       Re: Re: Refine, assumptions, domains

• To: mathgroup at smc.vnet.net
• Subject: [mg101750] Re: [mg101715] Re: Refine, assumptions, domains
• From: Andrzej Kozlowski <akozlowski at gmail.com>
• Date: Thu, 16 Jul 2009 08:17:38 -0400 (EDT)
• References: <h3hjkc\$1ue\$1@smc.vnet.net> <h3hqqr\$5l3\$1@smc.vnet.net> <200907151108.HAA16411@smc.vnet.net>

```On 15 Jul 2009, at 04:08, Richard Fateman wrote:

> Jon McLoone wrote:
>>
>> n Pi /. {ToRules[Reduce[0 < n*Pi < 10 && Element[n, Integers], n]]}
>>
>> On Jul 14, 10:37 am, Richard Fateman <fate... at cs.berkeley.edu> wrote:
>>> What I'm looking for is a simple way to obtain a finite list
>>>
>>> { Pi, 2 Pi, 3 Pi}
>>>
>>> from  this information:
>>>
>>>   0< n*Pi < 10,  Element[n,Integers]
>>>
>>> Refine doesn't do this, at least with Mathematica 6.0
>>> Suggestions? (cc to fate... at gmail.com would be nice.)
>>
>>
>
> Thanks for the suggestions!
>
> What I really wanted was a way of finding the solutions of any
> equation,
> say Sin[x]==0 that lie in a particular range, say the range 0 to 10.
> Naturally, I prefer that this be done symbolically and exactly rather
> than numerically, when possible.
>
> If we try
>
> Reduce[Sin[x]==0,x].
>
> After replacing the dummy name C by n, the answer condition is
> Element[n, Integers] && (x == 2*n*Pi || x == Pi + 2*n*Pi)
>
> which is correct but clumsy.
>
> Automatically mapping this into Jon's construction, is hairier.
> If we can manage to get this...
> (2 n Pi || Pi + 2 n Pi) /. {ToRules[
>    Reduce[((0 < 2*n*Pi < 10) || (0 < Pi + 2*n*Pi < 10)) &&
>      Element[n, Integers], n]]}
>
> {0 || Pi, 2*Pi || 3*Pi}
>
> which is pretty close;   % /. Or->List //Flatten  does the rest, given
> that particular form.
>
> Thanks.
>

It may be just me but I can't understand what you are trying to do at
all. If you just want:

> a way of finding the solutions of any equation,
> say Sin[x]==0 that lie in a particular range

then what's wrong with:

x /. {ToRules[Reduce[Sin[x] == 0 && 0 < x < 10, x]]}

{Pi, 2*Pi, 3*Pi}

???

Andrzej Kozlowski

```

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