what's wrong with this?!!
olfa
this is the system to solve:
Reduce[
Element[{iP, N}, Integers] &&
Not[(iP < N)] &&
Exists[{iPP}, (iPP < N) && iP == iPP + 1],
{iP}, Backsubstitution -> True]
the output is:
(N | iP) =E2=88=88 Integers && N =E2=89=A4 iP < 1 + N
but it should be:
iP==N
so what's the problem what is missing to get the correct output?
thank you very much.
Adam Strzebonski
the documentation page, more precisely, the basic design part
(above Examples) of the documentation page, was last modified.
The basic design of Reduce has not changed since Version 5.
New solving methods and bug fixes have been added to Reduce
in every version, and new documentation examples have been
added since version 5 as well.
The V5 output
In[1]:= Reduce[
Element[{iP, N}, Integers] && Not[(iP < N)] &&
Exists[{iPP}, (iPP < N) && iP == iPP + 1],
{iP}, Backsubstitution -> True]
Out[1]= (N | iP) \[Element] Integers && N <= iP < 1 + N
describes the correct solution set. It is not the simplest
possible description, since for integers N and iP,
N <= iP < 1 + N is equivalent to iP == N.
The result has been improved in V6.
Out[1]= (N | iP) \[Element] Integers && iP == N
Best regards,
Adam Strzebonski
Wolfram Research
Andrzej Kozlowski wrote:
> First, the information that Reduce has not been modified since version 5 (which indeed is what the documentation for Reduce suggests) is clearly false. In fact Reduce can do things in version 7 that it could not do in earlier ones, e.g
>
> Reduce[Sin[x] == x && Abs[x] < 1, x]
>
> x == 0
>
> $Version
>
> "8.0 for Mac OS X x86 (64-bit) (November 6, 2010)"
>
> Reduce[Sin[x] == x && Abs[x] < 1, x]
>
> During evaluation of In[1]:= Reduce::nsmet:This system cannot be \
> solved with the methods available to Reduce. >>
>
> Reduce[Sin[x] == x && Abs[x] < 1, x]
>
> $Version
>
> "6.0 for Mac OS X x86 (64-bit) (May 21, 2008)"
>
> So Reduce has certainly been "modified", at least between versions 6 and 7. In addition there have been bug fixes (which I presume does not count as "modification") and some functions called up by Reduce have been modified (which again perhaps does not count as a modification of Reduce).
>
> However, in this case Mathematica 6.03 gives the same answer as Mathematica 8. I can't check what happens with Mathematica 5 as I no longer have it installed.
>
> I am curious about the examples you claim work with Mathematica 5 but not with 8. Perhaps they only appear to work in 5 (or appear not to work in 8)?
>
> Andrzej Kozlowski
>
>
>
> On 22 Dec 2010, at 11:18, olfa mraihi wrote:
>
>> So what's wrong with Reduce in Mathematica 5 to get the output Element[n | iP, Integers] && n<= iP < 1 + n?
>>
>> knowing that Reduce has not been modified in Mathematica 8 so why does it give the correct output which is Element[n | iP, Integers] && iP == n?
>> I 'm using Mathematica 5 and I'm constrained to use it because some of my examples work on Mathematica 5 but not on Mathematica 8.
>>
>> Thank you.
>>
>> --- En date de : Mer 22.12.10, Andrzej Kozlowski <ak...@mimuw.edu.pl> a écrit :
>>
>> De: Andrzej Kozlowski <ak...@mimuw.edu.pl>
>> Objet: Re: what's wrong with this?!!
>> Ŕ: "olfa" <olfa....@yahoo.fr>
>> Cc: math...@smc.vnet.net
>> Date: Mercredi 22 décembre 2010, 9h48
>>
>> The output I get (with Mathematica 8) is:
>>
>> In[76]:= Reduce[Element[{iP, n}, Integers] && !iP < n &&
>> Exists[{iPP}, iPP < n && iP == iPP + 1], {iP},
>> Backsubstitution -> True]
>>
>> Element[n | iP, Integers] && iP == n
>>
>> Andrzej Kozlowski
Roland Franzius
> Hi mathematica community,
>
> this is the system to solve:
> Reduce[
> Element[{iP, N}, Integers]&&
> Not[(iP< N)]&&
> {iP}, Backsubstitution -> True]
>
> the output is:
>
> but it should be:
> iP==N
>
> so what's the problem what is missing to get the correct output?
>
> thank you very much.
>
0. commandment of Mathematica: You shall not make wrongful use of N
--
Roland Franzius
Andrzej Kozlowski
x == 0
$Version
$Version
Andrzej Kozlowski
> So what's wrong with Reduce in Mathematica 5 to get the output =
Element[n | iP, Integers] && n<= iP < 1 + n?
>
> knowing that Reduce has not been modified in Mathematica 8 so why does =
it give the correct output which is Element[n | iP, Integers] && iP ===
n?
> I 'm using Mathematica 5 and I'm constrained to use it because some of =
my examples work on Mathematica 5 but not on Mathematica 8.
>
> Thank you.
>
> --- En date de : Mer 22.12.10, Andrzej Kozlowski <ak...@mimuw.edu.pl> a =
=E9crit :
>
> De: Andrzej Kozlowski <ak...@mimuw.edu.pl>
> Objet: Re: what's wrong with this?!!
> =C0: "olfa" <olfa....@yahoo.fr>
> Cc: math...@smc.vnet.net
> Date: Mercredi 22 d=E9cembre 2010, 9h48
>
> The output I get (with Mathematica 8) is:
>
> In[76]:= Reduce[Element[{iP, n}, Integers] && !iP < n &&
> Exists[{iPP}, iPP < n && iP == iPP + 1], {iP},
> Backsubstitution -> True]
>
> Element[n | iP, Integers] && iP == n
>
> Andrzej Kozlowski
>
> On 22 Dec 2010, at 08:36, olfa wrote:
>
> > Hi mathematica community,
> >
> > this is the system to solve:
> > Reduce[
> > Element[{iP, N}, Integers] &&
> > Not[(iP < N)] &&
> > Exists[{iPP}, (iPP < N) && iP == iPP + 1],
> > {iP}, Backsubstitution -> True]
> >
> > the output is:
> > (N | iP) =E2=88=88 Integers && N =E2=89=A4 iP < 1 + N
olfa mraihi
So what's wrong with Reduce in Mathematica 5 to get the output Element[n
I 'm using Mathematica 5 and I'm constrained to use it because some of my examples work on Mathematica 5 but not on Mathematica 8.