Post a reply
Username:
Note:If not registered, provide any username. For more comfort, register here.
Subject:
Message body:
Enter your message here, it may contain no more than 60000 characters. 

Smilies
:D :) :( :o :shock: :? 8) :lol: :x :P :oops: :cry: :evil: :twisted: :roll: :wink: :!: :?: :idea: :arrow: :| :mrgreen:
Font size:
Font colour
Options:
BBCode is ON
[img] is ON
[flash] is OFF
[url] is ON
Smilies are ON
Disable BBCode
Disable smilies
Do not automatically parse URLs
Confirmation of post
To prevent automated posts the board requires you to enter a confirmation code. The code is displayed in the image you should see below. If you are visually impaired or cannot otherwise read this code please contact the %sBoard Administrator%s.
Confirmation code:
Enter the code exactly as it appears. All letters are case insensitive, there is no zero.
   

Topic review - symbolic elimination
Author Message
  Post subject:  symbolic elimination  Reply with quote
Hi,
I've recently started to use Singular, v. 3.1.0. The system I'm solving comes from the power flow analysis. I've computed "numerical" Groebner Basis in Mathematica and was very surprised by its very simple form, so I'm wondering if there is a way to find a symbolic form of the polynomial of one variable (Mathematica seems to be unable to do it). It's going to be a huge formula, but I'll deal with it.
The script I wrote:
//---
ring R = 0,(y12,y13,y23,w12,w13,w23,s1,s2,s3,s4,u3,u1,u2,u4,u5),(ds(11), lp(4));
ideal I=
-(y12+y13)*u1*u4 + y12*u2*u4 + y13*u3*u4 - s1,
y12*u1*u5 - (y12+y23)*u2*u5 + y23*u3*u5 - s2,
-(w12+w13)*u1*u4 + w12*u1*u5 + w13*u3*u1 - s3,
w12*u2*u4 - (w12+w23)*u2*u5 + w23*u3*u2 - s4;
//option(prot);
option(redSB);
ideal Istd = simplify(std(I),1);
list L = facstd(Istd);
L;
//----
The result is the basis of 43 polynomials, but I cannot find any equivalent to numerical computation. My unknowns are u1,u2,u4,u5.
I'd appreciate any help!

Thanks,
Eugene
Post Posted: Wed Dec 18, 2013 2:29 pm


It is currently Tue Jul 17, 2018 3:11 pm
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group