## 12605 Reputation

8 years, 249 days

## SubgroupLattice, IdentifySmallGroup...

with(GroupTheory):
S5:=Symm(5):
L:=SubgroupLattice(S5, output=list):
map([IdentifySmallGroup],{L[]});

{[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

## SubgroupLattice...

@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.

## typos...

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

## dsolve...

@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.

```restart;
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]
]:
print~(sys);
s:=NULL:
for i to 6 do  s:=s, dsolve((eval(sys[i], [s]))); print([s][-1]) od:
```

## eigenvectors...

@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.

```restart;
with(LinearAlgebra): with(plots):
n:=3:
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]]);
L:=10:
quad:=implicitplot3d(f, x[1]=-L..L, x[2]=-L..L, x[3]=-L..L, style=surface, scaling=constrained):
A:=VectorCalculus:-Hessian(1/2*f,[seq(x[i],i=1..n)]):
b:=eval(Vector( [ seq(diff(f,x[i]),i=1..n)]), [seq(x[i]=0,i=1..n)]):
c:=eval(f,[seq(x[i]=0,i=1..n)]):
X:=Vector([seq(x[i],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)],%) ):
J,Q:=Eigenvectors(A);
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
col:=[red,yellow,blue]:
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.

## Exhaustive check...

```restart;
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;
```

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

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

## ok...

```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 again...

```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;
```

## simplify...

@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!...

@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.

## R=0...

@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:=min(1/L1,1/L2);
```

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)

## seq...

@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.

## OK...

@stud_8013 You have changed the conditions. Then:

```restart;
A:={seq(0..2)}:  A0:=A minus {0}:
a:=Iterator:-CartesianProduct(A0,A\$3):