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Topic review - Resultant
Author Message
  Post subject:  Re: Resultant  Reply with quote
Sorry, option(prot); does not show anything for resultant.
You can use option(mem); to see the memory usage.
Post Posted: Sat Mar 06, 2010 4:37 am
  Post subject:  Re: Resultant  Reply with quote
The resultant computation is perhaps too long. To see that Singular indeed computes something, and to watch the progress of a computation, you should type option(prot); before the computation starts.

To your second question:
int n = 37; (n must be here a concete integer)
ring r = 0,(a(1..n),s),dp;
int ii;
poly x = s^n;
for( ii=n; ii>0; ii--)
{
x = x + (-1)^n*a(ii)*s^(n-ii);
}

GMG
Post Posted: Sat Mar 06, 2010 4:32 am
  Post subject:  Resultant  Reply with quote
I want to compute Resultant of two polynomial:

ring r = (0,t), (a(1..3),b(1..3),c(1..3),d(1..3),s) , lp;
poly x = s^3 + a(3)*s^2 + a(2)*s + a(3);
poly y = s^3 + b(3)*s^2 + b(2)*s + b(3);
poly z = s^3 + c(3)*s^2 + c(2)*s + c(3);
poly u = s^3 + d(3)*s^2 + d(2)*s + d(3);
poly f = x^2 - s*u^2 + s - (s^2*u^2-s)*y^2;
poly g = s*x^2 - 2*s^2*u^2 + 1 - s*(s^2*u^2 - s)*z^2;
resultant (f,g,s);

But it seems that Singular does not give me the answer, since there comes nothing. Can you help me please?
--------------
By the way i have a question, as above, but i want to have a(1..n), b(1..n), c(1..n), d(1..n), so i have to define

int n;

But then i have trouble with writting down polynomials, e.g. i want to define

x = s^n + \sum_{i=1}^{n} a_i * s^{n-i}

How can i represent it in SINGULAR?
Post Posted: Thu Mar 04, 2010 9:10 pm


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