use numerical solver automatically?

[certik]

bc.. This is taken from: http://trac.sagemath.org/sage_trac/ticket/1235 In [1]: e = exp(1)

In [2]: t = Symbol("t")

In [3]: a = 0.004_(8_e(-300_t) - 8_e(-1200t))(720000_e(-300_t)-11520000_e(-1200t)) + 0.004(9600_e(-1200_t) - 2400_e(-300t))*2

In [4]: a Out[4]: 2
⎛ -300_t -1200t⎞ ⎛
0.004000000000000000083266726847⎝- 2400_ℯ + 9600*ℯ ⎠ + ⎝- 115200

-1200*t           -300*t⎞ ⎛                                   -1200*t     

00_ℯ + 720000ℯ ⎠⎝- 0.03200000000000000066613381478_ℯ + 0.

                           -300*t⎞

03200000000000000066613381478*ℯ ⎠

In [5]: solve(a==0, t)

Not a polynomial equation. Can't solve it, yet.Traceback (most recent call last)

/home/ondra/sympy/

/home/ondra/sympy/sympy/solvers/solvers.py in solve(eq, syms, simplified) 73 solutions = list(set(roots(equ, syms[0]))) 74 except PolynomialException: ---> 75 raise "Not a polynomial equation. Can't solve it, yet." 76 77 if simplified == True:

Not a polynomial equation. Can't solve it, yet.: None

But it should call the iterative solver to return at least an approximative answer. Or it should return some kind of RootOf() or something, so that the user can manipulate with the answer further.

p. Original issue for "#3571":https://github.com/sympy/sympy/issues/3571: "http://code.google.com/p/sympy/issues/detail?id=472":http://code.google.com/p/sympy/issues/detail?id=472

p. Original author: "https://code.google.com/u/104039945248245758823/":https://code.google.com/u/104039945248245758823/

[smichr]

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 ?

**Labels:** Solvers  

Original comment: http://code.google.com/p/sympy/issues/detail?id=472#c1 Original author: https://code.google.com/u/117933771799683895267/

[asmeurer]

**Status:** Valid  

Original comment: http://code.google.com/p/sympy/issues/detail?id=472#c2 Original author: https://code.google.com/u/asmeurer@gmail.com/

[oscarbenjamin]

Still valid:

In [58]: a = (-2400*exp(-300*t) + 9600*exp(-1200*t))**2/250 + (4*exp(-300*t)/125 - 4*exp(-1200*t)/125)*(720000*exp(-300*t) - 11520000*exp(-1200*t))

In [59]: a
Out[59]: 
                                2                                                                
⎛        -300⋅t         -1200⋅t⎞    ⎛   -300⋅t      -1200⋅t⎞                                     
⎝- 2400⋅ℯ       + 9600⋅ℯ       ⎠    ⎜4⋅ℯ         4⋅ℯ       ⎟ ⎛        -300⋅t             -1200⋅t⎞
───────────────────────────────── + ⎜───────── - ──────────⎟⋅⎝720000⋅ℯ       - 11520000⋅ℯ       ⎠
               250                  ⎝   125         125    ⎠                                     

In [60]: solve(a, t)    # <----- hangs
^C---------------------------------------------------------------------------
KeyboardInterrupt

In [61]: solve(a.subs(t, z/300), z)
Out[61]: 
⎡   ⎛3 ___ 3 ____________⎞             ⎛3 ___ 3 ____________⎞             ⎛3 ___ 3 ____________⎞             ⎛3 ___ 3 __________
⎢   ⎜╲╱ 2 ⋅╲╱ 25 - 3⋅√41 ⎟   2⋅ⅈ⋅π     ⎜╲╱ 2 ⋅╲╱ 25 - 3⋅√41 ⎟   2⋅ⅈ⋅π     ⎜╲╱ 2 ⋅╲╱ 3⋅√41 + 25 ⎟   2⋅ⅈ⋅π     ⎜╲╱ 2 ⋅╲╱ 3⋅√41 + 2
⎢log⎜────────────────────⎟ - ─────, log⎜────────────────────⎟ + ─────, log⎜────────────────────⎟ - ─────, log⎜──────────────────
⎣   ⎝         2          ⎠     3       ⎝         2          ⎠     3       ⎝         2          ⎠     3       ⎝         2        

__⎞             ⎛    ____________⎞     ⎛    ____________⎞⎤
5 ⎟   2⋅ⅈ⋅π     ⎜   ╱ 25   3⋅√41 ⎟     ⎜   ╱ 3⋅√41   25 ⎟⎥
──⎟ + ─────, log⎜3 ╱  ── - ───── ⎟, log⎜3 ╱  ───── + ── ⎟⎥
  ⎠     3       ⎝╲╱   4      4   ⎠     ⎝╲╱     4     4  ⎠⎦