Christian Wolinski

MaplePrimes Activity

These are replies submitted by Christian Wolinski

@Carl Love What does `evala/toprof`(Pf); `evala/toprof`(indets(Pf,RootOf)); return? ANS2 should contain variable k. If it does not then what would explain this? Also I added a print command.

I am editing too much. The codes work flawlessly in Maple V R4. I wonder in which version the discontinuity occurs.

@Markiyan Hirnyk It appears _EnvExplicit:=true; is being used, converting all RootOfs into radicals. Lets see if _EnvExplicit:=false; would change that.

@Carl Love roots or evala@Roots or another? Which is better?

@Carl Love Thank you for your help. It appears roots command does not succeed. I'll change it to evala@Roots.

I ask, what command would you use to obtain roots of a factored polynomial?

@Carl Love I suppose split is the same thing as evala@AFactor. I tested it. It seems to work. I already edited my original answer.

@Dave L If you do that you should state the author of the statement you edit.

@Carl Love I have forgotten about this option completely, my apology. Unfortunately I have no access to the Groebner package. I can only suggest using method=buchberger and plex(X,Q). Quote from the help file for GB:

- GB is an implementation of Buchberger's algorithm for computing Grobner bases
  over Zp(a,b,...)[x1, ..., xn].
- GB(F,X,termorder) computes the reduced, minimal Grobner basis of the
  polynomials F with respect to the indeterminates X and the given term order-

No help otherwise.

..Also the basis you quoted is identical to tdeg basis.


Your definitions for x and q are:


L := 1+X+X^6+X^7+X^8, (X^3+X+1)*Q+Q^3+(X^7+X^6+X^4+1)*Q^2+X^7+X^6+X^4+X^3+X^2+X+1;

so the monic for Q is what you seek and it is obtained with:


GB([L], [X, Q], plex) mod 2;

@roman_pearce Thank You, that is precisely the function I needed. But also I was wondering if there is any support for this from side of Maple (even though the solution is so simple).




Perhaps you want to try it using symmetric polynomials:


A := [(a+b+c+d)*(a*b+a*c+a*d+b*c+b*d+c*d)-12*sqrt((a^2+b^2+c^2+d^2)*a*b*c*d), [a+b+c+d-1]];
G := [-a*b*c*d+p14, -a^2*b-a^2*c-a^2*d-a*b^2-a*b*c-a*b*d-a*c^2-a*c*d-a*d^2-b^2*c-b^2*d-b*c^2-b*c*d-b*d^2-c^2*d-c*d^2+p23, -a^2-a*b-a*c-a*d-b^2-b*c-b*d-c^2-c*d-d^2+p22, -a-b-c-d+p11], [a, b, c, d, p11, p22, p23, p14];


Look for symmetric polynomials package.

@Carl Love I was supposing this could work:


R := {Product(g-i, i=-4..-3), Product(f-i, i=0..2)};
#R := {product(g-i, i=-4..-3), product(f-i, i=0..2)};

But now I think it would work only for small cases.

@Carl Love Maple 16 seems nicer.

What would this produce, 25 answers?


isolve({x>3/2, x<13/2, y>3/2, y<13/2});



S := {-1/2 < 2/53*f+7/53*g, 7/53*f-2/53*g < 1/2, 2/53*f+7/53*g < -37/106, 3/106 < 7/53*f-2/53*g};

R := {(g+4)*(g+3), f*(f-1)*(f-2)};
isolve(S union R);

gives full solution. So a solution would be to first figure each variable's range. In observation it seems a false approach for isolve to produce over a cartesian product of ranges.

Does isolve accept inert Product expressions?

Dividing (c,d) plane with the polynomials involved this image is obtained. The range of x: -1..1 was not used.


@Carl Love The order of eliminations I employed:

{cos(p1), sin(p1), sin(p2), cos(p3), sin(p3), cos(p2)}, {cos(t3), sin(t3)}, {sin(t2)}, {cos(t2)}, {y}.

Eliminating cos(t2) gave the first occurence of RootOf. Eliminating y gave 3 cases, all quadratic in x. That is my previous post. Perhaps you could reproduce this.

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