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Edit detail for #431 exponentiation wrong in 'LODO' revision 2 of 6

1 2 3 4 5 6
Editor: kratt6
Time: 2008/08/29 05:23:13 GMT-7
Note:

changed:
-Dx: LODO(EXPR INT, f+->D(f,x)) := D()
-u := operator 'u
-L := Dx + u(x)
Dx: LODO(EXPR INT, f+->D(f,x)) := D();
u := operator 'u;
L := Dx + u(x);

added:
)cl co

changed:
-K := OREUP ( x, INT, 1, f)
-L:K :=x+1
K := OREUP(x, INT, 1, f);
L := x::K+1;

Submitted by : kratt6 at: 2008-08-29T05:18:31-07:00 (15 years ago)
Name :
Axiom Version :
Category : Severity : Status :
Optional subject :  
Optional comment :

axiom
Dx: LODO(EXPR INT, f+->D(f,x)) := D();
Type: LinearOrdinaryDifferentialOperator?(Expression Integer,theMap LAMBDA-CLOSURE(NIL,NIL,NIL,G1421 envArg,SPADCALL(G1421,QUOTE x,ELT(*1;anonymousFunction;0;initial;internal;MV,0))))
axiom
u := operator 'u;
Type: BasicOperator?
axiom
L := Dx + u(x);
Type: LinearOrdinaryDifferentialOperator?(Expression Integer,theMap LAMBDA-CLOSURE(NIL,NIL,NIL,G1421 envArg,SPADCALL(G1421,QUOTE x,ELT(*1;anonymousFunction;0;initial;internal;MV,0))))
axiom
L**2 = L*L
LatexWiki Image(1)
Type: Equation LinearOrdinaryDifferentialOperator?(Expression Integer,theMap LAMBDA-CLOSURE(NIL,NIL,NIL,G1421 envArg,SPADCALL(G1421,QUOTE x,ELT(*1;anonymousFunction;0;initial;internal;MV,0))))

or

axiom
)cl co
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. f: INT->INT:=x+->x+1
LatexWiki Image(2)
Type: (Integer -> Integer)
axiom
K := OREUP(x, INT, 1, f);
Type: Domain
axiom
L := x::K+1;
Type: UnivariateSkewPolynomial?(x,Integer,R -> R,theMap LAMBDA-CLOSURE(NIL,NIL,NIL,G1432 envArg,SPADCALL(G1432,1,ELT(*1;anonymousFunction;1;initial;internal;MV,0))))
axiom
L^2=L*L
LatexWiki Image(3)
Type: Equation UnivariateSkewPolynomial?(x,Integer,R -> R,theMap LAMBDA-CLOSURE(NIL,NIL,NIL,G1432 envArg,SPADCALL(G1432,1,ELT(*1;anonymousFunction;1;initial;internal;MV,0))))

Reason is, that exponentiation is not taken from Monoid, but from SUP.

Martin