Updates:
Labels: Solvers
Comment #1 on issue 472 by [email protected]: use numerical solver
automatically?
http://code.google.com/p/sympy/issues/detail?id=472
This shouldn't be necessary since the solver can solve this equation. But
probably a hint is needed to make it work:
from sympy.solvers.solvers import _tsolve as ts
e=exp(1)
a = 0.004*(8*e**(-300*t)
-8*e**(-1200*t))*(720000*e**(-300*t)-11520000*e**(-
1200*t)) +0.004*(9600*e**(-1200*t) - 2400*e**(-300*t))**2
ta=terms_gcd(a)
ta
(46080.0 - 576000.0*exp(-900*t) + 737280.0*exp(-1800*t))*exp(-600*t)
_.subs(e**(-600*t),u)
u*(-576000.0*u**(3/2) + 737280.0*u**3 + 46080.0)
solve(_,u)
[0, 0.201541162404989, 0.781429110349124]
[ts(w-e**(-600*t),t) for w in _]
[[zoo], [0.00266960273088699], [0.000411051404934985]]
ans = _[1:]
[a.subs(t,ai[0]).n(2) for ai in ans]
[-2.8e-11, 8.0e-11]
_tsolve has to be used or else several hundred results are going to come
back:
solve(e**(-t)-x,t)
[log(1/x)]
solve(e**(-2*t)-x,t)
[log(-sqrt(1/x)), log(sqrt(1/x))]
solve(e**(-3*t)-x,t)
[log(-(1/x)**(1/3)/2 - sqrt(3)*I*(1/x)**(1/3)/2), log(-(1/x)**(1/3)/2 +
sqrt(3)*
I*(1/x)**(1/3)/2), log((1/x)**(1/3))]
Should a hint like 'principle' or 'real' be used here to just get a simple
inversion of the e**(-600*t) - u -> t = log(u)/-600 ?
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