Show some respect for the people who read your Question. This might require some actual typing on your part. 

@mmcdara The inappropriately named command convert(..., polynom) does only one very specific thing: It converts a series representation to an ordinary algebraic expression. The most-obvious manifestation of the conversion is the absence of the series's O(...) term. A better name would be convert(..., from_series).

@Carl Love I said above that you can get back up and running by doing the three simple steps Save => Close => Reopen. There are occasionally cases where that doesn't work---the Save on the menu is grayed out. Here's a simplified version of the protocol that works even in those cases (in addition to working in the original cases):

1. Attempt to close the worksheet by using the X on the right side of its tab on the tabs bar (not the X that closes the whole Maple window!!).
2. You'll be interupted by a Save dialog, and you accept it.
3. Reopen the worksheet.

You'll also be interupted be the reappearance of the "Kernel Connection Lost" dialog. Just dismiss it.
 

@mmcdara The difference between in and = is purely syntactic; the difference between the set and list forms is not. When there are non-syntactic differences between two forms, the brevity has nothing to do with my choice. In that case my decision is based on efficiency.

@mmcdara It's 2 fewer characters; that's the only reason.

@Earl In my Answer above, note this command

plots:-implicitplot3d~([p1,p2], -12..5, (-7..7)$2, grid= [50$3], color=~ [red, blue])

It uses elementwise operators (~) to create two implicitplot3d commands with colors matched to p1 and p2 respectively.

@PatD Oddly enough, there is a built-in solution for the equivalent situation with sum and Sum, but none (at least none documented) for products. It is deeply buried in the Physics package. See ?Physics,Library subsection Add.

@Earl I don't see how it could be done with a "stock" Maple plotting command. But it's possible to write a brief "wrapper" procedure that calls a stock plotting command and passes it the function as well as other options, including color. Then this wrapper effectively becomes a new higher-level plotting command that does what you said. I've included several examples of these wrappers in this Post: "Numerically solving BVPs that have many parameters". (The Post is not about plot wrappers per se.)

@Axel Vogt You could try 4-sided dice, which would avoid the too-many-bins problem.

But I think that any robust method for this must address the run lengths. Note that my test decides overwhelmingly---with an odds ratio of more than a thousand---that sequence A is the one that's not truly random.

@Axel Vogt The chi-squared test of course uses a continuous approximation of a discrete distribution, essentially an elaboration of the normal approximation to the binomial. That approximation is usually considered worthless if the expected count in any bin is <= 5. Unfortunately, the Maple command does not check this. Your expected bin counts are (25 values / 5 bins = 5), so all of your bins are in violation of this criterion.

That is why my run-length chi-squared test has the line while E > 10 (where E is about to be divided by 2 before being used). Once E is less than 10 (but still greater than 5), all remaining run-length counts are lumped into the final bin.

@vv Yes, a string has several runs. Now imagine that we have no foreknowledge of the length of the string. Select one of the runs uniformly at random. What is the probability that it's length is k? It's given by the formula I gave. (Note that the sum k= 1..infinity of that formula is 1.) For p=1/2, the formula yields 1/2, 1/4, 1/8, etc. 

A small adjustment needs to be made because we do have foreknowledge of the length of the string. But some empirical testing with rand(0..1)() ​​​​on strings of lengths in the thousands shows that approximately 1/2 of all runs are length 1, 1/4 length 2, etc., and that those approximations are very close.

@janhardo The expression piecewise(f(x) < 0, 0, f(x)) is essentially equivalent to both of

1. max(f(x), 0)
2. if f(x)::realcons and is(f(x) < 0) then 0 else f(x) end if where f(x)::realcons is a relation that evaluates true iff f(x) is a real constant (i.e., contains no free symbols such as 'x').

Of course, 1 would be the preferable formulation, but symbolic integration doesn't handle it.

@vv I didn't read the paper. Is it not exactly true that we expect 1/2 of runs to be length 1, 1/4 length 2, etc. (assuming underlying null-hypothesis distribution is binomial with p=1/2)? Or do you mean that chi^2 does not give exact p-values for discrete cases? I totally agree with that.

Assuming that the underlying binomial has proportion p (1/2 or otherwise), my off-the-top-of-my-head computation is that the distribution of run lengths is

P(run length = k) = p^k*(1-p) + (1-p)^k*p. Do I need to make a finiteness correction to that?

Do you mean that you want to duplicate in Maple the free-body diagram shown in your notes?

@janhardo You asked for a way to check (or verify) your results for 1c. Maple's purely numeric integration is very robust. Do this:

evalf(Int(exp(-x^2)*(x^4 - x^2), x= -3..3));

By using the capital-I Int with evalf, you are requesting purely numeric integration (aka quadrature). The algorithms used for this are completely different than if you had used evalf(int(...)). Since the two answers agree, they are almost certainly correct.