Re: Refine, assumptions, domains
- To: mathgroup at smc.vnet.net
- Subject: [mg101765] Re: Refine, assumptions, domains
- From: Szabolcs <szhorvat at gmail.com>
- Date: Thu, 16 Jul 2009 08:20:37 -0400 (EDT)
- References: <h3hjkc$1ue$1@smc.vnet.net> <h3hqqr$5l3$1@smc.vnet.net>
On Jul 15, 2:08 pm, Richard Fateman <fateman at cs.berkeley.edu> wrote: > Jon McLoone wrote: > > How about... > > > 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[1] 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]]} > > The answer comes out > {0 || Pi, 2*Pi || 3*Pi} > > which is pretty close; % /. Or->List //Flatten does the rest, give= n > that particular form. > Try this: In[8]:= Reduce[Sin[x] == 0 && 0 < x < 10, x] Out[8]= x == \[Pi] || x == 2 \[Pi] || x == 3 \[Pi] In[9]:= x /. {ToRules[%]} Out[9]= {\[Pi], 2 \[Pi], 3 \[Pi]} In Mathematica 7 this works even for equation that can only solved numerically. In Mathematica 6, it will only work for those eqs that the system can handle symbolically.
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- From: Andrzej Kozlowski <akozlowski@gmail.com>
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