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19 years, 251 days

Split....

The roots of this polynomial can be obtained as follows:

poly:=x^6-3*x+3;
polyroots:=map2(op, 1, roots(poly, proc () local result, k1, k2, k3, k4, k5, k6, k7, k8, k9, k10, k11, k12, k13; global _Z; k6 := RootOf(_Z^6-3*_Z+3); k2 := 6*k6; k5 := k6^2; k9 := k5^2; k3 := k6*k9; k4 := k6*k5; k11 := -15+6*k3+10*k4+12*k5+14*k9; k10 := k2-k11; k8 := _Z^2; k7 := 17*k8; k12 := 3*k3+7*k9+5*k4+6*k5-3*k6; k13 := k7-33+k12; k1 := RootOf(_Z*k13+k8*k10+k12+18); result := {RootOf(k10*_Z+(17*_Z+17*k1+k10)*k1+k13), RootOf(k7+(11*k6+k11)*_Z+11*k3+3*k9+7*k4+5*k5+k2-36)} end proc()));

GAMMA....

Will this suffice?

'sum(
(A-B)
*(-1+binomial(N-2,i))
*(A)^(i)
*(B)^(N-2-i)
,i=0..N-2)';

simplify(evalindets('%',specfunc(anything, binomial), convert, GAMMA));

All results are as they should be....

g & h can access z because you passed the variable to them in the parameters.

Different equation....

You are trying to solve EQ rather than eq:

eq := sqrt(x^2-10*x+1) = sqrt(-8*x^2+9*x-1);
EQ := (lhs, rhs)(map(Im=0, eq)), eq;
sol__eq := [solve]({eq}, {x});
sol__EQ := [solve]({EQ}, {x});
map((radnormal@subs), sol__eq, [EQ]);
map2(map, is, %);

Specifically for this example this will ...

`type/thatsum`:='`+`'(anything);
f:=x->mul([op](x));
applyrule((A::thatsum)='f(A)', [a+b,c+d,a+b+c,c+e+f+g+h]);
applyrule((A::'''`+`'''(anything))='f(A)', [a+b,c+d,a+b+c,c+e+f+g+h]);

Try minimize:...

minimize(expr, a=0..infinity, b=0..infinity, c=0..infinity, d=0..infinity, location);

Gives the same 4 solutions.

Elementary polynomials....

As pointed out already SolveTools:-SemiAlgebraic may be of use here. First convert to elementary polynomials turning expr into (2*e2+2-e1^2)^2-(e1-1)^2+3+4*e4:

E := (2*e2+2-e1^2)^2-(e1-1)^2+3+4*e4,
[a+b+c+d = e1, a*b+a*c+a*d+b*c+b*d+c*d = e2, a*b*c+d*b*c+d*a*b+d*a*c = e3, a*b*c*d = e4];
S := SolveTools:-SemiAlgebraic([E[1], e1>0, e2>0, e4>0]);

Like this:...

N := zip(proc(L,n) if L[1]="H" then n fi end, L, [\$1..nops(L)]);
applyop('NULL', {op}(N), L);
remove(has, S, N);

Use rationalize....

A poorly posed radical form. Maple will answer the following:

int(rationalize(f), u);

See convert,disjcyc...

map(convert, Generators(g1), disjcyc);
PermutationGroup(%);

The following has the same outcome:
PermutationGroup(Generators(g1));

Also see:
g1;
map(convert, Generators(g1), disjcyc);
g2 := PermutationGroup(%);
g3 := PermutationGroup(Generators(g1));
g4 := PermutationGroup(g1);
evalb(g1 = g1), evalb(g1 = g2), evalb(g1 = g3), evalb(g1 = g4); #we are comparing objects

Element comparison....

Your comparison is not elementwise:

(a, b, c, d) := (1, -1, 1, -1);
c := NULL; d := 1, -1;
if (a, b, c, d) = (1, -1, 1, -1) then
print("foo");
end if;

Use series....

If it is a polynomial in x you are working with then use:
convert(series(P, x, n+1), polynom);

However, if P contains other series as coefficients then those too will be converted.

Replace the loop with this:...

Matrix(3, 1, proc(k, c) local i, j, r, C, R; r:=Row(DOFe, k); Matrix(6, 6, unapply('`if`(member(i, r, 'R') and member(j, r, 'C' ), (KTe[k, 1])[R, C], 0), i, j'))  end);

The above code requires Matrix initialization be executed sequentially, which I think it is.

This may be the code:...

restart;
with(ListTools);
with(LinearAlgebra);

assign(
KTe = Matrix(3, 1),
KTe[1,1] = Matrix(4, 4, [[216, -288, -216, 288], [-288, 384, 288, -384], [-216, 288, 216, -288], [288, -384, -288, 384]]),
KTe[2,1] = Matrix(4, 4, [[216, 288, -216, -288], [288, 384, -288, -384], [-216, -288, 216, 288], [-288, -384, 288, 384]]),
KTe[3,1] = Matrix(4, 4, [[500, 0, -500, 0], [0, 0, 0, 0], [-500, 0, 500, 0], [0, 0, 0, 0]]),
DOFe = Matrix(3, 4, [[1, 2, 3, 4], [3, 4, 5, 6], [5, 6, 1, 2]])
);

KG := Matrix(3, 1): for k from 1 to 3 do nn:=Row(DOFe,k); KG1 := Matrix(6, 6); for i from 1 to 4 do for j from 1 to 4 do KG1[nn[i],nn[j]]:=(KTe[k,1])[i,j]; end do; end do; KG[k,1]:=KG1; end do:

print('KG' = KG);

Use this:...

C := proc(E, T)
if type('E, T') then 1 elif hastype('E, T') then add(map(procname, [op]('E'), 'T')) else 0 fi;
end;

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