On Sunday, 17 February 2019 18:46:48 UTC+1, Martin R wrote:
>
> Hi Kurt!
>
> I'm not sure I understand the problem.  With the patch applied, I obtain:
>


You do, your patch is certainly a partial solution (now G and h are 
correct). What I actually wanted to say/ask is that/whether g and G ought 
to be identical.

Semantically g(x,y)=D(f(x,y), x) => g(x,x)=D(f(x,x), x), but is that really 
the answer the user expects? It might be a source of confusion. If you 
hadn't pointed out the issue, I certainly would have made the mistake using 
g instead of G ;)  
 
What is the general opinion? 
BTW thanks for the patch.
Kurt

>
> (1) -> f:=operator 'f
>
>    (1)  f
>                                                           Type: 
> BasicOperator
> (2) -> g(x,y) == D(f(x,y), x)
>                                                                    Type: 
> Void
> (3) -> G(x,y) == eval(D(f(t,s), t),[t=x,s=y])
>                                                                    Type: 
> Void
> (4) -> expr:=D(f(x,y), x, 1);
>
>                                                     Type: 
> Expression(Integer)
> (5) -> function(expr, h, [x,y]);
>
>                                                                  Type: 
> Symbol
> (6) -> [G(x,x), g(x,x), h(x,x)]
>
>    (6)  [f  (x,x), f  (x,x) + f  (x,x), f  (x,x)]
>           ,1        ,2         ,1        ,1
>                                               Type: 
> List(Expression(Integer))
>
> (7) -> [interpret(G(x,x)::INFORM), interpret(g(x,x)::INFORM), 
> interpret(h(x,x)::INFORM)]
>
>    (7)  [f  (x,x), f  (x,x) + f  (x,x), f  (x,x)]
>           ,1        ,2         ,1        ,1
>                                                               Type: 
> List(Any)
>
>
> I think that this is OK, let's look at it in detail.
>
> * G(x,y) == eval(D(f(t,s), t),[t=x,s=y])
>
> D(f(t,s), t) means "consider f as a bivariate function and compute the 
> derivative with respect to the first variable"
> G then evaluates this expression at t=x and s=x.
>
> * g(x,y) == D(f(x,y), x)
>
> g(x,x) means "compute D(f(x,x), x)"
>
> * function(expr, h, [x,y])
>
> I admit that I am not completely sure about the semantics of "function", 
> but I think it should be roughly equivalent to G.
>
> Maybe you could say what result you would expect?
>
> Best wishes,
>
> Martin
>

-- 
You received this message because you are subscribed to the Google Groups 
"FriCAS - computer algebra system" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to fricas-devel+unsubscr...@googlegroups.com.
To post to this group, send email to fricas-devel@googlegroups.com.
To view this discussion on the web visit 
https://groups.google.com/d/msgid/fricas-devel/c4c0c8d5-97a9-4e5f-a1ba-52d0e4e5a5f3%40googlegroups.com.
For more options, visit https://groups.google.com/d/optout.

Reply via email to