It can be done with a single command:

a2_||(pos,neg):= selectremove(t-> sign(t) = 1, a2);

Clearly you want a matrix whose entries are matrices rather than a block matrix because your sizes won't fit into a block matrix. Here's how to do it:

a1 := Matrix(3, [1, 2, 3, 7, 8, 9, 13, 14, 15]);
a2 := Matrix(3, 2, [5, 6, 11, 12, 17, 18]);
a3 := Matrix(2, [19, 20, 25, 26]);
a4 := Matrix(3, 2, [5, 6, 11, 12, 17, 18]);
A := Matrix(2, 2, (i,j)-> a||(2*(i-1)+j) );

Here's a complete example showing the method that I outlined in my Replies. I compute, by purely numeric computation, the surface area of the isosurface MF(x,y,z) = 11 for MF(x,y,z) = sin(y*z)+x+11. Then, for accuracy comparison, I compute the surface area by trivial numero-symbolic computation.

My worksheet could be made more efficient; there's some redundant calculation. I used the way that I did for clarity of exposition.

Download numeric_isosurface_surface_area.mw

Your immediate problem is that X is a formal parameter, and you can't assign to a formal parameter. It may appear to you that you are only assigning to the elements of X, but since it's a list, any assignment to it reassigns the whole list. This makes assigning to the elements of a list a very bad thing to do, even though Maple will sometimes allow you to get away with it. The solution is to convert the list to a Vector on input, do your sorting, and then convert back to a list on output.

Some other points:

1. Sorting the empty list should return the empty list; it shouldn't be an error. In the code below, the empty list just "falls through" because the outer for loop will just be skipped if n is 0 or 1.
2. You can switch two elements in a Vector in a single assignment without using an intermediate variable. Look carefully at the central assignment in my code below.
3. Don't use the print statement to return values. Either use a return statement, or just put the expression to be returned immediately before the end proc.

Here is your code with all of these recommendations implemented:

Bubble := proc(L::list(realcons))
local n := nops(L), X := Vector(L), i, j;
     for i to n-1 do
          for j to n-i do
               if X[j+1] < X[j] then
                    X[j..j+1]:= X[[j+1, j]]
               end if
          end do
     end do;
     convert(X, list)
end proc:

#Test it.
R:= rand(-99..99):
L:= ['R()'$99];

     output omitted

Bubble(L);

     output omitted

All in all, you did a good job! Your logic was perfect, and your only real problem was due to an idiosyncracy in the way that Maple implements lists.

I followed the link from the Wikipedia page to the paper that has the Matlab code. Then I meticulously translated that into efficient and compact Maple code. The totality of the Matlab code accounts for only the GRF:= ...; paragraph of my code! The result is a greyscale image. To add color, I put the high values on a blue scale and the low values on a red+green scale.

restart:

Digits:= 15:
GRF:= proc(n,r)
uses AT= ArrayTools, IT= ImageTools;
local
     GRF:= (IT:-FitIntensity@IT:-Convolution)(
          AT:-RandomArray(n, 'distribution'= 'normal'),
          Matrix(2*r+1$2, (i,j)-> `if`((i-r-1)^2 + (j-r-1)^2 <= r^2, 1, 0))        
     )
;     
     (IT:-Preview@IT:-CombineLayers)(
          IT:-FitIntensity(map[evalhf](x-> `if`(x >= .5, 0, x), GRF))$2,
          IT:-FitIntensity(map[evalhf](x-> `if`(x < .5, 0, x), GRF))
     )
end proc:

GRF(300, 13);

When the number of parameters of a BVP is greater than or equal to the number of conditions at one of the boundary points, it can be solved as an IVP with fsolve being used to determine the parameters. I like this idea because I think that the main IVP solver (rkf45) is more robust than any of the BVP solvers. The following worksheet produces a very high-accuracy solution in good time using little memory, although I haven't compared it timewise or memorywise to Preben's. Accuracywise, my value of omega computed at Digits = 10 has three more digits of accuracy than Preben's. (I don't mean to denigrate his solution, for he just used the default error control.)

Download BVP_as_IVP.mw

If I understand you correctly, you want to increase the voltage, V, by steps and do fsolve at the increased voltages, at each step using as initial values the solutions from the previous step. Is that correct? In the worksheet below, I've done all this using a step of 0.1. The keys to making this work is to leave V symbolic and to use unapply to turn f1 and f2 into procedures with parameter V. The other changes that I made are just to keep everything neat and symbolic until it needs to made numeric.

Download 2DOF_mod.mw

There are various uses for ~ in Maple. Your syntax is wrong. For elementwise operation, the ~ goes after the procedure name. So that should be int8~(round~(S)).

Simply take your first example, and change type~(r,anything) to evalb~(r).

@tomleslie 

I believe that the following graph shows what not satisfying the boundary conditions smoothly means. This picks up from the end of the worksheet that you gave.

plots:-display(seq(plots:-odeplot(sol[k], [eta, f(eta)], 0..1, color= col[k]), k= 1..2));

I think that the critics may be complaining about that bump in the red curve. If you simply change the method from bvp[middefer] to bvp[midrich], this artifact will disappear completely. I think that this answers the OP's Question "Is there any other way?"

Correcting the abserr problem pointed out by Preben will also correct the anomaly in the red curve.

You just need to include two t-> and remove t= from the domain specification:

plot(
     [
          t-> add(BS(bb[4], t, i, 3)*bb[1](i+1), i= 0..n),
          t-> add(BS(bb[4], t, i, 3)*bb[2](i+1), i= 0..n),
          0..1
     ],
     color= red, axes= normal, scaling= constrained, numpoints= 100, thickness= 2
);

Note that the last item in the square brackets is 0..1 rather than t= 0..1.

The above can be made more efficient by including option remember at the beginning of procedures BS and bb.

If you want to create a list of indeterminate length and the value of each list element can be computed independently of the others (including that the stopping criterion doesn't depend on what is computed), then you should probably use a seq command. It is faster than a for loop. But there are a great many situations where the computations are not independent or the stopping criterion must be re-evaluated at each step. In those cases, you must use a for loop (which includes do loops and while loops). The best way to create a list in a for loop is this:

MyList:= Vector();
for n from 1 to 10 do
     MyList(n):= n+1;
end do;
MyList:= convert(MyList, list);

Note that this creates a list of size 10, not 9, which is what you said.

To reiterate what Tom Leslie said, a "list in {} form" is a set. Lists and sets have many similarities, but they're not the same thing. In particular a set isn't a type of list. To create a set in a for loop, do as above, but change the last line to

MyList:= convert(MyList, set);

If you have a group of expressions e1, ..., e9, they can be made into a set of equations equated to 0 like this:

MyEqns:= {e||(1..9)} =~ 0;

The reason for your print problems is a feature of the old-style matrices and arrays which isn't worth explaining here because you shouldn't be using them anyway. Your code can be vastly simplified and modernized to the new Matrices and Arrays like this:

restart:

Der:=proc(A,n)
local i,j,k,l,C;
     eval(
          Matrix((n,n), symbol= C),
          solve({
               seq(seq(seq(add(
                    A[i,j,k]*C[k,l] = A[k,j,l]*C[i,k]+A[i,k,l]*C[j,k],
                    k= 1..n), l= 1..n), j= i+1..n), i= 1..n-1)
          })
     )
end proc:

A1:= Array((1..2)$3, {(1,1,2)= 1}, storage= sparse):
Der(A1,2);

The above code for procedure Der is functionally equivalent to Tom Leslie's. If the Array A is highly sparse, as in the above example, there'll several occurences of the escaped local variable C. This may be a nuisance, and you may wish to replace it with a global variable. In that case, remove C from the local variables, and change the procedure's other four occurences of C to _C.

That is done by converting the first ODE's solution function for f(x) into something that'll return unevaluated when passed a symbolic argument and return numeric when passed a numeric argument and then declaring that as known in the dsolve command for the second ODE. Like this:

Sol_f:= dsolve({diff(f(x),x)=f(x), f(0)=1}, numeric):
F:= x-> `if`(x::complexcons, eval(f(':-x'), Sol_f(x)), 'procname'(x)):
Sol_g:= dsolve({diff(g(x),x)=F(x)*g(x), g(0)=1}, numeric, known= [F]);

Kitonum's Answer works also, but there are times when you want to solve the two as separate systems. And the above (using option known) will work for any numeric procedure, not just one that comes from a prior dsolve.

By the way, your title is perfect. I don't know what you see wrong with it.

That is surprising that it can't do it, because it can do it immediately if you replace the floats with symbols:

int(x*(1+a*x^4)^e*(C1-C2*x^4/(1+a*x^4)), x= 0..1);

     (1/8)*(4*C2*a^(5/4)*GAMMA(3/2+e)*GAMMA(1-e)*a^(-3/4+e)*hypergeom([1-e, -1/2-e], [1/2-e],                      -1/a)+4*C1*a^(1/4+e)*hypergeom([-e, -1/2-e], [1/2-e], -1/a) * a^(5/4)*GAMMA(3/2+e)*GAMMA(-e) * e-(4*        (1/2+e))*Pi^(3/2)*sec(Pi*e)*(C1*a*e+(1/2)*C2)) / (e*a^(3/2)*GAMMA(3/2+e)*(2*e+1)*GAMMA(-e))

evalf(eval(%, [C1= 2.906663106, a= 1.38, e= -0.4311594203, C2= 3.458929096]));

      1.00000000009749

This is the same (except for a little round-off error in the last digit) as what you'll get if you apply numeric integration (evalf(Int(...))) to your original integral.