x == 1
In[2]:=
ToRules[%]
Out[2]=
Sequence[{x -> -1}, {x -> 1}]
In[3]:=
{%}
Out[3]=
{{x -> -1}, {x -> 1}}
In[4]:=
x /. %
Out[4]=
{-1, 1}
Best regards,
/J.M.