Unexpected results w.r.t. exponential operation of LODO (with july2008 release)
"Liu Xiaojun" <[email protected]> Wed, 27 Aug 2008 23:34:47 +0800
| Newsgroups | gmane.comp.mathematics.axiom.user |
|---|---|
| Message-ID | <[email protected]> |
Hi,
Does exponential operation (^ or **) of a differential operator, say L,
means repeating multiplications of L in Axiom ? If so, it seems the
following code produced an unexpected result:
(1) -> Dx: LODO(EXPR INT, f+->D(f,x)) := D()
(1) D
Type: LinearOrdinaryDifferentialOperator(Expression Integer,theMap
LAMBDA-CLOSURE(NIL,NIL,NIL,G1402 envArg,SPADCALL(G1402,QUOTE
x,ELT(*1;anonymousFunction;0;frame0;internal;MV,0))))
(2) -> u := operator 'u
(2) u
Type:
BasicOperator
(3) -> L := Dx + u(x)
(3) D + u(x)
Type: LinearOrdinaryDifferentialOperator(Expression Integer,theMap
LAMBDA-CLOSURE(NIL,NIL,NIL,G1402 envArg,SPADCALL(G1402,QUOTE
x,ELT(*1;anonymousFunction;0;frame0;internal;MV,0))))
(4) -> L**2 = L*L
2 2 2 , 2
(4) D + 2u(x)D + u(x) = D + 2u(x)D + u (x) + u(x)
L^2 produces the same result as l.h.s. of (4) shows. However, it does not
equal to the r.h.s. generally, and r.h.s. should be correct answer. I also
tried similar things w.r.t. the parent domain of LODO, which is
UnivariateSkewPolynomial, it produces similar unexpected result:
(1)-> f: INT->INT:=x+->x+1
(2) -> K := OREUP ( x, INT, 1, f)
Type:
Domain
(3) -> x:K
Type:
Void
(4) -> L:=x+1
(4) ->
(4) x + 1
(5) -> L^2=L*L
2 2
(5) x + 2x + 1= x + 4x + 3
Clearly, the r.h.s. is correct too. I checked the .spad file, it seems the
exponential operation of LODO (or its parent OREUP) is missed (or I guess it
is directly inherited as usual polynomial operation). However I am not
clever enough to modify the code. Please check this, thanks.
Best Regards,
Xiaojun Liu
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