My problem is related to recovering orbits from invariant polynomials, and their ideal of relations.

The invariant polynomials I obtained are:

u = x² - x y + y²,
v = 2 x⁶ - 6 x⁵ y + 15 x⁴ y² - 20 x³ y³ + 15 x² y⁴ - 6 x y⁵ +  2 y⁶ ,
w = x⁶ - 4 x⁵ y + 10 x⁴ y² - 10 x³ y³ + 5 x² y⁴ - 2 x y⁵ + y⁶ .

Using the logic from the Cox et al. book I got that the algebraic relation (ideal of relations) between the invariants, which is:

11 u⁶ - 10 u³ v + 3 (v² - v w + w²) = 0

Then, using Reduce[] (exact symbolic solver which uses cylindrical algebraic decomposition) from Mathematica I solved for x and y under the assumption x > y > 0, u > 0, v > 0 and w > 0, but I got rather a complex solution.

My question is whether there is a way to try getting something more straightforward than the solution given by Mathematica.
 

I tried to use Maple's solve function, but it immediately stops without any result of an error message.

My input for Reduce is:

Reduce[{x² - x y + y² == u, 2 x⁶ - 6 x⁵ y + 15 x⁴ y² - 20 x³ y³ + 15 x² y⁴ - 6 x y⁵ + 2 y⁶ == v,  x⁶ - 4 x⁵ y + 10 x⁴ y² - 10 x³ y³ + 5 x² y⁴ - 2 x y⁵ + y⁶ == w, 11 u⁶ - 10 u³ v + 3 (v² - v w + w²) == 0, x > y > 0, u > 0, v > 0, w > 0}, {x, y}, Complexes]

and my input for solve:

solve({u = x^2 - x*y + y^2, v = 2*x^6 - 6*x^5*y + 15*x^4*y^2 - 20*x^3*y^3 + 15*x^2*y^4 - 6*x*y^5 + 2*y^6, w = x^6 - 4*x^5*y + 10*x^4*y^2 - 10*x^3*y^3 + 5*x^2*y^4 - 2*x*y^5 + y^6, 11*u^6 - 10*u^3*v + 3*v^2 - 3*v*w + 3*w^2 = 0, 0 < u, 0 < v, 0 < w, 0 < x, 0 < y, y < x}, {x, y})

Do you know what I am doing wrong, or what else could I try?

Hello guys!

Could you tell me what are the state-of-the-art algorithms to compute Hermite and Smith normal forms (well or just Hermite since the later can be computed by applying Hermite twice)? I am interested in these algorihtms which outputs also the respective unimodular matrices.

Let say I have an array

A := Array([[1], [2], [3]])
Normally if I would like to extend it I will write, eg:

Extend(A, [[4], [5]])

But because this is not 1D array Maple will rise an error. Is there a way to go over this limitation?

I have the system:

Update:

{-1/2 < 2*f*(1/53)+7*g*(1/53), 3/106 < 7*f*(1/53)-2*g*(1/53), 2*f*(1/53)+7*g*(1/53) < -37/106, 7*f*(1/53)-2*g*(1/53) < 1/2}

which I wish to solve over integers but isolve() gives me "Warning, solutions may have been lost and no solutions". The solutions exist and are {[f =0, g = -3] || [f = 1, g = -4], [f = 1, g = -3] || [f = 2, g = -4]}, but I cannot obtain them with Maple. Could you tell me what is wrong and how I should treat this kind of problems in the future, please.

Mathematica 10.0

Reduce[{-1/2 < 2*f*(1/53) + 7*g*(1/53), 3/106 < 7*f*(1/53) - 2*g*(1/53), 2*f*(1/53) + 7*g*(1/53) < -37/106, 7*f*(1/53) - 2*g*(1/53) < 1/2}, {f, g}, Integers]

(f == 0 && g == -3) || (f == 1 && g == -4) || (f == 1 &&
   g == -3) || (f == 2 && g == -4)

Maple

isolve({-1/2 < 2*f*(1/53)+7*g*(1/53), 3/106 < 7*f*(1/53)-2*g*(1/53), 2*f*(1/53)+7*g*(1/53) < -37/106, 7*f*(1/53)-2*g*(1/53) < 1/2});
Warning, solutions may have been lost

Maybe this is trivial but could somebody tell me how to get bounds of interval returned by shake? For instance,

shake(sqrt(2)) gives INTERVAL(1.41421356167 .. 1.41421356308), and I would like to store the upper and lower bounds as rational numbers in two variables.