Groups > comp.soft-sys.math.mathematica > #1244

Re: About C[i]

From	Peter <petsie@dordos.net>
Newsgroups	comp.soft-sys.math.mathematica
Subject	Re: About C[i]
Date	2011-03-30 09:08 +0000
Organization	Steven M. Christensen and Associates, Inc and MathTensor, Inc.
Message-ID	<imuru2$hr5$1@smc.vnet.net> (permalink)
References	<imshf8$5r8$1@smc.vnet.net>

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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

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Re: About C[i] Peter <petsie@dordos.net> - 2011-03-30 09:08 +0000

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