Hi Martin
I suppose this problem is even more involved:
f:=operator 'f
g(x,y) == D(f(x,y), x)
G(x,y) == eval(D(f(t,s), t),[t=x,s=y])
expr:=D(f(x,y), x, 1)
function(expr, h, [x,y])
Then G(q,q) is what I had expected, but neither g nor h may be considered
as functions.
The problem seems to be that we here deal with a kind of "macro", i.e.
G,g,h are not functions as long as 'f' is not concrete.
Even unparse(G(q,q)::INFORM) yields "D(f(q,q),q::Symbol)" :(
I guess this requires some fundamental changes?
Best
Kurt
(8) -> [G(x,x),g(x,x),h(x,x)]
Compiling function g with type (Variable(x), Variable(x)) ->
Expression(Integer)
Compiling function h with type (Variable(x), Variable(x)) ->
Expression(Integer)
(8) [f (x,x), f (x,x) + f (x,x), f (x,x) + f (x,x)]
,1 ,2 ,1 ,2 ,1
Type:
List(Expression(Integer))
On Saturday, 16 February 2019 23:41:35 UTC+1, Martin R wrote:
>
> Hi there!
>
> I'm afraid that there is a slight bug in the InputForm of formal
> derivatives in 1.3.5.
>
> I guess the problem is that evaluation does not commute with
> differentiation.
>
> Best wishes,
>
> Martin
>
>
>
>
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