I have found a simple workaround for this problem. I implemented a function
that performs a hardcore substitution of the expression:
def parse_imaginary_real(expression):
"""
It takes an expression and parse correctly the imaginary and real part
of first derivatives of a
real or imaginary function.
This function works only for first derivatives.
It could be extended easily to work with second or higher order
derivatives as well.
"""
all_variables = globals()
# Get the variables that are functions
all_functions = [x for x in all_variables.values() if isinstance(x,
sy.Function)]
new_exp = expression
for f in all_functions:
if f.is_real:
new_exp = expression.subs(sy.im(sy.diff(f)),
0).subs(sy.re(sy.diff(f)), sy.diff(f))
elif f.is_imaginary:
new_exp = expression.subs(sy.im(sy.diff(f)),
sy.diff(f)).subs(sy.re(sy.diff(f)), 0)
return new_exp
This function can be used as simplify to an expression. Indeed, it has
still some issues: it does not check the type of the argument, and it
simplifies only first derivatives (and I think it could be very slow if
many variables have been defined, I do not know very well how the globals()
can work if the function is used on a library).
However, for my simple case, it does the job and allows me to go over,
simplifying the imaginary and real parts of first derivatives that are zero
by definition. I hope it can be useful for other users.
Bests,
Lorenzo
Il giorno venerdì 28 febbraio 2020 20:02:51 UTC+1, Aaron Meurer ha scritto:
>
> There is an open issue about this
> https://github.com/sympy/sympy/issues/11868.
>
> Aaron Meurer
>
> On Fri, Feb 28, 2020 at 12:00 PM Lorenzo Monacelli
> <[email protected] <javascript:>> wrote:
> >
> > Dear all,
> > I have a complex expression and I want to separate imaginary and real
> part, however I noticed that when I have derivatives on a function, the
> real and imaginary part separation does not work properly:
> >
> > t = sy.Symbol("t", real = True)
> > f = sy.Function("f", real = True)(t)
> >
> > Then if I ask sympy the imaginary part of f, it correctly returns 0,
> however if I ask the immaginary part of the derivative of f, it is not able
> to see that it is zero:
> >
> > sy.im(f) # this is zero (ok!)
> > sy.im(sy.diff(f, t)) # does not simplify to zero
> >
> > How can I force sympy to set automatically the derivative of f to real
> numbers?
> > Thanks in advance for the help,
> > Bests,
> > Lorenzo
> >
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>
>
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