On Feb 21, 8:06 pm, Mike Hansen <mhan...@gmail.com> wrote:
> Hello,
>
> On Sun, Feb 21, 2010 at 12:03 PM, bsdz <blai...@googlemail.com> wrote:
> > I used to be able to obtain a function name in Sage 3+ using something
> > like the following: -
>
> > fu = function('fu', x)
> > print fu._f._name
> > # would print "fu"
>
> > I notice that this does not work in Sage 4 and "fu" is no longer a
> > SymbolicFunctionEvaluation but an Expression.
>
> > Is there a way to obtain the function's name in the above?
>
> Note that "fu" as defined above creates the function "fu" and then
> applies it to x. If you leave out the x, then you get the function
> itself which has a name method. If you have fu(something), then you
> can use the operator method on that to get the function back:
>
> sage: fu = function('fu')
> sage: fu.name()
> 'fu'
> sage: fu(x).operator().name()
> 'fu'
>
> --Mike
Ah, I see my mistake. Thanks Mike :)
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