On Nov 20, 12:48 am, Chris Smith <smi...@gmail.com> wrote:
> On Sun, Nov 20, 2011 at 10:41 AM, litghost <litgh...@gmail.com> wrote:
> > I am new to sympy, and trying to utilize solve to solve systems of
> > linear equations. However there a couple things I have been unable to
> > figure out.
>
> > The most important is how get diagnostic information about why the
> > solve call failed (returned None). The system of equations I am
> > solving for are pretty simple, I am problem just miss using the
> > interface. However with the only output of the system being returning
> > None, I have no way to figure out my mistake.
>
> > So here is an example I am trying:
>
> > from sympy import Symbol, solve
> > C = Symbol('C', constant=True, real=True)
> > R = Symbol('R', constant=True, real=True)
> > Rf = Symbol('Rf', constant=True, real=True)
> > Ri = Symbol('Ri', constant=True, real=True)
> > s = Symbol('s')
> > V1 = Symbol('V1')
> > Vminus = Symbol('Vminus')
> > Vplus = Symbol('Vplus')
> > Vout = Symbol('Vout')
>
> The assumption 'constant=True' doesn't have a meaning to sympy. Also,
> since you are not solving this with any values (everything is
> symbolic) you don't have to put assumptions about the answer being
> real on your symbols. Just create your variables...
>
> >>> var('R, C, Ri, Vout, V1, Rf, Vminus, Vplus, s')
>
> (R, C, Ri, Vout, V1, Rf, Vminus, Vplus, s)
>
> define your equations...
>
> >>> eqs = [C*V1*s + Vplus*(-2*C*s - 1/R),
>
> ... Vminus*(-1/Ri - 1/Rf) + Vout/Rf,
> ... C*Vplus*s + V1*(-C*s - 1/R) + Vout/R,
> ... -Vminus + Vplus]
> ...
>
>
>
> Now, if you just want *some* solution, don't list any variables:
>
> >>> solve(eqs)
>
> [{Vminus: Vplus, Vout: (V1**2 - V1*Vplus - Vplus**2)/(V1 - 2*Vplus), R:
> Vplus/(C
> *s*(V1 - 2*Vplus)), Ri: Rf*Vplus*(V1 - 2*Vplus)/(V1 - Vplus)**2}]
>
> If you want to solve some equation for a single variable, use 'manual'
>
> >>> solve(eqs, Vout, manual=True)
>
> [(Vminus*(Rf + Ri)/Ri,)]
>
> > What I want to see is:
> > Vout = Vplus*(s**2*C**2 + 3*s*C/R + 1/R^2)/(s*C/R)
>
> Ahh... then that means you want the solution (lhs) to *not* contain
> those values. So what *do* you want to allow on the lhs?
>
> Here are all the variables in the equations...
>
> >>> vall=reduce(set.union, [e.free_symbols for e in eqs])
> >>> vall
>
> set([R, C, Ri, Vout, V1, Rf, Vminus, Vplus, s])
>
> You *don't* want the variables in the rhs of what you just indicated
> to be on the lhs of the answers...
>
> >>> dont = (Vplus*(s**2*C**2 + 3*s*C/R + 1/R**2)/(s*C/R)).free_symbols
> >>> dont
>
> set([R, Vplus, s, C])
>
> So what you will accept on the lhs is anything else...
>
> >>> want = vall - dont
> >>> want
>
> set([Vminus, Vout, V1, Rf, Ri])
>
> So try that:
>
> >>> solve(eqs,want)
>
> [{Vminus: Vplus, Vout: Vplus*(C**2*R**2*s**2 + 3*C*R*s + 1)/(C*R*s), V1:
> Vplus*(
> 2*C*R*s + 1)/(C*R*s), Rf: 0}, {Vminus: Vplus, Vout: Vplus*(C**2*R**2*s**2 +
> 3*C*
> R*s + 1)/(C*R*s), V1: Vplus*(2*C*R*s + 1)/(C*R*s), Rf: Ri*(C*R*s +
> 1)**2/(C*R*s)
>
> }]
>
> And there you go:
>
> >>> for k, v in _[0].items():
>
> ... print k,
> ... print '\t',v
> ...
> Vminus Vplus
> Vout Vplus*(C**2*R**2*s**2 + 3*C*R*s + 1)/(C*R*s)
> V1 Vplus*(2*C*R*s + 1)/(C*R*s)
> Rf 0
>
> > When I worked it by hand there was another solution
> > Vout = Vminus*(1/Ri + 1/Rf)*Rf
> > so that is likely the problem. However is there any way for sympy to
> > give me both solutions? Or at least indicate that was the problem?
>
> You have to give it direction...what do you want your solution in
> terms of. We might have gotten this other solution by saying we don't
> want any of those rhs variables (this is going to get ugly, but bear
> through til the end):
>
> Like before, define what you don't want solved for:
>
> >>> dont=(Vminus*(1/Ri + 1/Rf)*Rf).free_symbols
> >>> want = vall - dont
> >>> solve(eqs,want)
>
> Traceback (most recent call last):
> File "<stdin>", line 1, in <module>
> ...more error messages
> raise CoercionFailed("can't convert %s of type %s to %s" % (a, K0, K1))
> sympy.polys.polyerrors.CoercionFailed: can't convert
> DMF(([[[[[[PythonRationalTy
> pe(-1, 1)], []]], [[[PythonRationalType(2, 1)], []], [[PythonRationalType(-1,
> 1)
> , PythonRationalType(0, 1)]]]]]], [[[[[[PythonRationalType(2, 1)]]]]]]), QQ)
> of
> type QQ(s,C,Rf,Ri,Vminus,sqrt(Rf**2*Vminus**2 - 4*Rf*Ri*Vminus**2)) to
> QQ[s,C,Rf
> ,Ri,Vminus,sqrt(Rf**2*Vminus**2 - 4*Rf*Ri*Vminus**2)]
>
> So go ahead and use the manual option:
>
> >>> solve(eqs,want,manual=True)
>
> [{Vplus: Vminus, Vout: Vminus*(Rf + Ri)/Ri, V1: (Rf*Vminus + 2*Ri*Vminus -
> sqrt(
> Rf**2*Vminus**2 - 4*Rf*Ri*Vminus**2))/(2*Ri), C: Vminus/(-2*R*Vminus*s +
> R*s*(Rf
> *Vminus + 2*Ri*Vminus - sqrt(Rf**2*Vminus**2 - 4*Rf*Ri*Vminus**2))/(2*Ri))},
> {Vp
> lus: Vminus, Vout: Vminus*(Rf + Ri)/Ri, V1: (Rf*Vminus + 2*Ri*Vminus +
> sqrt(Rf**
> 2*Vminus**2 - 4*Rf*Ri*Vminus**2))/(2*Ri), C: Vminus/(-2*R*Vminus*s +
> R*s*(Rf*Vmi
> nus + 2*Ri*Vminus + sqrt(Rf**2*Vminus**2 - 4*Rf*Ri*Vminus**2))/(2*Ri))}]
>
> And again, there you go:
>
> >>> _[0][Vout]
>
> Vminus*(Rf + Ri)/Ri
>
> Write if you have more questions. I'm not sure why the None is
> returned in this case, buthopefully you have a better idea to get what
> you want and what you 'dont',
>
> Chris
This is pretty much exactly the kind of information I was looking
for. I figured those kinds of features existed, I was just unable to
locate them with the documentation. Is there a documentation page
that expands on what you are talking about, so in the future I know
where to start?
Keith
--
You received this message because you are subscribed to the Google Groups
"sympy" group.
To post to this group, send email to sympy@googlegroups.com.
To unsubscribe from this group, send email to
sympy+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/sympy?hl=en.
