@nm Your questions are very good, but I do not have the answers. I do use some trial-and-error. I have a good feeling for what expand, combine, and evalc do. But when I use one-argument simplify or convert(..., form) (where form is a function name or function class name), then I am using trial-and-error. One thing that makes one-argument simplify hard to understand is that it is not idempotent, i.e., simplify(simplify(expr)) is not necessarily the same as simplify(expr).

The reason that I did not answer in the other thread is that I think that what you ultimately want is a procedure that determines whether a line segment intersects a triangle. This idea about the end point of the line segment satisfying 2 or 3 of the inequalities is not enough for that.

@adel-00 From your previous posts, I am guessing that you simply want the original plot shifted up by 1. So, suppose that P is the original plot. Then use

P2:= plottools:-translate(P, 0, 1);
plots:-display([P,P2]);

@nm You are computing the (partial) derivatives of f with respect to x. What is wanted is the derivatives of y with respect to x, also known as the implicit derivatives.

@nm Numeric integration shows that 1/2/I/Pi is correct.

@nm Your specs are almost identical to mine except that I am using Windows 8. I doubt that that could account for the difference. I don't understand it, but I'll try playing with it some more.

Show me the result of the following: After the above, give the command op(rhs(%)).

@Hayno So, we have the line segment L joining the origin and some other point that we're calling the vector. Let's call the triangle T. Determing which of these is the goal:

1. whether L intersects T?
2. whether L lies completely inside T?
3. whether the terminal point of L lies inside T?

2 and 3 are easy. 1 seems harder than just determing whether a line intersects the triangle.

@Hayno That doesn't make sense to me. What does it mean for a vector to intersect something? Give me a specific example of a vector intersecting something simple, like a square; and an example where the vector does not intersect the square. A vector is not a line, nor does it determine a specific line.

I am making this Reply primarily to promote your Question to the top of the Active stack, because I think that it is a good question and I hope that it gets seen. This is by far the most active day that I've ever seen on MaplePrimes, and the Active stack is getting pushed out of the immediately visible range (the top seven) every few hours.

In the meantime, read the help pages ?plots,surfdata and ?plots,listplot3d . These commands are the key to turning 3d point data into a surface plot.

How do you expect anyone to help given so little information? Can you provide a detailed specification of your original problem? I don't mean an upload of your C++ code.

@Hayno What is your ultimate goal here? Do you want a procedure which when given a line and a triangle decides whether the line intersects the triangle? Or do you want a procedure which when given a point and a triangle decides whether the point is inside the triangle?

@danlun You are asking me questions about how the GUI works, which I really don't know much about. Hopefully someone else can answer.

You can take any Maple output and see it with all the proper symbols by assigning that output to a variable, say Sol, and then giving the command lprint(Sol). If you intend to cut-and-paste the output in a form that is acceptable for later re-input, then you should use lprint.

What is the actual range of dates for your plot? That is, tell me the exact lowest date and the exact highest date. Whatever those dates are, they are being translated into the range 36907 - 36937. So I guess it's 31 days.

We can probably make the correction just by relabelling the tickmarks on your plot.

You have 16 equations with 11 unknowns. You are solving for only 8 of the unknowns. There are too many equations.

@adel-00 Of course it doesn't start at 1. You just divided a curve by its maximum. It should end at 1.

As for the speed, your function is very complicated. You may be able to speed it up some by changing sum to add.