[Axiom-mail] Unexpected results w.r.t. exponential operation of LODO (with july2008 release)

[Liu Xiaojun]

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