@Kitonum 

If the polygon is not "convex", the order of the points is essential.
The triangles are colored in the order A1A2A3, A1A3A4, A1A4A5,...
so,  see:
PolyOnUnitSphere([ <1,2,1>, <1,0,0>, <0,1,0>, <0,0,1>]);

Probably inserting numbers near vertices as textplot could be useful.

Best regards,
V.A.

Indeed, e.g.

restart;
PolyOnUnitSphere:=proc(A::list(Vector[column](3)),pt:=true,sph:=true) #A[i] = directions (or points on sphere)
uses LinearAlgebra,plots,plottools;
local a:=map(Normalize,A,2), n:=nops(A), u,v,i, L,P:=NULL;
L:= seq( plot3d(1.01*Normalize(a[1]+v*(a[i]+u*(a[i+1]-a[i])-a[1]),2), u=0..1,v=0..1,color=red,style=surface,_rest), i=2..n-1);
if pt=true then P:= pointplot3d(eval(1.01*~a),symbolsize=20,color=blue,symbol=solidsphere)  fi;
display(`if`(sph=true,sphere(style=surface),NULL), L, P);
end:

PolyOnUnitSphere([ <0,0,1> , <1,0,0>, <1,1,-1.5>, <1,2,1>, <0,1,0>]);

@Markiyan Hirnyk 

I know the definitions (it's a part of my job to know them). I tried to explain the Maple point of view (that the argument of f is supposed to be an integer); it is mathematically OK. It would be easy in Maple to define f(x)=0 for x in R \ N but probably the actual implementation has some advantages (e.g. speed).

@sand15athome 

So, we have different opinions.

Let P be a discrete probability in N = {0,1,....}.
This means that P is defined on 2^N --> [0,1], (it is a measure).
Its probability function is
f : N --> R,  f(n) = P({n}).
Here f(5.1) does not make sense.
Maple has an extension F of f and returns F(5.1) instead. That's all.

In the case of a continuous distribution we have the ProbabilityDensityFunction.
For the Normal(0,1) distribution it is
f(t) = (1/2)*sqrt(2)*exp(-(1/2)*t^2)/sqrt(Pi),  t in R.
But here also, t=I (=sqrt(-1)) does not produce an error, f(I) is actually a real number, and I think that nobody sees this as a bug.

As a rule, Maple tries to use the largest possible domain of definition for a function.

We all know that Maple has bugs. But any CAS has.
It would be nice if Maple solves the bugs faster and maintains a list of known bugs and their status.
In your other thread (about Probability) the bugs are clear.
Here, ProbabilityFunction(X, 1.1) = 0 is OK: 1.1 is not in the sample (within current precision). Note that X is not supposed to take only integer values (unlike Geometric). The situation is not perfect but acceptable in my opinion (as long as it is documented).

I don't see any bug.

ProbabilityFunction(Geometric(p), t);
is defined clearly as piecewise(t < 0, 0, p*(1-p)^t)  for 0 < p <=1,  t::integer.
The fact that it returns values instead of an error when t is not integer should not be a problem, and it is easier then to work with floats (no need for round). [Think of n! for noninteger n].

max( maximize(a-b,x=0..1), maximize(b-a,x=0..1) );
convert(%, radical);

@Markiyan Hirnyk 

But it's not complicated. If alpha is algebraic and a[k] are its conjugates (k=1..n) considered by Norm, then the conjugates of x - alpha are x - a[k].
So,  mul(x - a[k], k=1..n)  = Norm(x - alpha) = the minimal polynomial of alpha.

@Markiyan Hirnyk 

f1 is an algebraic number (or element) over the field Q(z).
The conjugates of f1 are the roots of the minimal polynomial; ==> the mentioned expression of the minimal polynomial.

@Markiyan Hirnyk 

If alpha is an algebraic number (over a field)  expressed with RootOfs then

evala( Norm(x - alpha) )

is the minimal polynomial for alpha (by definition).
See ?evala,Norm

@Markiyan Hirnyk 

z>1 was considered implicitly because the ode was considered in an interval containing z=2 and z=3, corresponding to x=1 and x=2.
So, all we need to know is that the hypergeometric function
hypergeom([1/3, 2/3], [3/2], w)
exists (and is C^2) for w<=0.

I prefer to work in cartesian coordonates

restart;
with(plots):with(plottools):
f:=r^2 *cos(theta)+r*sin(theta):
r1:= 0.3+0.1*cos(theta):
r2:= 0.5+0.1*cos(theta):
p:=densityplot(f, theta=0..2*Pi, r=r1..r2,  
colorstyle = HUE, style = patchnogrid):
T:=transform( (u,v) -> [v*cos(u),v*sin(u)]):
display(T(p));

@arman 

When coords=polar, the axes are cartesian.
Why don't you plot directly in cartesian coordinates? Simply use x=r*cos(t), y=r*sin(t).
P.S. Your image is missing.

 

@one man 

Try this simple one which is obvious by hand:

x1^4 + x2^4 + x3^4 - 1 = 0, x3 = 0.

You must be careful with the numbering (0..n) and the fact that in matices indices start at 1.
You should also check the formulae in wiki. Note that the wiki formulae for TSP are completely wrong (they talk there about index 0 which does not exist) and I had to correct them before coding.

Happy New Year!