Download code_solve_system_of_equations_vv.mw

A219954list := nmax -> 
[seq[scan=`+`](ifelse(n=1, 0, 3^numboccur(convert(n-1, 'base', 2), 1) - ifelse(2^ilog2(n)=n, n/2, 0)), n = 1..nmax)]:

A219954list(33);
 [0, 2, 5, 12, 15, 24, 33, 56, 59, 68, 77, 104, 113, 140, 167,  240, 243, 252,
261, 288, 297, 324, 351, 432, 441, 468, 495,   576, 603, 684, 765, 992, 995]

Download solution_pde_check-exact.mw

Technically, you must use:

U,S,Vt := SingularValues(A, output=[':-U',':-S',':-Vt']);

(Of course, in your example, this does not make much sense because U, S, Vt are not needed/used.)

restart;
a:=floor(33*sqrt(3)):
R2:=33^2*3:  # the radius^2
L2:=R2*16/6: # the edge^2
nr:=0: W:=Array(1..0):
for  x from 0 to a do
for  y from x to a do
for  z from y to a do
if x^2+y^2+z^2=R2 then nr++: W(nr):=[x,y,z] fi
od od od:
f:=u -> ([u[1],u[2],u[3]], [-u[1],u[2],u[3]],[u[1],-u[2],u[3]], [u[1],u[2],-u[3]],
        [-u[1],-u[2],u[3]], [-u[1],u[2],-u[3]],[u[1],-u[2],-u[3]], [-u[1],-u[2],-u[3]]):
P:=f~([seq](W)): # the integer points on the sphere
n:=nops(P);
#                            n := 96

d2 :=(u,v) -> inner(u-v,u-v):
A:=Matrix( n, (i,j)-> `if`(d2(P[i],P[j])=L2, 1, 0) ):
with(GraphTheory):
G:=Graph(A):
K4 := CompleteGraph(4):
IsSubgraphIsomorphic(K4,G, isomorphism);
#             true, {1 = 95, 2 = 94, 3 = 93, 4 = 89}

T := P[[rhs~(%[2])[]]]; # A desired tetrahedron
# T := [ [33, 33, 33], [-33, -33, 33], [-33, 33, -33], [33, -33, -33]  ]

plots:-display(plottools:-tetrahedron(T), plottools:-sphere([0, 0, 0], sqrt(R2), transparency=0.8), color="Blue");

The floor method cannot work because NLPSolve finds only local extrema and almost any point is a local max,
actually any point in ( (-5, oo) \ Z )^9.

The DirectSearch package (in Application Center) works, but the solution is of course not guaranteed.

GlobalOptima(add(x[k],k=1..9), 
   [seq(x[k]::integer, k=1..9),seq(x[k]<=x[k+1], k=1..8), x[1]>=-5, x[9]<=50, add(x[k]^3,k=1..9)=0],
   evaluationlimit=50000, maximize);

     [12., [x[1] = -4, x[2] = 2, x[3] = 2, x[4] = 2, x[5] = 2, x[6] = 2, x[7] = 2, x[8] = 2, x[9] = 2], 2169]

No, such polynomial would have been found by solve.
(You have a Hermite interpolation problem here!)

ex:=F(u^(1/n), v^(1/n))^n - F(u, v):
ex1:=eval(ex, [u=exp(-A), v=exp(-B)]):
expand(simplify(ex1)) assuming positive;  # 0

You just need a pencil for this.

c=5, a=12/2=6, b^2 = a^2 - c^2 = 11 

So, x^2/36 + y^2/11 = 1.

The system is not very simple, but I agree that SolveTools:-SemiAlgebraic is very slow in general, and I had not the patience for a result.

Anyway, the result (for eq1) seems to be (and is confirmed by MMA):

a := RootOf(_Z^3 - 4*_Z^2 - 27, 5.054 .. 5.063):
sol := {z=0, y = x, a < x}, {y=0, z = x, a < x}, {x=0, y = z, a < z};

plots:-spacecurve([[t,t,0], [t,0,t], [0,t,t]], t=a..3*a, color=[red,blue,green], thickness=5);

The procedure ALG works only for finite sets! Actually, the support for infinite sets in Maple is very limited, see ?SetOf.

[0,2] is a list, not an interval. [0,1) is a nonsense in Maple (for 1D input).

Note also that ALG computes the algebra generated by C, not the sigma-algebra; of course, X, C must be finite, so this is the same thing.

You cannot solve such theoretical problems using Maple. You cannot even formulate them within Maple, because a CAS does other things!

But your problem is very simple if you know basic measure theory:

If A is a Borel subset in R^n and  g : A --> R  and  h : R^n \ A  --> R are continuous
then the function f : R^n --> R,  f(x) = g(x), if x in A, and f(x) = h(x), if x in R^n \ A
is a Borel function.

In your example, A = {(0,0)}; in your previous version A = {0} x R.