Björn's profile - activity

It works for me after updating SageMath from 10.0 to 10.1 via Homebrew. I still have no clue why this didn't work in 10.

Good point. sage0_version says "'SageMath version 10.0, Release Date: 2023-05-20'". I'll update and report back.

Symbolic matrix inversion broken? The result of the following is a bit unexpected. sage: n=3; sage: D = matrix.diagona

It is a partial workaround, but does not solve the original question, really. The sage: is a bit in the way, and this doesn't seem to support block code like function definitions or loops. Sorry for not noticing your answer earlier, I haven't logged in in a while.

Is there a version of \begin{sageblock} my multiline code \end{sageblock} that has code highlighting and line numbering like the sagecommandline environment and code execution?

Of course typesetting with lstlistings and execution with sagesilent could be a workaround, but it would seem to require me to have the same code in my document twice. And that I find unacceptable.

That's just perfect, thanks! latex(sexp2), latex(f) instead of sexp2 ; f really demonstrates the achievement of your recursive coercion. Very nice, I especially like that the code does not require a priori knowledge of the function to be substituted. This really should be built into sage's simplify() at the stage when it processes what comes back from maxima.

I guess will have to live with that ;) Thanks again for giving me the starting point, there wasn't much left to do after that.

Not only that, to get the operator name, one somehow needs to find it by location in the sage expression, at least I am not aware of a better way to find a reference to the "maxima" a(). Maybe one could loop over all operators in the expression and somehow filter by a feature that identifies them as maxima operators... but that's messy.

Thanks to kcrisman I figured the remaining bits out myself. Complete code as follows:

# [...]
print latex(sexp)
sexp2=simplify(sexp) # expand doesn't hurt it, so omit that
print latex(sexp2)
# observe they are different due to sexp2 being processed 
# via Maxima, as explained by kcrisman
# maxima replaces a() by a new a() that doesn't have the 
# print function:

print sexp2.operands()[0].operator().__class__ # maxima a()
print a.__class__ # the a() we have defined

# to coerce back, we need to replace the former by the latter:
print latex(sexp2.substitute_function(sexp2.operands()[0].operator(),a))

(And that does the intended.)

Aha, I see... At the end of the day I don't really mind Maxima, as long as I can coerce back, which I cannot seem to figure out how to achieve. Obvious (to me) candidates would seem to be latex(sexp2.subs({a(n):a(n)})) or even latex(sexp2.subs({a: a})). But they do not work as expected or throw errors when I use keywords in combination with arguments to a().

print sexp2.operands()[0].operator().__class__ # maxima a()
print a.__class__ # my a()

Reveals the functions a() are indeed different, the former function_factory.NewSymbolicFunction, the latter expression.Expression. Can't I just somehow retrospectively set the print_latex_func on the function_factory.NewSymbolicFunction?

Use bool() like so

,var n
f1=((-1)^n-2*n-1)/4/(2*n+1)
f2=(-1)^n/4/(2*n+1)-1/4
show(f1==f2)
bool(f1==f2)

I know it is simplify() and not expand() that causes the problem. But why and how?

Here's a minimal example of what I am talking about. Both print statements should return the same output in this scenario.

,var n x
def my_latex_print(self, *args): 
    return "a_{%s}" %(', '.join(map(latex, args)))
a=function('a',nargs=1,print_latex_func=my_latex_print)(n)

sexp=a(n=n+1)+a(n=n)
print latex(sexp)
sexp=simplify(expand(sexp))
print latex(sexp)

However, the outputs differ:

a_{n + 1} + a_{n}
a\left(n + 1\right) + a\left(n\right)

Of course simplify-expand may lead to a different expression, but that's not what I am concerned about. I am concerned about the representation of a(n) instead of a_n in the output.

My main question is: How can I recover a_n out of sexp at the end of my code?