fun_coeff:= proc(p::algebraic, ux::function(name), n::nonnegint)
local P:= convert(p, :-D), u:= op(0,ux), x:= op(ux), k;
     [seq(coeff(P, (D@@k)(u)(x)), k= 0..n)]
end proc:    

fun_coeff(P, u(x), 4);

Use the arrow operator -> to define functions. By combining this with `if`, the definition can be specialized to the first argument being 0.5:

g:= (x,y)-> `if`(x=0.5, 200*y, 'procname'(args)):

y:= 4:
g(0.5, 4);

     800

g(x,y);

     g(x, 4)

You simply need the elementwise operator ~:

EQ=~0;

If you end your intermediary commands with a semicolon rather than a colon, you'll be able to track down the error more easily. Upon doing this, I quickly see that at least one of the intermediary results, the one that you called s[2], contains the value Float(infinity). This indicates some invalid operation, akin to dividing by 0, was attempted.

Regarding the use of colon or semicolon inside loops: All that matters is which is after the end do (or equivalent) of the whole loop. So you might as well use semicolons for the statements inside loops because it's easier to type.

You can't treat a variable (aka a symbol or a name), such as s, and its indexed subscripted forms, such as s[2], as if they were separate and independent variables. This can cause trouble if you assign to the subscripted form and then attempt to use the unsubscripted form as a free variable (which is exactly what you do here), and it'll definitely cause trouble if you assign to both of them. I'm not sure if this is the source of all or any of your trouble, but it must be corrected before we can move forward.

This caution about subscripted names only applies to indexed subscripted names, i.e., those formed with square brackets. If you form the subscript by using double underscore __ or double vertical bar || or a combination of those, then you can treat the resulting names as separate and independent variables.

Okay, after letting it rest in my mind for a few days, I understand what you're trying to do. I think that I have some vastly simplified procedures for it. Let me know if these do the job for you.

The first thing is to simplify F and Finv. This is what I got:

restart:

(a,b):= (4,1):

Fp:= x-> (U-> x+2*U*b-a*U^2)(1+floor((2*b-a+sqrt((2*b-a)^2+8*a*x))/2/a)):
Fm:= x-> (U-> x+2*U*b+a*U^2)(ceil((-2*b-a+sqrt((2*b+a)^2-8*a*x))/2/a)):
F:= proc(x) option remember; Fm(Fp(x)) end proc:

Fpinv:= x-> (U-> x+2*U*(a-b)-a*U^2)(1+floor((a-2*b+sqrt((2*b-a)^2+8*a*x))/2/a)):
Fminv:= x-> (U-> x+2*U*(a-b)+a*U^2)(ceil((-3*a+2*b+sqrt((3*a-2*b)^2-8*a*x))/2/a)):
Finv:= proc(x) option remember; Fminv(Fpinv(x)) end proc:  

I do believe, as you do, that Finv is the inverse of F. However, I didn't need to use it to write procedure MEC.

MEC:= proc(n::nonnegint)
local N:= {$0..n}, ic, Escape, Capture;
     for ic from 0 to n do
          if F(ic) > ic then Escape[F(ic)]:= true
          elif F(ic) < ic then Capture[ic]:= true
          end if
     end do;
     Escape:= {indices(Escape, 'nolist')};
     Capture:= {indices(Capture, 'nolist')};
     N minus Escape minus Capture, Escape intersect N, Capture
end proc:

Using it on your example:

MEC(12);

The key to understanding the above procedure is that Escape and Capture are Maple tables rather than Boolean arrays. The true that I used is just a dummy; an assignment of anything would do. The Boolean information is conveyed simply by the fact that an entry is assigned. Then the indices command is used to extract all assigned entries.

Let me know if the above does the job for you.

You do it by using x(t) instead of x and y(t) instead of y:

f:= x(t)^2 + y(t)^2:
diff(f, t);

The option maxmesh is an option for the dsolve command, not for the odeplot command. So, if you do

sol:= dsolve({ode,ics}, numeric, maxmesh= 1000):
plots:-odeplot(sol, [x, y(x)], x=0..1);

then it'll work.

You say that you increased the precision in the Options menu. What did you increase it to? If I increase it to 40, then I get accurate results (for the plot in question). Rather than using the options menu, it is better to use the Digits environment variable.

Digits:= 40:

You can test the accuracy of the computation by using the shake command. For example, let's apply it to the last point in your plot. The second argument to shake is the number of Digits of precision.

expr:= eval(m(450, (1/250)*x, 250), x= 100): #Make sure to use exact rationals in this line.
shake(expr, 10);

     INTERVAL(-3.44691114182*10^28 .. 3.45185884337*10^28)

shake(expr, 40);

    INTERVAL(1.96553109914429154721697440528449547786023 ..
    2.03451842008920754181567678008881546299414)

shake(expr, 50);

     INTERVAL(2.000000043280774985094379521387152201289707358740187 ..  
     2.000000043287679818113471795018001692671257692766963)

The above indicates that at 10 digits precision (the default), the computation is total garbage; at 40 digits precision, there are two accurate digits; and at 50 digits precision, there are 11 accurate digits.

In Maple's 2D input, it's illegal to put a space between a function's name and its arguments. You have, for example, printf ("*"). That needs to be changed to printf("*"). 

After you correct that, you still have a small logic error that prevents the first line from being printed. I'll let you figure that out. Let me know if you can't.

plots:-inequal(
     {cos(x) <= y and y <= sin(x) or sin(x) <= y and y <= cos(x)},
     x= 0..4*Pi, y= -1..1, xtickmarks= "piticks", scaling= constrained, color= red
);

nprintf("%5.3f", a);

It's possible, although a bit tricky, to program this so that the 5 and the 3 are not hardcoded. I'll work on it if you need that.

Dr Venkat Subramanian: Here's my Lobatto procedure. The amount by which I increase Digits (essentially, by n^(2/3)) may be excessive, but high-degree polynomial evaluations are notoriously prone to round-off error. Nevertheless, this gives accurate results with very few integrand evaluations.

Download Lobatto.mw

Your model is a linear function of the unknown parameters a, b, and c, so you should use LinearFit. The fact that it's a nonlinear function of the independent variable t is irrelevant.

Statistics:-LinearFit([1, t, t^2], X, Y, t);

Nonetheless, NonlinearFit will still work. Unlike the other respondents, I was able to load and execute your entire worksheet (in Maple 16), including the NonlinearFit command, without error and without making any changes. I do however agree with the other respondents that you should never embed a large dataset like this in a worksheet. It should be read in from a file.

The result was

One way would be to use three colors, like this:

plot([seq([cos(t), sin(t), t= 2*Pi/3*(k-1)..2*Pi/3*k], k= 1..3)], color= [green, yellow, red]);

x=1;
map(`*`, %, 2);
map(`^`, %, 2); or map2(`^`, 2, %); (It's not clear which you meant.)
map(`^``, %, 1/3);

All the above work the same way for inequalities (even when they are not valid for inequalities---be careful!).

The most important commands for manipulating equations and inequalities are rhs and lhs, which extract the right side and the left side, respectively.