Note that each solution of the cubic has an interval of validity. E.g. compare the plot of s[1] with that of cos(theta/3).

Indeed, the current behavior seems inconsistent (fortunatelly, I do not have such "bad habit"). But the question stands if it is desirable that the behavior became consistent. If so, in which way?: 8(17) -->8 and Pi(17)-->Pi?, or 8(17)-->8(17) and Pi(17)-->Pi(17) ?

Indeed, the current behavior seems inconsistent (fortunatelly, I do not have such "bad habit"). But the question stands if it is desirable that the behavior became consistent. If so, in which way?: 8(17) -->8 and Pi(17)-->Pi?, or 8(17)-->8(17) and Pi(17)-->Pi(17) ?

I observe that in recent days, short periods of unresponsivness (time out trying to connect to the server) appeared again.

I observe that in recent days, short periods of unresponsivness (time out trying to connect to the server) appeared again.

This error message is confusing as the expression "typed procedures" does not seem to be defined anywhere in the Maple documentation (c.f. this thread).

I wonder why Erik does not seem to get this error message.

Anyways, in regards to function calls with a constant as 0-operand:

8(17);
                                  8

2^(1/3)(17);

                                 (1/3)
                                2

2^(1/2)(17);

                                  1/2
                                 2

sqrt(2)(17);

                                  1/2
                                 2

This error message is confusing as the expression "typed procedures" does not seem to be defined anywhere in the Maple documentation (c.f. this thread).

I wonder why Erik does not seem to get this error message.

Anyways, in regards to function calls with a constant as 0-operand:

8(17);
                                  8

2^(1/3)(17);

                                 (1/3)
                                2

2^(1/2)(17);

                                  1/2
                                 2

sqrt(2)(17);

                                  1/2
                                 2

Clearly the curves in that picture were chosen with some aesthetic criterion:  for each (stable) equilibrium point  they are kind of equispaced when crossing a line that passes through that point. So, for reproducing this picture, it might be better to fix ICs on such line and integrate backwards in time.

Clearly the curves in that picture were chosen with some aesthetic criterion:  for each (stable) equilibrium point  they are kind of equispaced when crossing a line that passes through that point. So, for reproducing this picture, it might be better to fix ICs on such line and integrate backwards in time.

For instance:

printf("%.5Zf %.5e %g",-0.123456-0.123456*I, 0.1334423423*10^(-15),3);
-0.12346-0.12346I 1.33442e-16 3

For instance:

printf("%.5Zf %.5e %g",-0.123456-0.123456*I, 0.1334423423*10^(-15),3);
-0.12346-0.12346I 1.33442e-16 3

In this simple case just by "looking" at the form of x and the form of the three terms. Their order 0 is 1 and the linear term is that of x. So, at first order exp(x) is fine. Now, at second order it holds:

series(exp(x),e=0,3);
                                               2
              c[2]         4 c[3] c[1] - 5 c[2]   2      3
          1 + ---- e + 1/2 --------------------- e  + O(e )
              c[1]                     2
                                   c[1]

So, that the coefficient 5/2 of (c[2]/c[1])^2 has to be changed. And as it is the linear term of x squared, it is just an issue of adding (3/2)x^2 as -5/2+3/2=-1.

Certainly, any analytic function G(x) whose expansion starts at x^3 can be added without changing the first three terms.

In this simple case just by "looking" at the form of x and the form of the three terms. Their order 0 is 1 and the linear term is that of x. So, at first order exp(x) is fine. Now, at second order it holds:

series(exp(x),e=0,3);
                                               2
              c[2]         4 c[3] c[1] - 5 c[2]   2      3
          1 + ---- e + 1/2 --------------------- e  + O(e )
              c[1]                     2
                                   c[1]

So, that the coefficient 5/2 of (c[2]/c[1])^2 has to be changed. And as it is the linear term of x squared, it is just an issue of adding (3/2)x^2 as -5/2+3/2=-1.

Certainly, any analytic function G(x) whose expansion starts at x^3 can be added without changing the first three terms.

I understand that multinational companies frequently invest a lot of money investigating those issues before launching a global trade mark. Even a "monstrously" elsewhere succesful company like Mac Donalds had to mutate its shops here into coffehouses many years ago for surviving.

Indeed, "hot" is an ambiguous word as it refers to quite different properties like high temperature and strong taste. Neither of them seem "good" for me.