On 01/20/2014 12:19 PM, Alasdair wrote:
> Try this:
Thanks... I've shortened it a bit and somehow believe that must be a bug.
)cl completely
f:=operator 'f
y:=operator 'y
f1:=D(f(x,y(x)),x)
f2:=D(f1,x)
f3:=D(f2,x)
kxy:=eval(D(f(x,z),[x,z]),z=y(x))
In the end I get...
f3:=D(f2,x)
(5)
,,, , 3
f (x,y(x)) + f (x,y(x))y (x) + y (x) f (x,y(x))
,1,1,1 ,2 ,2,2,2
+
, 2 , , 2
y (x) f (x,y(x)) + y (x)f (x,y(x)) + y (x) f (x,y(x))
,2,2,1 ,2,1,1 ,2,1,2
+
, , 2 ,
y (x)f (x,y(x)) + y (x) f (x,y(x)) + y (x)f (x,y(x))
,1,1,2 ,1,2,2 ,1,2,1
+
, ,,
(3y (x)f (x,y(x)) + f (x,y(x)) + f (x,%B) + f
(x,y(x)))y (x)
,2,2 ,2,1 ,2,1 ,1,2
Type:
Expression(Integer)
kxy:=eval(D(f(x,z),[x,z]),z=y(x))
(6) f (%A,y(x))
,1,2
Type:
Expression(Integer)
That looks to me like even two bugs. Why would evaluating f3:=D(f2,x)
have influence on kxy:=eval(D(f(x,z),[x,z]),z=y(x))? Note that the
ordinary result should be like this:
(7) -> )cl completely
All user variables and function definitions have been cleared.
All )browse facility databases have been cleared.
Internally cached functions and constructors have been cleared.
)clear completely is finished.
(1) -> f:=operator 'f
(1) f
Type:
BasicOperator
(2) -> y:=operator 'y
(2) y
Type:
BasicOperator
(3) -> kxy:=eval(D(f(x,z),[x,z]),z=y(x))
(3) f (x,y(x))
,1,2
Type:
Expression(Integer)
But I guess that is a follow-up error from the computation of
f3:=D(f2,x). Also there I wouldn't want to see %B to appear in the output.
I faintly remember that Waldek introduced that local variables to
indicate "derivation wrt the n-th variable", because there was a bug for
another expression. Looks to me like this code must be revisited.
But I'm somehow clueless why evaluation of D(f2,x) leads to a side effect.
Ralf
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