ASKSAGE: Sage Q&A Forum - Latest question feedhttp://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 27 Apr 2015 21:45:24 -0500numerical solutions from a for-loop solve()http://ask.sagemath.org/question/26661/numerical-solutions-from-a-for-loop-solve/I am trying to extract a list of 3-tuples which solve a system of equations. For the test I am using a simply equation whose solutions go from negative to positive around one, so
n, u = var('n, u')
sltns=solve([n + u==0, n + c*u - 1==0], n, u)
L = [0, 1/4, 1/2, 3/4, 1, 5/4, 6/4, 7/4, 2]
for c in L:
print (c,sltns[0].rhs(),sltns[1].rhs())
does not work.
What I am trying to get is, for example, if c = 1/2, then
(1/2, 2, -2]
because if c = 1/2 then n==2 and u==-2. And so on for all the elements of the list.
()
()
...
()
If I omit defining sltns, but use the solve() directly, I get out of range errors when for some c, u goes from positive to negative or vice versa for n, and I use [0].rhs(). Isn't sltns[2] the way to select, for example, [d==4] in [a==2, b==3, d==4], while sltns[0] and sltns[1] select the first two? Is it limited to some range positive or negative?
My question is:
1. What is the correct way to code the calculation above?
2. How would I alternatively code the loop if I needed the () s to be tuples in a matrix [(),(), ..., ()] for further calculations or plotting?gottfriedMon, 27 Apr 2015 21:45:24 -0500http://ask.sagemath.org/question/26661/