Hi,
Consider the following code with Sage 9.0.beta7:
sage: x, y = var('x y', domain='real')
sage: phi = function('phi', latex_name=r'\varphi')
sage: s = phi(x, y).simplify()
sage: s
phi(x, y)
It looks good, but it is not:
sage: s.subs({phi(x, y): 1})
phi(x, y)
The failure of the substitution is due to the fact that the phi that
appears in s is different from the original phi:
sage: s.operator()
phi
sage: s.operator() == phi
False
A LaTeX printiing would have revealed the difference:
sage: latex(s)
\phi\left(x, y\right)
sage: latex(phi(x,y))
\varphi\left(x, y\right)
What is weird is that if one permutes the first two lines of the above
code, i.e. if one starts with
sage: phi = function('phi', latex_name=r'\varphi')
sage: x, y = var('x y', domain='real')
Then everything is fine:
sage: s = phi(x, y).simplify()
sage: s
phi(x, y)
sage: s.subs({phi(x, y): 1})
1
sage: s.operator()
phi
sage: s.operator() == phi
True
sage: latex(s)
\varphi\left(x, y\right)
Other data points:
- if one omits domain='real' for x,y, then everything is fine
- if one uses a single variable for phi, I.e. phi(x), then everything is
fine
- if one omits latex_name in the declaration of phi, then everything is fine
- if one replaces latex_name by another argument like
sage: phi = function('phi', nargs=2)
then the same error appears
Is this a known bug?
Eric.
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