No. But remember that FAIL just means that testeq couldn't decide if the equation holds or not!

No. But remember that FAIL just means that testeq couldn't decide if the equation holds or not!

@acer Yes, this is quite a lot faster. OK, you also started the loop with 800. The statements after 'else' and 'then' are not visited for the specific equations dealt with here, though.

@acer Yes, this is quite a lot faster. OK, you also started the loop with 800. The statements after 'else' and 'then' are not visited for the specific equations dealt with here, though.

It looks very strange that n*n is not automatically simplified to n^2 and that two apparently equal terms in the result don't cancel each other automatically, so I suggest that you tell us what vergl_2, X1, and V1 are. Please give us just text or upload a worksheet.

Be aware that (quoting from the help page for assume)
"When the assume function is used to make an assumption about an expression x, all previous assumptions on x are removed."

But you can do assume(n::posint);  instead.

@J4James That would correspond to

plot3d(p(x)(y),x=0..1,y=-1.1..1.1,axes=boxed);

@J4James That would correspond to

plot3d(p(x)(y),x=0..1,y=-1.1..1.1,axes=boxed);

@Axel Vogt Your latest worksheet proves that the result is exactly zero, and that that fact was due to symmetries in a only.

I tried to simplify a little. The integrand considered as a function of 'a' really only depends on sin(2*a) and sin(a).

So the integrand can be written

u:=g(sin(2*a),sin(a));

where g is a function of two variables.
#Since sin is periodic with period 2*Pi we may integrate over -Pi..Pi instead of 0..2*Pi:
A:=Int(u,a=-Pi..Pi);
#Changing variables:
IntegrationTools:-Change(A,a=t-Pi);
#Changing the name of the variable:
subs(t=a,%);
#Changing interval again:
subsop(2=(a=-Pi..Pi),%);
#Thus 2*A is equal to
%+A;
#Which results in
   Int(g(sin(2*a), sin(a)), a = -Pi .. Pi)+Int(g(sin(2*a), -sin(a)), a = -Pi .. Pi)

#and that is zero if g is antisymmetric in its second variable, i.e. g(x,-y) = -g(x,y).

And our function in the concrete case satisfies that.

@Axel Vogt Your latest worksheet proves that the result is exactly zero, and that that fact was due to symmetries in a only.

I tried to simplify a little. The integrand considered as a function of 'a' really only depends on sin(2*a) and sin(a).

So the integrand can be written

u:=g(sin(2*a),sin(a));

where g is a function of two variables.
#Since sin is periodic with period 2*Pi we may integrate over -Pi..Pi instead of 0..2*Pi:
A:=Int(u,a=-Pi..Pi);
#Changing variables:
IntegrationTools:-Change(A,a=t-Pi);
#Changing the name of the variable:
subs(t=a,%);
#Changing interval again:
subsop(2=(a=-Pi..Pi),%);
#Thus 2*A is equal to
%+A;
#Which results in
   Int(g(sin(2*a), sin(a)), a = -Pi .. Pi)+Int(g(sin(2*a), -sin(a)), a = -Pi .. Pi)

#and that is zero if g is antisymmetric in its second variable, i.e. g(x,-y) = -g(x,y).

And our function in the concrete case satisfies that.

@J4James You are restricted to point style only if you plot M as is. And that would only be relevant if you imported M from a file. But then, as I said, you can recover the sequence dm, which was constructed directly from the plot data.

@J4James You are restricted to point style only if you plot M as is. And that would only be relevant if you imported M from a file. But then, as I said, you can recover the sequence dm, which was constructed directly from the plot data.

@Markiyan Hirnyk combine works too:

combine(t^n/t^k);

@Markiyan Hirnyk combine works too:

combine(t^n/t^k);

Could you elaborate?

Just one more:

evalindets(current_marks,list(integer),x->`+`(op(x))/nops(x));