Re: About C[i]
- To: mathgroup at smc.vnet.net
- Subject: [mg117690] Re: About C[i]
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Wed, 30 Mar 2011 04:07:50 -0500 (EST)
olfa wrote:
> Hi Mathematica community,
>
> I have this solution set:
> ((C[1] | C[2]) \[Element] Integers && i == C[2] && k == 7 C[1] &&
> kP == 0 && iP == C[1] + C[2]) ||((C[1] | C[2]) \[Element] Integers
> &&
> i == C[2] && k == 2 + 7 C[1] && kP == 2 && iP == C[1] + C[2])
>
> it means that it is an infinite solution set.
> I need to expess it without C[1] and C[2] since C[1]==k/7||(k-2)/7
> and C[2]==i . consequently the solution set would be expressed as
> (kP==0 && iP==k/7+i)||(kP==2&&iP==(k-2)/7+i)
> is it possible? how?can we say that is now transformed into finite
> set?
>
> thank you very much for your help.
In[89]:= Eliminate[
Reduce[(i == C[2] && k == 7 C[1] && kP == 0 &&
iP == C[1] + C[2]) || (i == C[2] && k == 2 + 7 C[1] && kP == 2 &&
iP == C[1] + C[2]), {i, k, kP, iP, C[1], C[2]}], {C[1], C[2]}]
Out[89]= (7 i == 7 iP - k && kP == 0) || (7 i == 2 + 7 iP - k &&
kP == 2)
To answer the last question, no, it does not look like a finite set.
Both k and i can take on infinitely many values, and they parametrize
the solution set.
Daniel Lichtblau
Wolfram Research