I think I am in agreement with Vinzent, this would probably solve the
issue I am having now...
I am trying to obtain the solutions using the linex below. When I
solve for the constants assigned to the variables, I get complex
solutions with very small imaginary components, that should be real
solutions. I assume this happens due to some precision settings. And
when I try to solve it symbolically, it hangs....
The equation to solve is the numerator of function I got using the
_.as_numer_denom()
>>> from sympy import var,solve
>>> #where 'a', 'e', 'c', 'x', 'o', 'v', 'b', 's', 'w', 'd', 'f' are constants,
>>> and want to solve it for 'r'
>>> a=5.75
>>> e=8e-6
>>> c=0.25
>>> x=20.
>>> o=0.02
>>> v=0.005
>>> b=0.12
>>> s=0.23
>>> w=0.23
>>> d=0.
>>> f=20.
>>>
>>> var('r')
r
>>>
>>> myEqConstants=-4*a**2*r**2*e**2*(0.002*c*x*(o + v) + 2*a*b*r*e + 2*a*s*r*e
>>> + 2*a*w*r*e + a*e*r**2)**2 -4*a**2*r**2*e**2*(0.002*c*x*(o + v) -
>>> (2*a*b*r*e + 2*a*s*r*e + 2*a*w*r*e + a*e*r**2))**2 +16*a**4*d**2*r**4*e**4
>>> + 16*a**4*r**4*f**2*e**4
>>>
>>> mySols=solve(myEqConstants,r)
>>> print [aa.evalf() for aa in mySols]
[27.1235615040284 - 2.75e-13*I, -29.4436689512413 - 2.53e-13*I,
0.192368597088186 + 2.7527e-13*I, 0, -0.192261149875308 +
2.5278e-13*I]
>>>
>>> var('a r e c x o v b s w d f')
(a, r, e, c, x, o, v, b, s, w, d, f)
>>>
>>> myEqSymbolic=-4*a**2*r**2*e**2*(0.002*c*x*(o + v) + 2*a*b*r*e + 2*a*s*r*e +
>>> 2*a*w*r*e + a*e*r**2)**2-4*a**2*r**2*e**2*(0.002*c*x*(o + v) - (2*a*b*r*e +
>>> 2*a*s*r*e + 2*a*w*r*e + a*e*r**2))**2+16*a**4*d**2*r**4*e**4 +
>>> 16*a**4*r**4*f**2*e**4
>>>
>>> mySols=solve(myEqSymbolic,r)
Thanks for your help!
regards, anartz
On Aug 6, 6:26 pm, Vinzent Steinberg
<vinzent.steinb...@googlemail.com> wrote:
> On 6 Aug., 19:57, smichr <smi...@gmail.com> wrote:
>
>
>
> > On Aug 5, 9:29 pm, Vinzent Steinberg
>
> > <vinzent.steinb...@googlemail.com> wrote:
> > > On Aug 2, 12:14 am, smichr <smi...@gmail.com> wrote:
>
> > > > >>> num,den = (1/(.001+a)**3-6/(.9-a)**3).as_numer_denom()
> > > > >>>nsolve(num,a,.3) # no need for sympy now
>
> > > Maybe solve()/nsolve() should do this for you. What do you think?
>
> > > Vinzent
>
> > I was thinking the same thing, but you would have to watch out that
> > you didn't introduce spurious roots: e.g.
>
> > >>> (x/(x-1)).diff(x)
>
> > -1/(1 - x) - x/(1 - x)**2>>> n,d=_.as_numer_denom();n/d
>
> > (-x*(1 - x) - (1 - x)**2)/(1 - x)**3
>
> > Unless we cancel out the common factors (something that isn't working
> > in general yet but will hopefully work with the new polys module) we
> > will get a root of x=1.
>
> It should be easy to check for false roots numerically, but yeah, it
> has to be implemented.
>
> > I was also wondering if the routine shouldn't
> > try to solve the expression symbolically to see if it's possible
> > before trying to solve it numerically...or should sympy just trust the
> > user and not waste the time checking symbolic solutions if a numeric
> > solution has been requested?
>
> It would do it this way: The user wants to solve() an equation, if we
> can't do it symbolically, we return an instance of RootOf(), which can
> be evalf()'ed to get a numerical approximation.nsolve() should solve
> it always numerically, possibly doing some symbolic optimizations,
> which can be disabled.
>
> BTW: It's now fixed to find the root of 1/(0.001+a)**3-6/(0.9-a)**3.
>
> Vinzent
>
> Vinzent
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