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smichr on master
caveat about autowrapping blind… Update sympy/utilities/autowrap… Merge pull request #21639 from … (compare)
smichr on smichr-patch-1
smichr on smichr-patch-1
Update sympy/utilities/autowrap… (compare)
smichr on master
correct int(TribonacciConstant)… Update test_numbers.py Update sympy/core/tests/test_nu… and 1 more (compare)
smichr on smichr-patch-2
I was just setting up my environment. I used git fetch upstream in order to sync my local repo with the master but I am getting this error: git@github.com: Permission denied (publickey).
fatal: Could not read from remote repository.
Please make sure you have the correct access rights
and the repository exists.
@Ashwin-Shenoy-NITK How exactly are you specifying the upstream?
'Coz git remote add upstream git@github.com:sympy/sympy.git
might throw that error as you don't own the repository and you won't have the access rights.
If that is the case, you could simply use git remote add upstream https://github.com/sympy/sympy
there.
linsolve(Matrix(np.c_[V,o]))
.
{(-1.0*tau0 - 0.916666666666667, 5.0*tau0 + 0.708333333333333, -10.0*tau0 + 0.375, 10.0*tau0 - 0.208333333333333, -5.0*tau0 + 0.0416666666666667, tau0)}
, seems is autonamed tau0
tau0 = 0
, but without know the variable name before hand.
lin(Matrix(np.c_[V, o])) | all_free_variables = 0
r.evalf(subs={tau0: 0})
doesn't works, and also ) don't want to specify the variable name, since I will not know beforehand
r.subs(tau0, 0)
, both complain about tau0
not being defined
doing:tau0 = symbols('tau0');
r.subs(tau0,0)
works. But I must know before hand each symbol. That I don't. =/
mysolution
is a numpy array as you could notice.
If no symbols are given, internally generated symbols will be used.
tau0
.
solution_as_list
, didn't returns a list per se, but a set with a tuple, as in: {(-0.916666666666667, 0.708333333333333, 0.375, -0.208333333333333, 0.0416666666666667, 0)}
of type sympy.sets.sets.FiniteSet
.
flatten
anyway... Never mind then
Actually,
solution = np.array(*[e for e in r_tmp]).astype(float)
should flatten it as well
wow. :-( Given the matrix PI
array([[ -0.917, 0.042, 0. , 0. , -1. ],
[ 0.708, -1.125, 0.042, 0. , 5. ],
[ 0.375, 1.125, -1.125, 0.042, -10. ],
[ -0.208, -0.042, 1.125, -1.125, 10. ],
[ 0.042, 0. , -0.042, 1.125, -5. ],
[ 0. , 0. , 0. , -0.042, 1. ]])
and the vector c
array([-1. , 0. , 0. , -0.042, 1.083, -0.042])
the solution given by
linsolve(Matrix(np.c_[PI,c]))
returns an EmptySet()
while the same operation in Maple 16 returns non empty, and exactly
np.array([649/576, 143/192, 75/64, 551/576, -25/13824])
that in float is around
array([ 1.127, 0.745, 1.172, 0.957, -0.002])
Any ideas?
PI = np.array([
[-11/12, 1/24, 0, 0, -1],
[ 17/24, -9/8, 1/24, 0 ,5],
[3/8, 9/8, -9/8, 1/24, -10],
[-5/24, -1/24, 9/8, -9/8, 10],
[1/24, 0, -1/24, 9/8, -5],
[0, 0, 0, -1/24, 1]
])
It seems that your system is over-determined. It has 6 equations in 5 variables. It can have solutions only if the determinant of the extended system is 0.
>>> P = Matrix([[ -0.917, 0.042, 0. , 0. , -1. ],
... [ 0.708, -1.125, 0.042, 0. , 5. ],
... [ 0.375, 1.125, -1.125, 0.042, -10. ],
... [ -0.208, -0.042, 1.125, -1.125, 10. ],
... [ 0.042, 0. , -0.042, 1.125, -5. ],
... [ 0. , 0. , 0. , -0.042, 1. ]])
>>> c = Matrix([-1. , 0. , 0. , -0.042, 1.083, -0.042])
>>> Pc = P.row_join(c)
>>> Pc.det()
0.000998125750125132
In this example, the determinant not 0 because of the inaccuracy of floating point numbers. Hence there is no solution.