Hello. I have the following equations (all equal to zero) which I am
able to solve using SymPy for the Ep,De and La subscripted variables
after supplying the DeltaP() DeltaQ() and mu values:
2*Ep(0)/7 + Ep(1)/7 + 2*Ep(2)/35 + Ep(3)/70 - La(3)
Ep(0)/7 + 6*Ep(1)/35 + 9*Ep(2)/70 + 2*Ep(3)/35 + La(2)
I don't think there is a way to get back the matrices for the input
equations, but I think the following will give you what you want. It can be
streamlined, but you can see the steps below:
>>> eqs=Tuple(*eqs)
>>> eqs.atoms(Function)
set([La(3), Ep(0), DeltaQ(1, 0), DeltaQ(0, 0), Ep(2), Ep(3), De(
If you could take a moment to look over
https://github.com/sympy/sympy/pull/1682 you can solve for indexed
quantities without having to do the dummy substitution yourself.
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On Mon, Dec 3, 2012 at 12:28 PM, Chris Smith wrote:
> If you could take a moment to look over
> https://github.com/sympy/sympy/pull/1682 you can solve for indexed
> quantities without having to do the dummy substitution yourself.
Hey nice! Do I understand correctly this mean that pulling the late
On Mon, Dec 3, 2012 at 4:25 PM, Shriramana Sharma wrote:
> Hey nice! Do I understand correctly this mean that pulling the latest
> Git (yes I've become a cloner) will allow me to use IndexedBase
> objects like this?
OK now pulled latest git but I understand this is not into trunk yet.
BTW from th
>
> OK now pulled latest git but I understand this is not into trunk yet.
> BTW from the example in the code:
>
> >>> from sympy import Indexed, IndexedBase, Tuple
> >>> A = IndexedBase('A')
> >>> eqs = Tuple(A[1] + A[2] - 3, A[1] - A[2] + 1)
> >>> solve(eqs, eqs.atoms(Indexed))
> {A[1]: 1, A[2]: 2
On Mon, Dec 3, 2012 at 4:45 PM, Shriramana Sharma wrote:
> A = IndexedBase('A')
> >>> eqs = Tuple(A[1] + A[2] - 3, A[1] - A[2] + 1)
>
After the latest commit "it just works":
>>> solve((A[1] + A[2] - 3, A[1] - A[2] + 1))
[{A[2]: 2, A[1]: 1}]
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numVars = ( n + 1 ) * 3
> for i in range ( numVars ) :
> eqnList [ i ] = eqnList [ i ] . expand ()
> print ( eqnList [ i ] )
>
> # the linear equation system is expressed in matrix form as: eqnCoeffs
> * eqnVars = eqnConstants
>
> eqnCoeffs = zeros ( numVars,
On Mon, Dec 3, 2012 at 11:25 PM, smichr wrote:
> That's the verbose way to say eqnCoeffs =
> Matrix(eqn).jacobian(Matrix(eqnVars)) and eqnConstants = eqnCoeffs -
> Matrix(eqn). Here's a toy system:
> (though not a lot, but th
eqns = (2*x+y+3,x-y-2)
E = Matrix(eqns)
X = Matrix([x, y]
