How would you visualize a four-dimensional field (assuming F(t)=0)?

The equation tan(x)=-2*x is transcendental, and has to be solved numerically. Hence, solve is not the right tool in this case.

When solve finds no solution, it returns NULL. And NULL  prettyprints as nothing. But you can see it using the line printer:

lprint(solve({sin(x^2)+2*x^2*cos(x^2) = 0, -1<=x, x<=3}, x));
NULL

This is not a bug in solve, but may be a bad design of the prettyprinter.

The equation tan(x)=-2*x is transcendental, and has to be solved numerically. Hence, solve is not the right tool in this case.

When solve finds no solution, it returns NULL. And NULL  prettyprints as nothing. But you can see it using the line printer:

lprint(solve({sin(x^2)+2*x^2*cos(x^2) = 0, -1<=x, x<=3}, x));
NULL

This is not a bug in solve, but may be a bad design of the prettyprinter.

It seems that verify is called, and:

restart:
verify( 0, -2..2, 'interval'('closed'..'closed') );

                                 true

interval:=-2..2:
verify( 0, -2..2, 'interval'('closed'..'closed') );

                                false

Yes, this is strange as the computation is strightforward:

dsolve( ode2 ):
solve(eval(eval(%,x=0),IC),{_C1}):
simplify(eval(%%,%));
odetest(%,[ode2,IC]);


                                      y0
     y(x) = - --------------------------------------------------
                          x (-x + 2 a)          x (-x + 2 a)
              -y0 - exp(- ------------) + exp(- ------------) y0
                               2                     2


                                [0, 0]

 

By keeping a history of the thread edits (as wikies do) there should not be a big problem about the loss of the original sense of the exchange.

?implicitplot3d, noted above by Alec, is not working now.

Like ?PDEtools[diff_table]

There are many more addon packages. E.g. the FourierTrigSeries package, GRTensor, etc.

Indeed, syntactically different objects should typeset differently. The current situation is a bit of a Pandora box.

For instance, different fonts (or colored backgrounds) could be used for function calls and type `*` expressions.

The same outputs labeled "document" occur in Standard worksheet from 2D input.

Given the circumstances, I would prefer the divergence route suggested by Jacques. 

Volume 0: Axiom Jenks and Sutor > 2.1.1 Domain Constructors, states:

In Axiom, domains are objects. You can create them, pass them to functions, and, [...], test them for certain properties.

It sounds to me as saying that Axiom domains are also first class objects.

I do not know whether it is nice, but as it has a series development:

s:=sum((-((n+1)/n)^(1/3))^r,r=0..infinity);
                                    1
                        s := ----------------
                                 /n + 1\(1/3)
                             1 + |-----|
                                 \  n  /

and, inserting this summand into the initial summand:

((n+1)/n-1)*(12*n^2+12*n+1)/
((1+((n+1)/n)^(1/3))*(12*n^3+18*n^2+6*n)):
f1:=simplify(%):
subs(s=((-((n+1)/n)^(1/3))^r),f1):
f2:=simplify(%) assuming r::posint;
                        2                 r /n + 1\(r/3)
                   (12 n  + 12 n + 1) (-1)  |-----|
                                            \  n  /
             f2 := -------------------------------------
                              2     2
                           6 n  (2 n  + 3 n + 1)

  produces terms whose sum can be done when r is multiple of 3:

eval(f2,r=3):
sum(%,n=1..infinity);
                          2
                      5 Pi
                8/3 - ----- - 1/6 Zeta(3) - 8/3 ln(2)
                       18
eval(f2,r=6):
sum(%,n=1..infinity);

                      2     4
                    Pi    Pi
              8/3 + --- + --- + 11/6 Zeta(3) - 8/3 ln(2)
                    18    540

So, it seems like there is a pattern here, and this might help to make this partial sum, at least.

I do not know whether it is nice, but as it has a series development:

s:=sum((-((n+1)/n)^(1/3))^r,r=0..infinity);
                                    1
                        s := ----------------
                                 /n + 1\(1/3)
                             1 + |-----|
                                 \  n  /

and, inserting this summand into the initial summand:

((n+1)/n-1)*(12*n^2+12*n+1)/
((1+((n+1)/n)^(1/3))*(12*n^3+18*n^2+6*n)):
f1:=simplify(%):
subs(s=((-((n+1)/n)^(1/3))^r),f1):
f2:=simplify(%) assuming r::posint;
                        2                 r /n + 1\(r/3)
                   (12 n  + 12 n + 1) (-1)  |-----|
                                            \  n  /
             f2 := -------------------------------------
                              2     2
                           6 n  (2 n  + 3 n + 1)

  produces terms whose sum can be done when r is multiple of 3:

eval(f2,r=3):
sum(%,n=1..infinity);
                          2
                      5 Pi
                8/3 - ----- - 1/6 Zeta(3) - 8/3 ln(2)
                       18
eval(f2,r=6):
sum(%,n=1..infinity);

                      2     4
                    Pi    Pi
              8/3 + --- + --- + 11/6 Zeta(3) - 8/3 ln(2)
                    18    540

So, it seems like there is a pattern here, and this might help to make this partial sum, at least.

You should install Aladjev's Library for Maple. See these threads:

partial derivatives

Aladjev library for Maple