Sorry...a more tidy path through the equations can look like this (but
you only know what "cherries to pick") by looking at the equations
yourself:
h[1] >>> var('a0 lam conc x y z')
(a0, lam, conc, x, y, z)
h[2] >>> neq=[w.subs(dict([(r,nsimplify(r, rational=True)) for r
in w.atoms(Real
)])).as_numer_denom()[0] for w in [z-(a0*(1 - x/2)*x),lam+2*y-
z-0.005*x/2*x, z-1
*y-0.743436700916726*y, x+y-conc]]
h[2] >>> ans=solve(neq[:3],[z,y,a0])
h[2] >>> xx=solve(neq[3].subs(ans),x)
h[2] >>> for k in ans:
... ans[k]=[ans[k].subs(x,xxi) for xxi in xx]
...
h[2] >>> ans[x]=xx
h[2] >>> for k in ans:
... print k, [ai.subs(lam,.1).subs(conc, .1).n(3) for ai in
ans[k]]
...
x [-103., 0.487]
a0 [-0.0332, -1.83]
y [103., -0.387]
z [180., -0.675]
The two values of x generate the two values for all the other
quantities. It's interesting that the denominator removal was not
necessary when done this time.
/c
--
You received this message because you are subscribed to the Google Groups
"sympy" group.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/sympy?hl=en.
