vv

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These are answers submitted by vv

The singularities of F1 are on the Jordan curve

plots:-implicitplot(1/F1, x=0..k - varepsilon,y=0..2*Pi, view=0..6.5, color=red);

Note that the integrand is even in x, so, one may integrate over x = 0 ... 1 - eps.

Unfortunately Maple cannot compute numerically a CPV integral, but it is easy to transform it into a regular one. If the singular point is c, then 

Int(f(x), x =c-r .. c+r, CPV) = Int(f(c+t) + f(c-t), t = 0 .. r)

Strange implementation (bug?) for numboccur.

restart;
L:=[1., HFloat(1.), 1., 0., 0.];
#                   L := [1., 1., 1., 0., 0.]

numboccur(L, 1.);
#                               2

numboccur(L, 1.0);
#                               0

numboccur(L, HFloat(1.0));
#                               1

evalb(L[1]=L[2]);
#                              true

select(`=`, L, 1.); select(`=`, L, 1.0);
#                          [1., 1., 1.]

#                          [1., 1., 1.]

dismantle(%);

#LIST(2)
#   EXPSEQ(4)
#      FLOAT(3): 1.
#         INTPOS(2): 1
#         INTPOS(2): 0
#      HFLOAT(2): 1.
#      FLOAT(3): 1.
#         INTPOS(2): 1
#         INTPOS(2): 0

 

                      

arctand ignores the second argument.

arctand(y,x) simplifies to arctand(y)

BTW, for y <> 0,

arctan(y,x) = (Pi/2)*signum(y)-arctan(x/y)

The fact that leadterm is not documented under asympt is probably because asympt(f, x, n) is actually equivalent to series(f, x=infinity, n), and here leadterm appears. By taking n=2, leadterm is not strictly necessary, but it is convenient because O(...) is removed.

There is an old worksheet at the Application Center https://www.maplesoft.com/applications/view.aspx?SID=6322

It is clear enough and includes an animation.


 

The Maple answer is correct for your question. It gives the two intervals on which the derivative is negative.
If you are interested in the (relative) extrema, use exprema.
Of course x=0 is not an extreme point.

A := x^2 + 400/x;

x^2+400/x

(1)

plot(A, x=-10..10, view=-500..500);

 

solve(evalf(diff(A, x) < 0));lprint([%]);

RealRange(-infinity, Open(0.)), RealRange(Open(0.), Open(5.848035476))

 

[RealRange(-infinity, Open(0.)), RealRange(Open(0.), Open(5.848035476))]

 

solve(diff(A, x) < 0);lprint(%);  # it is better not to approximate A

RealRange(-infinity, Open(0)), RealRange(Open(0), Open(2*5^(2/3)))

 

RealRange(-infinity, Open(0)), RealRange(Open(0), Open(2*5^(2/3)))

 

extrema(A, {}, x, 'S'); evalf(%);S=evalf(S);

{60*5^(1/3)}

 

{102.5985568}

 

{{x = 2*5^(2/3)}} = {{x = 5.848035476}}

(2)

 

Download extrema.mw

restart;
eq1:= f(x) + 2*f(1/x) + 3*f(x/(x-1)) = x:
eq:=eq1, simplify(eval(eq1, x=1/x)):
eq:=eq, simplify(eval(eq1, x=x/(x-1))):
eq:=eq, simplify(eval(eq1, x=-1/(x-1))):
eq:=eq, simplify(eval(eq1, x=1-x)):
eq:=eq, simplify(eval(eq1, x=(x-1)/x)):
solve([eq], indets([eq],function)): f(x) = eval(f(x), %);
simplify( eval(eq1,  f = unapply(rhs(%), x)) );   # check

             

             

 

So, you want the surface integral on the projection of a cuboid on a plane.

V:=<0,0,0>, <1,2,0>, <0,2,0>, <1,0,0>, <0,0,1>, <1,2,1>, <0,2,1>, <1,0,1>:
T:=<-1,0,0>, <0,-1,0>, <0,0,-1>: 
PM:=LinearAlgebra:-ProjectionMatrix([T[2]-T[1], T[3]-T[1]]):
Pr := v -> T[1]+PM.v:
Pr2:= v -> (PM.v)[[1,2]]:  # ConvexHull works in dimension 2.
P:=Pr~([V]):
P2:=Pr2~([V]):
ind:=ComputationalGeometry:-ConvexHull(Matrix(P2)^+):
C:=P[ind]: n:=nops(C):  # the vertices of the projection
Tri:=(v1,v2,v3) -> Surface( v1 + t*(v2+s*(v3-v2)-v1), s=0..1, t=0..1):
# the parametrization of the interior of a triangle
add(VectorCalculus:-SurfaceInt( x^2+y^2, [x,y,z] = Tri(C[1],C[i],C[i+1])), i=2..n-1);

                         

Maple solved your equation with respect to z.
Note that LambertW is a function just like sin or exp.

with(VectorCalculus):
SurfaceInt( x^4+y^4+z^4, [x,y,z] = Sphere( <0,0,0>, a ) );

(12/5)*Pi*a^6

(1)

# You may want to see the generated double integral
SurfaceInt( x^4+y^4+z^4, [x,y,z] = Sphere( <0,0,0>, a ), inert );

Int(Int(a^6*(2*cos(theta)^4*cos(phi)^4-4*cos(theta)^4*cos(phi)^2-2*cos(theta)^2*cos(phi)^4+2*cos(theta)^4+4*cos(theta)^2*cos(phi)^2+2*cos(phi)^4-2*cos(theta)^2-2*cos(phi)^2+1)*sin(phi), phi = 0 .. Pi), theta = 0 .. 2*Pi)

(2)

 

Just increase Digits (exp(-200) is very small), e.g.

Digits := 20;

 

1. A faster method is to use Iterator.

with(Iterator):
C:=Combination(5,3):
K:=[1,2,3,4,a]:
seq( K[[seq(c[]+1)]], c=C);

        [1, 2, 3], [1, 2, 4], [1, 3, 4], [2, 3, 4], [1, 2, a], [1, 3, a], [2, 3, a], [1, 4, a], [2, 4, a], [3, 4, a]

 

2. To print the body of a procedure, use

interface(verboseproc=2);
print(combinat:-subsets);

or, 

showstat(combinat:-subsets);

 

evalf( Int(h, Omega, epsilon=1e-4, method=_CubaCuhre));

gives 4.12122491078804

(To have a probability you forgot probably to divide by (2*Pi)^3.)

You cannot write like this. diff(z, x)  simplifies to 0, because z is interpreted as an independent variable (just like x).
If z depends on x, you must write z(x)
In this case,
diff(w(x,y,t,z(x)), x); 
==>  D[1](w)(x, y, t, z(x)) + D[4](w)(x, y, t, z(x))*diff(z(x), x) 

You may use e.g.
alias(z=z(x));
so that the argument x is omitted, but you must be careful (I personally do not use this facility in such situations).

 

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