x.is_real is True
True
var('x',positive=True)
x
(x0) is True
False
x0
True
type(_)
class 'sympy.logic.boolalg.BooleanTrue'
I understand that the above (x0) is True is False for the same reason that
S(1) is 1 is False, but why don't boolean expressions use BooleanTrue, too?
Is that ust
kernS is only to keep 2*(x+y)-like expressions from turning into 2*x+2*y.
What issues do you have when processing the result of parse_expr with
evaluate=False?
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Maybe https://github.com/sympy/sympy/blob/master/sympy/logic/boolalg.py#L81
can provide some insight here.
Aaron Meurer
On Thu, Jun 26, 2014 at 10:33 AM, Chris Smith smi...@gmail.com wrote:
x.is_real is True
True
var('x',positive=True)
x
(x0) is True
False
x0
True
type(_)
class
Above 0, the function is monotonic but is ill-behaved in that it is very
flat; this can cause troubles for different solver routines. A slow but
robust method is bisection
nsolve(eq,c2,(1,1000),solver='bisect')
mpf('22.964256014441664')
Using that, you don't need a very precise range for the
Not sure exactly what you mean, but can't you use collect? And you could
pull off the coefficients to know what are potential items that can be
grouped.
eq=f(x).diff(x)+x*f(x).diff(x)+f(x).diff(x,2)+x*f(x).diff(x,2)
collect(eq,x)
x*(Derivative(f(x), x) + Derivative(f(x), x, x)) +
With existing methods you can do the following:
First, to share a long expression, try filldedent.
from sympy.utilities.misc import filldedent
print filldedent(eq)
-2*g*psi^ss_1*conjugate(psi^ss_1) - 2*g*psi^ss_2*conjugate(psi^ss_2) +
omega_2 - (-k + k_2)**2/(2*m)
That can be sympified by
The code I'm working on in sympy.physics.mechanics often results in large
expressions (100,000 + operations, biggest I've seen was 31,000,000
operations). Once these expressions are obtained, we often need to
substitute in values/symbols (this is the operating point). However, due to
the
