vv

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7 years, 134 days

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These are replies submitted by vv

@jud As I said, you must check for isomorphism, not "=".

restart;
with(GroupTheory):
AreIsomorphic~(SubgroupLattice(Symm(5), output=list), CyclicGroup(6));

 

@ecterrab I don't understand how D[1$j, 2$k](f)(x,y); is correct

The denominator is infinity for j::nonnegint  and  produces an Error for our j=2. The GAMMA at numerator too.

@acer Unfortunately we have a bug in D (for symbolic orders) here:

restart;
f := (x,y) -> 1/(2+x*y^2):
D[1$j, 2$k](f)(0,0):
eval(%,[j=2, k=4]):  value(%);    # 0  ???
D[1$2,2$4](f)(0,0);               # 6

@jud   ~  is used to construct elementwise operators (see ?elementwise) which are applied for each element of a container (i.e. list, set, rtable).
f~([a,b,c], u)   ==>   [f(a,u), f(b,u), f(c,u)]

@jud You are not right. G itself is a permutation group. You can see this by typing
lprint(G)

It has 5 generators. The same group can be defined with 3 generators:
GG := PermutationGroup(gen);
lprint(GG);

P.S. If you want to get rid of Perm, use convert:

convert(Perm([[1,2],[4,5,6]]), disjcyc);
     
  [[1, 2], [4, 5, 6]]

So,

convert~(gen, disjcyc);

@jud 
 

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
 

@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.

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:

 

@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.

@C_R 

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)
 

@Cata 

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) ]);

@Cata 

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!

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