@Josolumoh Is your function a probability density function (pdf)? If so, what's the name? If it's a standard one, Maple likely already has a sampling procedure for it.

@JohnS If L is a list or set, then L[] extracts the underlying sequence. This is documented in the ninth paragraph of Description at ?list, which simply states

To extract the contents of a list or set, use the empty selection operator [ ].

That's not possible: The function is much too complicated. Perhaps if you describe why you want to invert it, we could come up with an alternative.

You can only solve an equation for the highest derivative of each function. What is the reason that you want to solve for the second derivative of f4 rather than the third derivative?

What's sys? How's Maple supposed to know the ODEs that you've written on the paper just from sys? And I don't see anything about y(t) on your paper.

@Markiyan Hirnyk Yes, I like your way better. To handle more-general cases, change the key to (c-> evalf(rhs(c[-1]))).

@patient It can be done like this

linspace:= (a,b,n)-> a +~ (b-a)/(n-1)*[$0..n-1]:
U1:= eval(U(x), res):
U2:= proc(alpha)
local x;
     U1(parameters= [alpha]);
     plots:-spacecurve([x, alpha, U1(x)], x= -1..-.02, color= black, thickness= 0)
end proc:
plots:-display(
     [seq(U2(alpha), alpha= linspace(0.1, 10, 23))],
     axes= frame, labels= [x, alpha, ``], orientation= [-86, 68]
);

@patient Here's how to do the 3-D plot that you want:

U1:= eval(U(x), res):
U2:= proc(x,alpha)
     U1(parameters= [alpha]);
     U1(x)
end proc:
plot3d(U2, -1..-0.02, 0.1..10, labels= [x, alpha, U], axes= boxed);

@Earl Better yet, any polygon can be partitioned into triangles whose vertices are vertices of the polygon.

@Carl Love I made an incorrect generalization from the n=2 to n=3 case. At first I thought that no group of three should occur more than once. Actually, it is much more strict: No pair can appear as part of one of the groups of three more than once. This seems to be a much more difficult problem, and the number of solutions is far less than 51 or 55. Brian's generalization of Kitonum's solution only applies to the more-liberal case: that no group of three occurs more than once.

@ben2015 It's easy. Use a Vector, not a list. Use add and mul rather than sum and product.

A:= <a||(1..10)>:
F:= unapply(add(A(k)*(1-A(k))^2*mul((1-2*A(l))^3, l= 1..k-1), k= 1..10), a||(1..10));

@Earl I'm not sure if this helps, but any polygon can be partitioned into triangles.

@brian bovril 

My argument is that 55 is an upper bound, not an upper and lower bound, for the g=12, n=3 case (binomial(12,3) / (12/3) = 55). You got 51 solutions, so 51 is a lower bound. Do you have reason to believe that there are more than 51 solutions?

@max125 By "parsed", vv means that in the 2-D input, Maple attempts to interpret your expression as it's being typed. With Maple Input (also called 1-D input), no interpretation is attempted until you press Enter, and only the characters that you actually typed and which appear on your screen are interpretted. The word "parse" is nearly synonymous with "interpret".

@wgarn 

In regular Maple, the command

ans := map(proc (x) options operator, arrow; expand(convert(convert(x, rational), expln)) end proc, {ans});

can be shortened to

ans:= (expand@convert)~({ans}, compose, rational, expln);

I don't know if this also works in Maple TA.