[sage-support] Re: symbolic substitution

2008-06-16 Thread vpv
Hello William, Thanks for your reply! I think I can use your idea to solve my problem. I tried to run your example in Sage for the case when x1,x2,x3,x5,x7 are in GF(2). I realized that in that case e1 and e2 should also be defined in GF(2). As I need to do a hundred substitutions in which all

[sage-support] Re: symbolic substitution

2008-06-16 Thread Marshall Hampton
There are probably more elegant ways to do this, but one way is: varstring = '' for i in range(1,101): varstring += 'e' + str(i)+ ',' bool100 = eval('BooleanPolynomialRing(100,' + varstring[0:-1] + ')') -M. Hampton On Jun 16, 10:43 am, vpv [EMAIL PROTECTED] wrote: Hello William, Thanks

[sage-support] Re: symbolic substitution

2008-06-16 Thread William Stein
On Mon, Jun 16, 2008 at 9:43 AM, vpv [EMAIL PROTECTED] wrote: Hello William, Thanks for your reply! I think I can use your idea to solve my problem. I tried to run your example in Sage for the case when x1,x2,x3,x5,x7 are in GF(2). I realized that in that case e1 and e2 should also be

[sage-support] Re: Why so slow?

2008-06-16 Thread Robert Bradshaw
I am sure this is the issue. Your phi here is probably going through maxima (and probably trying to simplify symbolically) because of the sqrt() and pi). If you do something like sage: p(n,y) = 1/(pi*sqrt(2*n-y^2)) sage: plot(p(5, x)^2, (x,-5,5)) it should be acceptable. On Jun 15, 2008,