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L:=SubgroupLattice(S5, output=list):

{[1, 1], [2, 1], [3, 1], [4, 1], [4, 2], [5, 1], [6, 1], [6, 2], [8, 3], [10, 1], [12, 3], [12, 4], [20, 3], [24, 12], [60, 5],   [120, 34]}

# [6,2] = C6, see: https://en.wikipedia.org/wiki/List_of_small_groups

@jud This simply means that this Mathematica command identifies (isomorphically) the subgroups.
The SubgroupLattice in Maple is also able to detect the subgroups. See also DrawSubgroupLattice.

@Carl Love  sys was copy-pasted from tomleslie's,  so he has typos too.

@Carl Love Nice, vote up.
dsolve can be made to work in a few seconds, but the last 2 solutions are expressed via Mathieuxxx functions.

sys:= [ 
[diff(v__0(t),t$2)+v__0(t)=0,  v__0(0)=1, D(v__0)(0)=0],
[diff(v__1(t),t$2)+v__1(t)=-v__0(t)^3,  v__1(0)=0,  D(v__1)(0)=0],
[diff(v__2(t),t$2)+v__2(t)=-v__0(t)^2*v__1(t),  v__2(0)=0,  D(v__2)(0)=0],
[diff(v__3(t),t$2)+v__3(t)=-3*v__0(t)^2*v__1(t)^2-3*v__0(t)^2*v__2(t),  v__3(0)=0, D(v__3)(0)=0],
[diff(v__4(t),t$2)+v__4(t)=-6*v__0(t)*v__1(t)*v__2(t)-3*v__0(t)^2*v__4(t)-v__1(t)^3,  v__4(0)=0,  D(v__4)(0)=0],
[diff(v__5(t),t$2)+v__5(t)=-6*v__0(t)*v__1(t)*v__3(t)-3*v__1(t)^2*v__2(t)-3*v__0(t)^2*v__4(t)-3*v__0(t)*v__2(t)^2,  v__5(0)=0, D(v__5)(0)=0]
for i to 6 do  s:=s, dsolve((eval(sys[i], [s]))); print([s][-1]) od:


@jalal The principal planes correspond to eigenvectors.
They are T[..,j] , j=1..3, (normalized).
But here there are infinitely many because there is a double eigenvalue.
I have included only the principal plane corresponding to the single eigenvalue. You may add other if you want.

with(LinearAlgebra): with(plots):
f:=-x^2 + 2*y^2 + 2*z^2 - 6*x + 4*x*y - 4*x*z - 8*y*z + 4*z - 12:
f:=eval(f, [x=x[1],y=x[2],z=x[3]]);
quad:=implicitplot3d(f, x[1]=-L..L, x[2]=-L..L, x[3]=-L..L, style=surface, scaling=constrained):
b:=eval(Vector( [ seq(diff(f,x[i]),i=1..n)]), [seq(x[i]=0,i=1..n)]):
solve([ seq(diff(f,x[i]),i=1..n)],{seq(x[i],i=1..n)}); # the center
X0:= Vector[column]( eval([seq(x[i],i=1..n)],%) ):
PJ:=sort(J, output=permutation);
T:=Matrix(GramSchmidt([seq( Q[..,PJ[j]],j=1..n)],normalized)): # T is orthogonal
fnew:=simplify( (T.X+X0)^+. A. (T.X+X0) + b.(T.X+X0) + c ); # ==> Hyperboloid of Two Sheets
ax:=seq(arrow(X0, T[..,j], length=10, width=0.3, color=col[j]), j=1..n): 
p1:=implicitplot3d((X-X0)^+ . Q[..,PJ[3]], x[1]=-L..L, x[2]=-L..L, x[3]=-L..L, style=surface, scaling=constrained, transparency=0.5):
display(quad, ax, p1, orientation=[175,63,21], caption="Hyperboloid of Two Sheets");

P.S. If you are using the document mode (which I don't recommend for programming), click the ">_" button to create a prompt, then press F5 and paste the code there.


f:=arctan(y,x) + arctan(-y,x):
# exhaustive check
simplify(f) assuming x>=0,y::real;
simplify(f) assuming x<0,y>0;
simplify(f) assuming x<0,y<0;
simplify(eval(f,y=0)) assuming x::real;

                    (1 - signum(0, x, 1)) Pi

So, for reals, (f <> 0)  <==>   (x<0 and y=0)


S:=[seq]([vq[1], [solve(V__out/vq[1] = 1/sqrt((-m*x^2 + m + 1)^2 + (vq[2]*(x - 1/x))^2), x, useassumptions)]], vq in `[]`~(v__line, Q__s)) assuming x::positive;

plot([ [seq]([u[1],min(u[2])], u=S),  [seq]([u[1],max(u[2])], u=S) ]);


seq(['vq'=vq, 'x'=solve(V__out/vq[1] = 1/sqrt((-m*x^2 + m + 1)^2 + (vq[2]*(x - 1/x))^2), x, useassumptions)], vq in `[]`~(v__line, Q__s)) assuming x::positive;


@Aliocha A simplify is enough for this.

For typesetting (fine-tuning) e.g. ((x-2)/3)^k  etc,  I prefer not to use Maple. LaTeX is easier and nicer here.

@Carl Love Nice catch!

You have no answers because the problem has not much sense. If you really need it, try to explain exactly what you want.

@Zeineb Yes, it was a mistake, the limit of (a_n)^(1/n) if obviously oo for this subsequence.
Actually Maple computes these (though not necessary):

a := n -> (n - sqrt(n*(n + 7))*signum(sin(n)) + 9)^n:
simplify((abs(a(n))^(1/n))) assuming sin(n)>0:
L1:=limit(%, n=infinity):
simplify((abs(a(n))^(1/n))) assuming sin(n)<0:
L2:=limit(%, n=infinity):

       R = 0

The two subsequences cannot be given explicitly. But we know that the sets

N_1 = {n in N: sin(n)>0},  N_2 = {n in N: sin(n)<0},  

are infinite, so the subsequences are ( |a_n|^(1/n) )_(n in N_1) ,  ( |a_n|^(1/n) )_(n in N_2).

The fact that N_1, N_2 are infinite follows from Kronecker's Approximation Theorem
because sin(n) = sin(n+2*k*PI) , Pi being irrational (k,n in N)

@stud_8013  You may assign the sequence to a variable using s:= %;  or  s:=seq(...);
then  s[k]   returns the k-th polynomial in the sequence.

@stud_8013 You have changed the conditions. Then:

A:={seq(0..2)}:  A0:=A minus {0}:
seq(add(v[i]*x^(i-1)+x^4,i=1..4), v=a);


@ecterrab Mathematics has also fundamental problems which wait to be solved, for example the Riemann  hypothesis. Unfortunately, in Maple, the Zeta function cannot be computed efficiently with high precision for large arguments. Maybe in the future, such problems will be supported by Maple too.

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