You asked a very important question that perplexes many new Maple programmers. The best way to create a list element by element when you do not have foreknowledge of the final number of elements is to temporarily store the elements in a Maple hash table and then convert that table to a list when you're done. Like this:

Base3:= proc(n::nonnegint)
local r, R:= table(), k, q:= n;
     for k while q > 0 do
          q:= iquo(q, 3, 'r');
          R[k]:= r
     end do;
     `if`(n=0, [0], convert(R, list))
end proc:

You can do the same thing with a dynamically sized rtable. The coding is almost identical, and, as far as I can tell, it is just as efficient:

Base3:= proc(n::nonnegint)
local r, R:= Vector(1), k, q:= n;
     for k while q > 0 do
          q:= iquo(q, 3, 'r');
          R(k):= r
     end do;
     `if`(n=0, [0], convert(R, list))
end proc:

In Maple, a list cannot be initialized; there would be no point since a list cannot be modified at all. (Any examples that you may have seen of lists being modified are faked by the interpreter and are highly inefficient.) A list can be created with its final (and only) entries by

mylist:= [seq(sumsquare(k), k= 1..n)];

An array (or Array) is different from a list; it can be initialized and modified. 

The following procedure uses StringTools:-RegSplit to achieve what you want. You said that you do not need the positions; however, my algorithm necessarily generates the positions, so I thought that I might as well return them also.

RegMatches:= proc(
     P::string, #Regular expression
     T::string  #Text to search
)
uses ST= StringTools;
local
     S:= [ST:-RegSplit(P,T)],
     s,           #one result of above
     p:= 0,       #current position in T
     R:= table(), #return
     m            #a match
;
     for s in S[1..-2] do
          p:= p+length(s);
          ST:-RegMatch(P, T[p+1..-1], 'm');
          R[p+1]:= m;
          p:= p+length(m)
     end do;
     eval(R)
end proc:

Your example:

T:= "variable %1 needs to be of the form %3+2.9 "
    "in order to use routine %2 to find the fizturl":

P:= "%[0-9]+":

RegMatches(P,T);

Also, before you go and re-invent the wheel, check if StringTools:-FormatMessage will do what you need.

Here's another way, not necessarily better, using functional-programming style and, naturally, LinearAlgebra:-RandomMatrix.

P:= (n::posint, m::posint)->
     (x-> (d-> `if`(d[1]=d[-1], x, ``))(convert(x, base, 10)))~
          (LinearAlgebra:-RandomMatrix(n,m, generator= rand(100..999))):

P(10,10);

Despite what Dr Lopez said (which I agree with), you may have your own good reasons for wanting to stick with the VectorCalculus package and for wanting to represent tensors as Maple Matrices. So, I just wanted to show you that the less elegant way that you described is not so bad. Here it is:

<seq(Divergence(VectorField(M[k,..])), k= 1..3)>;

IntTutor is not designed to handle symbolic constants; however, if you hold its hand at each step, you can get an answer for this problem. But what's the point? Are you saying that you want to learn Laplace transforms, yet you don't even know how to integrate exp((a-s)*t) from 0 to infinity? For IntTutor to work, you need to replace the infinity with a finite symbolic constant. I chose gamma below. Then, in the final answer, replace gamma with infinity or a use a limit expression to that effect.

Your syntax has numerous errors: lack of parentheses, using e instead of exp, and others. Here's the corrected syntax:

f:= t-> exp(a*t):
IntTutor(Int(exp(-s*t)*f(t), t= 0..gamma);

If you use disassemble to take apart the expression, you will see that dismantle is showing a truer representation of how the expression is actually stored than ToInert is.

L1:= disassemble(addressof(3*a));

     L1 := 16, 18446744080214109662, 18446744073709551621

kernelopts(dagtag= L1[1]);

     SUM

Note that a product of variables, or a single variable to a numeric power, is a PRODUCT. It is the presence of a numeric coefficient that turns it into a SUM.

Your simplification is only valid if you assume that the parameters such as theta are real. The command evalc makes those assumptions.

simplify(evalc(Re(A)));

It looks (vaguely) like your data is stored as row Vectors. The following procedure takes as input two Vectors (row or column), Array slices, or lists (of the same length, of course) and a boolean-valued filter procedure which when passed a point (x,y) returns true iff the point should be included. It returns the filtered output as column Vectors.

Select:= proc(f, A::~Vector[column], B::~Vector[column])
local M:= convert(select(f@op, convert(<A|B>, listlist)), Matrix);
     M[..,1], M[..,2]
end proc:

Example of use: Suppose that we want to select those points whose first coordinate is even.

A:= <1|2|3|4|5>;  B:= <6|7|8|9|10>;

     A:= [1 2 3 4 5]
     B:= [6 7 8 9 10]

f:= (x,y)-> x::even:
(AA,BB):= Select(f, A, B):
% ^~ %T;

                                        [2 4], [7 9]

Download Select.mw

In Maple, a line break has the same syntactic significance as a space; unlike Matlab, the end of line is not considered a statement terminator. So, just find a place where you could put a space and continue on the next line. Preferably, the continuation lines will be indented farther to the right than the statement's initial line. If you are having trouble finding a place to put a space, you should probably rethink your code, and if you post it, I'll help with that.

In the case of a very long constant, string, or name---where you can't use an extra space---you can end a line with backslash \ and continue at the beginning of the next line. I consider this sloppy coding, and in 16 years of Maple usage, I've never had to use it.

It's trivial:

solve({f[k] $ k= 1..n}, {x||(1..n)}):
a||(1..n):= eval([x||(1..n)], %[1])[]:
a||(1..n);

Yes: Use eliminate instead of solve. Your situation is precisely what it's for.

The following paragraph is from the help page ?dsolve,details:

Integrals appearing in answers returned by dsolve are expressed using the inert Int and Intat (not int or intat). These integrals appear when int is not able to calculate them or when it appears to be convenient not to evaluate them during the solving process. You can request the evaluation of these integrals using the value command.

So, try applying the value command to dsolve's results. If you still get an integral, then Maple simply can't do the integral. If you have parameters, sometimes making assumptions on the parameters will make the integral doable. See ?assuming.

If you have further questions about this, please post your differential equation and any initial or boundary conditions.

The $ is the repetition operator (it also has a mode where it acts like seq). It is a basic operator of Maple, not something special to permute. For example, [1$4] returns [1, 1, 1, 1]. In many ways, $ is to seq as sum is to add and as product is to mul. For example, $ will take a symbolic second argument. (It's a shame that seq and $ are not switched: The former is more commonly used.)

To get all 2^n sequences of 0s and 1s of length n, use

combinat:-permute([1$n, 0$n], n);

While the $ does accept symbolic arguments, permute does not. To run your command, you'll need to supply a nonnegative integer for n.

Pi is not considered numeric. So, you have to do something like this

for x[2] from 15/180 to 75/180 by 5/180 do    
     a[i,j]:= Pi*x[2];
     j:= j+1
end do;