Re: About C[i]
- To: mathgroup at smc.vnet.net
- Subject: [mg117693] Re: About C[i]
- From: Peter <petsie at dordos.net>
- Date: Wed, 30 Mar 2011 04:08:22 -0500 (EST)
- References: <imshf8$5r8$1@smc.vnet.net>
Am 29.03.2011 13:57, schrieb olfa:
> 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.
>
Hi,
this is far from being finite. You've put the property "being integer"
from C[1],C[2] to (i,k/7) and (i,(k-2)/7) respectively. It is just a bit
less effort to calculate e.g.:
Table[{{c2, 7 c1, 0, c1 + c2}, {c2, 2 + 7 c1, 2, c1 + c2}}, {c1, -5,
5}, {c2, -5, 5}]
than a table where you have to test for either k/7 or (k-2)/7 to be Integer.
hth,
Peter